ch09

9-1 Fully developed, laminar flow of a viscous fluid (µ = 2.17 N·s/m2) flows between horizontal parallel plates 1-m long that are spaced 3.0-mm apart. The pressure drop is 1.25 kPa. Determine the volumetric flow rate (per unit width) through the channel (in m3/s m).

Approach:

Pressure drop can be calculated for this non-circular duct with the combination of Eq. 9-31 and Eq. 9-32 and using the hydraulic diameter. The velocity is unknown; once it is determined the volume flow rate can be calculated. We need to evaluate the friction factor. We combine all the expressions and solve for velocity.

Assumptions:

1. The system is steady. Solution:

Volume flow rate is or per unit width xV A HW= =V V V W H= V . Pressure drop can be calculated using the hydraulic diameter:

2

2Lh

P Lh fg D gρ

∆= =

V

For fully develop laminar flow between infinite parallel plates, 96 96 hf Re Dµ ρ= = V . Substituting this into the pressure drop equation, simplifying, solving for the average velocity V:

22

96hD P

Lµ∆

=V

The hydraulic diameter with W , →∞ ( ) ( )4 4 2 2 2 0.003m =0.006mh x wettedD A p HW H W H= = + = =⎡ ⎤⎣ ⎦

( ) ( )( )

( )( )

2 2

2

2 0.006m 1.25kN m 1000 N kN m=0.000432s96 2.17 Ns m 1m

=V

V A VHW= =V

( )2 3

-6 -6m m mms

0.000432 0.006m =2.59×10 =2.59×10s s

V HW

⎛ ⎞= = ⎜ ⎟⎝ ⎠

V Answer

9- 1

9-2 Journal bearings are constructed with concentric cylinders with a very small gap between the two cylinders; the gap is filled with oil. Because of the very small gap, the flow in the gap is laminar. Consider a sealed journal bearing with inner and outer diameters of 50- and 51-mm, respectively, and a length of 75 mm. The shaft (inner cylinder) rotates at 3000 RPM. At start-up the torque needed to turn the shaft is 0.25 N-m. Determine the viscosity of the oil (in N·s/m2). After an hour of operation will the torque have increased or decreased? Explain.

Approach:

Because the gap is small compared to the diameter, we can analyze the flow as if it were between infinite parallel plates. Viscosity is defined with Newton’s law of viscosity, Eq. 9-2.

Assumptions:

1. The system is steady. 2. The flow is fully developed between infinite parallel plates. 3. Properties are constant.

Solution:

Shear stress at a solid wall, using Newton’s law of viscosity, is:

ddy

τ µ= V

Because we can analyze this flow as between infinite parallel plates, we know that with the inner shaft rotating and the outer shaft stationary, the velocity profile is linear, so the velocity gradient is;

( )2 1

2 1

r rr r

ττ µ µ

−= → =

−V

V

Velocity is 1rω=V . Torque is 1 1Fr Arτℑ = = . Substituting these expressions into the shear stress equation and solving for visocity:

( ) ( )

( )( )2 1 2 1 2 1

2 21 1 12 2

r r r r r rAr r L r Lr

µ 31ω π ω π

ℑ − ℑ − ℑ −= = =

ω

( )( )( )

( ) ( )( )( )3 2

0.25 N m 0.0255m-0.025m 60s 1min Ns=0.054m2π 0.025m 0.075m 3000 rev min 2π rad rev

= Answer

Comments:

After an hour of operation, torque will decrease. Assuming that the bearing has little heat loss, the viscous friction will raise the oil temperature. Because viscosity for a liquid decreases with increasing temperature, torque will decrease with time.

9- 2

9-3 Consider laminar water flow at 20 °C between two very large horizontal plates. The lower plate is stationary and the upper plate moves to the right at a velocity of 0.25 m/s. For a plate spacing of 2 mm, determine the pressure gradient and its direction required to produce zero net flow at a cross section.

Approach:

Because the plates are very large, we assume the flow is fully developed between infinite parallel plates. For this flow, we can perform an analysis similar to what was done in Section 9.3 for flow in a circular tube to find the relationship between flow and pressure gradient.

Assumptions:

1. The system is steady. 2. The flow is fully developed between infinite

parallel plates. 3. Properties are constant.

Solution:

For steady, fully developed flow, conservation of momentum in the x-direction reduces to: 0xF =∑Evaluating the forces on the differential element shown above: 1 1 2 2 0T T B BP A P A A Aτ τ− + − = With length, dx, along the channel and a width W: ( ) ( ) ( ) ( )1 2 0T BP Wdx P Wdy Wdx Wdxτ τ− + − = The width cancels.

Shear stress at a solid wall, using Newton’s law of viscosity, is: ddy

τ µ= V

Therefore, TT

ddy

τ µ=V

and BB

ddy

τ µ=V

Substituting into the force balance and simplifying: ( )1 2 0T Bd dP P dy dx dx

dy dyµ µ− − +

V V=

( ) 2

1 22

T Bd dy d dyP P dP ddx dy dx dy

µ µ−−

= →V V V

=

Recognizing that for fully developed flow, constant ,dP dx = we separate variables and integrate twice:

21 2

12

dP y C y Cdxµ

= +V +

The boundary conditions are: 1) at y = 0, V = 0; 2) at y = b, V = VwApplying the boundary conditions:

1) so 2) 20 0 0 C= + + 2 0C = 21

12W

dP b C bxµ

= +V so 1 2W b dPC

b xµ= −

V

Substituting into the general velocity equation and simplifying: ( )212 W

dP yy bydx bµ

= − +V V

Flow rate is obtained by integration:

( )3

2

0

12 12

b WW

bWdP y b W dPV dA Wdy y by Wdydx b dxµ µ

⎛ ⎞ −= = = − + = +⎜ ⎟

⎝ ⎠∫ ∫ ∫

VV V V

2

Solving this equation for zero flow ( ) and from Appendix A-6 for water at 20 ºC, 0V = 4 29.85 10 Ns mµ −= ×

( )( )

( )

-4 2

2 2 2

6 9.85×10 Ns m 0.25m s6 N=369m m0.002m

WdPdx b

µ= =

V Answer

Pressure must increase in the x-direction to obtain zero flow.

9- 3

9-4 In the ¾-in. pipe shown below, oil flows downward at 6 gal/min. The oil has a specific gravity of 0.87 and a dynamic viscosity of 0.4 lbm/ft·s. The specific gravity of the manometer fluid is 2.9. Determine the manometer defection, h (in ft).

Approach:

The pressure drop between points 1 and 2 can be calculated with the steady, incompressible flow energy equation. The manometer equation is used to calculate the manometer deflection once the pressure drop is determined.

Assumptions:

1. The system is steady. 2. The flow is fully developed. 3. Properties are constant.

Solution:

The steady, incompressible flow energy equation is: 2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The areas at 1 and 2 are equal, so . There is no pump or turbine, so 1 =V V2 0P Th h= = , and minor losses are

neglected. Therefore, ( )2

1 2 2 1 2LP P g z z fD

ρ ρ− = − +V

The velocity is ( )( )( )

( )

3

2 2

4 6gal min 0.1337ft gal 1min 60s4 f=4.36sπ 0.0625ft

V VA Dπ

= = =V t

The Reynolds number is: ( )( )( )( )30.87 62.4ft lbm 4.36ft s 0.0625ft

370.4 lbm fts

DRe ρµ

= = =V

For fully developed laminar flow in a straight circular tube: 64 64 37 1.73f Re= = = Substituting this into the pressure drop equation:

( ) ( ) ( ) ( ) ( )2

1 2 3 2 3

22

2 2

4.26ft slbm ft 15 lbm0.87 62.4 32.2 -15ft + 1.73 0.87 62.40.0625 2ft s ft

lbm lbf s 1 lbf178,300 =38.532.2ft lbm 12in.fts in.

P P ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

For the manometer, stepping through the various legs: 1 0AP P− = A B o ABP P ghρ− = − 0B CP P− = C D m CDP P ghρ− = + 0D EP P− = E F oP P ghEFρ− = − 2 0FP P− =

( )1 2 o AB m CD o EF o AB EF m CDP P gh gh gh g h h ghρ ρ ρ ρ ρ− = − + − = − + + From the geometry of the manometer and pipe: AB CD EF AB EF CDL h h h h h L h= − + → + = +

Substituting into the manometer equation: ( )1 2 o CD mP P g L h ghρ ρ− = − + + CD

( )( )( )( )( )

( )( )( )

2 3 21 2

2 3

178,300lbm fts + 0.87 62.4 lbm ft 32.2ft s 15ft=50.1ft

32.2ft s 62.4 lbm ft 2.9-0.87o

CDm o

P P gLh

gρ

ρ ρ− +

= =−

Answer

9- 4

9-5 In Problem P 9-4 if the flow is upward instead of downward, determine the manometer deflection, h (in ft). Approach:

The pressure drop between points 1 and 2 can be calculated with the steady, incompressible flow energy equation. The manometer equation is used to calculate the manometer deflection once the pressure drop is determined.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V


neglected. Therefore, ( )2

1 2 2 1 2LP P g z z fD

ρ ρ− = − +V

The velocity is ( )( )( )

( )

3

2 2

4 6gal min 0.1337 ft gal 1min 60s4 f=4.36sπ 0.0625ft

V VA Dπ

= = =V t

The Reynolds number is: ( )( )( )( )30.87 62.4ft lbm 4.36ft s 0.0625ft

370.4 lbm fts

DRe ρµ

= = =V

For fully developed laminar flow in a straight circular tube: 64 64 37 1.73f Re= = = Substituting this into the pressure drop equation:

( ) ( ) ( ) ( ) ( )2

1 2 3 2 3

22

2 2

4.26ft slbm ft 15 lbm0.87 62.4 32.2 15ft + 1.73 0.87 62.40.0625 2ft s ft

lbm lbf s 1 lbf230,700 =49.832.2ft lbm 12in.fts in.

