Art Montemayor August 21, 2006 Rev: 0 Page 1 of 345 FileName: document.xls WorkSheet: Introduction I received my first copy of Crane's Technical Paper No. 410, titled " Flow of Flui fittings, and pipe", in October 1962. It was given to me free of charge by the Cr the business account at Liquid Carbonic Division of General Dynamics in Chicago, w I had just completed my first year practicing engineering at Liquid Carbonic's aff Jamaica Oxygen and Acetylene Ltd. and Jamaica Carbonics. I had spent a year in Ja Manager, replacing Alf Newton who had gone to Barbados and Trinidad to erect a sma Barbados and an industrial gas facility (Oxygen and Acetylene) at Biljah Road in T My original copy was the 1957 copyrighted version, sixth printing. It had a stat This was my first experience in dealing with fluid flow problems using the Darcy e corresponding Moody Chart. At Texas A&M we were taught the Fanning equation and i friction factor (the Darcy friction factor = 4 x the Fanning friction factor). We had never been exposed to such practical and detailed fluid flow problems at Te was not only interesting, but it also taught the young engineer how to cope with a plant fluid problems. I went through all 27 example problems which were given in Later, during my tenure at Quaker Oats Chemical Division in Chicago (1968 - 1973), of the 1965 Crane version (9th printing). The example problems were basically the the same manner as in the 1957 version. The 1965 version had a stated price of $ When I worked for Allstates Engineering on DuPont projects (1989 -1994) I obtained printing) and it had a stated price of $8.00. This is the version that this Work I still retain the original 1957 version copy, although the original hard, orange have cracked and disintegrated from the stainless steel spiral hinge. I have transcribed the Example problems given in the 1979 (18th printing) Edition added those problems that were given in the 1957 Edition but were not included in I have included in each of the 1979 Example problem solutions those solutions that and that were resolved in a different manner. This enables the reader to see the solving these problems and how the methods have increased the accuracy of the answ My reason for transcribing these Example problems into the Spreadsheet format is t method by which an engineer can quickly detect the methodology and follow the math Emphasis is put on the logic and reasoning employed rather than worrying about the The spreadsheet allows the reader to insert a variety of different input values an in absorbing the manner in which the ultimate answer is affected. Art Montemayor
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Art Montemayor August 21, 2006Rev: 0
Page 1 of 287 FileName: document.xlsWorkSheet: Introduction
I received my first copy of Crane's Technical Paper No. 410, titled " Flow of Fluids through Valves, fittings, and pipe", in October 1962. It was given to me free of charge by the Crane salesman that took care of the business account at Liquid Carbonic Division of General Dynamics in Chicago, where I worked.
I had just completed my first year practicing engineering at Liquid Carbonic's affiliated companies in Jamaica,Jamaica Oxygen and Acetylene Ltd. and Jamaica Carbonics. I had spent a year in Jamaica as ProductionManager, replacing Alf Newton who had gone to Barbados and Trinidad to erect a small CO2 plant in Barbados and an industrial gas facility (Oxygen and Acetylene) at Biljah Road in Trinidad.
My original copy was the 1957 copyrighted version, sixth printing. It had a stated price of $10.00.
This was my first experience in dealing with fluid flow problems using the Darcy equation together with the corresponding Moody Chart. At Texas A&M we were taught the Fanning equation and its corresponding friction factor (the Darcy friction factor = 4 x the Fanning friction factor). We had never been exposed to such practical and detailed fluid flow problems at Texas A&M. This booklet was not only interesting, but it also taught the young engineer how to cope with and resolve practicalplant fluid problems. I went through all 27 example problems which were given in Section 4 of the booklet.
Later, during my tenure at Quaker Oats Chemical Division in Chicago (1968 - 1973), I would receive a copy of the 1965 Crane version (9th printing). The example problems were basically the same and resolved in the same manner as in the 1957 version. The 1965 version had a stated price of $2.00.
When I worked for Allstates Engineering on DuPont projects (1989 -1994) I obtained the 1979 version (18th printing) and it had a stated price of $8.00. This is the version that this Workbook's examples are based on.
I still retain the original 1957 version copy, although the original hard, orange carboard front and back coverhave cracked and disintegrated from the stainless steel spiral hinge.
I have transcribed the Example problems given in the 1979 (18th printing) Edition into this workbook and alsoadded those problems that were given in the 1957 Edition but were not included in the 1979 Edition. Additionally,I have included in each of the 1979 Example problem solutions those solutions that were given in the 1957 Editionand that were resolved in a different manner. This enables the reader to see the difference in the technology of solving these problems and how the methods have increased the accuracy of the answer.
My reason for transcribing these Example problems into the Spreadsheet format is to achieve a rapid and efficient method by which an engineer can quickly detect the methodology and follow the mathematical computations. Emphasis is put on the logic and reasoning employed rather than worrying about the mathematical mechanics.The spreadsheet allows the reader to insert a variety of different input values and thereby facilitates the engineerin absorbing the manner in which the ultimate answer is affected.
Art Montemayor
Art Montemayor August 21, 2006Rev: 0
Page 2 of 287 FileName: document.xlsWorkSheet: Example 4-1
Example 4-1 (18th printing)
Given:of 50 gallons per minute.
Find: The Reynolds Number and the friction factor.
Solution: The Reynolds Number is defined as:
Where,Q = 50 gallons/min
62.22d = 2.067 inches
m = 0.85733 cP
Re = 88,830f = 0.0182 (from the Moody Chart, for smooth flow)
Water at 80 oF is flowing through 70 feet of 2-inch standard wall plastic pipe (smooth wall) at a rate
r = lb/ft3
Re=50. 6 Q ρ
d μ
Art Montemayor August 21, 2006Rev: 0
Page 3 of 287 FileName: document.xlsWorkSheet: Example 4-2
Example 4-2 (18th printing)
Given: A 6-inch Class 125 iron Y-pattern globe valve has a flow coefficient, Cv , of 600.
Find: Resistance coefficent, K, the the equivalent lengths L/D and L for flow in the zone of complete turbulence.
olution: K, L/D, and L should be given in terms of 6-inch Schedule 40 pipe; When the resistance coefficient K is used in flow equations, the velocity and internal diameter dimensions used in the equation must be based on the dimensions of the basis Schedule numbers regardless of the pipe with which the valve may be installed.
The values in the "K" Factor Table are associated with the internal diameter of the following pipe schedule numbers for the various ANSI Classes of valves and fittings.
Class 300 and lower Schedule 40Class 400 and 600 Schedule 80Class 900 Schedule 120Class 1500 Schedule 160Class 2500 (sizes 1/2 to 6") XXSClass 2500 (sizes 8" & up) Schedule 160
where,d = 6.065 inches = 0.5054 ft
K = 3.36 (based on 6", Schedule 40 pipe)
f = 0.015 (from Moody Chart; for 6.065" ID pipe in fully turbulent flow range)
L / D = 224
L = (L / D) (D) = 113 feet
In the 1957 6th printing edition, an alternate solution is offered:
L = 113 feet
K=894 . 01 d4
(CV )2
LD=
Kf
L=74 .3 d5
f CV2
Art Montemayor August 21, 2006Rev: 0
Page 4 of 287 FileName: document.xlsWorkSheet: Example 4-3
Example 4-3 (18th printing)
Given: A 4-inch Class 600 steel conventional angle valve with full area seat.
Find: Resistance coefficient K, flow coefficient CV, and equivalent lengths L / D and L for flow in the zone of complete turbulence.
Solution: K, L / D, and L should be given in terms of 4-inch Schedule 80 pipe; When the resistance coefficient K is used in flow equations, the velocity and internal diameter dimensions used in the equation must be based on the dimensions of the basis Schedule numbers regardless of the pipe with which the valve may be installed.
The values in the "K" Factor Table are associated with the internal diameter of the following pipe schedule numbers for the various ANSI Classes of valves and fittings.
Class 300 and lower Schedule 40Class 400 and 600 Schedule 80Class 900 Schedule 120Class 1500 Schedule 160Class 2500 (sizes 1/2 to 6") XXSClass 2500 (sizes 8" & up) Schedule 160
From the "K" Factor tables,
d = 3.826 inches
0.017 (from table on page A-26)
K = 2.55
274
L / D = 150
L = 47.8 feet
fT =
CV =
K=150 f T
CV=29 .9 d2
√K
K=fLD
; orLD=
Kf T
Art Montemayor August 21, 2006Rev: 0
Page 5 of 287 FileName: document.xlsWorkSheet: Example 4-3
In the 1957 6th Printing edition, the problem is worded and resolved differently:
Given: A 4-inch 600-pound conventional angle valve with no obstruction in flat seat.
Find: The resistance coefficient K, flow coefficient Cv, and the equivalent lengths L / D and L for fully turbulent flow.
Solution: K, L / D, and L should be given in terms of 4-inch Schedule 80 pipe;
L/D = 145 (from L/D tables for valves)
d = 3.826 inches
L = 46
K = 2.4 (from Nomograph for equivalent lengths and K)
Cv = 283
Art Montemayor August 21, 2006Rev: 0
Page 6 of 287 FileName: document.xlsWorkSheet: Example 4-4
Example 4-4 (18th printing)
Given: A 6 x 4-inch Class 600 steel gate valve with inlet and outlet ports conically tapered from back of body rings to valve ends. Face-to-face dimensions is 22" and back of seat ring to back of seat ring is about 6".
Find:
Solution: K, L/D, and L should be given in terms of 6-inch Schedule 80 pipe; When the resistance coefficient K is used in flow equations, the velocity and internal diameter dimensions used in the equation must be based on the dimensions of the basis Schedule numbers regardless of the pipe with which the valve may be installed.
The values in the "K" Factor Table are associated with the internal diameter of the following pipe schedule numbers for the various ANSI Classes of valves and fittings.
Class 300 and lower Schedule 40Class 400 and 600 Schedule 80Class 900 Schedule 120Class 1500 Schedule 160Class 2500 (sizes 1/2 to 6") XXSClass 2500 (sizes 8" & up) Schedule 160
(from K Factor Table)
3.826 inches
5.761 inches
0.015
0.66
0.1209
1.40
L / D = 93 diameters of 6" Schedule 80 pipe
L = 45 feet of 6" Schedule 80 pipe
K2 for any flow condition, and L / D and L for flow in the zone of complete turbulence.
K1 = 8 fT
d1 =
d2 =
fT =
b =
tan (q / 2) = = sin (q / 2) --approximately
K2 =
22"
5.761"
6"
3.826"
K2=K1+sin
θ2
[0 . 8 (1−β2)+2 . 6 (1−β2)2 ]β4
K=fLD
; orLD=
Kf T
β=d1
d2
Art Montemayor August 21, 2006Rev: 0
Page 7 of 287 FileName: document.xlsWorkSheet: Example 4-4
In the 1957 6th Printing edition, the problem is worded differently:
Given: A 6 x 4-inch 600-pound steel gate valve.
Find: The valve resistance coefficient K, and the equivalent lengths L / D and L, for fully turbulent flow of Reynolds numbers indicated on the Moody Friction Factor diagram.
Solution: K, L / D, and L should be given in terms of 6-inch Schedule 80 pipe;
L/D = 13 This value is for a 4" constant diameter port gate valve
66.8 This value is for 6" Schedule 80 pipe
0.0151 This is the fully turbulent friction factor for 6" Schedule 80 pipeas seen in the Moody chart
This is the definition of the resistance coefficient, K
K = 1.01 This is based on the 6" Schedule 80 pipe
L = (L/D) D = 32.1 feet This equivalent pipe length is based on the 6" Schedule 80 pipe
d4 = in4
d4 = in4
(L/D)a =
fT =
( LD )
a=( L
D )b (
d a
db)
4
=( LD )
b
(da )4
(db )4
K=f ( LD )
Art Montemayor August 21, 2006Rev: 0
Page 8 of 287 FileName: document.xlsWorkSheet: Example 4-5
Example 4-5 (18th printing)
Given:
Find: The proper size check valve and the pressure drop. The valve should be sized so that the disc is fully lifted at the specified flow.
Solution: For all practical purposes, it can be assumed that the pressure drop or head loss due to the flow of
The minimum velocity required to lift a check valve's disc to the full-open and stable position has been determined by tests for numerous types of check and foot valves, and is given in the "K" Factor Table. It is expressed in terms of a constant times the square root of the specific volume of the fluid being handled, making it applicable for use with any fluid.
(from K factor table)
(the "continuity" equation)
(From K Factor Table)
(From K Factor Table)
(Definition of the "Beta" ratio)
2.469 inches (for 2-1/2" Schedule 40 pipe)
3.068 inches (for 3" Schedule 40 pipe)
0.01605
62.305
0.018 (for fully turbulent flow in 2-1/2" or 3" pipe)Q = 80 gpm
5.1 ft/sec
v = 3.5 ft/sec (for the 3" check valve)
Note that the mean flow velocity of the 3" check valve is less than the recommended.Try a smaller, 2-1/2" check valve instead.
v = 5.4 ft/sec (for the 2-1/2" check valve)
Based on the above, a 2-1/2" check valve is recommended to be installed in the 3" pipeusing concentric reducers.
A globe type, lift check valve with a wing-guided disc is required in a 3-inch Schedule 40 horizontal
pipe carrying 70 oF water at the rate of 80 gallons per minute.
fluids in the turbulent range through valves and fittings varies as the square of the velocity (DP = k v2 ).
d1 =
d2 =
ft3/lb (the specific volume of water at 70 oF)
r = lb/ft3 (the density of water at 70 oF)
fT =
vmin =
vmin=40 √V̄
v=0. 408 Q
d2
ΔP=17 . 99 x 10−6 K ρ Q2
d4
K1=600 f T
K2=K1+β [0. 5 (1−β2)+(1−β2)2]
β 4
β=d1
d2
V̄ =
Art Montemayor August 21, 2006Rev: 0
Page 9 of 287 FileName: document.xlsWorkSheet: Example 4-5
0.80 (this is the Beta ratio between the check valve and the pipe)
26
2.0 psi (pressure drop of 80 gpm through the 2-1/2" check valve)
In the 1957 6th Printing edition, the problem is worded differently:
Given:
Find: The proper size check valve and the pressure drop. The valve should be sized so that the disc is fully lifted under normal flow conditions; see page 2-7 for discussion and page A-30 for minimum flow.
Solution:
Calculate the Reynolds Number to determine the friction factor based on flow in the 3-inch pipe.
Q = 100 gpm2.0 psi This is the minimum pressure drop required across the globe lift check valve in
order to provide sufficient flow to lift the disc fully. This is given in page A-30.L/D = 450 This is the same as for a globe valve; also given in page A-30.
62.305d = 3.068 inches This is the ID of the 3-inch Schedule 40 pipe.
0.975 cP
Re = 105,396 This places the flow in the Transition Zone within the Moody Chart.
f = 0.021 The friction factor taken from the Moody Chart
53.0
88.597 while thatfor a 2-1/2 inch size similar check valve is 37.161It is evident that for this flow condition the pressure drop through a wide-open valve is more than 2.0 psi for the
b =
K2 is the value of the resistance coefficient in terms of the larger pipe size and is determined by
basically dividing K1 by b4.
K2 =
DP =
A globe type, lift check valve with a wing-guided disc is required in a 3-inch Schedule 40 horizontal
pipe carrying 70 oF water at the rate of 100 gallons per minute.
Solve the Darcy equation for the value of d4 :
DP =
r = lb/ft3 This is the density of water at 70 oF.
m = This is the viscosity of the 70 oF water.
d4 =
From the pipe tables, it is seen that the value of the d4 parameter for the 3-inch pipe is
ΔP=0 . 00001799 f (L D) ρ Q2
d 4
d 4=0. 00001799 f (L D) ρ Q2
ΔP
Re=50. 6 Q ρ
d μ
Art Montemayor August 21, 2006Rev: 0
Page 10 of 287 FileName: document.xlsWorkSheet: Example 4-5
2-1/2 inch size and less than 2.0 psi for the 3-inch size. Therefore, the 3-inch valve would not be fully lifted and
450
88.597
37.161
1,073
2.85 psi
a 2-1/2 inch size check valve should be used in this flow application.
(L/D)b =
(da)4 =
(db)4 =
(L/D)a =
DP =
( LD )
a
=( LD )
b ( (da)4
(db)4 )
Art Montemayor August 21, 2006Rev: 0
Page 11 of 287 FileName: document.xlsWorkSheet: Example 4-6
Example 4-6 (18th printing)
Given:
Quantity Item200 feet 3" Schedule 40 pipe
6
1
entrance is flush with the inside of the tank
Find: The water velocity in the pipe and the rate of discharge in gallons per minute.
Solution:
K = 0.5 (for pipe entrance at the tank)
K = 1.0 (for pipe exit at the end)
0.018 (for fully turbulent flow in 3" pipe)
2.375 inches (bore of the reduced port ball valve)
3.068 inches (inside diameter of the 3" pipe)
For the K of a ball valve, the normal formula must be expanded in order to compensate for the different inlet and outlet angles:
0.77 (Beta ratio of the valve's bore to the pipe ID)
sin(16/2)= 0.14 (this sine value is for the valve's inlet half angle)
sin(30/2)= 0.26 (this sine value is for the valve's outlet half angle)
0.054
0.58 (This is the K for the ball valve)
K = 0.54 (This is the K for one screwed elbow)K = 3.24 (This is the K for the 6 elbows)
K = f (L./ D) = 14.08 (This is the K for the 200 feet of 3" straight pipe)
Water at 60 oF is discharged from a tank with 22-feet of average head to the atmosphere through:
3" standard 90o threaded elbows
3" flanged ball valve having a 2-3/8" diameter seat, 16o
conical inlet, and 30o conical outlet end. Sharp-edged
fT =
d1 =
d2 =
b =
Sin q/2 =
Sin q/2 =
K1 = 3 fT =
K2 =
30 fT =
22'
hL=Kv2
2 gor , v=√ 2 g hL
K
v=0 .408Q
d2or , Q=2. 451 v d2
K2=K1+0 . 8 sin
θ2
(1−β2 )+2 . 6 sinθ2
(1−β2 )2
β4
Art Montemayor August 21, 2006Rev: 0
Page 12 of 287 FileName: document.xlsWorkSheet: Example 4-6
The K for the system is composed of the Ks for the entrance, pipe, elbows, ball valve, and exit.
K (total) = 19.4
v = 8.5 ft/sec (This is the mean velocity of the water in the 3" pipe)
Q = 197 gpm (This is the water flowrate exiting the system)
In order to verify that the assumption that the flow is in the fully turbulent zone,
Reynolds Number = 180,742 (Reynolds number in the 3" pipe)
From the Moody Chart, the corresponding f = 0.0195 and this is not in the fully turbulent zone.Although this flow is in the transition zone, the difference is small enough to forego any correctionof K for the pipe.
The solution is valid from a practical point.
123.9 d v r / m =
Art Montemayor August 21, 2006Rev: 0
Page 13 of 287 FileName: document.xlsWorkSheet: Example 4-7
Example 4-7 (18th printing)
Given:head.
Find: The oil velocity in the pipe and the rate of flow in gallons per minute.
Solution:
(Friction factor for laminar flow)
K = 0.5 (for pipe entrance at the tank)
K = 1.0 (for pipe exit at the end)
0.58 (This is the K for the ball valve as developed in Example 4-6)
K = 3.24 (This is the K for the 6 elbows as developed in Example 4-6)
54.64 lb/ft3
100 cP
22 ft of oil (static pressure head developed by oil)
Note that the resistance coefficient K is considered as being independent of the Friction Factor or the ReynoldsNumber, and is treated as a constant for any valve or fitting under all condtions of flow (including laminar flow) regardless of the fluid handled.It is left to calculate the K for the 200 feet of 3" pipe and a velocity must be assumed in order to generate the pipe K. The velocity can be checked through trial-and-error methods until convergence is reached.
Assume v = 5.26 ft/sec
Re = 1,093 (Flow is laminar)
f = 0.059 (Friction factor for the 3" pipe)
K = 45.8 (K for 3" pipe)
K (total) = 51.1 (K for the entire system)
v = 5.26 ft/sec (Oil mean velocity in 3" pipe -- this should equal the assumed value)
Q = 121 gpm (Oil flowrate in the 3" pipe)
S.A.E. 10 Lube Oil at 60 oF flows through the system described in Example 4-6 at the same differential
K2 =
r = (oil density @ 60 oF)
m = (oil absolute viscosity @ 60 oF)
hL =
22'
hL=Kv2
2 gor , v=√ 2 g hL
K
v=0 .408Q
d2or , Q=2 . 451 v d2
Re=123. 9d v ρ
μ
f=64Re
Art Montemayor August 21, 2006Rev: 0
Page 14 of 287 FileName: document.xlsWorkSheet: Example 4-8
Example 4-8 (18th printing)
Given:Schedule 40 pipe in which an 9-inch conventional globe valve with full bore is installed.
Find: The pressure drop due to flow through the pipe and valve.
Solution:
S = 0.916
S = 0.900
B = 600 barrels/hr
d = 7.981 inches (8" Schedule 40 pipe inside diameter)
470 cP
0.014 (Friction factor for fully turbulent flow in 8" pipe)
56.13 lb/ft3
Re = 318 (Oil flow is very Laminar)
f = 0.20 (Friction factor for the laminar flow in the 8" pipe)
4.76 (K factor for the full-bore 8" globe valve)
60.55 (K factor for the 200 feet of 8" pipe)
K (Total) = 65.31 (Total K for the system)
2.87 psi
S.A.E. 70 Lube Oil at 100 oF is flowing at the rate of 600 barrels per hour through 200 feet of 8-inch
at 60 oF (Specific Gravity of the oil relative to water at 60 oF)
at 100 oF (Specific Gravity of the oil relative to water at 60 oF)
m = (Oil viscosity at 100 oF)
fT =
r = (Oil density at 100 oF)
K1 =
K =
DP =
ΔP=8 . 82 x 10−6 K ρ B2
d4
Re=35. 4 ρ B
d μ
K1=340 f T
K=fLD
f=64Re
Art Montemayor August 21, 2006Rev: 0
Page 15 of 287 FileName: document.xlsWorkSheet: Example 4-9
Example 4-9 (18th printing)
Given:minute, as shown in the following sketch.