P P ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

For the manometer, stepping through the various legs: 1 0AP P− = A B o ABP P ghρ− = + 0B CP P− = C D m CP P gh Dρ− = − 0D EP P− = E F oP P ghEFρ− = + 2 0FP P− =

( )1 2 o AB m CD o EF o AB EF m CDP P gh gh gh g h h ghρ ρ ρ ρ ρ− = − + = + − From the geometry of the manometer and pipe: AB CD EF AB EF CDL h h h h h L h= − + → + = +

Substituting into the manometer equation: ( )1 2 o CD mP P g L h ghρ ρ− = + − CD

( )( )( )( )( )

( )( )( )

2 3 21 2

2 3

230,700lbm fts - 0.87 62.4lbm ft 32.2ft s 15ft=-50.1ft

32.2ft s 62.4lbm ft 0.87-2.9o

CDm o

P P gLh

gρ

ρ ρ− +

= =−

Answer

Comments: The minus sign indicates the manometer deflection is in the opposite direction than what is shown in the figure.

9- 5

9-6 Data are read from and written to spinning computer disks (3600 rpm) by small read-write heads that float above the disk on a thin (0.5µm) film of air. Consider a 10-mm by 10-mm head located 55 mm from the disk centerline. For air at 25 °C, and assuming the flow is similar to that between infinite parallel plates, determine:

a. the Reynolds number based on the gap dimension b. the power required to overcome the viscous shear (in W).

Approach:

Reynolds number can be obtained from its definition. Power is torque times rotational speed, and torque is forces times distance. The force is caused by the shear stress, which can be determined with Newton’s law of viscosity, Eq. 9-2.

Assumptions:

1. The system is steady. 2. The flow is fully developed between infinite

parallel plates. 3. Properties are constant.

Solution:

a) The Reynolds number is defined as:

b bRe ρµ υ

= =V V

Velocity is rω=V , and air viscosity from Appendix A-7 at 25 ºC is 6 215.4 10 m sυ −= × , 5 21.83 10 Ns mµ −= ×

( )( )( )( )( )-7

-6 2

0.055m 3600 rev min 1min 60s 2π rad rev 5×10 m0.67

15.4×10 m sRe = =

b) Power is defined as: W ω= ℑ Torque is Fr Ar LDrτ τℑ = = = . Assuming Newtonian flow between parallel walls with one wall moving and using Newton’s law of viscosity:

d rdy b b

ωτ µ µ µ= = =V V

Substituting all the expressions into the power expression and simplifying:

( )( ) ( )

222 2 -5

2 -7

0.01m 0.01mNs 3600 rad 1Ws1.83×10 0.055m 2π60 s Nmm 5×10 m

LDW rb

µ ω ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛= = ⎜ ⎟ ⎜ ⎟ ⎜⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝⎣ ⎦

⎞⎟⎠

=1.57W Answer Comments:

This is only part of the power required to spin the disk. Shear forces on other parts of the rotating disk would increase the power required to spin the disk.

9- 6

9-7 Skimmers are used to remove viscous fluids, such as oil, from the surface of water. As shown on the diagram below, a continuous belt moves upward at velocity Vo through the fluid and the more viscous liquid (with density ρ and viscosity µ) adheres to the belt. A film with thickness h forms on the belt. Gravity tends to drain the liquid, but the upward belt velocity is such that net liquid is transported upward. Assume the flow is fully developed, laminar, with zero pressure gradient, and zero shear stress at the outer film surface where air contacts it. Determine an expression for the velocity profile and flow rate. Use a differential analysis similar to that used for fully developed laminar flow through an inclined pipe. Clearly state the velocity boundary conditions at the belt surface and at the free surface.

Approach:

The amount of oil that is picked up by the moving belt is a result of a force balance between gravity, which causes the oil to flow downward, and shear stress, which causes the oil to flow upward. We apply a force balance in the x-direction to obtain the velocity profile, and then we integrate it to find volume flow rate.

Assumptions:

1. The system is steady. 2. The flow is fully developed between infinite parallel plates. 3. Properties are constant.

Solution:

A force balance in the x-direction on the differential element shown above, assuming fully developed laminar flow with constant properties, is: 0x T T B BF A A dτ τ= = − − W∑ where the weight is W Vgρ= . Note there is no net pressure force. Atmospheric pressure acts all along the surface of the oil, and the infinitesimal difference in hydrostatic pressure is negligible. Letting D be the depth of the plane into the page: 0T BDdx Ddx Ddxdygτ τ ρ− − =

Canceling Ddx and rearranging: T B gdy

τ τρ

−=

Recognizing the left hand side is a derivative: d gdyτ ρ=

Using Newton’s law of viscosity ddy

τ µ= V → 2

2

d d d d gdy dy dy dyτ µ µ ρ

⎛ ⎞= =⎜ ⎟

⎝ ⎠

V V=

Separating variables and integrating twice: 2

1 12d g g yy C C y Cdy

ρ ρµ µ 2= + → = + +

V V

Boundary conditions are: 1) at y = 0, B=V V 2) at y h= there is no shear, so 0d dy =V Applying the boundary conditions:

2 2

1 1

0 0

0

B BC Cg gh C C hρ ρµ µ

= + + → =

= + → = −

V V

Therefore: 2 2

2 2B Bg y g g yhy hyρ ρ ρµ µ µ

⎛ ⎞= − + = − +⎜ ⎟

⎝ ⎠V V V

Volume flow rat is

2

0 0 2h h

BA

g yV dA Ddy hy Ddyρµ

⎡ ⎤⎛ ⎞= = = − +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦∫ ∫ ∫V V V

3

3 BV gh hD

ρµ

= − + V Answer

9- 7

9-8 Consider a fully developed laminar flow of 20 °C water down an inclined plane that is 20° to the horizontal. The water thickness is 1-mm. The water is exposed to atmosphere everywhere, and the air exerts zero shear on the water. Using a differential analysis similar to that used for fully developed laminar flow through an inclined pipe, determine the volume flow rate per unit width (in m3/s m).

Approach:

The water flows down the inclined plane due to the influence of gravity. A force balance between gravity and shear forces is solved to find the velocity profile, and then integrated to find the volume flow rate.

Assumptions:


Solution:

A force balance in the x-direction on the differential element shown above, assuming fully developed laminar flow with constant properties, is: 0x T T B BF A A dW sinτ τ= = − + θ∑ where the weight is W Vgρ= . Note there is no net pressure force. Atmospheric pressure acts all along the surface of the oil, and the infinitesimal difference in hydrostatic pressure is negligible. Letting D be the depth of the plane into the page: sin 0T BDdx Ddx Ddxdygτ τ ρ θ− + =

Canceling Ddx and rearranging: sinT B gdy

τ τρ θ

−= −

Recognizing the left hand side is a derivative: sind gdyτ ρ θ= −

Using Newton’s law of viscosity d dyτ µ= V

2

2 sind d d gdy dy dy

µ µ ρ⎛ ⎞

= = −⎜ ⎟⎝ ⎠

V V θ

Separating variables and integrating twice: 2

1 1sin sin2

d g g yC Cdy

ρ ρθ θµ µ 2y C= − + → = − + +

V V

Boundary conditions are: 1) at y = 0, 0=V 2) at y h= there is no shear, so 0d dy =V Applying the boundary conditions:

2 2

1 1

0 0 0 0

0 sin sin

C Cg gh C C hρ ρθ θµ µ

= + + → =

= − + → =

Therefore: 2 2

sin sin sin2 2

g y g g yhy hyρ ρ ρθ θ θµ µ µ

⎛ ⎞= − + = −⎜ ⎟

⎝ ⎠V

Volume flow rate is

2

0 0 2h h

A

g yV dA Ddy ty Ddρµ

⎡ ⎤⎛ ⎞= = = −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦∫ ∫ ∫V V y

3sin

3V g hD

ρ θµ

=

For water at 20 ºC from Appendix A-6, 4 29.85 10 Ns m , 998.2kg mµ ρ−= × = 3

( )( ) ( )( ) ( )

( )

33 2 o 2 3

-4 2

998.2kg m 9.81m s sin 20 0.001m Ns kgm m=0.00113sm3 9.85×10 Ns m

VD= Answer

9- 8

9-9 A biomedical device start-up company is developing a liquid drug injection device. The device uses compressed air to drive the plunger in a piston-cylinder assembly that will push the drug (viscosity and density similar to water at 10 °C) through the hypodermic needle (inside diameter 0.25 mm and length 50 mm). If the flow must remain laminar in the hypodermic needle, determine:

a. the maximum flow possible (in cm3/s) b. the required air pressure for the maximum flow if the pressure at the end of the needle must be 105

kPa (in kPa). (Assume fully developed flow.) Approach:

Using a transition Reynolds number of 2100 and the Reynolds number definition, we can calculate the maximum allowable velocity and, hence, flow rate. We can use Eq. 9-13, which relates pressure drop and velocity, to calculate the required air pressure.

Assumptions:

1. The system is steady. 2. The flow is fully developed laminar flow. 3. Properties are constant.

Solution:

a) Volume flow rate is defined as: 2 4xV A Dπ= =V V Reynolds number is defined as:

DRe ρµ

=V

Using for the transition between laminar and turbulent flow, and water properties from Appendix A-6 at 10 ºC,

2100Re =4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 :

( ) ( )( )

( )( )

4 2 2

3

12.9 10 Ns m 2100 kgm Ns m=10.8s999.6kg m 0.00025m

ReD

µρ

−×= =V

( )2 3

-7π 0.00025mm ms

10.8 =5.30×10s 4

V ⎛ ⎞= ⎜ ⎟⎝ ⎠

Answer

b) From Eq. 9-13 for a horizontal flow:

2 2

8 32L LPR Dµ µ

∆ = =V V

( )( )( )( )

( )

4 2

2 1 2 2

32 12.9 10 Ns m 0.05m 10.8m s 1kN 1000N32 105kPa+0.00025m

LP PDµ

−×= + =

V

105kPa+357kPa=462kPa= Answer

9- 9

9-10 The viscosity of liquids is measured with a capillary viscometer, in which a laminar flow is maintained in a small diameter tube, and the pressure drop and flow rate are measured. If the flow is fully developed, then Eq. 9-13 can be used to calculate the liquid viscosity. However, entrance effects often are present. Consider the flow of a liquid (SG = 0.92) through a tube 450-mm long and 0.75-mm in diameter. A flow of 1 cm3/s is obtained when the pressure drop is 65 kPa. Determine:

a. the viscosity if the flow is fully developed (in Ns/m2) b. the viscosity if the pressure drop in the entrance length is twice that for the same length of fully

developed flow (in N·s/m2). Approach:

With the given information, Eq. 9-13 is used to calculate the viscosity. If entrance effects are present, a correction to the effective length of the tube must be made, and that corrected length used in Eq. 9-13. Laminar entrance length can be estimated with Eq. 9-39.