Find:
Solution:
(This is the pressure loss in the system due to flow)
(This is the pressure loss in the system due to elevation change)
d = 5.047 inches
S = 0.916
S = 0.900
470 cP
56.1 lb/ft3
0.016 (Friction factor for oil in fully turbulent flow)
Q = 600 gpm (Oil flow rate)
First, establish if the flow is laminar or turbulent.
Re = 723 (This flow is Laminar)
f = 64/Re = 0.089 (This is the fricition factor for the oil in laminar flow)
S.A.E. 20 Lube Oil at 100 oF is flowing through 5-inch Schedule 40 pipe at a rate of 600 gallons per
The velocity in feet per second and the pressure difference between gauges P1 and P2.
(Oil Specific Gravity @ 60 oF referenced to water @ 60 oF)
(Oil Specific Gravity @ 100 oF referenced to water @ 60 oF)
m = (Oil absolute viscosity @ 100 oF)
r = (Oil density @ 100 oF)
fT =
P2
P1
Flow
75' 50'
175'
5" Class 150 steel gate valve with full area seat - wide open.
5" Class 150, steel angle valve with full area seat - wide open
5" R
5" weld elbow
v=0. 408 Q
d2
Re=50. 6 Q ρ
d μ
ΔP=18 x 10−6 K ρ Q2
d4
ΔP=hL ρ
144
Art Montemayor August 21, 2006Rev: 0
Page 16 of 287 FileName: document.xlsWorkSheet: Example 4-9
This K is for the 5" gate valve and = 0.13
This K is for the 5" angle valve and = 2.40
This K is for the 5" elbow and is = 0.32
This K is for the 5" pipe and is = 63.17
The K for the entire system is the sum of the Ks = 66.01
v = 9.6 This is the oil mean velocity in the pipe.
37.0 psi (This is Pressure drop due to flow)
19.5 psi (This is Pressure drop due to elevation increase)
56.5 psi This is the total pressure drop between the gauges
DP =
DP =
DP (Total) =
K1=8 f T
K1=150 f T
K=20 f T
K=fLD
Art Montemayor August 21, 2006Rev: 0
Page 17 of 287 FileName: document.xlsWorkSheet: Example 4-10
Example 4-10 (18th printing)
Given: 600 psig steam at 850 oF flows through 400 feet of horizontal 6-inch Schedule 80 pipe at a rate of 90,000 lb/hr.
Class 600 venturi gate valve as described in example 4-4, and one 6-inch Class 600 y-pattern globe .valve. The latter has a seat diameter equal to 0.9 of the inside diameter of Schedule 80 pipe, disc fully lifted
Find: The pressure drop through the system.
Solution:
0.90 (This is the given Beta ratio for the 6" globe valve)
(This is the K for each 6" welded elbow)
(This is the K for the 6" pipe)
W = 90,000 lb/hrd = 5.761 inches (This is the 6" Schedule 80 pipe ID)
1.22 ft3/lb (This is the Specific Volume of the superheated steam)0.026727 cP (This is the viscosity of the superheated steam)
0.015 (This the friction factor for the 6" pipe in fully turbulent flow)
0.825 (for Globe valve)
1.41 (for Globe valve)
1.40 (for Gate valve, as calculated in example 4-4)
Re = 3,688,279
f = 0.015 (The pipe friction factor as found in the Moody Chart)
K = 12.5 (for the 6" Schedule 80 pipe)
K = 0.63 (for the 3 - 6" welded elbows)
K (Total) = 15.94 (The total K for globe & gate valves, pipe, and elbows)
39.9 psi This is the pressure drop for the system
The system contains three 90o weld elbows having a relative radius of 1.5, one fully-open 6x4-inch
(This is the K2 for the 6" globe valve)
(This is the K1 for the 6" globe valve)
b =
m =
fT =
K1 =
K2 =
K2 =
DP =
ΔP=28 x 10−8 K W 2 V̄d4
K2=K1+β [0. 5 (1−β2)+(1−β2)2]
β 4
K1=55 f T
K=14 f T
K=fLD
Re=6 .13W
d μ
V̄ =
This exercise compares the results and answers found in Crane Tech Paper #410 with those generated at:
http://www.engineeringpage.com/cgi-bin/dp/dpf.pl
Crane Tech Paper #410; example 4-10
CALCULATION INPUT
FLUID DATA PIPE DATA
Medium 600 psia Steam (@ 850 oF) ANSI B36.10
Flowrate 109,440 ft3/h Size
Density 0.8223 lb/ft3 Schedule
Dynamic Viscosity 0.027 mPa.s [ = cP] Pipe length
FITTINGS AND VALVES
Number of fittings were manually put in
Elbows 90 Number R/D = Type Connection
3 R/D = 1.5 Long Radius All Types
Valves Number Type Special
1 Gate, Ball, Plug Full line size ß = 1
1 Globe Angle or Y-type
CALCULATION RESULTSExternal Diameter 168.28 mm 6.6252 inch
Wall Thickness 10.97 mm 0.4319 inch
Internal Diameter 146.34 mm 5.7614 inch
Flowrate 0.8608 m3/s 13,644.46 gpm (U.S.)
3098.9957 m3/h 109,440 ft3/h
Density 13.172 kg/m3 0.8223 lb/ft3
Fluid velocity 51.1803 m/s 167.9143 ft/s
Reynolds Number 3,653,869 [ - ] turbulent flow
Wall Roughness 0.05 mm 0.002 inch
Relative roughness factor 0.0003 [ - ]
Moody Friction Factor 0.0156 [ - ]
Pressure Drop straight line 737,060 Pa 7.3706 bar 106.90 psi
Pressure Drop fittings 54,681 Pa 0.5468 bar 7.93 psi
Total Pressure Drop 791,741 Pa 7.9174 bar 114.83 psi
Art Montemayor August 21, 2006Rev: 0
Page 19 of 287 FileName: document.xlsWorkSheet: Example 4-11
Example 4-11 (18th printing)
Given:
Find: The pressure drop from Point A to Point B.
Solution:
(This is the radius of the pipe bends
60.569
0.34 cPd = 1.049 inches (1" Schedule 40 pipe ID)
0.023 (Friction factor for 1" pipe in fully turbulent flow)Q = 15 gpm (Hot water flow rate)
Re = 127,131 This reveals that the water is in the transition zonef = 0.024 This is the pipe friction factor at the flowing conditions
K = 4.94 This is the K for the 18 feet of straight 1" pipe
K = 0.64
3.89K(Total) = 9.47 This is the total K for the entire system
1.92 psi This is the total system pressure drop
Art's Note:For practical purposes, it is very important to know the orientation, although the pressure drop will
This air will create a 2-phase flow region at these sites and the pressure drop will increase. It is
initially vented to the atmosphere and all air be expelled from there in order to ensure that the system is started up with 100% water-filled condtions. It is in this manner that the system can "recover" the energy spent in going vertically up 2 feet four times in the coil run. If the system is not 100% water-filled, then total recovery of this energy cannot be done because of the air's
Water at 180 oF is flowing through a flat heating coil, shown in the sketch below, at a rate of 15 gpm.
(This is the K for each of the two 90o bends
(K for each 180o bend)
r = lb/ft3 (Water density at 180 oF)
m = (Water viscosity at 180 oF)
fT =
This is the K for the 2 -90o bends
KB = This is the K for the 7-180o bends
DP =
This example problem doesn't mention whether the flat coil is in a horizontal or a vertical orientation.
be the same --- as long as the entire 1" pipe is 100% water filled.
If the coil is vertically oriented, then air will initially be trapped at the top of the 4 - 180o returns.
very important in an actual, industrial application that the top of each of the 4 top 180o returns be
A B
1" Schedule 40 pipe
1' 1'
2'
4" Radius
4" Radius
4" Radius
ΔP=18 x 10−6 K ρ Q2
d4
Re=50. 6 Q ρ
d μ
K=fLD
rd=4
K90=14 f T
KB= (n−1 ) (0.25 π f Trd+0 .5 K90)+K90
Art Montemayor August 21, 2006Rev: 0
Page 20 of 287 FileName: document.xlsWorkSheet: Example 4-11
compressibility.
This exercise compares the results and answers found in Crane Tech Paper #410 with those generated at:
http://www.engineeringpage.com/cgi-bin/dp/dpf.pl
Crane Tech Paper #410; example 4-11
CALCULATION INPUTFLUID DATA PIPE DATA
Medium Water at 180 oF ANSI B36.10
Flowrate 15 GPM Size 1 inch
Density 60.57 lb/ft3 Schedule Sch 40
Dynamic Viscosity 0.34 mPa.s [= cP] Pipe length 18 ft
FITTINGS AND VALVES
Number of fittings were manually put in
Elbows 90 Number R/D = ... Type Connection
2 R/D = 5 Pipe Bend
Return 180 Number R/D = ... Type Connection
7 R/D = 1.5 Long Radius All Types
CALCULATION RESULTSExternal Diameter 33.4 mm 1.315 inch
Wall Thickness 3.38 mm 0.1331 inch
Internal Diameter 26.64 mm 1.0488 inch
Flowrate 0.0009 m3/s 15 gpm (U.S.)
3.4069 m3/h 120.3125 ft3/h
Density 970.2383 kg/m3 60.57 lb/ft3
Fluid velocity 1.6978 m/s 5.5703 ft/s
Reynolds Number 129,071 turbulent flow
Wall Roughness 0.05 mm 0.002 inch
Relative roughness factor 0.0019
Moody Friction Factor 0.0247
Pressure Drop straight line 23,318 Pa 0.2332 bar 3.382 psi
Pressure Drop fittings 6,923 Pa 0.0692 bar 1.004 psi
Total Pressure Drop 30,241 Pa 0.3024 bar 4.386 psi
Art Montemayor August 21, 2006Rev: 0
Page 22 of 287 FileName: document.xlsWorkSheet: Example 4-12
Example 4-12 (18th printing)
Given: A 12-inch Schedule 40 steel pipe 60 feet long, containing a standard gate valve 10 feet from the
the reservoir and its center line is 12 feet below the water level in the reservoir.
Find: The diameter of a thin-plate orifice that must be centrally installed in the pipe to restrict the velocity of flow to 10 feet per second when the gate valve is wide open.
Solution:
K = 0.78 (K for pipe entrance)
K = 1.00 (K for pipe exit)
(K for the 12" gate valve)
(K for the 12" pipe)
v = 10 ft/sec
d = 11.938 inches
62.371
1.12 cP
0.013 (This the friction factor for the 12" pipe in fully turbulent flow)
12.0 ft (This is the system head; assume the reservoir's level to remain constant)
Re = 822,889 flow is in the transition zone
f = 0.014 The pipe's friction factor, using the Moody Chart
K = 7.73 This is the system's K required for the desired velocity of 10 ft/sec.
0.104 This is gate valve's K
K = 0.84 This is the 12" pipe's K
K(Total) = 2.728 + the orifice's K
K(orifice)= 5.00 (This is the K of the orifice that satisfies the system's needs)
From the graph on page A-20 showing C versus Reynolds Number at varying Beta values, an assumed Betayields a C with which the K(orifice) can be calculated. Trial-and-Error method is used as follows:
At the Reynolds Number of 822,889, different values of Beta ratio are assumed.
entrance, discharges 60 oF water to atmosphere from a reservoir. The entrance projects inward into
K1 = 8 fT
(this is the relationship between an orifice K and its b ratio)
r = lb/ft3
m =
fT =
hL =
K1 =
RO
12' 60'
30'10'
hL=Kv2
2 gor , System K=
2 g hL
v2
Re=123. 9 d v ρ
μ
K=fLD
KOrifice≃1−β2
C2 β 4
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Page 23 of 287 FileName: document.xlsWorkSheet: Example 4-12
C value K0.70 0.70 4.330.65 0.67 7.210.67 0.682 5.88
8.12 inches (This is the bore diameter of the required orifice)
Assumed b valueThe assumed b is too large; use a smaller oneThe assumed b is too small; use an intermediateThis is close enough; use b = 0.68
d1 =
Art Montemayor August 21, 2006Rev: 0
Page 24 of 287 FileName: document.xlsWorkSheet: Example 4-13
Example 4-13 (18th printing)
Given: Fuel oil with a density of 0.815 grams per cubic centimeter and a kinematic viscosity of 2.7 centistokes is flowing through 50 millimeter I.D. steel pipe, 30 meters long at a rate of 7.0 liters per second.
Find:
Solution: Define the symbols in SI units as follows:
A = 0.001963D = Pipe internal diameter, in meters = 0.0500
g = Acceleration of gravity = 9.80665
head loss, in meters of fluidL = Pipe length, in meters = 30
q = 0.007v = mean fluid velocity = 3.565 meter/sec (mean velocity by the continuity equation)
0.815
fluid pressure drop, in barsfluid pressure drop, in megapascals
1.0 meter = 3.28 feet = 39.37 inches
1.0 0.98067 bar = 14.22334 psi
1.0 0.098067 Mpascal = 14.22334 psi
A column of fluid one square centimeter in cross-sectional area and one meter high is equal to a pressure of
Re = where, d = 1.9685 inches
Re = 65,986 v = 11.6934 ft/sec2.7 centistokes (Kinematic viscosity)
f = 0.023 (This is the friction factor from the Moody Chart)
= 8.94 meters (This is the head loss through the 50 mm pipe)
0.729
0.715 bar
0.071 MegaPascal
Head loss in meters of fluid and pressure drop in kg/cm2, bar, and megapascal (MPa)
Pipe cross-sectional flow area, in meters2 =
meters/sec2
hL =
flow rate, in meters3/sec =
r = fluid density, in grams/centimeter3 =
DP(kpc) = fluid pressure drop, in kilograms/centimeter2
DP(bar) =DP(MPa)=
kg/cm2 =
kg/cm3 =
0.1 r kg/cm2; therefore:
7740 d v/ n
n =
DP(kpc) = kg/cm2
DP(bar) =
DP(MPa) =
hL=fLD
v2
2 g
Art Montemayor August 21, 2006Rev: 0
Page 25 of 287 FileName: document.xlsWorkSheet: Example 4-14
Example 4-14 (18th printing)
Given:
Find: The velocity in both the 4 and 5-inch pipe sizes and the pressure differential between the pressure
Solution:
Use Daniel Bernoulli's theorem, which states:
(This is the K for the 4" pipe, in terms of the 5" pipe)
This is the K for the 4" x 5" expanding elbow & is the sum of a straight-sized elbow + a sudden enlargement
62.371
1.121 cP
4.026 inches
Water at 60 oF is flowing through the piping system, shown in the sketch below, at a rate of 400 gpm.
gauges P1 and P2.
Assume that the hydraulic head of 75 feet has negligible effect on the water's density; therefore, r1 = r2 :
(This is the K for the 5" 90o elbow
r = lb/ft3
m =
d1 =
P1
P2
Flow
110'
150'75'
4" Schedule 40 pipe
5" Schedule 40 pipe
5" Schedule 40 pipe
5"x4" Reducing weld elbowwith 7-1/2" radius
5" weld elbow with 7-1/2" radius
Z1+144 P1
ρ1
+v1
2
2 g=Z2+
144 P2
ρ2
+v2
2
2 g+hL
(P1−P2)=( ρ144 ) [(Z2−Z1)+
(v22−v1
2 )2 g
+hL]hL=
0 . 00259 K Q2
d4
Re=50. 6 Q ρ
dμ
K=fLD
K=f L
D β4
K=14 f T
K=14 f T+(1−β2)
β4
2
Art Montemayor August 21, 2006Rev: 0
Page 26 of 287 FileName: document.xlsWorkSheet: Example 4-14
5.047 inches
Q = 400 gpm
0.016
0.80
75 feet
10.08 ft/sec (This is the mean velocity in the 4" pipe by the continuity equation)
6.41 ft/sec (This is the mean velocity in the 5" pipe by the continuity equation)
-0.94 feet
Re = 279,689 (This is the Reynolds Number for the 4" pipe; it falls within the Transition Zone)
Re = 223,108 (This is the Reynolds Number for the 5" pipe; it falls within the Transition Zone)
f = 0.018 (This is the friction factor for both 4" and 5" pipe)
K = 9.6 (This is the K for the 225 feet of 5" pipe)
K = 5.9 (This is the K for the 110 feet of 4" pipe)
14.6 (This is the K for the 4" pipe in terms of the 5" pipe)
K = 0.22 (This is the K for the 5" common elbow)
K = 0.55 (This is the K for the 4" x 5" expander elbow)
K (Total) = 24.98 This is the total K for the entire system of pipe & elbows
16.0 feet of fluid
39.0 psi
d2 =
fT =
b =
(Z1 - Z2) =
v1 =
v2 =
(v22 - v2
1)/2g =
K2 =
hL =
DP = (P1 - P2) =
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Page 27 of 287 FileName: document.xlsWorkSheet: Example 4-15
Example 4-15 (18th printing)
Given:
Find: The total discharge head (H) at flowing conditions and the brake horsepower (bhp) required for a pump
Solution:
Use Daniel Bernoulli's theorem, which states:
The head loss through the 3" discharge pipe system due to flow
The Reynolds Number through the 3" pipe system
The mean water velocity through the 3" pipe system
The Brake Horsepower required to pump the water throughthe 3" pipe system and up to the 400 feet of elevation.
K =
K1 = The K for the gate valve
K = f (L/D) The K for the 3" pipe
Water at 70 oF is pumped through the piping system shown in the sketch at the rate of 100 gpm.
having an efficiency (eP) of 70 per cent.
Assume that the hydraulic head of 400 feet has negligible effect on the water's density; therefore, r1 = r2.
Since the mean velocity at the pump discharge is the same as at the outlet (v1 = v2), the equation is:
30 fT The K for 90o elbows
8 fT
30'
70'100'
3" Std. Gate Valve3" Schedule 40 pipe
Four 3" Std. 90o threadedelbows
300'
2-1/2" Globe lift check valve with wing-guided disc & concentric reducers for 3" pipe.
Z1+144 P1
ρ1
+v1
2
2 g=Z2+
144 P2
ρ2
+v2
2
2 g+hL
144ρ
(P1−P2 )=(Z2−Z1)+hL
hL=0 . 00259 K Q2
d4
Re=123. 9d v ρ
μ
v=0. 408 Q
d2
bhp=Q H ρ
247 , 000 e P
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Page 28 of 287 FileName: document.xlsWorkSheet: Example 4-15
K = 1.00 This is the K for the 3" pipe fluid exit lossd = 3.068 inches This is the ID of the 3" Schedule 40 pipe
62.305
0.97 cP
0.018 This is the friction factor for water in fully turbulent flow in the 3" pipeQ = 100 gpm
v = 4.33 ft/sec
Re = 105,294 This Reynolds Number identifies the flow as in the Transition zone.
f = 0.021 This is the 3" pipe's friction factor from the Moody Chart
K = 2.16
0.14 This is the K for the gate valve
K = 26.33 This is the K for the Check Valve with reducers as per Example 4-5
K = 41.07 This is the K for the 500 feet of 3" Schedule 40 pipe
K (Total) = 70.70 This is the Total K for the system, including the exit K
21 ft of water This is the head loss for the 3" pipe system due to flow
H = 421 ft of water
bhp = 15.2 This is the Pump's Brake Horsepower, assuming 70% efficiency.
r = lb/ft3 This is the density of the 70 oF water
m = This is the absolute viscosity of 70 oF water
fT =
This is the K for the four 90o elbows
K1 =
hL =
This is the head loss for the 3" pipe system including static head
Art Montemayor September 30, 2006Rev: 0
Page 29 of 287 FileName: document.xlsWorkSheet: Example 4-16
Example 4-16 (18th printing)
Given:standard cubic feet per minute (scfm).
Find: The pressure drop in psi and the velocity in ft/min at both upstream and downstream gauges.
Solution:Refer to the Table for the Flow of Air through Schedule 40 steel pipe.
2.21 psiof 1", Schedule 40 pipe.
2.61 psi
100 scfm The air flow measured at Std. conditions
20.2 acfm
20.9 acfm
V = 3,369 ft/min Mean air velocity at upstream conditions
V = 3,483 ft/min Mean air velocity at downstream conditions
Art's Note:1.) Standard Conditions were not defined for this problem. This should be mandatory.2.) The 65 psig pressure is not defined as to location; it was assumed that this is the initial pressure.3.) Note that it is not mentioned that it is assumed that the temperature stays constant (Isothermal).4.) Since the pressure drop is less than 10% of the initial pressure, the conventional Darcy equation
could have been used to calculate the pressure drop:
where,f = the friction factor for flow in the 1" pipe
L = the 100 feet of 1" pipe
V = the mean air velocity at the initial conditionsd = the internal diameter of the 1" pipe.
Note that the Reynolds Number would also have to be calculated to obtain f; this means that the viscosity of the air at the initial conditions would also have to be known.
The density of air at standard conditions would have to be known in order to find the mass of air flowing. This mass air flow is constant, so the volumetric flow can be calculated using the densityvalues at the initial as well as at the final points.
The air tables have already taken the air density into consideration and this is the reason they are useful for calculating this type of problem.
Air at 65 psig and 110 oF is flowing through 75 feet of 1-inch Schedule 40 pipe at a rate of 100
(Standard Conditions are defined here as: 60 oF and 14.7 psia)
DP = for 100 psi, 60 oF air at a flow rate of 100 scfm through 100 ft
DP = This is the DP corrected for length, pressure, & temperature.
q'm =
qm(1) = The air flow measured at 65 psig & 110 oF (Upstream)
qm(2) = The air flow measured at (65 - 2.61) psig & 110 oF (Downstream)
r = the density of the air, lb/ft3, at the initial conditions
ΔP=(0 . 000000359 ) f L ρ V 2
dpsi
Art Montemayor September 30, 2006Rev: 0
Page 30 of 287 FileName: document.xlsWorkSheet: Example 4-17
Example 4-17 (18th printing)
Given:pumped through a 12-inch Schedule 30 steel pipe at a rate of 1,900 barrels per hour. The pipe line is 50 miles long with discharge at an elevation of 2,000 feet above the pump inlet. Assume the pump has an efficiency of 67%.
Find: The brake horsepower of the pump.
Solution:
t = 15.6 60.1 Temperature of the crude oil; it remains constantL = 50 miles = 264,050 ft
54.65 lb/ft3 This is the degrees API converted to density0.8762 This is the density converted to Specific Gravity
B = 1,900 bbls/hr = 1,330 gpmd = 12.090 inches
14.2 cP This is the kinematic viscosity converted to absolute viscosityh = 2,000 feet This is the height that the crude is elevated by the pump
Re = 21,442 This Reynolds Number identifies the flow as in the Transition Zone.
f = 0.025 This is the friction factor as read from the Moody Chart
533 psi This is the oil's pressure drop over 50 miles of pipeline
1,406 ft This is the pressure drop converted to head loss
H = 3,406 ft This is the total head developed by the pump to pump the oil for 50 miles and raise it to an elevation of 2,000 feet.
The pump's Brake Horsepower = 1,496
Crude oil with 30 degrees API at 15.6 oC with a viscosity of 75 Universal Saybolt seconds is being
oC = oF
r =S =
m =
DP =
hL =
ΔP=(0 . 0001058 ) f L ρ B2
d5
Re=35. 4B ρd μ
hL=144 Δ P
ρ
brake horsepower=Q H ρ
247 , 000 e P
Art Montemayor September 30, 2006Rev: 0
Page 31 of 287 FileName: document.xlsWorkSheet: Example 4-18
Example 4-18 (18th printing)
Given: A 14-inch, Schedule 20, natural gas pipe line is 100 miles long. The inlet pressure is 1,300 psia, the
Find: The flowrate in millions of standard cubic feet per day (MMscfd).
Solution: (Art's Note: First and foremost, note that the basis of the gas composition is totally lacking; this is a
Three solutions to this example are presented for the purpose of illustrating the variations in results obtained by use of:
1. the Simplified Compressible Flow forumula;2. the Weymouth formula; and,3. the Panhandle formula.
Simplified Compressible Flow forumula
d = 13.376 inches (the pipeline internal diameter)0.011 cP (absolute viscosity estimated from gas graphs)
f = 0.0128 (assumed friction factor for fully turbulent flow; from the Moody Chart)
T = 500 (the absolute gas temperature)
1,300 psia (initial pipeline pressure)
300 psia (final pipeline pressure)
100 miles (pipeline length)
Component mole %
16 75 12.00
30 21 6.30
44 4 1.76Mixture Molecular Weight = 20.06
The Specific Gravity of gases is defined as the ratio of the molecular weight of the gas to that of air.
MW(air) = 28.964
0.693
4,489,682 scfh = 107.8 MMscfd
outlet pressure is 300 psia, and the average temperature is 40 oF. The gas analysis is 75% CH4,
21% C2H6, and 4% C3H8.
gross error on the part of any engineer. Without a basis, this problem can't be solved. A mole % basis is therefore assumed. This is the same as a volumetric basis.)
m =
oR
P'1 =
P'2 =
Lm =
molecular weight
MW x mole frac.
CH4
C2H6
C3H8
Sg =
q'h =
qh' =114. 2 √[ (P1
' )2−(P2' )2
f Lm T Sg] d5
Re=0 .482 qh
' Sg
d μ
Art Montemayor September 30, 2006Rev: 0
Page 32 of 287 FileName: document.xlsWorkSheet: Example 4-18
Re = 10,186,290 This Reynolds Number identifies the flow as fully turbulent
f = 0.0128 This friction factor is equal to that of the assumed
Since the assumed friction factor is correct, the correct flow rate is that calculated. Note that if the assumedfriction factor were found to be incorrect, it would have necessary to repeat the exercise with a new assumptionuntil the assumed friction factor was in reasonable agreement with that based upon ithe calculated ReynoldsNumber.
Weymouth formula
4,379,360 scfh = 105.1 MMscfd
Panhandle formula
E = 0.92 Assume a flow efficiency for average operation conditions
5,575,520 scfh = 133.8 MMscfd
q'h =
q'h =
qh' =28 . 0 d2. 667 √[ (P1
' )2−( P2' )2
Sg Lm] (520
T )
qh' =36 . 8 E d2 .6182 [ ( P1
' )2−(P2' )2
Lm]0. 5394
Art Montemayor September 30, 2006Rev: 0
Page 33 of 287 FileName: document.xlsWorkSheet: Example 4-19
Example 4-19 (18th printing)
Given:constant head of 11.5 feet.
Find: The flow rate in gallons per minute.
Solution:
K = 0.5 K for entrance loss
K = K for the Mitre elbow
K for the gate valveK = f (L/D) K for straight pipe runs
This is the K for the sudden contraction between the 3" and 2" pipes
This is K for the 2" pipe in terms of the 3" pipe
K = This is the K for the exit from the 2" pipe in terms of the 3" pipe
11.5 ft of water the head above the water outlet
3.068 inches the large pipe inside diameter
2.067 inches the smaller pipe inside diameter
1.1211 cP62.371 lb/ft3 the density of water at 60 oF
0.018 the friction factor for fully turbulent flow in the 3" pipe
0.019 the friction factor for fully turbulent flow in the 2" pipe0.67 this is the Beta ratio between the two pipes
K = 0.50 This K is for the water entering from the tank into the outletK = 1.08 This K is for the 3" mitered elbow
0.14 This K is for the 3" gate valveK = 0.70 This K is for the 10' of 3" pipeK = 10.71 This K is for the 20' of 2" pipe in terms of the 3" pipeK = 4.85 This K is for the 2" exit in terms of the 3" pipe
1.33 This K is for the sudden contraction from 3" to 2"K (Total) = 19.31 This is the Total K for the system
Q = 143 gpm This solution has assumed the pipe flow is in the fully turbulent zone
Water at 60 oF is flowing from a reservoir through the piping system below. The reservoir has a
60 fT
K1 = 8 fT
1/b4
hL =
d1 =
d2 =
m = the absolute viscosity of water at 60 oFr =
fT =
fT =b =
K1 =
K2 =
and has used the corresponding fT to calculate the individual pipe Ks.
10'd1 20'
11.5'
Water @ 60o F 3" Sch 40 Pipe
3" Std. gate valve; wide open
2" Sch 40 Pipe
Q=19.65 d2 √ hL
K
Re=50. 6 Q ρ
d μ
K2=0 .5 (1−β2)
β4
K=f L
D β4
Art Montemayor September 30, 2006Rev: 0
Page 34 of 287 FileName: document.xlsWorkSheet: Example 4-19
In order to verify that the pipe flow is verily in the fully turbulent zone, the Reynolds Numbers must be calculated
Re = 130,953 This is the Reynolds Number for the 3" pipef = 0.020 This is the friction factor for flow in the 3" pipe
Re = 194,371 This is the Reynolds Number for the 2" pipef = 0.021 This is the friction factor for flow in the 2" pipe
The friction factors used for the straight pipe are not in agreement with the fully turbulent flow values used inobtaining the approximate flow rate. Therefore, the K factors for the two pipes should be corrected accordingly:
K = 0.78 This corrected K is for the 10' of 3" pipe
K = 11.83 This corrected K is for the 20' of 2" pipe in terms of the 3" pipe
K (Total) = 20.52 This is the corrected Total K for the system
Q = 138 gpm This the correct solution.
and the corresponding friction factors found and compared with the fT used.
Art Montemayor September 30, 2006Rev: 0
Page 35 of 287 FileName: document.xlsWorkSheet: Example 4-20
Example 4-20 (18th printing)
Given: A header with 170 psia saturated steam is feeding a pulp stock digester through 30 feet of 2-inchSchedule 40 pipe which includes one standard 90 degree elbow and a fully-open conventional plug type disc globe valve. The initial pressure in the digester is atmospheric.
Find: The initial flow rate in pounds per hour, using both the modified Darcy formula and the sonic velocity and continuity equations.
Solution:Using the Modified Darcy Formula
K = 0.5 K for Entrance from header
K =
K for 2" globe valveK = 1.0 K for exit from 2" pipe into digester
k = Cp/Cv = 1.297 Ratio of specific heats for saturated steam vapor; from tabular datad = 2.067 inches Internal diameter of 2" pipeL = 30 ft Straight length of 2" pipe
2.6748 Specific volume of saturated steam at 170 psia
0.019 The friction factor for fully turbulent flow in the 2" pipeg = 32.20 ft/sec2 The acceleration of gravity
K = 0.50 This is the K for the header steam entering the 2" pipeK = 0.57 This is the K for the 2", 90o elbow
6.46 This is the K for the 2" globe valve, fully openedK = 1.00 This is the K for the steam exiting the 2" pipe and entering the digesterK = 3.31 This is the K for the 2" pipe (assumed @ fully turbulent flow)
K (Total) = 11.84 This is the total K for the system (assuming fully turbulent flow)
0.914 This is the ratio of the overall pressure drop to the initial pressure
133.5 psi This is the pressure drop that is fixed by the steam sonic flow
Y = 0.710 Y net expansion factor for adiabatic flow (isenthalpic expansion)and used to compensate for the changes in fluid properties due tofluid expansion.
W = 11,776 lb/hr This is the steam flow rate occurring at sonic conditions and is
30 fT K for 2", 90o Elbow
K1 = 340 fT
ft3/lb
fT =
K1 =
DP/P1 =
From tabular data, it is seen that the maximum DP/P1 for sonic flow to exist is 0.785 when the system K = 11.84.
This means that, since this value is less than the calculated DP/P1, sonic ("choked") flow is taking placeat the end of the 2" pipe as it enters the digester. Therefore, the pressure drop for these flow conditions is:
DP =
20'
Spherical Digester @ 14.7 psia
10'
W=1891 Y d2 √ ΔPK V̄
K=fLD
V̄ =
Art Montemayor September 30, 2006Rev: 0
Page 36 of 287 FileName: document.xlsWorkSheet: Example 4-20
conditions.the maximum mass flow rate that can occur under these pressure
Art Montemayor September 30, 2006Rev: 0
Page 37 of 287 FileName: document.xlsWorkSheet: Example 4-20
Using Sonic Velocity and Continuity Equations
This is the sonic velocity for an expanded gas
This is the Continuity equation
170 psia The initial saturated steam pressure133.5 psi The pressure drop @ sonic conditions as developed above
1,197.2 btu/lb The enthalpy of the saturated steam entering the 2" pipe
36.5 psia This is the final (maximum) pressure that will yield sonic flow;constant, sonic flow will continue even though the final pressure is dropped to a lower value - such as 14.7 psia.
The steam expansion that takes place across the globe valve is assumed to be adiabatic and, therefore, the
1,197.2 btu/lb The enthalpy of the saturated steam exiting the 2" pipe;
The enthalpy of the saturated steam exiting the 2" pipe;
12.425 This is the steam's specific volume at 36.6 psia and H= 1,197.2 but/lb
1,653 ft/sec
W = 11,164 lb/hr
Art's Note: is the downstream pressure when the initial downstream pressure will be, by definition, 14.7 psia.
P'1 =DP =
hg =
P'2 =
enthalpy conditions are equal on both sides of the valve - DH = 0.0. Therefore,
hg =This is the same as the steam entering the pipe (DH = 0)
ft3/lb
The resulting temperature = 317 oF and superheated.
vS = This is the steam's sonic velocity 36.6 psia and 317 oF
The problem asks for "the initial flow rate"; the DP has been calculated assuming that 36.6 psia
vS=√144 g k P2' V̄
W= v d2
0 .0509 V̄
V̄ =
Art Montemayor September 30, 2006Rev: 0
Page 38 of 287 FileName: document.xlsWorkSheet: Example 4-21
Example 4-21 (18th printing)
Given: Coke oven gas with a specific gravity of 0.42, a header pressure of 125 psig, and a temperature
Assume that the gas's ratio of specific heats, k = 1.4.
Find: The flow rate in standard cubic feet per hour (scfh).
Solution:
139.7 psia header absolute pressure
0.0175 The Reynolds Number is not necessary to identify f since the discharged gasto the atmosphere will have a very large Re, and flow will always be in the fully turbulent range, in which the friction factor is a horizontal line on the Moody Chart - and constant.
d = 3.068 inches the discharge pipe ID
0.42 the specific gravity of the gas
600 the initial, absolute temperature of the gas in the header.L = 20 feet length of 3" discharge pipe
K = 0.5 This is the entrance loss K K = 1.369 This is the 3" pipe KK = 1.0 This is the K for the exit loss out of the 3" pipe
K (Total) = 2.87 This is the K for the Total System
0.895 This is the ratio of the overall pressure drop to the initial pressure
91.8 psi
Y = 0.637 Interpolated from the tabulated data for Y
1,027,686 scfh
of 140 oF is flowing through 20 feet of 3-inch Schedule 40 pipe before discharging to atmosphere.
P'1 =
f = fT =
Sg =
T1 = oR
DP/P'1 =
From tabular data, it is seen that the maximum DP/P1 for sonic flow to exist is 0.657 when the system K = 2.87.
Since the indicated DP/P'1 is less than that calculated above, sonic velocity occurs at the end of the 3" pipeand the DP for the corresponding flow equation is:
DP =
q'h = gas flow rate measured at 60 oF & 14.7 psia
125 psigqh
' =40 , 700 Y d2 √ ΔP P1'
K T1 Sg
K=fLD
Art Montemayor September 30, 2006Rev: 0
Page 39 of 287 FileName: document.xlsWorkSheet: Example 4-21
Source: http://www.cheresources.com/indexzz.shtml
Taran Baker
Dec 1 2004 TP-410 includes an exit loss when calculating the equivalent length of pipe. However, I struggle to comprehend why this is required. If we are including an exit loss, aren't we saying that the flow is choked not at the end of the pipe, but some distance from the pipe discharge? The reason I am asking is that I am trying to calculate the reaction force from a pipeline vent. As such, I require the choked pressure at the end of the vent pipe. My gut feeling tells me that I shouldn't include an exit loss. Any assistance with this would be most appreciated.
rxnarang In adibatic flow of gas through a conduit of constant cross section, the sonic velocity is always reached at the end of the pipe. You are correct.
However, there is always energy loss involved due to friction, including entrance and exit losses, which are a form of energy loss. All these energy losses influence the energy profile of the fluid. Given that sonic velocity occurs at the end of the line, higher the losses, lesser the amount of fluid flow. Hence, taking the exit losses into account is important, as it will influence the amount of fluid which can flow in the pipe.
Crane Example 4-21 is correct in computing all the losses which occur.
Hope this helps.Regards
Sonic Flow, Crane TP-410 example 4-21
My question concerns sonic gas flow from a pipe. Example 4-21 (pg 4-14) of Crane
Art Montemayor September 30, 2006Rev: 0
Page 40 of 287 FileName: document.xlsWorkSheet: Example 4-21
gut feeling tells me that I shouldn't include an exit loss. Any assistance with this would be
Art Montemayor September 30, 2006Rev: 0
Page 41 of 287 FileName: document.xlsWorkSheet: Example 4-22
Example 4-22 (18th printing)
Given:the outlet of a 1/2-inch Schedule 80 pipe discharging to atmosphere.
Find: The air flow rate in standard cubic feet per minute (scfm).
Solution:
The modified Darcy flow equation for compensating for the changeof fluid properties due to free, adiabatic expansion of the fluid.
34.0 psia The initial, absolute pressure in the 1/2" pipe19.3 psi The pressure difference between the pipe initial pressure and atm.
d = 0.546 inches The internal diameter of the 1/2" pipe
0.0275 The friction factor under fully turbulent flowL = 10 ft The straight length of the 1/2" pipe
560 The initial, absolute temperature of the air in the pipe
K = 6.04 This is the K for the 10 feet of 1/2" pipeK = 1.00 This is the K for the pipe exit into the atmosphere
K (Total) = 7.04 This is the Total K for the system
0.568 This is the ratio of the overall pressure drop to the initial pressure
Y = 0.76 Y net expansion factor for adiabatic flow (isenthalpic expansion)and used to compensate for the changes in fluid properties due to fluid expansion.
q'm = 62.7 scfm
Art's Note: Note that fully turbulent flow is assumed to be occurring while the 1/2" pipe is venting to atmosphere.While this is more than probable, it hasn't been verified.Sub-sonic air flow is also assumed in this problem. This is another item that hasn't been verified.
Air at a pressure of 19.3 psig and a temperature of 100 oF is measured at a point 10 feet from
P'1 =DP =
fT =
T1 = oR
DP/P'1 =
qm' =678 Y d2 √ ΔP P1
'
K T 1 S g
K=fLD
Art Montemayor September 30, 2006Rev: 0
Page 42 of 287 FileName: document.xlsWorkSheet: Example 4-23
Example 4-23 (18th printing)
Given: A square-edged orifice with 2.0-inch diameter is installed in a 4-inch Schedule 40 pipe having amercury manometer connected between taps located 1 diameter upstream and 0.5 diameter downstream.
Find: a) the flow range where the orifice flow coefficient C is constant.
b)
Solution:
The flow equation for an orifice
The Reynolds number
The differential pressure across the orifice taps
the differential head in inches of Mercury (Hg)
The weight density of mercury under water =
62.371
13.568
1.000
783.88 Density of Mercury underwater
0.454 the differential pressure across the orifice taps
2.000 inches the orifice diameter
4.026 inches the pipe's I.D.0.497 The ratio of the orifice diameter to the I.D. of the pipe
C = 0.625
Q = 50.32 This is the calibration constant and solution "a)".
Q = 106 gpm
1.1211 cPRe = 73,800
flow rate 106 gpm through the 4" pipe is verified as correct. (this is solution "b)" )
Should the C factor prove to be incorrect for the Reynolds number based on the calculated flow, it must be adjusted by trial-and-error methodology until reasonable agreement is reached.
The theoretical calibration constant for the meter when used on 60 oF water and for
The flow rate of 60 oF water when the mercury manometer difference is 4.4 inches.
Dhm =
rW (SHg - SW)
rW = lb/ft3 Density of water at 60 oF
SHg = Specific Gravity of Mercury at 60 oF, referred to Water at 60 oF
SW = Specific Gravity of Water at 60 oF, referred to Water at 60 oF
rHg = lb/ft3
DP = Dhm
d1 =
d2 =b =
(Dhm)^0.5
m = this is the viscosity of the 60 oF water
The calculated value of C = 0.625 corresponds to the calculated Reynolds Number of 73,800; therefore, the
Flow
Q=236 d12 C √ ΔP
ρ
Re=50.6 Q ρ
d μ
ΔP=Δhm ρ
(12 ) (144 )
Example 4-24 (18th printing)
Given:pressure differential between the pipe taps of a 2.15-inch I.D. square-edged orifice.
Find: The oil flow rate in gallons per minute (gpm).
Solution:
SAE 10 Lube Oil at 90 oF is flowing through a 3-inch Schedule 40 pipe and produces 0.4 psi
Example 4-25 (18th printing)
Given: A rectangular concrete overflow aqueduct, 25 feet high and 16.5 feet wide, has an absolute
Find: The discharge rate in cubic feet per second when the liquid in the reservoir has reached the
Solution:
roughness (e) of 0.01 foot. Refer to the sketch below.
maximum height indicated in the sketch below. Assume the average water temperature is 60 oF.
Example 4-26 (18th printing)
Given: A cast iron pipe is two-thirds full of steady, uniform flowing water at 60 oF. The pipe has an insidediameter of 24 inches and a slope of 3/4-inch per foot. Note the sketch below.
Find: The water flow rate in gallons per minute (gpm).
Solution:
Example 4-27 (18th printing)
Given: A steam boiler operating at 300 psia has a maximum capacity of 100,000 pounds per hour of saturated steam.
Find: The boiler capacity in both kilo Btu per hour and in boiler horsepower.
Solution:
Example 4-6 in the 1957 (6th printing) version
Given: A 600-pound steel in-line ball check valve is required in a 2-inch pipe line carrying 48 degree APIcrude oil at 150 oF and a rate of 37 gallons per minute.
Find: The proper size check valve and the pressure drop with the valve installed in both vertical and horizontal positions. The valve should be sized so that the disc is fully lifted under normal flowconditions.
Solution:
Example 4-7 in the 1957 (6th printing) version
Given:rate of 400 gallons per minute; the face-to-face dimension of the valve is 10 inches.
Find: The pressure drop through the valve.
Solution:
Bunker C fuel oil at 90 oF is flowing through a wide open 5-inch, 150-pound steel gate valve at a
Example 4-12 in the 1957 (6th printing) version
Given: Saturated steam at 150 psig is flowing through a cylindrical pipe coil, as shown in the sketchbelow, at a rate of 2,000 pounds per hour. The coil is made of 2-inch Schedule 40 steel pipe.
Find: The pressure drop from Point A to Point B.