Assumptions:

1. The system is steady. 2. The flow is laminar. 3. Properties are constant.

Solution:

a) Volume flow rate is defined as: 2 4xV A Dπ= =V V Pressure drop for laminar flow of a Newtonian fluid in a circular tube is obtained from Eq. 9-13: 2 28 32P L R L Dµ µ∆ = =V V Noting that 2 4A Dπ= , combining these three equations, and solving for viscosity:

( ) ( )( ) ( )( )( )

4 324-3

23

π 0.00075m 65kN m 100cm m 1000 N 1kN Ns=1.122×10128 m128 0.45m 1cm s

D PLV

πµ ∆= = Answer

b) If entrance effects are taken into account (assuming 2ent withoutent

P P∆ = ∆ ) and from Eq. 9-39 0.065entL R= eD

tot ent fullydeveloped

P P P∆ = ∆ + ∆

( ) ( ) ( )2 2 2 2

3232 32 322 2ententtot ent ent ent

L LL V VP L L L L LD D D D

µµ µ µ−⎛ ⎞∆ = + = + − = +⎜ ⎟⎝ ⎠

VV( )

4

128tot

ent

D PL L V

πµ

∆=

+

Because the entrance length is a function of Reynolds number, which depends on viscosity, an iterative solution is required. The procedure is: guess the viscosity, calculate the Reynolds number, calculate the entrance length, and then calculate viscosity and compare to the guessed value. Guess 3 21.122 10 Ns mµ −= ×

Reynolds number is defined as: 2

4 4andD VRe ReDDVρ ρ ρ

µ πµπ= = →

V V =

( )( )( )( ) ( )

( )( )

33 3 2

-3 2

4 0.92 1000 kg m 1cm s 1m 100cm Ns kgm=1392

π 1.122×10 Ns m 0.00075mRe =

( )( )0.065 1392 0.00075m =0.0679mentL =

( ) ( )( ) ( )

( )( )

4 32-4

23

π 0.00075m 65kN m 100cm m 1000 N 1kN Ns=9.75×10m128 0.45m+0.0679m 1cm s

µ =

This does not match the guessed value, so continuing the iteration until it converges: 4 20.953 10 Ns m 1640 0.080mentRe Lµ −= × = = Answer Therefore, the error caused if entrance effects are not taken into account is:

3 4

4

1.122 10 9.53 10error= 100 17.7%9.53 10

− −

−

× − ×× =

× Answer

9- 10

9-11 A machine tool manufacturer is considering using gravity flow to supply cutting oil (SG = 0.87, µ = 0.003 N·s/m2) to the tool and workpiece. The vertical 5-mm diameter tube connecting the oil reservoir to the workpiece is very long so the flow can be assumed fully developed; in addition, the depth of oil in the reservoir is negligible compared to the tube length. The pressure is atmospheric at the exit of the tube and at the surface of the reservoir. Determine the volumetric flow rate of the oil (in cm3/s).

Approach:

Flow is caused by a balance between gravity, pressure, and shear stress/viscosity forces. Equation 9-13 was developed for this situation, so we will apply it directly. Because that equation uses the average velocity, volume flow rate is obtained easily from its definition.

Assumptions:

1. The system is steady. 2. The flow is fully developed and laminar. 3. Properties are constant.

Solution:

Volume flow rate is defined as: 2 4xV A Dπ= =V V Pressure drop for fully developed laminar flow of a Newtonian fluid in a circular tube is obtained from Eq. 9-13:

2 2

8 32sin sinL LP gL gLR Dµ µρ θ ρ∆ = + = +

V V θ

For the present problem . With atmospheric pressure at the surface of the reservoir and at the exit of the tube, . Using this information and solving for velocity:

o90θ =0P∆ =

2sin

32g Dρ θ

µ−

=V

Note that in the development of Eq. 9-13, the x-direction was in opposition to the gravity force, so the negative sign indicates a downward flow. Using the given information:

( )( ) ( )( )( )( )

23 o2

2 2

- 0.87 1000kg m sin 90 0.005msin m=0.22732 s32 0.003Ns m kgm Ns

g Dρ θµ

−= =V Answer

Checking the Reynolds number:

2

4 4andD VRe ReDDVρ ρ ρ

µ πµπ= = →

V V =

( )( )( )( )( )3 2

2

0.87 1000kg m 0.227 m s 0.005m Ns kgm=329

0.003Ns mRe =

This is laminar flow, so our assumption checks out. Therefore, the volume flow rate is:

( )2 3

-6π 0.005mm ms

0.227 =4.46×10s 4

V ⎛ ⎞= ⎜ ⎟⎝ ⎠

Answer

9- 11

9-12 A manometer, with pressure taps 25 ft apart, is used to measure the pressure drop of oil (SG = 0.82) flowing in a 1.5-in. pipe with a volumetric flow rate of 4 ft2/min. The manometer fluid is mercury (SG = 13.6). From the lower pressure tap to the surface of the mercury highest in the manometer is 2 ft, and the distance from the upper pressure tap to the same height in the mercury is 4 ft. For a manometer deflection of 4 in., determine:

a. the flow direction b. the friction factor c. whether the flow is laminar or turbulent d. the oil viscosity (in lbm/ft s).

Approach: Flow direction can be determined by inspection, and the friction factor can be calculated using the steady, incompressible flow energy equation. Whether the flow is laminar or turbulent must be determined by assuming one or the other, using a correlation for the friction factor, calculating the viscosity, and then checking the Reynolds number. The manometer equation is used to calculate the pressure drop.

Assumptions:


Solution:

a) Hydrostatic pressure increases farther down in a fluid. The manometer indicates that . The only way this could occur would be if flow was from point 1 to point 2 with friction losses.

2P P< 1

Answer b, c, d) The steady, incompressible flow energy equation is:

2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V


neglected. Therefore, ( ) ( )2

1 2 1 21 2 1 22

22o o

P P P PL Dgz z f f z zg D g gLρ ρ

⎡ ⎤− −+ − = → = + −⎢ ⎥

⎣ ⎦

VV

From the definition of volume flow rate, the velocity is ( )( )

( )

3

2 2

4 4ft min 1min 60s4 f=5.43sπ 0.125ft

V VA Dπ

= = =V t

A

For pressure drop, using the manometer equation and stepping through the various legs: 1 1A oP P ghρ− = − 0A BP P− = B C m BP P gh Cρ− = 2 2C oP P ghCρ− =

1 2 1o A m BC o CP P gh gh gh 2ρ ρ ρ− = − + + Substituting this into the friction factor expression:

( )( )( )( )( )

( )2

1 2 1 22 2

2 0.125ft 32.2ft s2 1-4.33ft+ 0.33ft +2ft+2ft =0.05620.8225ft 5.43ft s

mA BC C

o

Dgf h h h z zL

ρρ

⎡ ⎤ ⎡ ⎤= − + + + − =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦V3.6

Assuming the flow is fully developed laminar flow: ( )( )( )( )( )30.82 62.4ft lbm 5.43ft s 0.125ft 0.056264 0.0305lbm fts

64 64Dff

Reρµ= → = = =

V Answer

Check the Reynolds number: ( )( )( )( )30.82 62.4ft lbm 5.43ft s 0.125ft

11400.0305lbm fts

DRe ρµ

= = =V

which is laminar, so our assumption is valid.

9- 12

9-13 Develop an expression for the velocity profile for fully developed laminar flow between stationary infinite parallel plates. Use an approach similar to that applied in Section 9.3 for a circular tube.

Approach:

Flow through this infinite parallel plate channel results from a balance among viscous, pressure, and gravity forces. A force balance on a differential element is used to determine the velocity profile, similar to what was done for the circular tube.

Assumptions:


Solution:

A force balance in the x-direction on the differential element shown above, assuming fully developed laminar flow with constant properties, is: 1 1 2 20 2x sF P A P A A mg sinτ θ= = − − −∑ Let D be the depth of the plane into the page: Using Newton’s law of viscosity d dnτ µ= V where n is the direction from the wall. To put this in terms of y (from the channel centerline), ( )y H n dy d H n dn= − → = − = − . Substituting these expressions into the force balance, and letting 1 2 2 , , and 2sA A Dy A DL m D yLρ= = = = :

( )1 2 2 2 2 sindP P Dy DL L yDgdy

µ ρ θ− + −V 0=

Canceling 2D, and simplifying:

1 2 sinP P dg

L yµρ θ

−− = −

Vdy

Separating variables and integrating:

2

1 2 1 21

1 1sin sin2

P P P P yg ydy d g CL L

ρ θ ρ θµ µ

− −⎡ ⎤ ⎡ ⎤− − = → = − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦∫ ∫ V V +

The boundary condition is: at y = H, 0=V . Applying the boundary condition:

2

1 21

1 sin2

P P HC gL

ρ θµ

−⎡ ⎤= −⎢ ⎥⎣ ⎦

Therefore:

( )22

2 21 2 1 21 sin sin 12 2

P P P PH yH

g H y gL L

ρ θ ρ θµ µ

⎡ ⎤− −⎡ ⎤ ⎡ ⎤ ⎛ ⎞= − − − = − −⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎢ ⎥⎣ ⎦V Answer

Comments:

Note the parabolic velocity profile, which is similar to that in a circular tube.

9- 13

9-14 In an inclined 50-mm diameter pipe, a fluid (SG = 0.88) flows with a volumetric flow rate of 0.003 m3/s. The gage pressure at the pipe inlet is 720 kPa. The pipe outlet is at atmospheric pressure and is 15 m above the inlet. Determine the head loss between the inlet and outlet (in m).

Approach:

We can use Eq. 9-41 directly to find head loss. Assumptions:

1. The system is steady. 2. The flow is fully developed. 3. Properties are constant. 4. Neglect minor losses.

Solution:

The steady, incompressible flow energy equation is:

2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V


neglected. Therefore, ( )1 21 2L

P Ph z

gρ−

= + −∑ z

( )

( )( )

2 2

3 2

kN 1000kg m720-0m kNs

+ -15m =83.4-15=68.4mkg m0.88 1000 9.81m s

Lh

⎛ ⎞⎜ ⎟⎝ ⎠=

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∑ Answer

9- 14

9-15 The pipe exit in Problem P 9-14 is lowered to the same elevation as the inlet. Determine the inlet pressure for this new condition (in kPa).