Solution:
3'
30"
A
B
Art Montemayor U.S.A. Pipe Dimensions September 30, 2003Rev: 0
Page 56 of 287 FileName: document.xlsWorkSheet: Pipe Tables
Inside Diameter Functions
(in Inches)
Inches Inches Inches Feet
Commercial Wrought Steel Pipe Data Although pipe classification is common knowledge that is taken for granted among a lot of us old engineers, I have Schedule Wall Thickness - Per ASA B36.10 - 1950 found that young engineers are lacking this information because both academic professors and we experienced
engineers are both guilty of not passing on the information which used to be common and available when piping and
Sch
edul
e 10
14 14 0.250 13.500 1.1250 182.25 2,460.4 33,215.1 448,403.3 143.14 0.994 fitting catalogs like Vogt, Tube Turns, Walworth, etc. used to be freely available to us. Now, these valuable free 16 16 0.250 15.500 1.2917 240.25 3,723.9 57,720.1 894,661.0 188.69 1.310 catalogs have become a thing of the past…18 18 0.250 17.500 1.4583 306.25 5,359.4 93,789.1 1,641,308.6 240.53 1.670
20 20 0.250 19.500 1.6250 380.25 7,414.9 144,590.1 2,819,506.2 298.65 2.074 Because I regard this subject as very basic and important for all engineers to dominate, some years back I prepared 24 24 0.250 23.500 1.9583 552.25 12,977.9 304,980.1 7,167,031.5 433.74 3.012 the following explanation for young engineers working under me and with me in plant projects. I would like to share
30 30 0.312 29.376 2.4480 862.95 25,350.0 744,681.6 21,875,767.4 677.76 4.707 it with any one else who hasn't had the opportunity to find out this logical explanation of how pipe is classified.
Sch
edul
e 20
8 8.625 0.250 8.125 0.6771 66.02 536.4 4,358.1 35,409.3 51.85 0.360 Industrial pipe thicknesses follow a set formula, expressed as the “schedule number” as established by the 10 10.750 0.250 10.250 0.8542 105.06 1,076.9 11,038.1 113,140.8 82.52 0.573 American Standards Association (ASA) now re-organized as ANSI - the American National Standards Institute. 12 12.750 0.250 12.250 1.0208 150.06 1,838.3 22,518.8 275,854.7 117.86 0.818 Eleven schedule numbers are available for use: 5, 10, 20, 30, 40, 60, 80, 100, 120, 140, & 160. The most popular 14 14 0.312 13.376 1.1147 178.92 2,393.2 32,011.4 428,184.9 140.52 0.976 schedule, by far, is 40. Sch 5, 60, 100, 120, & 140 have rarely, if ever been employed by myself in over 40 years 16 16 0.312 15.376 1.2813 236.42 3,635.2 55,895.1 859,442.6 185.68 1.289 as a practicing engineer. The schedule number is defined as the approximate value of the expression:18 18 0.312 17.376 1.4480 301.93 5,246.3 91,158.9 1,583,977.6 237.13 1.647
10 10.750 0.307 10.136 0.8447 102.74 1,041.4 10,555.2 106,987.5 80.69 0.560 For example, the schedule number of ordinary steel pipe having an allowable stress of 10,000 psi for use at a 12 12.750 0.330 12.090 1.0075 146.17 1,767.2 21,365.1 258,304.2 114.80 0.797 working pressure of 350 psig would be:14 14 0.375 13.250 1.1042 175.56 2,326.2 30,822.2 408,394.0 137.89 0.958
20 20 0.500 19.000 1.5833 361.00 6,859.0 130,321.0 2,476,099.0 283.53 1.969 This would be the proper schedule for welded joints and steel fittings but not for threaded connections and cast-iron 24 24 0.562 22.876 1.9063 523.31 11,971.3 273,854.8 6,264,702.3 411.01 2.854 or malleable-iron fittings. In practice, schedule 40 would be used for welded construction and Sch 80 (about 2x the
30 30 0.625 28.750 2.3958 826.56 23,763.7 683,205.6 19,642,160.0 649.18 4.508 computed value) for iron fittings. The higher schedule is required because of weaknesses in the iron fittings and the metal lost in cutting the threads.
1/4 0.540 0.088 0.364 0.0303 0.13 0.0 0.0 0.0 0.10 0.001 For all pipe sizes below 10", Sch 40 pipe is identical with what was once called “standard” pipe, and Sch 80 is 3/8 0.675 0.091 0.493 0.0411 0.24 0.1 0.1 0.0 0.19 0.001 identical with the former “extra-strong” pipe. There is no equivalent schedule number for “double-extra-strong” 1/2 0.840 0.109 0.622 0.0518 0.39 0.2 0.1 0.1 0.30 0.002 pipe, and Sch 160 is the only other weight in which pipe smaller than 4" is available. 3/4 1.050 0.113 0.824 0.0687 0.68 0.6 0.5 0.4 0.53 0.004
1 1.315 0.133 1.049 0.0874 1.10 1.2 1.2 1.3 0.86 0.006 Temperature has no direct bearing on the schedule, except as it either weakens (or strengthens) the material's 1 1/4 1.660 0.140 1.380 0.1150 1.90 2.6 3.6 5.0 1.50 0.010 allowable stress. Stainless steels (304ELC & 316ELC), for example, yield a stronger allowable stress at the low
2 2.375 0.154 2.067 0.1723 4.27 8.8 18.3 37.7 3.36 0.023 I've used the rule of thumb that the softer the metal, the stronger it is at the lower temperatures.2 1/2 2.875 0.203 2.469 0.2058 6.10 15.1 37.2 91.7 4.79 0.033
3 3.500 0.216 3.068 0.2557 9.41 28.9 88.6 271.8 7.39 0.051 I hope this has helped you in explaining how pipe is classified.3 1/2 4.000 0.226 3.548 0.2957 12.59 44.7 158.5 562.2 9.89 0.069
Note:Stainless Steel Pipe Schedule 40S values are the same, size for size, as those shown above on the Standard Wall Pipe Table (heaviest weight on 8, 10, and 12-inch sizes).
Stainless Steel Pipe Schedule 80S values are the same, size for size, as those shown above on the Extra Strong Pipe Table.
Art Montemayor U.S.A. Pipe Dimensions September 30, 2003Rev: 0
Page 61 of 287 FileName: document.xlsWorkSheet: Pipe Tables
May 8, 2003
Although pipe classification is common knowledge that is taken for granted among a lot of us old engineers, I have found that young engineers are lacking this information because both academic professors and we experienced engineers are both guilty of not passing on the information which used to be common and available when piping and fitting catalogs like Vogt, Tube Turns, Walworth, etc. used to be freely available to us. Now, these valuable free catalogs have become a thing of the past…
Because I regard this subject as very basic and important for all engineers to dominate, some years back I prepared the following explanation for young engineers working under me and with me in plant projects. I would like to share it with any one else who hasn't had the opportunity to find out this logical explanation of how pipe is classified.
Industrial pipe thicknesses follow a set formula, expressed as the “schedule number” as established by the American Standards Association (ASA) now re-organized as ANSI - the American National Standards Institute. Eleven schedule numbers are available for use: 5, 10, 20, 30, 40, 60, 80, 100, 120, 140, & 160. The most popular schedule, by far, is 40. Sch 5, 60, 100, 120, & 140 have rarely, if ever been employed by myself in over 40 years as a practicing engineer. The schedule number is defined as the approximate value of the expression:
Schedule Number = (1,000)(P/S)
P = the internal working pressure, psigS = the allowable stress (psi) for the material of construction at the conditions of use.
For example, the schedule number of ordinary steel pipe having an allowable stress of 10,000 psi for use at a working pressure of 350 psig would be:
This would be the proper schedule for welded joints and steel fittings but not for threaded connections and cast-iron or malleable-iron fittings. In practice, schedule 40 would be used for welded construction and Sch 80 (about 2x the computed value) for iron fittings. The higher schedule is required because of weaknesses in the iron fittings and the metal lost in cutting the threads.
For all pipe sizes below 10", Sch 40 pipe is identical with what was once called “standard” pipe, and Sch 80 is identical with the former “extra-strong” pipe. There is no equivalent schedule number for “double-extra-strong” pipe, and Sch 160 is the only other weight in which pipe smaller than 4" is available.
Temperature has no direct bearing on the schedule, except as it either weakens (or strengthens) the material's allowable stress. Stainless steels (304ELC & 316ELC), for example, yield a stronger allowable stress at the low
I've used the rule of thumb that the softer the metal, the stronger it is at the lower temperatures.
I hope this has helped you in explaining how pipe is classified.
Art Montemayor
Schedule Number = (1,000)(350/10,000) = 35 (approx. 40)
temperatures near the cryogenic zone (-50 to -150 oF). Copper and Brass also exhibit the same behavior.
In Crane TP 410, K for a fitting is found by multiplying a number times f T. fT is called the friction factor. Is fT related to the roughness?
dtn6770 (Mechanical) 13 Dec 06 9:45
fT is related to roughness by way of the Moody Diagram and its use of relative roughness. Relative roughness is the materials absolute roughness divided by the inside diameter.
STYMIEDPIPER (Mechanical) 13 Dec 06 10:53
Crane 410 lists values for fT
wfn217 (Chemical) 13 Dec 06 10:57
I wanted to know whether the listed fT values are strictly due to geometry or also a function of roughness.
BigInch (Petroleum) 13 Dec 06 11:05
geometry
vzeos (Mechanical) 13 Dec 06 12:03
wfn217- In Crane TP 410 fT is the fully turbulent flow friction factor defined by the equation below. Dis the pipe I. D. and k is the pipe roughness.
1/√fT= 2Log{(3.7D/k)
Crane's equivalent length calculations assume fully developed turbulent flow. See Crane TP 410, Page26. The fT values provided by Crane imply a k value of about 0.00180".
BigInch (Petroleum) 13 Dec 06 12:16
vzeos,
You are totally correct. Seems I reversed the question. K is determined by geometry, but the equivaletlength is indeed based on 0.0018 roughness.
DLANDISSR (Mechanical) 13 Dec 06 14:45
fT is the friction factor in the turbulent region of flow. It is calculated from the Colebrook-White equation which is an iteration. The Moody Diagram is derived from the Colebrook-White equation.
katmar (Chemical) 13 Dec 06 15:00
BigInch nailed the answer with his first post, but this question highlights what I consider to be one of the weaker points of the Crane 410 manual so please forgive me a little rant here. I have seen Crane's treatment of the K value for pipe fittings cause so much confusion - it really is a great pity they chose to do it this way.
The Crane engineers noted that the K values for fittings generally decreased as the fitting size increased. But it all went wrong when they noticed that this rate of decrease was close to the same as the rate at which the friction factor for fully developed turbulent flow in commercial steel pipe decreased as the pipe size increased. The fatal mistake was to link the two. See Crane Fig 2-14 and associated commentary.
For example, on page A-29 the K value for a 90 degree butt-weld pipe bend with an r/d of 1.5 is given as 14fT. The values of fT are given at the top of page A-26 as a function of pipe size. The values may have been calculated using the function referenced by vzeos, but for the purposes of calculating K values they are constants for each pipe size.
This apparent link between the K value and the friction factor gives the impression that the K value is linked to the pipe roughness - but in fact it is not because fT is defined to be at a particular roughness. Even worse, it is possible to be misled into believing the Crane K values compensate for changes in Reynolds number because everyone knows that the friction factor is influenced by the Reynolds number. But again, it is not because fT is defined to be in a particular Reynolds regime (fully turbulent).
To use the example of the 90 degree bend I gave above, it would have been better for Crane to give theK value as 14J, where J is simply a fudge-factor and would still be given by the values in the table on Page A-26 but without any reference to the friction factor. (Note that I have selected J as my symbol simply because it has no prior definition in the Crane Nomenclature table.)
The upshot of all of this is that in Crane's treatment, the K value of a fitting is a function onlypipe size (or geometry to use the terms used by wfn217 and BigInch). This was an improvement over previous work where the K value had been assumed to be constant for all sizes of fittings, and at the time that Crane first published this method it was rightly acclaimed as an important advance but IMHOit was badly worded and newer editions of 410 have unfortunately done nothing to remove the confusion.
I have awarded a star to BigInch for his comment that if you want to convert the Crane K value to an equivalent length, you must use the fT value from Crane's table on page A-26 (which is based on a roughness of 0.0018") and NOT the actual friction factor of the pipe you are using. The Crane description of this on pages 2-8 to 2-11 is extremely confusing, and the example 4-7 is just plain wrong because the K values given in the 410 manual apply only to fully developed turbulent flow and should never be used for laminar flow.
If you are working with laminar flow it is much better to work with equivalent lengths than with fixed or even Crane K values. Resistance values for fittings increase rapidly at low Reynolds numbers, but so does the friction factor. This means that if you use fixed L/D values, which get multiplied by the friction factor in the Darcy-Weisbach equation, the high resistance values are automatically compensated for. Or even better, use the 2-K or 3-K methods proposed by Hooper and Darby.
katmar (Chemical) 13 Dec 06 15:00
BigInch nailed the answer with his first post, but this question highlights what I consider to be one of the weaker points of the Crane 410 manual so please forgive me a little rant here. I have seen Crane's treatment of the K value for pipe fittings cause so much confusion - it really is a great pity they chose to do it this way.
The Crane engineers noted that the K values for fittings generally decreased as the fitting size increased. But it all went wrong when they noticed that this rate of decrease was close to the same as the rate at which the friction factor for fully developed turbulent flow in commercial steel pipe decreased as the pipe size increased. The fatal mistake was to link the two. See Crane Fig 2-14 and associated commentary.
For example, on page A-29 the K value for a 90 degree butt-weld pipe bend with an r/d of 1.5 is given as 14fT. The values of fT are given at the top of page A-26 as a function of pipe size. The values may have been calculated using the function referenced by vzeos, but for the purposes of calculating K values they are constants for each pipe size.
This apparent link between the K value and the friction factor gives the impression that the K value is linked to the pipe roughness - but in fact it is not because fT is defined to be at a particular roughness. Even worse, it is possible to be misled into believing the Crane K values compensate for changes in Reynolds number because everyone knows that the friction factor is influenced by the Reynolds number. But again, it is not because fT is defined to be in a particular Reynolds regime (fully turbulent).
To use the example of the 90 degree bend I gave above, it would have been better for Crane to give theK value as 14J, where J is simply a fudge-factor and would still be given by the values in the table on Page A-26 but without any reference to the friction factor. (Note that I have selected J as my symbol simply because it has no prior definition in the Crane Nomenclature table.)
The upshot of all of this is that in Crane's treatment, the K value of a fitting is a function onlypipe size (or geometry to use the terms used by wfn217 and BigInch). This was an improvement over previous work where the K value had been assumed to be constant for all sizes of fittings, and at the time that Crane first published this method it was rightly acclaimed as an important advance but IMHOit was badly worded and newer editions of 410 have unfortunately done nothing to remove the confusion.
I have awarded a star to BigInch for his comment that if you want to convert the Crane K value to an equivalent length, you must use the fT value from Crane's table on page A-26 (which is based on a roughness of 0.0018") and NOT the actual friction factor of the pipe you are using. The Crane description of this on pages 2-8 to 2-11 is extremely confusing, and the example 4-7 is just plain wrong because the K values given in the 410 manual apply only to fully developed turbulent flow and should never be used for laminar flow.
If you are working with laminar flow it is much better to work with equivalent lengths than with fixed or even Crane K values. Resistance values for fittings increase rapidly at low Reynolds numbers, but so does the friction factor. This means that if you use fixed L/D values, which get multiplied by the friction factor in the Darcy-Weisbach equation, the high resistance values are automatically compensated for. Or even better, use the 2-K or 3-K methods proposed by Hooper and Darby.
katmar (Chemical) 13 Dec 06 15:00
BigInch nailed the answer with his first post, but this question highlights what I consider to be one of the weaker points of the Crane 410 manual so please forgive me a little rant here. I have seen Crane's treatment of the K value for pipe fittings cause so much confusion - it really is a great pity they chose to do it this way.
The Crane engineers noted that the K values for fittings generally decreased as the fitting size increased. But it all went wrong when they noticed that this rate of decrease was close to the same as the rate at which the friction factor for fully developed turbulent flow in commercial steel pipe decreased as the pipe size increased. The fatal mistake was to link the two. See Crane Fig 2-14 and associated commentary.
For example, on page A-29 the K value for a 90 degree butt-weld pipe bend with an r/d of 1.5 is given as 14fT. The values of fT are given at the top of page A-26 as a function of pipe size. The values may have been calculated using the function referenced by vzeos, but for the purposes of calculating K values they are constants for each pipe size.
This apparent link between the K value and the friction factor gives the impression that the K value is linked to the pipe roughness - but in fact it is not because fT is defined to be at a particular roughness. Even worse, it is possible to be misled into believing the Crane K values compensate for changes in Reynolds number because everyone knows that the friction factor is influenced by the Reynolds number. But again, it is not because fT is defined to be in a particular Reynolds regime (fully turbulent).
To use the example of the 90 degree bend I gave above, it would have been better for Crane to give theK value as 14J, where J is simply a fudge-factor and would still be given by the values in the table on Page A-26 but without any reference to the friction factor. (Note that I have selected J as my symbol simply because it has no prior definition in the Crane Nomenclature table.)
The upshot of all of this is that in Crane's treatment, the K value of a fitting is a function onlypipe size (or geometry to use the terms used by wfn217 and BigInch). This was an improvement over previous work where the K value had been assumed to be constant for all sizes of fittings, and at the time that Crane first published this method it was rightly acclaimed as an important advance but IMHOit was badly worded and newer editions of 410 have unfortunately done nothing to remove the confusion.
I have awarded a star to BigInch for his comment that if you want to convert the Crane K value to an equivalent length, you must use the fT value from Crane's table on page A-26 (which is based on a roughness of 0.0018") and NOT the actual friction factor of the pipe you are using. The Crane description of this on pages 2-8 to 2-11 is extremely confusing, and the example 4-7 is just plain wrong because the K values given in the 410 manual apply only to fully developed turbulent flow and should never be used for laminar flow.
If you are working with laminar flow it is much better to work with equivalent lengths than with fixed or even Crane K values. Resistance values for fittings increase rapidly at low Reynolds numbers, but so does the friction factor. This means that if you use fixed L/D values, which get multiplied by the friction factor in the Darcy-Weisbach equation, the high resistance values are automatically compensated for. Or even better, use the 2-K or 3-K methods proposed by Hooper and Darby.
Montemayor (Chemical) 13 Dec 06 18:59
And I select Harvey as the preferred Ranter of the day and worthy - as well - of recognition for bringing to everyone's attention the importance of really understanding and reading through what is putin front of our eyes. We, as professional engineers, are not being asked to believe or accept 100% of what we are offered or given - regardless of how “sacred” the Cow may seem. Everything in engineering is subject to scrutiny and improvements.
I have a lot of respect and gratitude for what has gone into putting together Crane's Tech Paper #410. However, everything Harvey has stated regarding the concept of their K values is not only valid, but 100% positive criticism that should be heard and applied. Major world-class engineering firms agree with what Harvey states - and so do some of the biggest chemical process companies. To quote one: “Until recently, the use of K coefficients for valves and fittings has been considered more accurate than the use of equivalent lengths of pipe, but recent research has disclosed that K coefficients are not constant for all sizes of any one type of valve or fitting; so the use of equivalent lengths, with some exceptions, is now preferred.” And this is in addition to the problems of understanding/interpreting what TP 410 is saying. We have a better option, as Harvey states, in the 2-K or 3-K methods.
Great comments and good engineering knowledge, Harvey!
wfn217 (Chemical) 14 Dec 06 11:39
The number that Crane multiplies by fT to obtain K is the same number listed as L/D in Cameron
katmar (Chemical) 14 Dec 06 13:45
wfn217,
the Darcy-Weisbach correlation for pressure drop is
?P = ( fL/D + K ) ?V2 / 2
So if you want to express the resistance of a fitting in terms of the equivalent length (i.e. L/D) instead of K then you have to calculate
L/D = K/f
and since Crane express their K's as (Constant) x fT you would get
L/D = (Constant) x fT / f
If f is evaluated at the same conditions as fT (which is what my rant above was all about!) then of course
L/D = (Constant)
which means that for commercial steel pipe in the fully turbulent regime the (Constant) used by Crane will indeed be the L/D value given by Cameron. You just have to remember that because Crane evaluated these (Constants) for commercial steel pipe in the fully turbulent regime, if you want to get back to the L/D values you must use fT and not the friction factor in your particular case.
the Darcy-Weisbach correlation for pressure drop is
?P = ( fL/D + K ) ?V2 / 2
So if you want to express the resistance of a fitting in terms of the equivalent length (i.e. L/D) instead of K then you have to calculate
L/D = K/f
and since Crane express their K's as (Constant) x fT you would get
L/D = (Constant) x fT / f
If f is evaluated at the same conditions as fT (which is what my rant above was all about!) then of course
L/D = (Constant)
which means that for commercial steel pipe in the fully turbulent regime the (Constant) used by Crane will indeed be the L/D value given by Cameron. You just have to remember that because Crane evaluated these (Constants) for commercial steel pipe in the fully turbulent regime, if you want to get back to the L/D values you must use fT and not the friction factor in your particular case.
It has been found experimentally that the resistance factor (i.e. K value) for a fitting depends on both the fitting size and the Reynolds number. In the Darcy-Weisbach equation (in which I managed to get the wrong symbols before)
ΔP = ( fL/D + K ) ρV2 / 2
you can see that if you use a K value the pressure drop is not influenced by the fitting size or the Reynolds number. When you use Crane K values the value is corrected (via the table on page A-26) for fitting size before you insert the value into the D-W equation, but many of the older references gave single K values for all fitting sizes.
On the other hand, you can see that if you use equivalent lengths, then the value is multiplied by the actual friction factor for your application, and because the friction factor does depend on the fitting size and the Reynolds number it automatically corrects the calculation for both these factors. This makes the equivalent length method both easier to use, and more accurate than K values. It is easier to use because you only have to remember a single L/D value for a given type of fitting.
As Montemayor and I said before, the very best way is to use the 2-K or 3-K method but these are computationally much more involved. In my software I use 3-K because once it is programmed it makes no extra work to use it, but if I am doing a hand calc then I use the equivalent length method.
Hope this makes it clear
Harvey
tickle (Chemical) 14 Dec 06 18:17
Thank you Katmar for your wonderful explanation.
I have questioned Crane myself, but learnt to accept that its findings were as practical as required.
I think that younger engineers blindly use equations and computer programs - because they are there. They do not understand the applicable circumstances or the limitations. In fact whilst reading the post one of the junior engineers questioned me regarding recommended pipe diameters. They had sized thesteam piping for maximum flow and minimum pressure. They were blindly going to use a calculated diameter because it was less than or equal to the recommended velocity without considering the application and whether the worst case conditions could occur at the same time, (they can't and a smaller pipe diameter was my recommendation).
katmar (Chemical) 14 Dec 06 15:15
wfn217,
It has been found experimentally that the resistance factor (i.e. K value) for a fitting depends on both the fitting size and the Reynolds number. In the Darcy-Weisbach equation (in which I managed to get the wrong symbols before)
ΔP = ( fL/D + K ) ρV2 / 2
you can see that if you use a K value the pressure drop is not influenced by the fitting size or the Reynolds number. When you use Crane K values the value is corrected (via the table on page A-26) for fitting size before you insert the value into the D-W equation, but many of the older references gave single K values for all fitting sizes.
On the other hand, you can see that if you use equivalent lengths, then the value is multiplied by the actual friction factor for your application, and because the friction factor does depend on the fitting size and the Reynolds number it automatically corrects the calculation for both these factors. This makes the equivalent length method both easier to use, and more accurate than K values. It is easier to use because you only have to remember a single L/D value for a given type of fitting.
As Montemayor and I said before, the very best way is to use the 2-K or 3-K method but these are computationally much more involved. In my software I use 3-K because once it is programmed it makes no extra work to use it, but if I am doing a hand calc then I use the equivalent length method.
Hope this makes it clear
Harvey
tickle (Chemical) 14 Dec 06 18:17
Thank you Katmar for your wonderful explanation.
I have questioned Crane myself, but learnt to accept that its findings were as practical as required.
I think that younger engineers blindly use equations and computer programs - because they are there. They do not understand the applicable circumstances or the limitations. In fact whilst reading the post one of the junior engineers questioned me regarding recommended pipe diameters. They had sized thesteam piping for maximum flow and minimum pressure. They were blindly going to use a calculated diameter because it was less than or equal to the recommended velocity without considering the application and whether the worst case conditions could occur at the same time, (they can't and a smaller pipe diameter was my recommendation).
pleckner (Chemical) 14 Dec 06 20:11
I invite everyone to read my article on this subject published at
http://www.cheresources.com/eqlength.shtml
I don't agree that using equivalent lengths is the proper way to perform a system hydraulcs calculation (my arguments are in the above mentioned article). You must not mix the friction factor for a fitting with the friction factor of a pipe because they are not the same, whether we are talking about CRANE'sfriction factor at fully developed turbulent flow or the two-K or three-K methods. The pressure loss of a fitting obtained from these methods is not the same nor nearly the same as the pressure loss of a fitting obtained from multiplying the pipe friction factor x the equivalent length of the fitting.
The most accurate way to perform the calculation is to use one of these methods (three-k perferred) and add the losses to those of the pipe, not to combine them all into an equivalent length of straight pipe and multiply by the pipe friction factor.
The two-K or three-K methods were developed to address the fact that fittings are indded somewhat dependent on Reynolds number, but this is still not the same as a pipe friction factor.
I will also argue that for fully developed turbulent flow, there is nothing wrong with CRANE's values.
katmar (Chemical) 15 Dec 06 5:15
pleckner,
There is a lot more that we agree on than on which we differ. It was never my intention to suggest that there is anything wrong with applying Crane's K values to fully developed turbulent flow. I have recommended MANY times here that people should get their hands on a copy of Crane 410. Probably 95% or more of our flow calcs are for fully developed turbulent flow, and Crane is ideal for this.
My main objection was to the confusion caused by Crane's wording and their linking of the K value to the turbulent friction factor. I believe the posts in this thread confirm that this confusion is widespread, and I have often seen engineers struggle to come to terms with it. Despite the shortcomings of Crane 410, my copy "lives" in the very front of the top drawer of the filing cabinet right next to my desk.
We all agree that the multi-K methods are better than the L/D method. However, I disagree with Crane's statement (which is echoed in your article) that “K is a constant under all flow conditions, including laminar flow”. In your example you give the K value for a long radius bend as 0.36. Applying the 3-K method at a Reynolds number of 100 in the same sized pipe gives a K value of 8.3. This means that a pressure drop calculation using the Crane value will be 90% understated, if we take 3-K as the benchmark. I accept that a similar calculation at Re = 100 using the L/D method will over-estimate the pressure drop but for most calculations this would be the conservative option.
I have one objection to the example in your article. 92% of the pressure drop is due to the reducer, but nobody would ever try to calculate the pressure drop through a reducer with the L/D method. Including the reducer exaggerates the negative aspects of the L/D method.
Neglecting the reducer, the calculated pressure drops using the various methods are
Who would put his neck on the block over which of these numbers is correct? The L/D and Crane answers are within 15% of the 3-K answer, and all three answers are probably adequate for practical purposes.Katmar SoftwareEngineering & Risk Analysis Software
BigInch (Petroleum) 15 Dec 06 6:12
That's my humble opinion also. With all the other inaccuracies inherent in hydraulics and more specifically how systems are usually operated at one or the other of the extreme ranges, if you design a system close enough so that you have to worry about the difference in pressure drops from laminar or turbulent flow in a few fittings, that's just plain tooooo close. If for some reason you need to examine an existing system, the actual data's there for the taking.
katmar (Chemical) 15 Dec 06 5:15
pleckner,
There is a lot more that we agree on than on which we differ. It was never my intention to suggest that there is anything wrong with applying Crane's K values to fully developed turbulent flow. I have recommended MANY times here that people should get their hands on a copy of Crane 410. Probably 95% or more of our flow calcs are for fully developed turbulent flow, and Crane is ideal for this.
My main objection was to the confusion caused by Crane's wording and their linking of the K value to the turbulent friction factor. I believe the posts in this thread confirm that this confusion is widespread, and I have often seen engineers struggle to come to terms with it. Despite the shortcomings of Crane 410, my copy "lives" in the very front of the top drawer of the filing cabinet right next to my desk.
We all agree that the multi-K methods are better than the L/D method. However, I disagree with Crane's statement (which is echoed in your article) that “K is a constant under all flow conditions, including laminar flow”. In your example you give the K value for a long radius bend as 0.36. Applying the 3-K method at a Reynolds number of 100 in the same sized pipe gives a K value of 8.3. This means that a pressure drop calculation using the Crane value will be 90% understated, if we take 3-K as the benchmark. I accept that a similar calculation at Re = 100 using the L/D method will over-estimate the pressure drop but for most calculations this would be the conservative option.
I have one objection to the example in your article. 92% of the pressure drop is due to the reducer, but nobody would ever try to calculate the pressure drop through a reducer with the L/D method. Including the reducer exaggerates the negative aspects of the L/D method.
Neglecting the reducer, the calculated pressure drops using the various methods are
Who would put his neck on the block over which of these numbers is correct? The L/D and Crane answers are within 15% of the 3-K answer, and all three answers are probably adequate for practical purposes.Katmar SoftwareEngineering & Risk Analysis Software
BigInch (Petroleum) 15 Dec 06 6:12
That's my humble opinion also. With all the other inaccuracies inherent in hydraulics and more specifically how systems are usually operated at one or the other of the extreme ranges, if you design a system close enough so that you have to worry about the difference in pressure drops from laminar or turbulent flow in a few fittings, that's just plain tooooo close. If for some reason you need to examine an existing system, the actual data's there for the taking.
pleckner (Chemical) 15 Dec 06 8:01
We are in agreement.
I put the reducers into the L/D equation only as a means for comparison. I agree with you (Katmar) onthat. But then again, I don't use the equivalent length method, ever.
Katmar, I'm one up on you, I keep CRANE on my desk, I'm too lazy to go into my file cabinet.
wfn217 (Chemical) 15 Dec 06 9:09
It should be noted that pleckner's article states that K for a fitting has little to do with friction, which answers my original question. I have seen someone try to scale the fT values in Crane from pipe to tubing based on roughness.
LeSabre (Petroleum) 15 Dec 06 10:07
pleckner,
Great article. I appreciate your logical presentation. I am curious, however, about your representation that in 1979 Crane “...discussed and used the two-friction factor method for calculating the total pressure drop in a piping system...(f for straight pipe and ft for valves and fittings)”. It is my understanding from reading the TP 410 Foreword (4th paragraph, quoted below) that Crane’s intent was to have the pipe friction factor, f, apply to the K factor calculation and that the K factor was intended to apply to the full range of flow regimes from laminar to full turbulence. Can you comment on this?
Quote (TP 410 FOREWORD):The fifteenth printing (1976 edition) presented a conceptual change regarding the values of EquivalentLength "L/D" and Resistance Coefficient "K" for valves and fittings relative to the friction factor in pipes. This change has relatively minor effect on most problems dealing with flow conditions that result in Reynolds numbers falling in the turbulent zone. However, for flow in the laminar zone, the change avoids a significant overstatement of pressure drop. Consistent with this conceptual revision, the resistance to flow through valves and fittings is now expressed in terms of resistance co-efficient
pleckner (Chemical) 15 Dec 06 8:01
We are in agreement.
I put the reducers into the L/D equation only as a means for comparison. I agree with you (Katmar) onthat. But then again, I don't use the equivalent length method, ever.
Katmar, I'm one up on you, I keep CRANE on my desk, I'm too lazy to go into my file cabinet.
wfn217 (Chemical) 15 Dec 06 9:09
It should be noted that pleckner's article states that K for a fitting has little to do with friction, which answers my original question. I have seen someone try to scale the fT values in Crane from pipe to tubing based on roughness.
LeSabre (Petroleum) 15 Dec 06 10:07
pleckner,
Great article. I appreciate your logical presentation. I am curious, however, about your representation that in 1979 Crane “...discussed and used the two-friction factor method for calculating the total pressure drop in a piping system...(f for straight pipe and ft for valves and fittings)”. It is my understanding from reading the TP 410 Foreword (4th paragraph, quoted below) that Crane’s intent was to have the pipe friction factor, f, apply to the K factor calculation and that the K factor was intended to apply to the full range of flow regimes from laminar to full turbulence. Can you comment on this?
Quote (TP 410 FOREWORD):The fifteenth printing (1976 edition) presented a conceptual change regarding the values of EquivalentLength "L/D" and Resistance Coefficient "K" for valves and fittings relative to the friction factor in pipes. This change has relatively minor effect on most problems dealing with flow conditions that result in Reynolds numbers falling in the turbulent zone. However, for flow in the laminar zone, the change avoids a significant overstatement of pressure drop. Consistent with this conceptual revision, the resistance to flow through valves and fittings is now expressed in terms of resistance co-efficient
katmar (Chemical) 15 Dec 06 11:22
LeSabre,
The example in pleckner's article shows how the "old" (i.e. pre-1976) L/D method over-estimates the pressure drop in fittings for low Reynolds numbers. All my calcs agree with and confirm pleckner's result. Crane were quite correct to make this claim in the 1976 foreword.
It is true that using the “new” K values avoids the overstatement of the pressure drop in the laminar regime, but my example (See 5 Dec 5:15) shows how the Crane K values now understate the pressure drop by 90% at a Reynolds number of 100. If you were designing a new pipeline would you rather estimate the pressure drop as 40% too high or 90% too low?
Your query, and the latest post by wfn217, are typical examples of the confusion that is caused by the Crane method, and which lead to my rant near the start of this thread.
I have probably stretched everyone's patience to the limit by going on and on about this problem so I will leave it at this now. Katmar SoftwareEngineering & Risk Analysis Software
pleckner (Chemical) 15 Dec 06 12:49
To LaSabre:
Thanks.
To me that statement in the Foreword just means that they made a change on how they relate the new K value to pipe friction factor. The statement obviously doesn't give any specifics as to how to apply the two, that is left to the rest of the document. Indeed, everywhere K for a valve or fitting is calculatedwithin TP410 they are very careful to note the friction factor as fT and NOT f.
CRANE notes that the friction factors (f and fT) will essentially be the same in the zone of complete turbulence.
But you are correct in that it appears CRANE was intent on having K applied throughout all flow zones and that is where they got it wrong as we've been pointing out in these Postings.
ccfowler (Mechanical) 16 Dec 06 2:03
katmar,
Thanks for your comments and their effects on this discussion.
BigInch,You have pointed toward one of my pet peeves, the dreadful misunderstandings brought about by the ability of computers and calculators to thrash around 10 digit numbers. If the accuracy to which most parameters are known is only 1%, 5%, 10%, and sometimes even worse, most of those many digits presented by the mighty computer or calculator are really just so much drivel. Unfortunately, there is never a shortage of people who want to believe that all of those digits are significant (and are willing to
BigInch (Petroleum) 16 Dec 06 3:37
The discussion has been very informative, but there is a certain logic for Crane to just continuing to publish equivalent lengths with conservative roughness at turbulent flows... its conservative and that's exactly how a lookup table intended for general application to all hydraulic systems should be. Somewhere in Tips, I recall seeing a comment I liked questioning how many digits were actually needed before the answer became believable, or something to that effect. Guess we should have stayed with 8 bit computers. Even if all variables were known to 1%, it seems to me that most systems would spend an excessive amount of time operating outside the range where those accuracies were valid. Lastly, risking repeating myself, sooner or later any given system reaches capacity at one extreme end of operational range or another, so over the long term is using Eq.Len at max turbulence really a bad thing? BigInch-born in the trenches. LeSabre (Petroleum) 18 Dec 06 10:40
katmar, Thanks for directing my attention to your “rant”. I agree that the ft factor depends on geometry. Your j-factor theory, however, leads to the unlikely result that two different pipe I.D.s, say, 4” sch 40 (I.D. = 4.0260”) and 4” sch 160 (I.D. = 3.4380”) could have the same K-factor. How do you reconcile this situation? Also, how would you get the K-factor for a 36" XS L.R. Ell.? Thanks. sailoday28 (Mechanical) 18 Dec 06 12:11
And the L/d or K for a run or branch of a "T" is? Regards
BigInch (Petroleum) 18 Dec 06 12:28
You could have a look at these that Weldbend gives for fittings, including straight tees and they even give a method for assemblies as well, http://www.weldbend.com/Technical%20Data/Flow%20Resistance%20For%20Fittings/flowresistance.htm and click on <next> for the following page BigInch-born in the trenches. LeSabre (Petroleum) 18 Dec 06 12:44
BigInch, Great site, thanks! But I didn't see any info on K-factors or ft.
katmar (Chemical) 15 Dec 06 11:22
LeSabre,
The example in pleckner's article shows how the "old" (i.e. pre-1976) L/D method over-estimates the pressure drop in fittings for low Reynolds numbers. All my calcs agree with and confirm pleckner's result. Crane were quite correct to make this claim in the 1976 foreword.
It is true that using the “new” K values avoids the overstatement of the pressure drop in the laminar regime, but my example (See 5 Dec 5:15) shows how the Crane K values now understate the pressure drop by 90% at a Reynolds number of 100. If you were designing a new pipeline would you rather estimate the pressure drop as 40% too high or 90% too low?
Your query, and the latest post by wfn217, are typical examples of the confusion that is caused by the Crane method, and which lead to my rant near the start of this thread.
I have probably stretched everyone's patience to the limit by going on and on about this problem so I will leave it at this now. Katmar SoftwareEngineering & Risk Analysis Software
pleckner (Chemical) 15 Dec 06 12:49
To LaSabre:
Thanks.
To me that statement in the Foreword just means that they made a change on how they relate the new K value to pipe friction factor. The statement obviously doesn't give any specifics as to how to apply the two, that is left to the rest of the document. Indeed, everywhere K for a valve or fitting is calculatedwithin TP410 they are very careful to note the friction factor as fT and NOT f.
CRANE notes that the friction factors (f and fT) will essentially be the same in the zone of complete turbulence.
But you are correct in that it appears CRANE was intent on having K applied throughout all flow zones and that is where they got it wrong as we've been pointing out in these Postings.
ccfowler (Mechanical) 16 Dec 06 2:03
katmar,
Thanks for your comments and their effects on this discussion.
BigInch,You have pointed toward one of my pet peeves, the dreadful misunderstandings brought about by the ability of computers and calculators to thrash around 10 digit numbers. If the accuracy to which most parameters are known is only 1%, 5%, 10%, and sometimes even worse, most of those many digits presented by the mighty computer or calculator are really just so much drivel. Unfortunately, there is never a shortage of people who want to believe that all of those digits are significant (and are willing to
BigInch (Petroleum) 16 Dec 06 3:37
The discussion has been very informative, but there is a certain logic for Crane to just continuing to publish equivalent lengths with conservative roughness at turbulent flows... its conservative and that's exactly how a lookup table intended for general application to all hydraulic systems should be. Somewhere in Tips, I recall seeing a comment I liked questioning how many digits were actually needed before the answer became believable, or something to that effect. Guess we should have stayed with 8 bit computers. Even if all variables were known to 1%, it seems to me that most systems would spend an excessive amount of time operating outside the range where those accuracies were valid. Lastly, risking repeating myself, sooner or later any given system reaches capacity at one extreme end of operational range or another, so over the long term is using Eq.Len at max turbulence really a bad thing? BigInch-born in the trenches. LeSabre (Petroleum) 18 Dec 06 10:40
katmar, Thanks for directing my attention to your “rant”. I agree that the ft factor depends on geometry. Your j-factor theory, however, leads to the unlikely result that two different pipe I.D.s, say, 4” sch 40 (I.D. = 4.0260”) and 4” sch 160 (I.D. = 3.4380”) could have the same K-factor. How do you reconcile this situation? Also, how would you get the K-factor for a 36" XS L.R. Ell.? Thanks. sailoday28 (Mechanical) 18 Dec 06 12:11
And the L/d or K for a run or branch of a "T" is? Regards
BigInch (Petroleum) 18 Dec 06 12:28
You could have a look at these that Weldbend gives for fittings, including straight tees and they even give a method for assemblies as well, http://www.weldbend.com/Technical%20Data/Flow%20Resistance%20For%20Fittings/flowresistance.htm and click on <next> for the following page BigInch-born in the trenches. LeSabre (Petroleum) 18 Dec 06 12:44
BigInch, Great site, thanks! But I didn't see any info on K-factors or ft.
pleckner (Chemical) 18 Dec 06 13:03
Let's remember that CRANE TP410 doesn't publish equivalent lenghts for valves and fittings, it provides K values or the means to calculate K values from which we can then calculate the equivalent length using the appropriate friction factor at fully developed turbulent flow.
Let us also not forget that we can obtain various fT values from the graph on Page A-23. This graph allows us to choose the fT for the type of pipe being used (and this translates into a constant absolute roughness for that type of pipe) and the diameter of that pipe. So I don't agree that CRANE TP410 necessarily publishes conservative values of roughness. It all depends on how you want to manipulate the piping material you are using, e.g. For pharmaceutical grade highly polished SS pipe I might choose to use the roughness for Drawn Tubing (an absolute roughness of 0.000005 feet) rather than clean commercial steel pipe (0.00015 feet). Note that with this table, I can also try to interpolate and obtain fT for various pipe diameters!
Let us also not forget that the published fT values given in the table at the top of Page A-26 are strictly for CLEAN COMMERCIAL STEEL PIPE and for the schedule of pipe listed on Page 2-10. If one desires the the K value for another fitting of a different schedule pipe, they should adjust it using Equation 2-5 on that same page or go to the Graph on Page A-23.
One last thing for now, one can still use published equivalent lenghts as per the reference given by BigInch but just remember to calculate the pressure loss through that fitting using fT for that fitting rather than the pipe friction factor as I've shown in my paper and my post above.
BigInch (Petroleum) 18 Dec 06 13:12
on the <next> page, they give a method for dP, from which you can backcalculate Ks, right?
Let's remember that CRANE TP410 doesn't publish equivalent lenghts for valves and fittings, it provides K values or the means to calculate K values from which we can then calculate the equivalent length using the appropriate friction factor at fully developed turbulent flow.
Let us also not forget that we can obtain various fT values from the graph on Page A-23. This graph allows us to choose the fT for the type of pipe being used (and this translates into a constant absolute roughness for that type of pipe) and the diameter of that pipe. So I don't agree that CRANE TP410 necessarily publishes conservative values of roughness. It all depends on how you want to manipulate the piping material you are using, e.g. For pharmaceutical grade highly polished SS pipe I might choose to use the roughness for Drawn Tubing (an absolute roughness of 0.000005 feet) rather than clean commercial steel pipe (0.00015 feet). Note that with this table, I can also try to interpolate and obtain fT for various pipe diameters!
Let us also not forget that the published fT values given in the table at the top of Page A-26 are strictly for CLEAN COMMERCIAL STEEL PIPE and for the schedule of pipe listed on Page 2-10. If one desires the the K value for another fitting of a different schedule pipe, they should adjust it using Equation 2-5 on that same page or go to the Graph on Page A-23.