Approach:

We can use the steady flow energy equation to find the head loss in the inclined pipe. The frictional head loss is the same in a horizontal pipe, so we can again use the steady flow energy equation, but this time we will determine the new inlet pressure.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V


neglected. Therefore, for when the outlet is 15 m above the inlet ( )1 21 2L

P Ph z

gρ−

= + −∑ z

( )

( )( )

2 2

3 2

kN 1000kg m720-0m kNs

+ -15m =83.4-15=68.4mkg m0.88 1000 9.81m s

Lh

⎛ ⎞⎜ ⎟⎝ ⎠=

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∑ Answer

If the pipe is in a horizontal orientation, 1z z2= , with the same flow rate, and assuming the same total head loss, solving the energy equation for P1:

( ) ( )( )( )( )3 2 21 2 0 68.4m 0.88 1000kg m 9.81m s kNs 1000kg m =591kPaLP P h gρ= + = +∑ Answer

9- 15

9-16 An air conditioning duct is 25-cm square and must convey 25 m3/min of air at 100 kPa, 25 °C. The duct is made of sheet metal that has a roughness of approximately 0.05 mm. Determine the pressure drop for 25-m of horizontal duct run (in kPa and mm of water).

Approach:

Pressure drop is calculated directly with the steady, incompressible flow energy equation

Assumptions:

1. The system is steady. 2. The flow is fully developed. 3. Properties are constant. 4. The duct is smooth.

Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The duct is constant area and horizontal, so 1 2=V V and 1z z2= There is no pump or turbine, so , and there are no minor losses:

0P Th h= =

2 2

1 21 22 2

P P L Lf P P fg D g D g

ρρ−

= → − =V V

The friction factor is a function of Reynolds number and roughness. For air from Appendix A-7 at 100 kPa, 25 ºC, 5 21.83 10 Ns m , 1.169kg mµ ρ−= × = 3 .

Velocity is : ( )( )

( )( )

325m min 1min 60s m=6.670.25m 0.25m s

VA

= =V

Because this is a non-circular duct, the hydraulic diameter must be used:

( )( )

( )4 0.25m 0.25m4

=0.25m4 0.25m

xh

wetted

AD

P= =

( )( )( )3

-5 2

1.169kg m 6.67 m s 0.25m=106,500

1.83×10 Ns mhD

Reρµ

= =V

This is turbulent flow so with a smooth duct:

1.11 1.111 6.9 0 6.91.8log 1.8log 0.0176

3.7 3.7 106,500D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Substituting known values into the pressure drop equation:

( ) ( )2 2

1 2 3

6.67 m s25 kg 1kN s0.0176 1.169 =0.0457kPa0.25 2 1000kg mm

P P⎛ ⎞⎛ ⎞⎛ ⎞− = ⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠ Answer

Using the manometer equation wP ghρ∆ =

( )( )( )( )

2 2

3 2

0.0457 kN m 1000kgm 1kNs=0.00466m of water = 4.66mm of water

1000kg m 9.81m sPw

Phgρ

∆= = Answer

9- 16

9-17 A manufacturer develops a new type of flow control valve. Before it can be advertised and sold, its loss coefficient must be determined. The valve is installed in a 6-in. pipe and 2 ft3/s of water flows through it. The pressure drop is measured with a manometer whose fluid has a specific gravity of 1.3. The manometer deflection is 7.5 in. Determine the loss coefficient for the valve.

Approach:

We can use the definition of head loss for minor losses to calculate the loss coefficient. The pressure drop (head loss) can be determined from the manometer equation.

Assumptions:


Solution:

The head loss due to minor losses is given by:

2

2L Lh Kg

=V

and 1 2L

P Ph

gρ−

=

Combining the expressions and solving for the loss coefficient:

( )1 2

2

2L

w

P PK

ρ−

=V

The velocity in the upstream pipe is ( )( )

3

2 2

4 2ft s4 f=10.2sπ 0.5ft

V VA Dπ

= = =V t

A

For the manometer, stepping through the various legs: 1 1A wP P ghρ− = − 0A BP P− = B C m BP P gh Cρ− = 2 2C wP P ghCρ− = +

( ) ( )1 2 2 2 2w AB m BC w C w BC C m BC w C m w BCP P gh gh gh g h h gh gh ghρ ρ ρ ρ ρ ρ ρ ρ− = − + + = − + + + = − Substituting this expression into the loss coefficient equation:

( ) ( )

2 2

2 2m w BC m BCL

w

1gh SG ghK

ρ ρρ− −

= =V V

( )( )( )

( )

2

2

2 1.3-1 32.2ft s 7.5 12ft0.116

10.2ft sLK = = Answer

9- 17

9-18 When pumping a fluid, the pressure at the entrance to the pump must never drop below the saturation pressure of the fluid. If the pressure does drop below the saturation pressure, cavitation (the forming of vapor bubbles) occurs which can damage the pump impeller. Consider the system shown below constructed of commercial steel pipe and threaded connections. For water at 10 °C, determine the maximum possible flow rate without cavitation occurring (in m3/s).

Approach:

Cavitation will occur if the pressure at point 2 falls below the saturation pressure of 10 ºC water. We can solve for the flow rate using the steady, incompressible flow energy equation. The pressure at 2 is the saturation pressure.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . We assume the area at 1 large, so 0T ph h= = 1 0≈V , and the pressure at 1 is

atmospheric, so P1 = 101.3 kPa. At point 2, from the saturated water table at 10 ºC, . The

losses include one entrance, two bends, and line loss:

( )o10 C 1.23kPasatP =2

22L ent bend

Lh f K KD g

⎛ ⎞= + +⎜ ⎟⎝ ⎠

∑ V . Therefore,

( )2

1 21 2 2 1

2ent bendP P Lz z f K K

g D gρ− ⎛ ⎞+ − = + + +⎜ ⎟

⎝ ⎠

V

The friction factor depends on flow, so an iterative solution is required. From Figure 9-15, for an zero area ratio at the entrance Kent = 0.5. From Table 9-3, for a regular 90º threaded bend, Kbend = 1.5. Velocity is:

( )( )

3

2 2

4 m s4 =226.4sπ 0.075m

VV V VA Dπ

= = =V m (1)

For commercial steel pipe =0.045mmε , and for water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 .

The friction factor is a function of Reynolds number and roughness.

( )( )( )3

64 2

999.6kg m 226.2 m s 0.075m1.316 10

12.9 10 Ns m

VDRe Vρµ −= = = ×

×V (2)

The volume flow will be large enough so that the flow will be turbulent, so

1.11 1.111 6.9 0.045 75 6.91.8log 1.8log

3.7 3.7D

Re Refε⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞= − + = − +⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢⎣ ⎦ ⎣

⎤⎥⎥⎦

(3)


( ) ( )

( )( ) ( ) ( ) ( )2 2 2

3 2 2

101.3-1.23 kN m 1000kg m kNs 13+ -5m = +0.5+2 1.5 +10.075999.6kg m 9.81m s 2 9.81m s

f⎡ ⎤⎢ ⎥⎣ ⎦

V

(4) ( 2102 4.5 173.3 f= + V)Iterating on the four equations given above, we obtain: 30.0147 m s 3.33m s 0.0271 19, 200V f Re= = = =V Answer

Comments: When the water enters the pump, it accelerates more. Hence, the actual minimum pressure at 2 would need to be greater than 1.23 kPa. The pump manufacturer would specify the minimum required pressure.

9- 18

9-19 Fire codes mandate that the pressure drop in horizontal runs of commercial steel pipe must not exceed 1.0 lbf/in.2 per 150-ft of pipe for flows up to 500 gal/min. For a water temperature of 50 °F, determine the minimum pipe diameter required (in in.). Is the number you calculated feasible?

Approach:

Pressure loss depends on velocity, which depends on pipe diameter. The friction factor is a function of velocity. Hence, an iterative solution is required. The steady, incompressible flow energy equation is used to determine the minimum pipe diameter.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The areas at 1 and 2 are equal, so . The pipe is horizontal, so 1 =V V2 21z z= . There is no pump or turbine, so

, and minor losses are neglected. Therefore, 0P Th h= =2

1 2

2a

P P Lfg D gρ

−=

V

Velocity is :( )( )( )

( ) ( )

3

2 2

4 500gal min 0.1337ft gal 1min 60s4 1=sπ Dft ft

V VA D Dπ

= = =V 2

.419 ft (1)

The friction factor is a function of Reynolds number and roughness. From Table 9-2 for commercial steel pipe, =0.00015ftε , and for water from Appendix B-6 at 50 ºF, 5 388 10 lbm fts , 62.4 lbm ftµ ρ−= × = :

( )( )( )3 2

5

62.4 lbm ft 1.419 ft s ft 100,60088 10 lbm fts

D DDReD

ρµ −= = =

×V (2)

For any reasonable size diameter, the flow will be turbulent, so

1.111 0.00015 6.91.8log

3.7D

Ref

⎡ ⎤⎛ ⎞= − +⎢⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

⎥ (3)


( )( )( ) ( )22 2 2 2 2

3

1lbf in. 32.2ft lbm lbf s 144in. ft ft s150 0.991262.4lbm ft

ffD D

⎛ ⎞= →⎜ ⎟⎝ ⎠

V V= (4)

An iterative solution can be found using the above four equations. The procedure to use is: assume a diameter D, calculate the velocity V, Reynolds number Re, and friction factor f, and then use equation 4 to calculate the diameter. Compare the calculated diameter with the guessed one; iterate until converged. Performing the iteration: 0.513ft = 6.2 in. 5.39ft s =196,000 =0.0175D Re f= =V Answer

Comments: The next large standard pipe size would be used.

9- 19

9-20 The owners of a luxurious mountain resort want to install a fancy water fountain. The artist’s initial design uses 75-m of 7.5-cm diameter commercial steel pipe ending in a nozzle with a diameter of 3.75-cm with a 40-kW pump to pull water from a lake above the resort at a flow rate of 0.05 m3/s. To save operating costs, the owners want to remove the pump and rely only on gravity head to power the fountain. Assuming the friction factor is 0.016 for both cases and neglecting minor losses, determine:

a. the flow rate if the pump is removed from the system (in m3/s) b. the height of the water jet with and without the pump if the nozzle is pointed vertically up (in m).