One last thing for now, one can still use published equivalent lenghts as per the reference given by BigInch but just remember to calculate the pressure loss through that fitting using fT for that fitting rather than the pipe friction factor as I've shown in my paper and my post above.
BigInch (Petroleum) 18 Dec 06 13:12
on the <next> page, they give a method for dP, from which you can backcalculate Ks, right?
Yes, indeed (assuming we are still talking LR bends) the Sch40 and Sch160 fittings would have the same K value using my J factor, or using the Crane method. The table at the top of page A-26 has a note "K is based on use of schedule pipe as listed on page 2-10". And page 2-10 seems to say that the K values apply to Schedules 40 to 160, but that the velocity that is used to calculate the velocity head must be based on the actual ID of the fitting. Makes sense to me. If you take a look at Figure 2-16 on page 2-13 of Crane 410 you will doubt every calculation you have ever made. This figure shows the variability in the experimental data on which the K values we use are based. A great deal of license has been used in getting to the "averages" we accept as gospel.
This question again highlights the reason I disagree with the Crane fT method. People want to start fiddling with the fT for their particular pipe, whereas Crane's intention was that if you have a 4" fitting you use an fT of 0.017 irrespective of the schedule or actual roughness of that pipe . That is why I suggested that we call it "J" and eliminate the false link to the friction factor that is confusing everybody. All the experimental work shows that the K value has almost no dependency on the roughness, and if you look at Fig 2-16 again you will see that there is no room for hair splitting here.
Using the Darby 3-K method to calculate K values for the Sch40 and Sch160 bends gives values on 0.28 and 0.29 respectively (at Re = 300,000). If I was doing a calculation that involved these fittings I would use these two different values for the two different schedules, but only for the sake of computational consistency and to allow anyone to later check my calcs using the same methods. In my heart I would know that in fact they are for all intents and purposes the same. Of course for the same flowrate the pressure drop is higher through the Sch160 bend because of the higher velocity, but not because of any real change in K value. To try to calculate the actual K value from the Crane method by interpolating fT between Sch40 and Sch160 is IMO like measuring the length of a football pitch with a micrometer.
For the 36" bend I would use 3-K and get a K value of about 0.18. The rate of decrease with size gets less as the sizes get bigger. You could use Crane fT of 0.0105 (interpolating from the figure on page A-24) because there is a substantial increment to 36" from the data on page A-26, but the fact that Crane neglected to give values on page A-26 for 36" pipe does not detract from my argument that theirprocedure is confusing.
@Sailoday28,Crane have neglected to give K values for welded or flanged Tee's. I have no idea why. As always I would use 3-K. A good reference for this type of data is the classic article by Larry Simpson and Martin Weirick (Chemical Engineering, April 3, 1978).
@pleckner,I am confused over what you are saying Phil. If we apply your example of polished SS with a roughness of 0.000005 inch to our 4" LR bend we would have to have a Reynolds number of over 100 million to get to full turbulence (See Crane A-24). Under these conditions you would get an fabout 0.007. This compares with an fT of 0.017 for a 4" fitting given on page A-26. Using the value of0.007 would make a 4" highly polished LR bend have a K value of 17x0.007 = 0.119 and not the 0.28 that I would calculate from Darby's 3-K. Is this what you are saying, or have I got the wrong end of the stick?
katmar (Chemical) 18 Dec 06 15:29
@LeSabre,
Yes, indeed (assuming we are still talking LR bends) the Sch40 and Sch160 fittings would have the same K value using my J factor, or using the Crane method. The table at the top of page A-26 has a note "K is based on use of schedule pipe as listed on page 2-10". And page 2-10 seems to say that the K values apply to Schedules 40 to 160, but that the velocity that is used to calculate the velocity head must be based on the actual ID of the fitting. Makes sense to me. If you take a look at Figure 2-16 on page 2-13 of Crane 410 you will doubt every calculation you have ever made. This figure shows the variability in the experimental data on which the K values we use are based. A great deal of license has been used in getting to the "averages" we accept as gospel.
This question again highlights the reason I disagree with the Crane fT method. People want to start fiddling with the fT for their particular pipe, whereas Crane's intention was that if you have a 4" fitting you use an fT of 0.017 irrespective of the schedule or actual roughness of that pipe . That is why I suggested that we call it "J" and eliminate the false link to the friction factor that is confusing everybody. All the experimental work shows that the K value has almost no dependency on the roughness, and if you look at Fig 2-16 again you will see that there is no room for hair splitting here.
Using the Darby 3-K method to calculate K values for the Sch40 and Sch160 bends gives values on 0.28 and 0.29 respectively (at Re = 300,000). If I was doing a calculation that involved these fittings I would use these two different values for the two different schedules, but only for the sake of computational consistency and to allow anyone to later check my calcs using the same methods. In my heart I would know that in fact they are for all intents and purposes the same. Of course for the same flowrate the pressure drop is higher through the Sch160 bend because of the higher velocity, but not because of any real change in K value. To try to calculate the actual K value from the Crane method by interpolating fT between Sch40 and Sch160 is IMO like measuring the length of a football pitch with a micrometer.
For the 36" bend I would use 3-K and get a K value of about 0.18. The rate of decrease with size gets less as the sizes get bigger. You could use Crane fT of 0.0105 (interpolating from the figure on page A-24) because there is a substantial increment to 36" from the data on page A-26, but the fact that Crane neglected to give values on page A-26 for 36" pipe does not detract from my argument that theirprocedure is confusing.
@Sailoday28,Crane have neglected to give K values for welded or flanged Tee's. I have no idea why. As always I would use 3-K. A good reference for this type of data is the classic article by Larry Simpson and Martin Weirick (Chemical Engineering, April 3, 1978).
@pleckner,I am confused over what you are saying Phil. If we apply your example of polished SS with a roughness of 0.000005 inch to our 4" LR bend we would have to have a Reynolds number of over 100 million to get to full turbulence (See Crane A-24). Under these conditions you would get an fabout 0.007. This compares with an fT of 0.017 for a 4" fitting given on page A-26. Using the value of0.007 would make a 4" highly polished LR bend have a K value of 17x0.007 = 0.119 and not the 0.28 that I would calculate from Darby's 3-K. Is this what you are saying, or have I got the wrong end of the stick?
katmar (Chemical) 18 Dec 06 16:33
@pleckner,
I got confused between inches and feet for your roughness in my working of your example with the polished SS. The fT is more like 0.010, making the K value 0.17. The numbers are different from my earlier calc but the principle is the same. My question is, are you proposing that we use the calculated fT value of about 0.010 rather than Crane's 0.017 to calculate the K value of a highly polished LR bend?
Sorry for the calculation error - it’s nearly midnight here!
Have any of you looked at the “Handbook of Hydraulic Resistance” by I. E. Idelchik? It was my impression that this book was the ultimate reference for calculating flow resistance through fittings. Perhaps someone who studied this text might have some comments on how Idelchik handles these issues.
pleckner (Chemical) 18 Dec 06 19:55
Yes, CRANE TP410, page 2-10 says that the K values apply to Schedules 40 to 160 but they are not constant at all these schedules. Look again and you will see that Schedule 40 pipe only applies to Class300 and less; Schedule 80 pipe for Class 400 and 600 and so on. The way I interpret this is that you would have to adjust the K value using equation 2-5 for the different internal diameter if you were using a schedule 80 pipe in a Class 300 or less application. For example, 2" pipe, schedule 80 has an ID of 1.939" but that same pipe in schedule 40 has an ID of 2.067". If you were using schedule 80 pipein a Class 150 application, the K value would have to be adjusted accordingly because the K values arepublished for the schedule 40 Class 150 pipe, not the schedule 80 Class 150 pipe; different velocities.
@katmar:My example is to show how one can manipulate, or try to manipulate absolute roughness to suite the type of piping material they have rather than just blindly following the table on A-26 that most people do. And yes, for this instance, I don't see anything wrong with using a lower K value for polished stainless steel pipe (actually tubing for bio-pharm use) over clean commercial steel pipe. The frictionallosses will be significantly less for pharma pipe/tubing than for the standard chemical/petrochemical industry steel pipe.
BTW, I am doing a project for vegatable oil tank farm expansion and we are using a smaller diameter 304 SS pipe than I would normally choose for our flow rates (asked for by the client) because in their experience, there is essentially very little effective friction at all, the stuff just glides right down the pipe!
katmar (Chemical) 18 Dec 06 16:33
@pleckner,
I got confused between inches and feet for your roughness in my working of your example with the polished SS. The fT is more like 0.010, making the K value 0.17. The numbers are different from my earlier calc but the principle is the same. My question is, are you proposing that we use the calculated fT value of about 0.010 rather than Crane's 0.017 to calculate the K value of a highly polished LR bend?
Sorry for the calculation error - it’s nearly midnight here!
Have any of you looked at the “Handbook of Hydraulic Resistance” by I. E. Idelchik? It was my impression that this book was the ultimate reference for calculating flow resistance through fittings. Perhaps someone who studied this text might have some comments on how Idelchik handles these issues.
pleckner (Chemical) 18 Dec 06 19:55
Yes, CRANE TP410, page 2-10 says that the K values apply to Schedules 40 to 160 but they are not constant at all these schedules. Look again and you will see that Schedule 40 pipe only applies to Class300 and less; Schedule 80 pipe for Class 400 and 600 and so on. The way I interpret this is that you would have to adjust the K value using equation 2-5 for the different internal diameter if you were using a schedule 80 pipe in a Class 300 or less application. For example, 2" pipe, schedule 80 has an ID of 1.939" but that same pipe in schedule 40 has an ID of 2.067". If you were using schedule 80 pipein a Class 150 application, the K value would have to be adjusted accordingly because the K values arepublished for the schedule 40 Class 150 pipe, not the schedule 80 Class 150 pipe; different velocities.
@katmar:My example is to show how one can manipulate, or try to manipulate absolute roughness to suite the type of piping material they have rather than just blindly following the table on A-26 that most people do. And yes, for this instance, I don't see anything wrong with using a lower K value for polished stainless steel pipe (actually tubing for bio-pharm use) over clean commercial steel pipe. The frictionallosses will be significantly less for pharma pipe/tubing than for the standard chemical/petrochemical industry steel pipe.
BTW, I am doing a project for vegatable oil tank farm expansion and we are using a smaller diameter 304 SS pipe than I would normally choose for our flow rates (asked for by the client) because in their experience, there is essentially very little effective friction at all, the stuff just glides right down the pipe!
vzeos (Mechanical) 22 Dec 06 11:49
There seems to be a good deal of confusion about the relationship and use of the friction factor and the K-factor. Inspection of equations 2-1, 2-2, 2-3 and 2-4 on page 2-8 of Crane TP 410 (see below) should clarify the issue.
Equation 2-1, hL = v2/2g, this is the velocity head of a flowing fluid.
Equation 2-2, hL = K v2/2g, this defines the K-factor as the number of velocity heads lost due to a valve or fitting.
Equation 2-3, hL = (f L/D) v2/2g, this is the Darcy equation.
Equation 2-4, K = (f L/D), this is the familiar K-factor equation.
It should be completely evident that the link between the friction factor and the K-factor is established by virtue of the Darcy equation and that f is the Darcy friction factor. The purpose for inventing a K-factor and doing this algebraic manipulation is to develop a dimensionless group to be used for model scaling and to generalize experimental findings based on a limited number of experimental results. It should be further noted that the friction factor, f, applies only to straight pipe; there is no friction factor associated with any valve or fitting.
The Darcy friction factor can be obtained using your favorite charts or equations. For the special case of fully developed turbulent flow, fT, which represents the maximum possible friction factor, the Karman Rough Pipe Law should be used, namely 1/√fT = 2Log (3.7D/k) or fT = 1/[2Log (3.7D/k)] where k is the pipe roughness. There is no special significance given to the fT values provided on page A-26. Indeed, the first paragraph on page 2-10 states that these fT values are provided for “convenience.”
K-factor values for valves and fittings are determined experimentally by measuring the head lost due tothe valve or fitting during a flow test. Once the K-factor value is determined for a valve or fitting it can be equated to a hypothetical pipeline (having particular values for the parameters f, L and, D) by using equation 2-4. The values for the parameters of the hypothetical pipe are not set in stone.
For example:A flow test for a 2” valve produced a pressure loss equal to 3 velocity heads under fully developed turbulent flow. What is the length of 2.067” ID hypothetical pipe that would cause a 3 velocity head loss if the hypothetical pipe roughness were 0.0015”?
Using the rough pipe law, fT = 1/[2Log (3.7*2.067/0.0015)] 2 = 0.01819
Using Equation 2-4, LT = (3* 2.067)/0.01819 = 340.9” or 28.4 feetLT is the equivalent length of the valve under fully developed turbulent flow.
wfn217 (Chemical) 22 Dec 06 12:01
vzeos (Mechanical) 22 Dec 06 11:49
There seems to be a good deal of confusion about the relationship and use of the friction factor and the K-factor. Inspection of equations 2-1, 2-2, 2-3 and 2-4 on page 2-8 of Crane TP 410 (see below) should clarify the issue.
Equation 2-1, hL = v2/2g, this is the velocity head of a flowing fluid.
Equation 2-2, hL = K v2/2g, this defines the K-factor as the number of velocity heads lost due to a valve or fitting.
Equation 2-3, hL = (f L/D) v2/2g, this is the Darcy equation.
Equation 2-4, K = (f L/D), this is the familiar K-factor equation.
It should be completely evident that the link between the friction factor and the K-factor is established by virtue of the Darcy equation and that f is the Darcy friction factor. The purpose for inventing a K-factor and doing this algebraic manipulation is to develop a dimensionless group to be used for model scaling and to generalize experimental findings based on a limited number of experimental results. It should be further noted that the friction factor, f, applies only to straight pipe; there is no friction factor associated with any valve or fitting.
The Darcy friction factor can be obtained using your favorite charts or equations. For the special case of fully developed turbulent flow, fT, which represents the maximum possible friction factor, the Karman Rough Pipe Law should be used, namely 1/√fT = 2Log (3.7D/k) or fT = 1/[2Log (3.7D/k)] where k is the pipe roughness. There is no special significance given to the fT values provided on page A-26. Indeed, the first paragraph on page 2-10 states that these fT values are provided for “convenience.”
K-factor values for valves and fittings are determined experimentally by measuring the head lost due tothe valve or fitting during a flow test. Once the K-factor value is determined for a valve or fitting it can be equated to a hypothetical pipeline (having particular values for the parameters f, L and, D) by using equation 2-4. The values for the parameters of the hypothetical pipe are not set in stone.
For example:A flow test for a 2” valve produced a pressure loss equal to 3 velocity heads under fully developed turbulent flow. What is the length of 2.067” ID hypothetical pipe that would cause a 3 velocity head loss if the hypothetical pipe roughness were 0.0015”?
Using the rough pipe law, fT = 1/[2Log (3.7*2.067/0.0015)] 2 = 0.01819
Using Equation 2-4, LT = (3* 2.067)/0.01819 = 340.9” or 28.4 feetLT is the equivalent length of the valve under fully developed turbulent flow.
wfn217 (Chemical) 22 Dec 06 12:01
Crane 410 fittingsthread378-173164
wfn217 (Chemical) 13 Dec 06 8:49
In Crane TP 410, K for a fitting is found by multiplying a number times f T. fT is called the friction factor. Is fT related to the roughness?
dtn6770 (Mechanical) 13 Dec 06 9:45
fT is related to roughness by way of the Moody Diagram and its use of relative roughness. Relative roughness is the materials absolute roughness divided by the inside diameter.
STYMIEDPIPER (Mechanical) 13 Dec 06 10:53
Crane 410 lists values for fT
wfn217 (Chemical) 13 Dec 06 10:57
I wanted to know whether the listed fT values are strictly due to geometry or also a function of roughness.
BigInch (Petroleum) 13 Dec 06 11:05
geometry
vzeos (Mechanical) 13 Dec 06 12:03
wfn217- In Crane TP 410 fT is the fully turbulent flow friction factor defined by the equation below. Dis the pipe I. D. and k is the pipe roughness.
1/√fT= 2Log{(3.7D/k)
Crane's equivalent length calculations assume fully developed turbulent flow. See Crane TP 410, Page26. The fT values provided by Crane imply a k value of about 0.00180".
BigInch (Petroleum) 13 Dec 06 12:16
vzeos,
You are totally correct. Seems I reversed the question. K is determined by geometry, but the equivaletlength is indeed based on 0.0018 roughness.
DLANDISSR (Mechanical) 13 Dec 06 14:45
fT is the friction factor in the turbulent region of flow. It is calculated from the Colebrook-White equation which is an iteration. The Moody Diagram is derived from the Colebrook-White equation.
katmar (Chemical) 13 Dec 06 15:00
BigInch nailed the answer with his first post, but this question highlights what I consider to be one of the weaker points of the Crane 410 manual so please forgive me a little rant here. I have seen Crane's treatment of the K value for pipe fittings cause so much confusion - it really is a great pity they chose to do it this way.
The Crane engineers noted that the K values for fittings generally decreased as the fitting size increased. But it all went wrong when they noticed that this rate of decrease was close to the same as the rate at which the friction factor for fully developed turbulent flow in commercial steel pipe decreased as the pipe size increased. The fatal mistake was to link the two. See Crane Fig 2-14 and associated commentary.
For example, on page A-29 the K value for a 90 degree butt-weld pipe bend with an r/d of 1.5 is given as 14fT. The values of fT are given at the top of page A-26 as a function of pipe size. The values may have been calculated using the function referenced by vzeos, but for the purposes of calculating K values they are constants for each pipe size.
This apparent link between the K value and the friction factor gives the impression that the K value is linked to the pipe roughness - but in fact it is not because fT is defined to be at a particular roughness. Even worse, it is possible to be misled into believing the Crane K values compensate for changes in Reynolds number because everyone knows that the friction factor is influenced by the Reynolds number. But again, it is not because fT is defined to be in a particular Reynolds regime (fully turbulent).
To use the example of the 90 degree bend I gave above, it would have been better for Crane to give theK value as 14J, where J is simply a fudge-factor and would still be given by the values in the table on Page A-26 but without any reference to the friction factor. (Note that I have selected J as my symbol simply because it has no prior definition in the Crane Nomenclature table.)
The upshot of all of this is that in Crane's treatment, the K value of a fitting is a function onlypipe size (or geometry to use the terms used by wfn217 and BigInch). This was an improvement over previous work where the K value had been assumed to be constant for all sizes of fittings, and at the time that Crane first published this method it was rightly acclaimed as an important advance but IMHOit was badly worded and newer editions of 410 have unfortunately done nothing to remove the confusion.
I have awarded a star to BigInch for his comment that if you want to convert the Crane K value to an equivalent length, you must use the fT value from Crane's table on page A-26 (which is based on a roughness of 0.0018") and NOT the actual friction factor of the pipe you are using. The Crane description of this on pages 2-8 to 2-11 is extremely confusing, and the example 4-7 is just plain wrong because the K values given in the 410 manual apply only to fully developed turbulent flow and should never be used for laminar flow.
If you are working with laminar flow it is much better to work with equivalent lengths than with fixed or even Crane K values. Resistance values for fittings increase rapidly at low Reynolds numbers, but so does the friction factor. This means that if you use fixed L/D values, which get multiplied by the friction factor in the Darcy-Weisbach equation, the high resistance values are automatically compensated for. Or even better, use the 2-K or 3-K methods proposed by Hooper and Darby.
katmar (Chemical) 13 Dec 06 15:00
BigInch nailed the answer with his first post, but this question highlights what I consider to be one of the weaker points of the Crane 410 manual so please forgive me a little rant here. I have seen Crane's treatment of the K value for pipe fittings cause so much confusion - it really is a great pity they chose to do it this way.
The Crane engineers noted that the K values for fittings generally decreased as the fitting size increased. But it all went wrong when they noticed that this rate of decrease was close to the same as the rate at which the friction factor for fully developed turbulent flow in commercial steel pipe decreased as the pipe size increased. The fatal mistake was to link the two. See Crane Fig 2-14 and associated commentary.
For example, on page A-29 the K value for a 90 degree butt-weld pipe bend with an r/d of 1.5 is given as 14fT. The values of fT are given at the top of page A-26 as a function of pipe size. The values may have been calculated using the function referenced by vzeos, but for the purposes of calculating K values they are constants for each pipe size.
This apparent link between the K value and the friction factor gives the impression that the K value is linked to the pipe roughness - but in fact it is not because fT is defined to be at a particular roughness. Even worse, it is possible to be misled into believing the Crane K values compensate for changes in Reynolds number because everyone knows that the friction factor is influenced by the Reynolds number. But again, it is not because fT is defined to be in a particular Reynolds regime (fully turbulent).
To use the example of the 90 degree bend I gave above, it would have been better for Crane to give theK value as 14J, where J is simply a fudge-factor and would still be given by the values in the table on Page A-26 but without any reference to the friction factor. (Note that I have selected J as my symbol simply because it has no prior definition in the Crane Nomenclature table.)
The upshot of all of this is that in Crane's treatment, the K value of a fitting is a function onlypipe size (or geometry to use the terms used by wfn217 and BigInch). This was an improvement over previous work where the K value had been assumed to be constant for all sizes of fittings, and at the time that Crane first published this method it was rightly acclaimed as an important advance but IMHOit was badly worded and newer editions of 410 have unfortunately done nothing to remove the confusion.
I have awarded a star to BigInch for his comment that if you want to convert the Crane K value to an equivalent length, you must use the fT value from Crane's table on page A-26 (which is based on a roughness of 0.0018") and NOT the actual friction factor of the pipe you are using. The Crane description of this on pages 2-8 to 2-11 is extremely confusing, and the example 4-7 is just plain wrong because the K values given in the 410 manual apply only to fully developed turbulent flow and should never be used for laminar flow.
If you are working with laminar flow it is much better to work with equivalent lengths than with fixed or even Crane K values. Resistance values for fittings increase rapidly at low Reynolds numbers, but so does the friction factor. This means that if you use fixed L/D values, which get multiplied by the friction factor in the Darcy-Weisbach equation, the high resistance values are automatically compensated for. Or even better, use the 2-K or 3-K methods proposed by Hooper and Darby.
katmar (Chemical) 13 Dec 06 15:00
BigInch nailed the answer with his first post, but this question highlights what I consider to be one of the weaker points of the Crane 410 manual so please forgive me a little rant here. I have seen Crane's treatment of the K value for pipe fittings cause so much confusion - it really is a great pity they chose to do it this way.
The Crane engineers noted that the K values for fittings generally decreased as the fitting size increased. But it all went wrong when they noticed that this rate of decrease was close to the same as the rate at which the friction factor for fully developed turbulent flow in commercial steel pipe decreased as the pipe size increased. The fatal mistake was to link the two. See Crane Fig 2-14 and associated commentary.
For example, on page A-29 the K value for a 90 degree butt-weld pipe bend with an r/d of 1.5 is given as 14fT. The values of fT are given at the top of page A-26 as a function of pipe size. The values may have been calculated using the function referenced by vzeos, but for the purposes of calculating K values they are constants for each pipe size.
This apparent link between the K value and the friction factor gives the impression that the K value is linked to the pipe roughness - but in fact it is not because fT is defined to be at a particular roughness. Even worse, it is possible to be misled into believing the Crane K values compensate for changes in Reynolds number because everyone knows that the friction factor is influenced by the Reynolds number. But again, it is not because fT is defined to be in a particular Reynolds regime (fully turbulent).
To use the example of the 90 degree bend I gave above, it would have been better for Crane to give theK value as 14J, where J is simply a fudge-factor and would still be given by the values in the table on Page A-26 but without any reference to the friction factor. (Note that I have selected J as my symbol simply because it has no prior definition in the Crane Nomenclature table.)
The upshot of all of this is that in Crane's treatment, the K value of a fitting is a function onlypipe size (or geometry to use the terms used by wfn217 and BigInch). This was an improvement over previous work where the K value had been assumed to be constant for all sizes of fittings, and at the time that Crane first published this method it was rightly acclaimed as an important advance but IMHOit was badly worded and newer editions of 410 have unfortunately done nothing to remove the confusion.
I have awarded a star to BigInch for his comment that if you want to convert the Crane K value to an equivalent length, you must use the fT value from Crane's table on page A-26 (which is based on a roughness of 0.0018") and NOT the actual friction factor of the pipe you are using. The Crane description of this on pages 2-8 to 2-11 is extremely confusing, and the example 4-7 is just plain wrong because the K values given in the 410 manual apply only to fully developed turbulent flow and should never be used for laminar flow.
If you are working with laminar flow it is much better to work with equivalent lengths than with fixed or even Crane K values. Resistance values for fittings increase rapidly at low Reynolds numbers, but so does the friction factor. This means that if you use fixed L/D values, which get multiplied by the friction factor in the Darcy-Weisbach equation, the high resistance values are automatically compensated for. Or even better, use the 2-K or 3-K methods proposed by Hooper and Darby.
Montemayor (Chemical) 13 Dec 06 18:59
And I select Harvey as the preferred Ranter of the day and worthy - as well - of recognition for bringing to everyone's attention the importance of really understanding and reading through what is putin front of our eyes. We, as professional engineers, are not being asked to believe or accept 100% of what we are offered or given - regardless of how “sacred” the Cow may seem. Everything in engineering is subject to scrutiny and improvements.
I have a lot of respect and gratitude for what has gone into putting together Crane's Tech Paper #410. However, everything Harvey has stated regarding the concept of their K values is not only valid, but 100% positive criticism that should be heard and applied. Major world-class engineering firms agree with what Harvey states - and so do some of the biggest chemical process companies. To quote one: “Until recently, the use of K coefficients for valves and fittings has been considered more accurate than the use of equivalent lengths of pipe, but recent research has disclosed that K coefficients are not constant for all sizes of any one type of valve or fitting; so the use of equivalent lengths, with some exceptions, is now preferred.” And this is in addition to the problems of understanding/interpreting what TP 410 is saying. We have a better option, as Harvey states, in the 2-K or 3-K methods.
Great comments and good engineering knowledge, Harvey!
wfn217 (Chemical) 14 Dec 06 11:39
The number that Crane multiplies by fT to obtain K is the same number listed as L/D in Cameron
katmar (Chemical) 14 Dec 06 13:45
wfn217,
the Darcy-Weisbach correlation for pressure drop is
?P = ( fL/D + K ) ?V2 / 2
So if you want to express the resistance of a fitting in terms of the equivalent length (i.e. L/D) instead of K then you have to calculate
L/D = K/f
and since Crane express their K's as (Constant) x fT you would get
L/D = (Constant) x fT / f
If f is evaluated at the same conditions as fT (which is what my rant above was all about!) then of course
L/D = (Constant)
which means that for commercial steel pipe in the fully turbulent regime the (Constant) used by Crane will indeed be the L/D value given by Cameron. You just have to remember that because Crane evaluated these (Constants) for commercial steel pipe in the fully turbulent regime, if you want to get back to the L/D values you must use fT and not the friction factor in your particular case.
the Darcy-Weisbach correlation for pressure drop is
?P = ( fL/D + K ) ?V2 / 2
So if you want to express the resistance of a fitting in terms of the equivalent length (i.e. L/D) instead of K then you have to calculate
L/D = K/f
and since Crane express their K's as (Constant) x fT you would get
L/D = (Constant) x fT / f
If f is evaluated at the same conditions as fT (which is what my rant above was all about!) then of course
L/D = (Constant)
which means that for commercial steel pipe in the fully turbulent regime the (Constant) used by Crane will indeed be the L/D value given by Cameron. You just have to remember that because Crane evaluated these (Constants) for commercial steel pipe in the fully turbulent regime, if you want to get back to the L/D values you must use fT and not the friction factor in your particular case.
It has been found experimentally that the resistance factor (i.e. K value) for a fitting depends on both the fitting size and the Reynolds number. In the Darcy-Weisbach equation (in which I managed to get the wrong symbols before)
ΔP = ( fL/D + K ) ρV2 / 2
you can see that if you use a K value the pressure drop is not influenced by the fitting size or the Reynolds number. When you use Crane K values the value is corrected (via the table on page A-26) for fitting size before you insert the value into the D-W equation, but many of the older references gave single K values for all fitting sizes.
On the other hand, you can see that if you use equivalent lengths, then the value is multiplied by the actual friction factor for your application, and because the friction factor does depend on the fitting size and the Reynolds number it automatically corrects the calculation for both these factors. This makes the equivalent length method both easier to use, and more accurate than K values. It is easier to use because you only have to remember a single L/D value for a given type of fitting.
As Montemayor and I said before, the very best way is to use the 2-K or 3-K method but these are computationally much more involved. In my software I use 3-K because once it is programmed it makes no extra work to use it, but if I am doing a hand calc then I use the equivalent length method.
Hope this makes it clear
Harvey
tickle (Chemical) 14 Dec 06 18:17
Thank you Katmar for your wonderful explanation.
I have questioned Crane myself, but learnt to accept that its findings were as practical as required.
I think that younger engineers blindly use equations and computer programs - because they are there. They do not understand the applicable circumstances or the limitations. In fact whilst reading the post one of the junior engineers questioned me regarding recommended pipe diameters. They had sized thesteam piping for maximum flow and minimum pressure. They were blindly going to use a calculated diameter because it was less than or equal to the recommended velocity without considering the application and whether the worst case conditions could occur at the same time, (they can't and a smaller pipe diameter was my recommendation).
katmar (Chemical) 14 Dec 06 15:15
wfn217,
It has been found experimentally that the resistance factor (i.e. K value) for a fitting depends on both the fitting size and the Reynolds number. In the Darcy-Weisbach equation (in which I managed to get the wrong symbols before)
ΔP = ( fL/D + K ) ρV2 / 2
you can see that if you use a K value the pressure drop is not influenced by the fitting size or the Reynolds number. When you use Crane K values the value is corrected (via the table on page A-26) for fitting size before you insert the value into the D-W equation, but many of the older references gave single K values for all fitting sizes.
On the other hand, you can see that if you use equivalent lengths, then the value is multiplied by the actual friction factor for your application, and because the friction factor does depend on the fitting size and the Reynolds number it automatically corrects the calculation for both these factors. This makes the equivalent length method both easier to use, and more accurate than K values. It is easier to use because you only have to remember a single L/D value for a given type of fitting.
As Montemayor and I said before, the very best way is to use the 2-K or 3-K method but these are computationally much more involved. In my software I use 3-K because once it is programmed it makes no extra work to use it, but if I am doing a hand calc then I use the equivalent length method.
Hope this makes it clear
Harvey
tickle (Chemical) 14 Dec 06 18:17
Thank you Katmar for your wonderful explanation.
I have questioned Crane myself, but learnt to accept that its findings were as practical as required.
I think that younger engineers blindly use equations and computer programs - because they are there. They do not understand the applicable circumstances or the limitations. In fact whilst reading the post one of the junior engineers questioned me regarding recommended pipe diameters. They had sized thesteam piping for maximum flow and minimum pressure. They were blindly going to use a calculated diameter because it was less than or equal to the recommended velocity without considering the application and whether the worst case conditions could occur at the same time, (they can't and a smaller pipe diameter was my recommendation).
pleckner (Chemical) 14 Dec 06 20:11
I invite everyone to read my article on this subject published at
http://www.cheresources.com/eqlength.shtml
I don't agree that using equivalent lengths is the proper way to perform a system hydraulcs calculation (my arguments are in the above mentioned article). You must not mix the friction factor for a fitting with the friction factor of a pipe because they are not the same, whether we are talking about CRANE'sfriction factor at fully developed turbulent flow or the two-K or three-K methods. The pressure loss of a fitting obtained from these methods is not the same nor nearly the same as the pressure loss of a fitting obtained from multiplying the pipe friction factor x the equivalent length of the fitting.
The most accurate way to perform the calculation is to use one of these methods (three-k perferred) and add the losses to those of the pipe, not to combine them all into an equivalent length of straight pipe and multiply by the pipe friction factor.
The two-K or three-K methods were developed to address the fact that fittings are indded somewhat dependent on Reynolds number, but this is still not the same as a pipe friction factor.
I will also argue that for fully developed turbulent flow, there is nothing wrong with CRANE's values.
katmar (Chemical) 15 Dec 06 5:15
pleckner,
There is a lot more that we agree on than on which we differ. It was never my intention to suggest that there is anything wrong with applying Crane's K values to fully developed turbulent flow. I have recommended MANY times here that people should get their hands on a copy of Crane 410. Probably 95% or more of our flow calcs are for fully developed turbulent flow, and Crane is ideal for this.
My main objection was to the confusion caused by Crane's wording and their linking of the K value to the turbulent friction factor. I believe the posts in this thread confirm that this confusion is widespread, and I have often seen engineers struggle to come to terms with it. Despite the shortcomings of Crane 410, my copy "lives" in the very front of the top drawer of the filing cabinet right next to my desk.
We all agree that the multi-K methods are better than the L/D method. However, I disagree with Crane's statement (which is echoed in your article) that “K is a constant under all flow conditions, including laminar flow”. In your example you give the K value for a long radius bend as 0.36. Applying the 3-K method at a Reynolds number of 100 in the same sized pipe gives a K value of 8.3. This means that a pressure drop calculation using the Crane value will be 90% understated, if we take 3-K as the benchmark. I accept that a similar calculation at Re = 100 using the L/D method will over-estimate the pressure drop but for most calculations this would be the conservative option.
I have one objection to the example in your article. 92% of the pressure drop is due to the reducer, but nobody would ever try to calculate the pressure drop through a reducer with the L/D method. Including the reducer exaggerates the negative aspects of the L/D method.
Neglecting the reducer, the calculated pressure drops using the various methods are
Who would put his neck on the block over which of these numbers is correct? The L/D and Crane answers are within 15% of the 3-K answer, and all three answers are probably adequate for practical purposes.Katmar SoftwareEngineering & Risk Analysis Software
BigInch (Petroleum) 15 Dec 06 6:12
That's my humble opinion also. With all the other inaccuracies inherent in hydraulics and more specifically how systems are usually operated at one or the other of the extreme ranges, if you design a system close enough so that you have to worry about the difference in pressure drops from laminar or turbulent flow in a few fittings, that's just plain tooooo close. If for some reason you need to examine an existing system, the actual data's there for the taking.
katmar (Chemical) 15 Dec 06 5:15
pleckner,
There is a lot more that we agree on than on which we differ. It was never my intention to suggest that there is anything wrong with applying Crane's K values to fully developed turbulent flow. I have recommended MANY times here that people should get their hands on a copy of Crane 410. Probably 95% or more of our flow calcs are for fully developed turbulent flow, and Crane is ideal for this.
My main objection was to the confusion caused by Crane's wording and their linking of the K value to the turbulent friction factor. I believe the posts in this thread confirm that this confusion is widespread, and I have often seen engineers struggle to come to terms with it. Despite the shortcomings of Crane 410, my copy "lives" in the very front of the top drawer of the filing cabinet right next to my desk.
We all agree that the multi-K methods are better than the L/D method. However, I disagree with Crane's statement (which is echoed in your article) that “K is a constant under all flow conditions, including laminar flow”. In your example you give the K value for a long radius bend as 0.36. Applying the 3-K method at a Reynolds number of 100 in the same sized pipe gives a K value of 8.3. This means that a pressure drop calculation using the Crane value will be 90% understated, if we take 3-K as the benchmark. I accept that a similar calculation at Re = 100 using the L/D method will over-estimate the pressure drop but for most calculations this would be the conservative option.
I have one objection to the example in your article. 92% of the pressure drop is due to the reducer, but nobody would ever try to calculate the pressure drop through a reducer with the L/D method. Including the reducer exaggerates the negative aspects of the L/D method.
Neglecting the reducer, the calculated pressure drops using the various methods are
Who would put his neck on the block over which of these numbers is correct? The L/D and Crane answers are within 15% of the 3-K answer, and all three answers are probably adequate for practical purposes.Katmar SoftwareEngineering & Risk Analysis Software
BigInch (Petroleum) 15 Dec 06 6:12
That's my humble opinion also. With all the other inaccuracies inherent in hydraulics and more specifically how systems are usually operated at one or the other of the extreme ranges, if you design a system close enough so that you have to worry about the difference in pressure drops from laminar or turbulent flow in a few fittings, that's just plain tooooo close. If for some reason you need to examine an existing system, the actual data's there for the taking.
pleckner (Chemical) 15 Dec 06 8:01
We are in agreement.
I put the reducers into the L/D equation only as a means for comparison. I agree with you (Katmar) onthat. But then again, I don't use the equivalent length method, ever.
Katmar, I'm one up on you, I keep CRANE on my desk, I'm too lazy to go into my file cabinet.
wfn217 (Chemical) 15 Dec 06 9:09
It should be noted that pleckner's article states that K for a fitting has little to do with friction, which answers my original question. I have seen someone try to scale the fT values in Crane from pipe to tubing based on roughness.
LeSabre (Petroleum) 15 Dec 06 10:07
pleckner,
Great article. I appreciate your logical presentation. I am curious, however, about your representation that in 1979 Crane “...discussed and used the two-friction factor method for calculating the total pressure drop in a piping system...(f for straight pipe and ft for valves and fittings)”. It is my understanding from reading the TP 410 Foreword (4th paragraph, quoted below) that Crane’s intent was to have the pipe friction factor, f, apply to the K factor calculation and that the K factor was intended to apply to the full range of flow regimes from laminar to full turbulence. Can you comment on this?
Quote (TP 410 FOREWORD):The fifteenth printing (1976 edition) presented a conceptual change regarding the values of EquivalentLength "L/D" and Resistance Coefficient "K" for valves and fittings relative to the friction factor in pipes. This change has relatively minor effect on most problems dealing with flow conditions that result in Reynolds numbers falling in the turbulent zone. However, for flow in the laminar zone, the change avoids a significant overstatement of pressure drop. Consistent with this conceptual revision, the resistance to flow through valves and fittings is now expressed in terms of resistance co-efficient
pleckner (Chemical) 15 Dec 06 8:01
We are in agreement.
I put the reducers into the L/D equation only as a means for comparison. I agree with you (Katmar) onthat. But then again, I don't use the equivalent length method, ever.
Katmar, I'm one up on you, I keep CRANE on my desk, I'm too lazy to go into my file cabinet.
wfn217 (Chemical) 15 Dec 06 9:09
It should be noted that pleckner's article states that K for a fitting has little to do with friction, which answers my original question. I have seen someone try to scale the fT values in Crane from pipe to tubing based on roughness.
LeSabre (Petroleum) 15 Dec 06 10:07
pleckner,
Great article. I appreciate your logical presentation. I am curious, however, about your representation that in 1979 Crane “...discussed and used the two-friction factor method for calculating the total pressure drop in a piping system...(f for straight pipe and ft for valves and fittings)”. It is my understanding from reading the TP 410 Foreword (4th paragraph, quoted below) that Crane’s intent was to have the pipe friction factor, f, apply to the K factor calculation and that the K factor was intended to apply to the full range of flow regimes from laminar to full turbulence. Can you comment on this?
Quote (TP 410 FOREWORD):The fifteenth printing (1976 edition) presented a conceptual change regarding the values of EquivalentLength "L/D" and Resistance Coefficient "K" for valves and fittings relative to the friction factor in pipes. This change has relatively minor effect on most problems dealing with flow conditions that result in Reynolds numbers falling in the turbulent zone. However, for flow in the laminar zone, the change avoids a significant overstatement of pressure drop. Consistent with this conceptual revision, the resistance to flow through valves and fittings is now expressed in terms of resistance co-efficient
katmar (Chemical) 15 Dec 06 11:22
LeSabre,
The example in pleckner's article shows how the "old" (i.e. pre-1976) L/D method over-estimates the pressure drop in fittings for low Reynolds numbers. All my calcs agree with and confirm pleckner's result. Crane were quite correct to make this claim in the 1976 foreword.
It is true that using the “new” K values avoids the overstatement of the pressure drop in the laminar regime, but my example (See 5 Dec 5:15) shows how the Crane K values now understate the pressure drop by 90% at a Reynolds number of 100. If you were designing a new pipeline would you rather estimate the pressure drop as 40% too high or 90% too low?
Your query, and the latest post by wfn217, are typical examples of the confusion that is caused by the Crane method, and which lead to my rant near the start of this thread.
I have probably stretched everyone's patience to the limit by going on and on about this problem so I will leave it at this now. Katmar SoftwareEngineering & Risk Analysis Software
pleckner (Chemical) 15 Dec 06 12:49
To LaSabre:
Thanks.
To me that statement in the Foreword just means that they made a change on how they relate the new K value to pipe friction factor. The statement obviously doesn't give any specifics as to how to apply the two, that is left to the rest of the document. Indeed, everywhere K for a valve or fitting is calculatedwithin TP410 they are very careful to note the friction factor as fT and NOT f.
CRANE notes that the friction factors (f and fT) will essentially be the same in the zone of complete turbulence.
But you are correct in that it appears CRANE was intent on having K applied throughout all flow zones and that is where they got it wrong as we've been pointing out in these Postings.
ccfowler (Mechanical) 16 Dec 06 2:03
katmar,
Thanks for your comments and their effects on this discussion.
BigInch,You have pointed toward one of my pet peeves, the dreadful misunderstandings brought about by the ability of computers and calculators to thrash around 10 digit numbers. If the accuracy to which most parameters are known is only 1%, 5%, 10%, and sometimes even worse, most of those many digits presented by the mighty computer or calculator are really just so much drivel. Unfortunately, there is never a shortage of people who want to believe that all of those digits are significant (and are willing to
BigInch (Petroleum) 16 Dec 06 3:37
The discussion has been very informative, but there is a certain logic for Crane to just continuing to publish equivalent lengths with conservative roughness at turbulent flows... its conservative and that's exactly how a lookup table intended for general application to all hydraulic systems should be. Somewhere in Tips, I recall seeing a comment I liked questioning how many digits were actually needed before the answer became believable, or something to that effect. Guess we should have stayed with 8 bit computers. Even if all variables were known to 1%, it seems to me that most systems would spend an excessive amount of time operating outside the range where those accuracies were valid. Lastly, risking repeating myself, sooner or later any given system reaches capacity at one extreme end of operational range or another, so over the long term is using Eq.Len at max turbulence really a bad thing? BigInch-born in the trenches. LeSabre (Petroleum) 18 Dec 06 10:40
katmar, Thanks for directing my attention to your “rant”. I agree that the ft factor depends on geometry. Your j-factor theory, however, leads to the unlikely result that two different pipe I.D.s, say, 4” sch 40 (I.D. = 4.0260”) and 4” sch 160 (I.D. = 3.4380”) could have the same K-factor. How do you reconcile this situation? Also, how would you get the K-factor for a 36" XS L.R. Ell.? Thanks. sailoday28 (Mechanical) 18 Dec 06 12:11
And the L/d or K for a run or branch of a "T" is? Regards
BigInch (Petroleum) 18 Dec 06 12:28
You could have a look at these that Weldbend gives for fittings, including straight tees and they even give a method for assemblies as well, http://www.weldbend.com/Technical%20Data/Flow%20Resistance%20For%20Fittings/flowresistance.htm and click on <next> for the following page BigInch-born in the trenches. LeSabre (Petroleum) 18 Dec 06 12:44
BigInch, Great site, thanks! But I didn't see any info on K-factors or ft.
katmar (Chemical) 15 Dec 06 11:22
LeSabre,
The example in pleckner's article shows how the "old" (i.e. pre-1976) L/D method over-estimates the pressure drop in fittings for low Reynolds numbers. All my calcs agree with and confirm pleckner's result. Crane were quite correct to make this claim in the 1976 foreword.