Approach:

The elevation difference between the lake surface and the nozzle exit is not given and must be determined. The steady, incompressible flow energy equation can be used to calculate the elevation difference and the height of the fountain.

Assumptions:

1. The system is steady. 2. The flow is fully developed. 3. Properties are constant and evaluated at 10 ºC.

Solution:

a) The steady, incompressible flow energy equation is: 2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no turbine, so . At points 1 and 2 the pressure is atmospheric and the area at 1 is very large, so . We will assume no minor losses. Therefore,

0Th =

1 10 and P P=V 2=

2 2

1 2 2 2pipe nozzle

P

V VLz z h fD g g

− = − + +

The pump head can be evaluated from P PW mgh Vghρ= = P . For water from Appendix A-10 at 10 ºC, 3999.6kg mρ =

( )( )( )( )( )

2

3 3 2

40,000W N s kg m=81.6m

999.6kg m 0.05m s 9.81m sP

PW

hVgρ

= =

The velocity at the nozzle exit is: ( )( )

3

2 2

4 0.05m s4 m=45.3sπ 0.0375m

nozzlenozzle

V VA Dπ

= = =V

Likewise, for the velocity in the pipe, ( )( )

3

2 2

4 0.05m s4 m=11.3sπ 0.075m

pipepipe

V VA Dπ

= = =V

Note that . Therefore, 4nozzle pipe=V V

( )( ) ( ) ( )

( )2 2

1 2 2 2

45.3m s 11.3m s7581.6m 0.0160.0752 9.81m s 2 9.81m s

81.6 104 104 127.4m

z z ⎛ ⎞− = − + + ⎜ ⎟⎝ ⎠

= − + + =

Now using the energy equation without the pump:

2 22

1 2 162 2 2

pipe pipenozzleL Lz z f fD g g D g

⎛ ⎞− = + = +⎜ ⎟⎝ ⎠

V VV

Solving for the velocity in the pipe:

( ) ( )( )

( )( )

0.50.5 21 2

2 9.81m s 127.4m2 m=8.8316 16+ 0.016 75 0.075 spipe

g z zf L D

⎡ ⎤−⎡ ⎤⎢ ⎥= =⎢ ⎥+ ⎢ ⎥⎣ ⎦ ⎣ ⎦

V

Therefore, the flow rate without the pump is: ( )( )( )2 38.83m s π 4 0.075m =0.039m sV A= =V Answer

9- 20

b) To determine the height of the fountain, apply the energy equation between points 2 and 3. The pressure is the same and the velocity at 3 is zero, so 3 20 and P P3= =V :

22 2

32 22 3 3 22 2

z z z z2g g g

+ = + → − =VV V

With the pump: ( )( )

2

3 2 2

45.3m s=104.6m

2 9.81m sz z− = Answer

Without the pump, the velocity at the nozzle exit is

( )( )

3

2

4 0.039 m s m=35.3sπ 0.0375m

nozzle =V

( )( )

2

3 2 2

35.3m s=63.6m

2 9.81m sz z− = Answer

Comments: Even without the pump, the fountain still reaches an impressive height.

9- 21

9-21 At an oil tank farm, a vandal opens a valve at the end of a 5-cm diameter, 50-m long horizontal pipe from the bottom of a large diameter oil tank. The oil tank is open to the atmosphere, and the oil depth is 6.5 m. The oil has a SG = 0.85 and a kinematic viscosity of 6.8×10-4 m2/s. Neglecting minor losses, determine the initial flow rate from the tank (in m3/s).

Approach:

The flow rate depends on a balance between the head supplied by the oil depth and the head loss in the drain pipe. The steady, incompressible flow energy equation can be used to determine the flow. An iterative solution may be required because the friction factor is a function of the flow (velocity).

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The areas at 1 and 2 are large, so . The pressures at 1 and 2 are both atmospheric, so . There is no pump or turbine, so , and minor losses are neglected. Therefore,

1 2 0= ≈V V 1P P= 2

0P Th h= =

2

1 2 2Lz z fD g

− =V

The friction factor is a function of Reynolds number and roughness. Assuming a smooth pipe:

( ) ( )

-4 2

m s 0.05m73.5

6.8×10 m sD DRe ρµ ν

= = = =VV V V

For any reasonable velocity, the flow will be laminar (check this). Therefore, for fully developed laminar flow in a straight circular tube: 64 64f Re Dν= = V Substituting this into the pressure drop equation and solving for velocity:

( ) 22

1 21 2

2642 64

g z z DLz zD D g Lν

ν−

− = → =V V

V

( )( )( )

( )( )

22

-4 2

2 9.81m s 6.5m 0.05m m=0.147s64 6.8×10 m s 50m

=V Answer

Checking the Reynolds number:

( ) ( )

-4 2

0.147m s 0.05m10.8

6.8×10 m sRe = =

Laminar flow, so our assumption is correct.

Comments: Note that the velocity calculated above is the initial velocity, assuming that the transient start-up of the flow is brief and the flow is quasi-steady. As the oil level drops, then the velocity will decrease.

9- 22

9-22 In a large convention center, heated air at 85 °F must be conveyed from the furnace room to the display rooms through a 500-ft smooth duct. The required flow rate is 7500 ft3/min. If the pressure loss must not exceed 2.5 in. of water, determine:

a. the minimum diameter required (in in.) b. pumping power required (in hp).

Approach:

Pressure loss depends on velocity, which depends on pipe diameter. The friction factor is a function of velocity. Hence, an iterative solution is required. The steady, incompressible flow energy equation is used to determine the minimum pipe diameter.

Assumptions:

1. The system is steady. 2. The flow is fully developed with constant properties. 3. Neglect minor losses. 4. Air is an ideal gas.

Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The areas at 1 and 2 are equal, so . The pipe is horizontal, so 1 =V V2 21z z= . There is no pump or turbine, so

P Th h= , and minor losses are neglected. Therefore, 2

1 2

2a

P P Lfg D gρ

−=

V

The pressure drop can be obtained from the manometer equation:

wa a w w a

a a

PP g h g h h hg

ρρ ρ

ρ ρ∆

∆ = ∆ = ∆ → = ∆ = ∆ w

Assuming air is an ideal gas at one atmosphere:

( )( )( )

( ) ( )

2 2 2

3

14.7 lbf in. 28.97 lbm lbmol 144in. ft lbm=0.07281545ft lbf lbmR 85+460 R fta

PMRT

ρ = =

( )3

3

62.4 lbm ft 1ft2.5in. =178.6ft12in.0.0728lbm fta

Pgρ

∆ ⎛ ⎞= ⎜ ⎟⎝ ⎠


( ) ( )

3

2 2

4 7500ft min 1min 60s4 =sπ ft ft

V VA D D Dπ

= = =V 2

159.2 ft (1)

The friction factor is a function of Reynolds number and roughness. Assuming a smooth pipe, and for air from Appendix B-7 at 85 ºF, 51.259 10 lbm ftsµ −= × :

( )( )( )3 2

5

0.0728ft lbm 159.2 ft s ft 920,6001.259 10 lbm fts

D DDReD

ρµ −= = =

×V (2)

For any reasonable size diameter, the flow will be turbulent, so with 0ε =

1.111 1.8log

3.7D

Refε⎡ ⎤⎛ ⎞= − +⎢⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

6.9⎥ (3)


( )( )

2 2

2

ft s500178.623.02 32.2ft sfft f D

D⎛ ⎞= →⎜ ⎟⎝ ⎠

V V= (4)

An iterative solution can be found using the above four equations. The procedure to use is: assume a diameter D, calculate the velocity V, Reynolds number Re, and friction factor f, and then use equation 4 to calculate the

9- 23

diameter. Compare the calculated diameter with the guessed one; iterate until converged. Performing the iteration: 1.70ft 55.1ft s =541,500 =0.0129D Re f= =V Answer The next large standard duct size would be used. b) The pumping power is a aW mgh Vhρ= = a

( )3 3 2

2

ft ft 1min ft lbfs 1hp0.0728 7500 32.2 178.6ftlbm min 60s 32.2ft lbm 550ftlbf ss

W⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠

=2.96hp Answer

9- 24

9-23 Consider a heat exchanger that has 1000 2.5-cm diameter smooth tubes in parallel, each 6-m long. The total water flow of 1 m3/s at 10 °C flows through the tubes. Neglecting entrance and exit losses, determine:

a. the pressure drop (in kPa) b. the pumping power required (in kW) c. the pumping power for the same flow rate if solid deposits from the water build up on the inner

surface of the pipe with a thickness of 1-mm and an equivalent roughness of 0.4 mm. Approach:

Because the tubes operate in parallel, all tubes have the same pressure drop, so we only need to calculate the pressure drop in one tube. The steady, incompressible flow energy equation is used.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . At points 1 and 2 the pipe areas are the same, so .

Therefore,

0T Ph h= = 1 2=V V2

1 2 2VLP P f

Dρ− =

The friction factor is a function of Reynolds number and roughness. For smooth pipe =0ε , and for water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 .

The velocity is in one tube is: ( )

( ) ( )

3

2 2

4 1.0 m s4 m=2.04s1000 π 0.025m

V VA N Dπ

= = =V

( )( )( )3

4 2

999.6kg m 2.04m s 0.025m39,500

12.9 10 Ns mDRe ρµ −= = =

×V

This is turbulent, so 1.11 1.111 6.9 0 6.91.8log 1.8log 0.0219

3.7 3.7 39,500D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Therefore, ( ) ( )2 2

1 2 3

2.04m s6 kg Ns 1kN0.0219 999.6 =10.9kPa0.025 2 kg m 1000Nm

P P⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞− = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ Answer

b) Pumping power for the total flow rate is: 3

2

m kN kW s1 10.9 =10.9kWs kN mmPW V P

⎛ ⎞⎛ ⎞⎛ ⎞= ∆ = ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

Answer

c) The new pipe diameter is ( )0.025m-2 0.001m =0.023mD = . Therefore, 0.4 23 0.0174Dε = =

( )22.04 25 23 2.41m s= =V → ( )( )( ) ( )4999.6 2.41 0.023 12.9 10 43,000Re −= × = 1.11 1.111 6.9 0.0174 6.91.8log 1.8log 0.0472

3.7 3.7 43,000D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( ) ( )2 2

1 2 3

2.41m s6 kg Ns 1kN0.0472 999.6 =35.7kPa0.023 2 kg m 1000Nm

P P⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞− = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

3

2

m kN kW s1 35.7 =35.7kWs kN mmPW V P

⎛ ⎞⎛ ⎞⎛ ⎞= ∆ = ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

Answer

Comments: Fouling can have a significant effect on flow rate. Note, however, that with centrifugal pumps, the motor power is fixed, so the flow rate would decrease.

9- 25

9-24 The piping system that connects one reservoir to a second reservoir consists of 150-ft of 3-in. cast iron pipe that has four flanged elbows, a well-rounded entrance, and sharp-edged exit, and a fully open gate valve. For 75 gal/min of water at 50 °F, determine the elevation difference between the two reservoirs (in ft).

Approach:

The flow rate is set by the balance between the head caused by the difference in reservoir elevations and the frictional losses in the pipe. The steady, incompressible flow energy equation applied between points 1 and 2 is used to determine the elevation difference.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . We assume the surface of the reservoirs are large, so 0T ph h= = 1 2 0≈ ≈V V , and the pressure at 1 and 2 are atmospheric, so P1 = P2. Therefore, taking into account all the minor losses:

2

1 2 42ent valve exit bend

Lz z f K K K KD g

⎛ ⎞− = + + + +⎜ ⎟⎝ ⎠

V

From Table 9-3, Figure 9-13 and Figure 9-14, Kent = 0.04, Kbend = 0.3, and Kvalve = 0.15. For cast iron pipe =0.00085ftε , and for water from Appendix B-6 at 50 ºF, 5 388 10 lbm ft s , 62.4 lbm ftµ ρ−= × = .

The velocity is determined from conservation of mass for incompressible flow:

( ) ( )( )

( )

3 3

2 2

4 75gal ft 0.1337ft gal 1min 60s4 f=3.40s0.25ft

V VA Dπ π

= = =V t


( )( )( )3

5

62.4 lbm ft 3.40ft s 0.25ft60, 400

88 10 lbm ft sDRe ρµ −= = =

×V

The flow is turbulent, so

1.11 1.111 6.9 0.00085 0.25 6.91.8log 1.8log 0.0289

3.7 3.7 60, 400D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( ) ( ) ( )( )

2

1 2 2

3.40ft s1500.0289 0.04 0.15 1.0 4 0.3 3.54ft0.25 2 32.2ft s

z z ⎡ ⎤− = + + + + =⎢ ⎥⎣ ⎦ Answer

9- 26

9-25 Vandals open the drain valve on a water tower that is 10-m in diameter with a water depth of 8 m. The water flows out a sharp-edged opening into a horizontal 30-m long pipe, both of which are 10-cm in diameter; the gate valve in the pipe is half opened. Assume the friction factor is 0.016. Determine:

a. the time required for the tank to drain (in min) b. the time required for the tank to drain if only the sharp-edged opening and the valve are present (in

min) c. the appropriateness of the friction factor value used.

Approach:

Conservation of mass is needed to determine the time to drain the pool. The driving pressure head decreases with time, so the head as a function of time is needed. The open system conservation of mass equation and the steady, incompressible flow energy equation are used to calculate the time.

Assumptions:


Solution:

a) The open system conservation of mass equation is for the control volume defined as the water in the pool: in outdm dt m m= − . Water only leaves, and expressing the mass in terms of volume and flow rate in terms of

velocity:

( ) ( )1

2 2

d V d A zA

dt dtρ ρ

ρ= = − V

Density cancels so rearranging the equation:

2

2 22 2

1 1

A Ddzdt A D

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠V V

We can obtain an expression for velocity by assuming (at any instant in time) that the steady, incompressible flow energy equation is applicable with a constant friction factor:

2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

At points 1 and 2 the pressure is atmospheric, so 1P P2= . There is no pump or turbine, so . Let (the datum) and drop the subscript on z

0P Th h= =

2 0z = 1, the depth of water in the tank. The losses include an entrance, a

valve, and line loss, so 2 2

1 2

2 2 ent valveLz K K f

22

2g g D⎛ ⎞+ = + + +⎜ ⎟⎝ ⎠

V Vg

V

The tank velocity can be described in terms of the pipe velocity by conservation of mass. Any mass flow rate from the tank must equal the mass flow rate through the pipe: ( )2

1 2 1 1 2 2 1 2 1m m A A D Dρ ρ= → = → =V V V 2V Substituting this into the energy equation and solving for velocity:

( )

0.5

2 42 1

21 ent valve

gzD D K K f L D

⎡ ⎤= ⎢ ⎥

− + + +⎢ ⎥⎣ ⎦V

Substituting this in the conservation of mass equation and separating variables:

( )

0.52

20.5 4

1 2 1

21 ent valve

Ddz g dtDz D D K K f L D

⎡ ⎤⎛ ⎞= − ⎢ ⎥⎜ ⎟

− + + +⎢ ⎥⎝ ⎠ ⎣ ⎦

We integrate this from initial height zo to 0:

9- 27

( )

( )

( )

0.5202

0.5 410 2 1

0.520.5 2

41 2 1

0.52 42 10.5 1

2

21

221

12

2

o

t

z ent valve

oent valve

ent valveo

Ddz g dtDz D D K K f L D

D gz tD D D K K f L D

D D K K f L DDt z

D g

⎡ ⎤⎛ ⎞= − ⎢ ⎥⎜ ⎟

− + + +⎢ ⎥⎝ ⎠ ⎣ ⎦

⎡ ⎤⎛ ⎞− = − ⎢ ⎥⎜ ⎟

− + + +⎢ ⎥⎝ ⎠ ⎣ ⎦

⎡ ⎤− + + +⎛ ⎞= ⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠ ⎣ ⎦

∫ ∫

From Figure 9-14, Knit = 1.0, and from Table 9-3, Kvalve = 2.1.

( ) ( ) ( )( )( )

0.5420.5

2

1- 0.1 10 +1.0+2.1+ 0.016 30 0.10102 8m =38,100 s=635 min=10.6hr0.1 2 9.81m s

t⎡ ⎤⎛ ⎞ ⎢ ⎥= ⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦ Answer

b) If the 30 m long pipe is not taken into account:

( ) ( )( )

0.5420.5

2

1- 0.1 10 +1.0+2.1102 8m =25,900 s=432 min=7.2 hr0.1 2 9.81m s

t⎡ ⎤⎛ ⎞ ⎢ ⎥= ⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦

c) We can evaluate whether the chosen friction factor is appropriate by calculating an “average” velocity, Reynolds number, and friction factor. The velocity is:

( ) ( )

( )( )

22 31

3

2 2 22

8m 10m 44 m0.016538,100s s

4 0.0165m s4 m2.1s0.10m

tot oV z DV

t t

V VA D

ππ

π π

≈ = = =

≈ = = =V

The friction factor is a function of Reynolds number and roughness. For a commercial steel pipe =0.045mmε , and for water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 . The Reynolds numbers are:

( )( )( )3

4 2

999.6kg m 2.1m s 0.10m162,700

12.9 10 Ns mDRe ρµ −= = =

×V


1.111 0.045 100 6.91.8log 0.0187

3.7 162,700A

ff

⎡ ⎤⎛ ⎞= − + → =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

Using this friction factor, we obtain t = 39,800 s = 11.1 hr. Answer

Comments: Because an “average” flow rate was used, there is some uncertainty in this answer.

9- 28

9-26 An oil transporter truck is filled from the top with 15 m3 of fuel oil (SG = 0.86, µ = 5.3 × 10-2 N·s/m2) from a reservoir that is 4 m below the truck top. A 10-m long flexible hose 6-cm in diameter whose surface roughness is equivalent to that of galvanized iron connects the truck to the reservoir. A one-third closed ball valve and two bends that are equivalent to 90° threaded elbows are in the hose. For a filling time of 15 min, and a pump mechanical efficiency of 75%, determine the required pump power (in kW).

Approach:

The steady, incompressible flow energy equation is used to determine the required pump head. Once that is known, the power can be calculated.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no turbine, so . At points 1 and 2 the areas are large and the pressure is atmospheric, so . Therefore,

0Th =

1 2 10 and P P= = =V V 2

2

2 1 22P exit ent valve bendVLh z z f K K K K

D g⎛ ⎞= − + + + + +⎜ ⎟⎝ ⎠

where, from Table 9-3 and Figure 9-13: 1.0exitK = , Kent = 1.0, 1.5, 5.5bend valveK K= = . The friction factor is a function of Reynolds number and roughness. The volume flow rate is:

( )( )

3 315m m=0.016715min 60s min s

VVt

= =

The velocity is: ( )( )

3

2

4 0.0167 m s m=5.89sπ 0.06m

VA

= =V

( )( )( )( )3

2 2

0.86 1000kg m 5.89 m s 0.06m5740

5.3 10 Ns mDRe ρµ −= = =

×V

The flow is turbulent, so using =0.15 mmε for galvanized iron.

1.11 1.111 6.9 0.15 60 6.91.8log 1.8log 0.0387

3.7 3.7 5740D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Therefore, ( ) ( ) ( )( )

2

2

5.89m s104m+ 0.0387 +1.0+1.0+5.5+2 1.5 34.0m0.06 2 9.81m sPh ⎡ ⎤⎛ ⎞= =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Power is

( ) ( )

3 2

3 2kg m m kN s0.86 1000 0.0167 9.81 34.0m

s 1000kg mm s0.75

P PP

P P

mgh VghW

ρη η

⎛ ⎞ ⎛⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝= = =

⎞⎟⎠

=6.37kW Answer

9- 29

9-27 Large office buildings use circulating hot water systems to ensure that hot water is available instantly in all restrooms. Consider a system that consists of 200-m of 2.5-cm commercial steel pipe. It has 15 90° regular threaded elbows, two fully open gate valves, three half open gate valves and one three-quarter closed gate valve. For water at 50 °C and a pump with a mechanical efficiency of 75%, determine:

a. the power required if the water velocity is 2 m/s (in kW) b. the power required if the water velocity is 1 m/s (in kW).

Approach:

The required pump head can be determined with the steady, incompressible flow energy equation must be used. Once the head is known, the power can be calculated.