It is true that using the “new” K values avoids the overstatement of the pressure drop in the laminar regime, but my example (See 5 Dec 5:15) shows how the Crane K values now understate the pressure drop by 90% at a Reynolds number of 100. If you were designing a new pipeline would you rather estimate the pressure drop as 40% too high or 90% too low?
Your query, and the latest post by wfn217, are typical examples of the confusion that is caused by the Crane method, and which lead to my rant near the start of this thread.
I have probably stretched everyone's patience to the limit by going on and on about this problem so I will leave it at this now. Katmar SoftwareEngineering & Risk Analysis Software
pleckner (Chemical) 15 Dec 06 12:49
To LaSabre:
Thanks.
To me that statement in the Foreword just means that they made a change on how they relate the new K value to pipe friction factor. The statement obviously doesn't give any specifics as to how to apply the two, that is left to the rest of the document. Indeed, everywhere K for a valve or fitting is calculatedwithin TP410 they are very careful to note the friction factor as fT and NOT f.
CRANE notes that the friction factors (f and fT) will essentially be the same in the zone of complete turbulence.
But you are correct in that it appears CRANE was intent on having K applied throughout all flow zones and that is where they got it wrong as we've been pointing out in these Postings.
ccfowler (Mechanical) 16 Dec 06 2:03
katmar,
Thanks for your comments and their effects on this discussion.
BigInch,You have pointed toward one of my pet peeves, the dreadful misunderstandings brought about by the ability of computers and calculators to thrash around 10 digit numbers. If the accuracy to which most parameters are known is only 1%, 5%, 10%, and sometimes even worse, most of those many digits presented by the mighty computer or calculator are really just so much drivel. Unfortunately, there is never a shortage of people who want to believe that all of those digits are significant (and are willing to
BigInch (Petroleum) 16 Dec 06 3:37
The discussion has been very informative, but there is a certain logic for Crane to just continuing to publish equivalent lengths with conservative roughness at turbulent flows... its conservative and that's exactly how a lookup table intended for general application to all hydraulic systems should be. Somewhere in Tips, I recall seeing a comment I liked questioning how many digits were actually needed before the answer became believable, or something to that effect. Guess we should have stayed with 8 bit computers. Even if all variables were known to 1%, it seems to me that most systems would spend an excessive amount of time operating outside the range where those accuracies were valid. Lastly, risking repeating myself, sooner or later any given system reaches capacity at one extreme end of operational range or another, so over the long term is using Eq.Len at max turbulence really a bad thing? BigInch-born in the trenches. LeSabre (Petroleum) 18 Dec 06 10:40
katmar, Thanks for directing my attention to your “rant”. I agree that the ft factor depends on geometry. Your j-factor theory, however, leads to the unlikely result that two different pipe I.D.s, say, 4” sch 40 (I.D. = 4.0260”) and 4” sch 160 (I.D. = 3.4380”) could have the same K-factor. How do you reconcile this situation? Also, how would you get the K-factor for a 36" XS L.R. Ell.? Thanks. sailoday28 (Mechanical) 18 Dec 06 12:11
And the L/d or K for a run or branch of a "T" is? Regards
BigInch (Petroleum) 18 Dec 06 12:28
You could have a look at these that Weldbend gives for fittings, including straight tees and they even give a method for assemblies as well, http://www.weldbend.com/Technical%20Data/Flow%20Resistance%20For%20Fittings/flowresistance.htm and click on <next> for the following page BigInch-born in the trenches. LeSabre (Petroleum) 18 Dec 06 12:44
BigInch, Great site, thanks! But I didn't see any info on K-factors or ft.
pleckner (Chemical) 18 Dec 06 13:03
Let's remember that CRANE TP410 doesn't publish equivalent lenghts for valves and fittings, it provides K values or the means to calculate K values from which we can then calculate the equivalent length using the appropriate friction factor at fully developed turbulent flow.
Let us also not forget that we can obtain various fT values from the graph on Page A-23. This graph allows us to choose the fT for the type of pipe being used (and this translates into a constant absolute roughness for that type of pipe) and the diameter of that pipe. So I don't agree that CRANE TP410 necessarily publishes conservative values of roughness. It all depends on how you want to manipulate the piping material you are using, e.g. For pharmaceutical grade highly polished SS pipe I might choose to use the roughness for Drawn Tubing (an absolute roughness of 0.000005 feet) rather than clean commercial steel pipe (0.00015 feet). Note that with this table, I can also try to interpolate and obtain fT for various pipe diameters!
Let us also not forget that the published fT values given in the table at the top of Page A-26 are strictly for CLEAN COMMERCIAL STEEL PIPE and for the schedule of pipe listed on Page 2-10. If one desires the the K value for another fitting of a different schedule pipe, they should adjust it using Equation 2-5 on that same page or go to the Graph on Page A-23.
One last thing for now, one can still use published equivalent lenghts as per the reference given by BigInch but just remember to calculate the pressure loss through that fitting using fT for that fitting rather than the pipe friction factor as I've shown in my paper and my post above.
BigInch (Petroleum) 18 Dec 06 13:12
on the <next> page, they give a method for dP, from which you can backcalculate Ks, right?
Let's remember that CRANE TP410 doesn't publish equivalent lenghts for valves and fittings, it provides K values or the means to calculate K values from which we can then calculate the equivalent length using the appropriate friction factor at fully developed turbulent flow.
Let us also not forget that we can obtain various fT values from the graph on Page A-23. This graph allows us to choose the fT for the type of pipe being used (and this translates into a constant absolute roughness for that type of pipe) and the diameter of that pipe. So I don't agree that CRANE TP410 necessarily publishes conservative values of roughness. It all depends on how you want to manipulate the piping material you are using, e.g. For pharmaceutical grade highly polished SS pipe I might choose to use the roughness for Drawn Tubing (an absolute roughness of 0.000005 feet) rather than clean commercial steel pipe (0.00015 feet). Note that with this table, I can also try to interpolate and obtain fT for various pipe diameters!
Let us also not forget that the published fT values given in the table at the top of Page A-26 are strictly for CLEAN COMMERCIAL STEEL PIPE and for the schedule of pipe listed on Page 2-10. If one desires the the K value for another fitting of a different schedule pipe, they should adjust it using Equation 2-5 on that same page or go to the Graph on Page A-23.
One last thing for now, one can still use published equivalent lenghts as per the reference given by BigInch but just remember to calculate the pressure loss through that fitting using fT for that fitting rather than the pipe friction factor as I've shown in my paper and my post above.
BigInch (Petroleum) 18 Dec 06 13:12
on the <next> page, they give a method for dP, from which you can backcalculate Ks, right?
Yes, indeed (assuming we are still talking LR bends) the Sch40 and Sch160 fittings would have the same K value using my J factor, or using the Crane method. The table at the top of page A-26 has a note "K is based on use of schedule pipe as listed on page 2-10". And page 2-10 seems to say that the K values apply to Schedules 40 to 160, but that the velocity that is used to calculate the velocity head must be based on the actual ID of the fitting. Makes sense to me. If you take a look at Figure 2-16 on page 2-13 of Crane 410 you will doubt every calculation you have ever made. This figure shows the variability in the experimental data on which the K values we use are based. A great deal of license has been used in getting to the "averages" we accept as gospel.
This question again highlights the reason I disagree with the Crane fT method. People want to start fiddling with the fT for their particular pipe, whereas Crane's intention was that if you have a 4" fitting you use an fT of 0.017 irrespective of the schedule or actual roughness of that pipe . That is why I suggested that we call it "J" and eliminate the false link to the friction factor that is confusing everybody. All the experimental work shows that the K value has almost no dependency on the roughness, and if you look at Fig 2-16 again you will see that there is no room for hair splitting here.
Using the Darby 3-K method to calculate K values for the Sch40 and Sch160 bends gives values on 0.28 and 0.29 respectively (at Re = 300,000). If I was doing a calculation that involved these fittings I would use these two different values for the two different schedules, but only for the sake of computational consistency and to allow anyone to later check my calcs using the same methods. In my heart I would know that in fact they are for all intents and purposes the same. Of course for the same flowrate the pressure drop is higher through the Sch160 bend because of the higher velocity, but not because of any real change in K value. To try to calculate the actual K value from the Crane method by interpolating fT between Sch40 and Sch160 is IMO like measuring the length of a football pitch with a micrometer.
For the 36" bend I would use 3-K and get a K value of about 0.18. The rate of decrease with size gets less as the sizes get bigger. You could use Crane fT of 0.0105 (interpolating from the figure on page A-24) because there is a substantial increment to 36" from the data on page A-26, but the fact that Crane neglected to give values on page A-26 for 36" pipe does not detract from my argument that theirprocedure is confusing.
@Sailoday28,Crane have neglected to give K values for welded or flanged Tee's. I have no idea why. As always I would use 3-K. A good reference for this type of data is the classic article by Larry Simpson and Martin Weirick (Chemical Engineering, April 3, 1978).
@pleckner,I am confused over what you are saying Phil. If we apply your example of polished SS with a roughness of 0.000005 inch to our 4" LR bend we would have to have a Reynolds number of over 100 million to get to full turbulence (See Crane A-24). Under these conditions you would get an fabout 0.007. This compares with an fT of 0.017 for a 4" fitting given on page A-26. Using the value of0.007 would make a 4" highly polished LR bend have a K value of 17x0.007 = 0.119 and not the 0.28 that I would calculate from Darby's 3-K. Is this what you are saying, or have I got the wrong end of the stick?
katmar (Chemical) 18 Dec 06 15:29
@LeSabre,
Yes, indeed (assuming we are still talking LR bends) the Sch40 and Sch160 fittings would have the same K value using my J factor, or using the Crane method. The table at the top of page A-26 has a note "K is based on use of schedule pipe as listed on page 2-10". And page 2-10 seems to say that the K values apply to Schedules 40 to 160, but that the velocity that is used to calculate the velocity head must be based on the actual ID of the fitting. Makes sense to me. If you take a look at Figure 2-16 on page 2-13 of Crane 410 you will doubt every calculation you have ever made. This figure shows the variability in the experimental data on which the K values we use are based. A great deal of license has been used in getting to the "averages" we accept as gospel.
This question again highlights the reason I disagree with the Crane fT method. People want to start fiddling with the fT for their particular pipe, whereas Crane's intention was that if you have a 4" fitting you use an fT of 0.017 irrespective of the schedule or actual roughness of that pipe . That is why I suggested that we call it "J" and eliminate the false link to the friction factor that is confusing everybody. All the experimental work shows that the K value has almost no dependency on the roughness, and if you look at Fig 2-16 again you will see that there is no room for hair splitting here.
Using the Darby 3-K method to calculate K values for the Sch40 and Sch160 bends gives values on 0.28 and 0.29 respectively (at Re = 300,000). If I was doing a calculation that involved these fittings I would use these two different values for the two different schedules, but only for the sake of computational consistency and to allow anyone to later check my calcs using the same methods. In my heart I would know that in fact they are for all intents and purposes the same. Of course for the same flowrate the pressure drop is higher through the Sch160 bend because of the higher velocity, but not because of any real change in K value. To try to calculate the actual K value from the Crane method by interpolating fT between Sch40 and Sch160 is IMO like measuring the length of a football pitch with a micrometer.
For the 36" bend I would use 3-K and get a K value of about 0.18. The rate of decrease with size gets less as the sizes get bigger. You could use Crane fT of 0.0105 (interpolating from the figure on page A-24) because there is a substantial increment to 36" from the data on page A-26, but the fact that Crane neglected to give values on page A-26 for 36" pipe does not detract from my argument that theirprocedure is confusing.
@Sailoday28,Crane have neglected to give K values for welded or flanged Tee's. I have no idea why. As always I would use 3-K. A good reference for this type of data is the classic article by Larry Simpson and Martin Weirick (Chemical Engineering, April 3, 1978).
@pleckner,I am confused over what you are saying Phil. If we apply your example of polished SS with a roughness of 0.000005 inch to our 4" LR bend we would have to have a Reynolds number of over 100 million to get to full turbulence (See Crane A-24). Under these conditions you would get an fabout 0.007. This compares with an fT of 0.017 for a 4" fitting given on page A-26. Using the value of0.007 would make a 4" highly polished LR bend have a K value of 17x0.007 = 0.119 and not the 0.28 that I would calculate from Darby's 3-K. Is this what you are saying, or have I got the wrong end of the stick?
katmar (Chemical) 18 Dec 06 16:33
@pleckner,
I got confused between inches and feet for your roughness in my working of your example with the polished SS. The fT is more like 0.010, making the K value 0.17. The numbers are different from my earlier calc but the principle is the same. My question is, are you proposing that we use the calculated fT value of about 0.010 rather than Crane's 0.017 to calculate the K value of a highly polished LR bend?
Sorry for the calculation error - it’s nearly midnight here!
Have any of you looked at the “Handbook of Hydraulic Resistance” by I. E. Idelchik? It was my impression that this book was the ultimate reference for calculating flow resistance through fittings. Perhaps someone who studied this text might have some comments on how Idelchik handles these issues.
pleckner (Chemical) 18 Dec 06 19:55
Yes, CRANE TP410, page 2-10 says that the K values apply to Schedules 40 to 160 but they are not constant at all these schedules. Look again and you will see that Schedule 40 pipe only applies to Class300 and less; Schedule 80 pipe for Class 400 and 600 and so on. The way I interpret this is that you would have to adjust the K value using equation 2-5 for the different internal diameter if you were using a schedule 80 pipe in a Class 300 or less application. For example, 2" pipe, schedule 80 has an ID of 1.939" but that same pipe in schedule 40 has an ID of 2.067". If you were using schedule 80 pipein a Class 150 application, the K value would have to be adjusted accordingly because the K values arepublished for the schedule 40 Class 150 pipe, not the schedule 80 Class 150 pipe; different velocities.
@katmar:My example is to show how one can manipulate, or try to manipulate absolute roughness to suite the type of piping material they have rather than just blindly following the table on A-26 that most people do. And yes, for this instance, I don't see anything wrong with using a lower K value for polished stainless steel pipe (actually tubing for bio-pharm use) over clean commercial steel pipe. The frictionallosses will be significantly less for pharma pipe/tubing than for the standard chemical/petrochemical industry steel pipe.
BTW, I am doing a project for vegatable oil tank farm expansion and we are using a smaller diameter 304 SS pipe than I would normally choose for our flow rates (asked for by the client) because in their experience, there is essentially very little effective friction at all, the stuff just glides right down the pipe!
katmar (Chemical) 18 Dec 06 16:33
@pleckner,
I got confused between inches and feet for your roughness in my working of your example with the polished SS. The fT is more like 0.010, making the K value 0.17. The numbers are different from my earlier calc but the principle is the same. My question is, are you proposing that we use the calculated fT value of about 0.010 rather than Crane's 0.017 to calculate the K value of a highly polished LR bend?
Sorry for the calculation error - it’s nearly midnight here!
Have any of you looked at the “Handbook of Hydraulic Resistance” by I. E. Idelchik? It was my impression that this book was the ultimate reference for calculating flow resistance through fittings. Perhaps someone who studied this text might have some comments on how Idelchik handles these issues.
pleckner (Chemical) 18 Dec 06 19:55
Yes, CRANE TP410, page 2-10 says that the K values apply to Schedules 40 to 160 but they are not constant at all these schedules. Look again and you will see that Schedule 40 pipe only applies to Class300 and less; Schedule 80 pipe for Class 400 and 600 and so on. The way I interpret this is that you would have to adjust the K value using equation 2-5 for the different internal diameter if you were using a schedule 80 pipe in a Class 300 or less application. For example, 2" pipe, schedule 80 has an ID of 1.939" but that same pipe in schedule 40 has an ID of 2.067". If you were using schedule 80 pipein a Class 150 application, the K value would have to be adjusted accordingly because the K values arepublished for the schedule 40 Class 150 pipe, not the schedule 80 Class 150 pipe; different velocities.
@katmar:My example is to show how one can manipulate, or try to manipulate absolute roughness to suite the type of piping material they have rather than just blindly following the table on A-26 that most people do. And yes, for this instance, I don't see anything wrong with using a lower K value for polished stainless steel pipe (actually tubing for bio-pharm use) over clean commercial steel pipe. The frictionallosses will be significantly less for pharma pipe/tubing than for the standard chemical/petrochemical industry steel pipe.
BTW, I am doing a project for vegatable oil tank farm expansion and we are using a smaller diameter 304 SS pipe than I would normally choose for our flow rates (asked for by the client) because in their experience, there is essentially very little effective friction at all, the stuff just glides right down the pipe!
vzeos (Mechanical) 22 Dec 06 11:49
There seems to be a good deal of confusion about the relationship and use of the friction factor and the K-factor. Inspection of equations 2-1, 2-2, 2-3 and 2-4 on page 2-8 of Crane TP 410 (see below) should clarify the issue.
Equation 2-1, hL = v2/2g, this is the velocity head of a flowing fluid.
Equation 2-2, hL = K v2/2g, this defines the K-factor as the number of velocity heads lost due to a valve or fitting.
Equation 2-3, hL = (f L/D) v2/2g, this is the Darcy equation.
Equation 2-4, K = (f L/D), this is the familiar K-factor equation.
It should be completely evident that the link between the friction factor and the K-factor is established by virtue of the Darcy equation and that f is the Darcy friction factor. The purpose for inventing a K-factor and doing this algebraic manipulation is to develop a dimensionless group to be used for model scaling and to generalize experimental findings based on a limited number of experimental results. It should be further noted that the friction factor, f, applies only to straight pipe; there is no friction factor associated with any valve or fitting.
The Darcy friction factor can be obtained using your favorite charts or equations. For the special case of fully developed turbulent flow, fT, which represents the maximum possible friction factor, the Karman Rough Pipe Law should be used, namely 1/√fT = 2Log (3.7D/k) or fT = 1/[2Log (3.7D/k)] where k is the pipe roughness. There is no special significance given to the fT values provided on page A-26. Indeed, the first paragraph on page 2-10 states that these fT values are provided for “convenience.”
K-factor values for valves and fittings are determined experimentally by measuring the head lost due tothe valve or fitting during a flow test. Once the K-factor value is determined for a valve or fitting it can be equated to a hypothetical pipeline (having particular values for the parameters f, L and, D) by using equation 2-4. The values for the parameters of the hypothetical pipe are not set in stone.
For example:A flow test for a 2” valve produced a pressure loss equal to 3 velocity heads under fully developed turbulent flow. What is the length of 2.067” ID hypothetical pipe that would cause a 3 velocity head loss if the hypothetical pipe roughness were 0.0015”?
Using the rough pipe law, fT = 1/[2Log (3.7*2.067/0.0015)] 2 = 0.01819
Using Equation 2-4, LT = (3* 2.067)/0.01819 = 340.9” or 28.4 feetLT is the equivalent length of the valve under fully developed turbulent flow.
wfn217 (Chemical) 22 Dec 06 12:01
vzeos (Mechanical) 22 Dec 06 11:49
There seems to be a good deal of confusion about the relationship and use of the friction factor and the K-factor. Inspection of equations 2-1, 2-2, 2-3 and 2-4 on page 2-8 of Crane TP 410 (see below) should clarify the issue.
Equation 2-1, hL = v2/2g, this is the velocity head of a flowing fluid.
Equation 2-2, hL = K v2/2g, this defines the K-factor as the number of velocity heads lost due to a valve or fitting.
Equation 2-3, hL = (f L/D) v2/2g, this is the Darcy equation.
Equation 2-4, K = (f L/D), this is the familiar K-factor equation.
It should be completely evident that the link between the friction factor and the K-factor is established by virtue of the Darcy equation and that f is the Darcy friction factor. The purpose for inventing a K-factor and doing this algebraic manipulation is to develop a dimensionless group to be used for model scaling and to generalize experimental findings based on a limited number of experimental results. It should be further noted that the friction factor, f, applies only to straight pipe; there is no friction factor associated with any valve or fitting.
The Darcy friction factor can be obtained using your favorite charts or equations. For the special case of fully developed turbulent flow, fT, which represents the maximum possible friction factor, the Karman Rough Pipe Law should be used, namely 1/√fT = 2Log (3.7D/k) or fT = 1/[2Log (3.7D/k)] where k is the pipe roughness. There is no special significance given to the fT values provided on page A-26. Indeed, the first paragraph on page 2-10 states that these fT values are provided for “convenience.”
K-factor values for valves and fittings are determined experimentally by measuring the head lost due tothe valve or fitting during a flow test. Once the K-factor value is determined for a valve or fitting it can be equated to a hypothetical pipeline (having particular values for the parameters f, L and, D) by using equation 2-4. The values for the parameters of the hypothetical pipe are not set in stone.
For example:A flow test for a 2” valve produced a pressure loss equal to 3 velocity heads under fully developed turbulent flow. What is the length of 2.067” ID hypothetical pipe that would cause a 3 velocity head loss if the hypothetical pipe roughness were 0.0015”?
Using the rough pipe law, fT = 1/[2Log (3.7*2.067/0.0015)] 2 = 0.01819
Using Equation 2-4, LT = (3* 2.067)/0.01819 = 340.9” or 28.4 feetLT is the equivalent length of the valve under fully developed turbulent flow.