Assumptions:


Solution:

a) Because this is a closed loop, we can use the steady, incompressible flow energy equation from point 1 to point

1: 2 2

1 1 1 11 12 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + + +∑V V

There is no turbine, so , and the types of losses are given in the problem statement, so that: 0Th =

2

2 15 3 12P valve bend valve half valve quarter

open open open

Lh f K K K KD g

⎛ ⎞= + + + +⎜ ⎟⎝ ⎠

V

From Table 9-2 for the pipe =0.045mε , and for water from Appendix B-6 at 50 ºC, 4 25.29 10 Nm s ,µ −= × 3988kg mρ = . From Table 9-3, Kvalve open = 0.15, Kvalve half open = 2.1, Kvalve quarter open = 17, Kbend = 1.5.

( )( )( )3

4 2

988kg m 2m s 0.025m93, 400

5.29 10 Nm sDRe ρµ −= = =

×V


1.11 1.111 6.9 0.045 25 6.91.8log 1.8log 0.0245

3.7 3.7 93, 400D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( ) ( ) ( ) ( ) ( ) ( )( )

2

2

2m s2000.0245 +2 0.15 +15 1.5 +3 2.1 +1 17 =49.4m0.025 2 9.81m sPh ⎡ ⎤= ⎢ ⎥⎣ ⎦

Answer

( )( )( )( ) ( )( )( )23 2988kg m 2m s 4 0.025m 9.81m s 49.4m Ns kgm=626W

0.75P

P

mghWπ

η= = Answer

b) If the velocity is 1 m/s, the same calculations are made as in part (a): 46,700 0.0260Re f= =

12.9mPh = Answer 82.1WW = Answer

9- 30

9-28 To ensure adequate water supplies to a town, a municipal water department developed a second reservoir and wants to connect the new reservoir to the old one using a concrete pipe. The reservoirs are 1.5 miles apart with a difference in surface elevations of 25 ft. Determine the minimum pipe diameter needed to carry 10 ft3/s of water at 50 °F?

Approach:

The head caused by the difference in reservoir elevations is balance by the frictional head loss, which depends on the velocity and diameter. Applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the pipe diameter with an iterative solution.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . The areas of the reservoirs are large, and the pressure are atmospheric, so . Taking into account line losses only:

0P Th h= =

1 2 10 and P P= = =V V 2

2

1 2 2Lz z fD g

− =V

The friction factor is a function of Reynolds number and roughness. For concrete pipe =0.005 ftε (estimated), and for water from Appendix B-6 at 50 ºF, 5 388 10 lbm ft s , 62.4 lbm ftµ ρ−= × = .


3

2 2

4 10ft s4 12.73 ft=sπ ft

V VA D DDπ

= = =V 2 (1)

( )( )( )3 2

5

62.4 lbm ft 12.73 ft s ft903,000

88 10 lbm ft s

D DDRe Dρµ −= = =

×V (2)

For any reasonable size pipe, the flow is turbulent, so

1.111 1.8log

3.7D

Refε⎡ ⎤⎛ ⎞= − +⎢⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

6.9⎥ (3)

Substituting known quantities into the pressure drop equation:

( )( ) ( )

( )2

2

1.5mi 5280ft mi ft s25ft

ft 2 32.2ft sf

D⎡ ⎤

= ⎢ ⎥⎣ ⎦

V

2

0.2033 fD

=V (4)

The above four equations are solved iteratively. Doing so: 2.64ftD = Answer 4.81ft s=V 0.0232f =The next large standard size of concrete pipe would be used.

9- 31

9-29 The reservoir behind a dam is connected to a hydroelectric power plant with a penstock (a large pipe to convey the water). At a particular plant, the elevation difference between the reservoir surface and the hydroturbine is 50 m, and the penstock is constructed of 150-m of 1-m diameter cast iron pipe. The turbine has a mechanical efficiency of 78% and the electric generator has an efficiency of 94%. For a 1 m3/s flow of 10 °C water, determine:

a. the power output from the plant (in kW) b. the power output if a fully open gate valve and two long radius 45° flanged elbows also are in the

pipe (in kW). Approach:

The steady, incompressible flow energy equation is used to determine the turbine head. Once that is known, the power can be calculated.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump, so . At points 1 and 2 the areas are large and the pressure is atmospheric, so . Including the exit loss where,

0Ph =

1 2 10 and P P= = =V V 2 1.0exitK = .:

2

1 2 2T eVLh z z f K

D g⎛ ⎞= − − +⎜ ⎟⎝ ⎠

xit

The friction factor is a function of Reynolds number and roughness, and from Table 9-2 =0.26 mmε . For water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 .


3

2 2

4 1m s4 m=1.27s1m

V VA Dπ π

= = =V

( )( )( )3

4 2

999.6kg m 1.27 m s 1m987,000

12.9 10 Ns mDRe ρµ −= = =

×V


1.11 1.111 6.9 0.26 1000 6.91.8log 1.8log 0.0152

3.7 3.7 987,000D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Therefore, ( ) ( )( )

2

2

1.27 m s15050m- 0.0152 +1.0 49.7m1 2 9.81m sTh ⎡ ⎤⎛ ⎞= =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Power extracted from the water is

( )3 2

3 2

kg m m kN s999.6 1 9.81 49.7m =488kWs 1000kg mm sT T TW mgh Vghρ

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

Taking into account the turbine and electric generator efficiencies: ( )( )( )0.78 0.94 488kW =358kWelec T gen TW Wη η= = Answer b) Adding a gate valve, , and 2 long radius 45º bends, 0.15gateK = 0.2bendK =

( ) ( ) ( )( )

2

2

1.27 m s15050m- 0.0152 +1.0+0.15+2 0.2 49.7m1 2 9.81m sTh ⎡ ⎤⎛ ⎞= =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

358kWelecW = Answer Comments:

Other losses dominate the process so adding a gate valve has essentially no effect.

9- 32

9-30 The drain at the bottom of a swimming pool (10-m in diameter and 2-m deep) is well rounded and is connected to a 5-cm diameter, 20-m long plastic pipe. The water is at 20 °C. For a friction factor of 0.021, determine:

a. the time required to drain the pool (in min) b. if this value of friction factor is appropriate.

Approach:

Conservation of mass is needed to determine the time to drain the pool. The driving pressure head decreases with time, so the head as a function of time is needed. The open system conservation of mass equation and the steady, incompressible flow energy equation are used to calculate the time.

Assumptions:


Solution:

a) The open system conservation of mass equation is for the control volume defined as the water in the pool: in outdm dt m m= − . Water only leaves, and expressing the mass in terms of volume and flow rate in terms of

velocity:

( ) ( )1

2 2

d V d A zA

dt dtρ ρ

ρ= = − V

Density cancels so rearranging the equation:

2

2 22 2

1 1

A Ddzdt A D

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠V V

We can obtain an expression for velocity by assuming (at any instant in time) that the steady, incompressible flow energy equation is applicable with a constant friction factor:

2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

At points 1 and 2 the pressure is atmospheric, so 1P P2= . There is no pump or turbine, so . Let (the datum) and drop the subscript on z

0P Th h= =

2 0z = 1, the depth of water in the tank. The losses include an exit and line

loss, so 2 2

1 2

2 2 entLz K f

22

2g g D⎛ ⎞+ = + +⎜ ⎟⎝ ⎠

V Vg

V

The pool velocity can be described in terms of the pipe velocity by conservation of mass. Any mass flow rate from the pool must equal the mass flow rate through the pipe: ( )2

1 2 1 1 2 2 1 2 1m m A A D Dρ ρ= → = → =V V V 2V Substituting this into the energy equation and solving for velocity:

( )

0.5

2 42 1

21 ent

gzD D K f L D

⎡ ⎤= ⎢ ⎥

− + +⎢ ⎥⎣ ⎦V

Substituting this in the conservation of mass equation and separating variables:

( )

0.52

20.5 4

1 2 1

21 ent

Ddz g dtDz D D K f L D

⎡ ⎤⎛ ⎞= − ⎢ ⎥⎜ ⎟

− + +⎢ ⎥⎝ ⎠ ⎣ ⎦

We integrate this from initial height zo to 0:

9- 33

( )

( )

( )

0.5202

0.5 410 2 1

0.520.5 2

41 2 1

0.52 42 10.5 1

2

21

221

12

2

o

t

z ent

oent

ento

Ddz g dtDz D D K f L D

D gz tD D D K f L D

D D K f L DDt z

D g

⎡ ⎤⎛ ⎞= − ⎢ ⎥⎜ ⎟

− + +⎢ ⎥⎝ ⎠ ⎣ ⎦

⎡ ⎤⎛ ⎞− = − ⎢ ⎥⎜ ⎟

− + +⎢ ⎥⎝ ⎠ ⎣ ⎦

⎡ ⎤− + +⎛ ⎞= ⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠ ⎣ ⎦

∫ ∫

From Figure 9-13, Kent = 0.04

( ) ( ) ( )( )( )

0.5420.5

2

1- 0.05 10 +0.04+ 0.021 20 0.05202 2m =314,000s=87.2hr0.05 2 9.81m s

t⎡ ⎤⎛ ⎞ ⎢ ⎥= ⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦ Answer

b) We can evaluate whether the chosen friction factor is appropriate by calculating an “average” velocity, Reynolds number, and friction factor. The velocity is:

( ) ( )

( )( )

22 31

3

2 2 22

2m 10m 44 m0.00050314,000 s

4 0.00050m s4 m0.255s0.05m

tot oV z DV

t t s

V VA D

ππ

π π

≈ = = =

≈ = = =V

The friction factor is a function of Reynolds number and roughness. For smooth pipe =0ε , and for water from Appendix A-6 at 20 ºC, 4 29.85 10 Ns m , 998.2kg mµ ρ−= × = 3 . The Reynolds numbers are:

( )( ) ( )3

-4 2

998.2kg m 0.255m s 0.05m12,900

9.85×10 Ns mDRe ρµ

= = =V


1.111 0 6.91.8log 0.0288

3.7 12,900A

ff

⎡ ⎤⎛ ⎞= − + → =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

Using this friction factor, we obtain t = 362,000 s and 100.6 hr. Answer

Comments: The answer for part (b) is significantly different than that in (a), so the new estimate is probably better.

9- 34

9-31 If the pool in Problem P 9-30 has a sharp-edged entrance and two 90° regular threaded elbows, determine the time required to drain the pool (in min).

Approach:

We use the same equations as derived in Problem P 9-30. Additional minor losses are included.

Assumptions:


Solution:

The equation for the draining time, including the additional minor losses is:

( )

0.52 42 10.5 1

2

1 22

2ent bend

o

D D K K f L DDt z

D g

⎡ ⎤− + + +⎛ ⎞= ⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠ ⎣ ⎦

From Table 9-3, Kbend = 1.5 and from Figure 9-13, Kent = 1.0. Using the friction factor from part (b) in Problem 9-31:

( ) ( ) ( ) ( )( )( )

0.5420.5

2

1- 0.05 20 +1+2 1.5 + 0.0288 20 0.05202 2m0.05 2 9.81m s

t⎡ ⎤⎛ ⎞ ⎢ ⎥= ⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦

=415,300 s=6922 min=115.4hr Answer Comments:

As can be seen the addition of a small amount of minor loss increased the draining time from about 100 hr to 115 hr. Note, however, we used “average” values of velocity and friction factor, so there is uncertainty in this answer.

9- 35

9-32 A pipe connects two reservoirs at different elevations. The pipe is constructed of 12-in. diameter commercial steel with flanged fittings. The gate valve is one-fourth closed. The water temperature is 50 °F. Determine the required elevation difference between the two reservoirs to produce a water flow rate of 10 ft3/s (in ft).

Approach:

The head caused by the difference in reservoir elevations is balance by the frictional head loss. Applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the required elevation difference.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . The areas of the reservoirs are large, and the pressure are atmospheric, so . Taking into account all losses:

0P Th h= =

1 2 10 and P P= = =V V 2

2

1 2 22ent valve exit bend

Lz z f K K K KD g

⎛ ⎞− = + + + +⎜ ⎟⎝ ⎠

V

The friction factor is a function of Reynolds number and roughness. For commercial steel pipe =0.00015 ftε , and for water from Appendix B-6 at 50 ºF, 5 388 10 lbm ft s , 62.4 lbm ftµ ρ−= × = .


3

2 2

4 10ft s4 f=12.7sπ 1ft

V VA Dπ

= = =V t

( ) ( )( )3

5

62.4 lbm ft 12.7 ft s 1ft903,000


×V


1.11 1.111 6.9 0.00015 1 6.91.8log 1.8log 0.0141

3.7 3.7 903,000D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

From Table 9-3 and Figures 9-13 and 9-14, 0.6, 0.26, 1.0, 0.3ent valve exit bendK K K K= = = = Therefore,

( ) ( ) ( )( )

2

1 2 2

12.7 ft s3100.5 0.26 1.0 2 0.3 0.0141 16.9ft1 2 32.2ft s

z z ⎡ ⎤⎛ ⎞− = + + + + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ Answer

9- 36

9-33 A liquid (SG = 0.93, µ = 0.00068 N·s/m2) is contained in a vertical 2-cm diameter pipe. At one elevation the fluid pressure is 230 kPa; at an elevation 10-m higher, the pressure is 110 kPa. Determine:

a. if the flow is moving and in what direction b. the flow velocity if it is flowing (in m/s).

Approach:

If the fluid is moving, there will be frictional losses. The sign on the frictional head loss will tell us the flow direction. The frictional head loss can be determined with the steady, incompressible flow energy equation.

Assumptions:


Solution:

a) Assuming the flow is from point 1 to point 2, the steady, incompressible flow energy equation is:

2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + + +∑V V

There is no pump or turbine, so , the pipe diameter is constant, 0P Th h= = 1 2=V V , and there are no minor losses:

( )( ) ( )( )( )( ) ( )

2 21 2

1 2 3 2

110-230 kN m 1000kg m kNs+ 10m =-3.15m

0.93 1000kg m 9.81m sLP P

h z zgρ−

= + − =

The frictional head loss is negative, the assumed flow direction is incorrect. The flow is upward. Answer

b) The head loss is 2

2LLh fD g

=V . We do not know the velocity or the friction factor, so we will need to iterate

to find a solution. The friction factor depend on Reynolds number, so

( )( )( )( )3

2

0.93 1000kg m m s 0.020m27,400

0.00068 Nm sDRe ρµ

= = =VV V (1)

For any reasonable velocity the flow will be turbulent. We assume the flow is turbulent and that the pipe is smooth, =0ε , so

1.111 1.8log

3.7D

Refε⎡ ⎤⎛ ⎞= − +⎢⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

6.9⎥ (2)

Substituting known values into the head loss equation:

( )( )

0.52

2

m s10m 0.0928 m3.15m0.015m s2 9.81m s

ff

⎛ ⎞⎛ ⎞= → = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

VV (3)

The iterative procedure is to: guess a velocity, calculate the Reynolds number and the friction factor, solve equation (3) for velocity, and then compare to the guessed value. Continue until converged.

V (m/s) Re f V (m/s)

0.5 13,700 0.0284 1.81 1.81 49,600 0.0208 2.11 2.11 57,800 0.0201 2.15 2.15 58,900 0.0200 2.16 Answer

9- 37

9-34 The designers of a large shopping mall install 18-in. diameter smooth concrete storm sewers to channel away runoff after heavy rainstorms. Each storm sewer will need to carry a flow of 10 ft3/s. The pressures at the entrance and exit of the sewer are atmospheric. If the sewers are 200 ft long before they join with larger pipes, determine the required elevation change per 100 ft of pipe (in ft).

Approach:

Because we can solve for elevation change directly, the steady, incompressible flow energy equation can be used to determine the required slope.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

Let locations 1 and 2 be just outside the pipe ends, and both are at atmospheric pressure, so . The duct is constant area, so . There is no turbine or pump, so

1P P= 2

21 =V V 0T Ph h= = . For the minor losses at the entrance and exit, assume the area ratios are zero, so from Figure 9-15, Kent = 0.5 and Kexit = 1.0.

2

1 2 2ent exitLz z K K fD g

⎛ ⎞− = + +⎜ ⎟⎝ ⎠

V

The friction factor is a function of Reynolds number and roughness. For water from Appendix B-6 at 50 ºF, 5 388 10 lbm fts , 62.4 lbm ftµ ρ−= × = .


3

2 2

4 10ft s4 f=5.66s1.5ft

V VA Dπ π

= = =V t

( ) ( ) ( )3

5

62.4 lbm ft 5.66ft s 1.5ft602,000

88 10 lbm ftsDRe ρµ −= = =

×V

This is turbulent flow. From Table 9-2 for smooth concrete 0.001ft, 0.00067Dε ε= = :

1.11 1.111 6.9 0.00067 6.91.8log 1.8log 0.0184

3.7 3.7 602,000D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( ) ( )( )

2

1 2 2

5.66ft s2000.5+1.0+ 0.0184 1.97ft1.5 2 32.2ft s

z z ⎡ ⎤− = =⎢ ⎥⎣ ⎦

Therefore, the required negative slope is 1.97 200 -0.985 100ft of pipe− = Answer Comments:

Note that the angle is . ( )1 osin 0.00985 0.56θ −= − = −

9- 38

9-35 In mountainous regions, tunnels often are used for cars, trucks, and trains. If the tunnel is too long, ventilation air must be supplied to dilute and purge vehicle exhaust gases from the tunnel. Consider a 3-ft diameter, 2500-ft long duct constructed of commercial steel pipe that carries air at 45 °F, 14.1 psia with a flow rate of 10,000 ft3/min. Determine:

a. the pressure drop (in in. of water) b. the power required (in hp).

Approach:

The pressure drop is determined with the steady, incompressible flow energy equation. The pump power can be calculated with the same equation but applied just across the pump.

Assumptions:

1. The system is steady. 2. The flow is fully developed. 3. Properties are constant. 4. Air is an ideal gas.

Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The pipe diameter and elevation are constant, so 1 2 1and z z2= =V V . There is no pump or turbine, so

. There are no minor losses. Therefore, 0P Th h= =2

1 2

2aP P Lh f

g D gρ−

= =V

Assuming air is an ideal gas: ( )( )( )

( )( )

2 2 2

3

14.1lbf in. 28.97 lbm lbmol 144in. ft lbm=0.07541545ft lbf lbmR 45+460 R ft

PMRT

ρ = =

The friction factor is a function of Reynolds number and roughness. For air from Appendix B-7 at 45 ºF, 51.188 10 lbm ftsµ −= × . For commercial steel pipe, 0.00015ft, 0.00015 3 0.00005Dε ε= = = .


( )

3

2 2

4 10000ft min 1min 60s4 f=23.6sπ 3ft

V VA Dπ

= = =V t

( )( )( )3

5-5

0.0754lbm ft 23.6ft s 3ft4.49 10

1.188×10 lbm ftsDRe ρµ

= = = ×V

This is turbulent flow so

1.11 1.11

5

1 6.9 0.00005 6.91.8log 1.8log 0.01393.7 3.7 4.49 10

D fRef

ε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ×⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


( ) ( )( )

2

2

23.6ft s25000.0139 100ft of air3 2 32.2ft sah ⎛ ⎞= =⎜ ⎟

⎝ ⎠ Answer

For pressure in inches of water:

( )0.0754 12in.100ft =1.45in. of water62.4 1ft

aw w a a w a

w

P gh gh h hρ

ρ ρρ

⎛ ⎞ ⎛ ⎞∆ = = → = = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Answer

b) Applying the same energy equation between points 1 and 2: 2, 1, 2,g g gP

P P Ph

g gρ ρ−

= =

The pumping power is ( ) a aW Vg P g V P V ghρ ρ ρ= ∆ = ∆ =

( )3 2

3 2

ft lbm ft 1min lbfs 1hp10000 0.0754 32.2 100ft =2.28hpmin 60s 32.2ftlbm 550ft lbf sft s

⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

Answer

9- 39

9-36 Water at 10 °C flows from a lake at a flow rate of 0.1 m3/s. A 15-cm diameter, 100-m long galvanized iron pipe connects the lake to a building in which either a pump or a turbine is located. The elevation difference between the lake surface and the building is 10 m. Determine:

a. if the device in the building is a pump or a turbine b. the power of the device (in W).

Approach:

The steady, incompressible flow energy equation has terms for pump and turbine power. We will assume the device inside the building is a pump. Applying the energy equation, if the power calculated is positive, then our assumption is correct. If it is negative, then our assumption was wrong, and the device actually is a turbine.

Assumptions:

1. The system is steady. 2. The flow is fully developed. 3. Properties are constant. 4. The device is a pump.

Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

We assume the device is a pump, so . At points 1 and 2 the areas are large and the pressure is atmospheric, so . The losses are line, exit, and entrance losses. Therefore,

0Th =

1 2 10 and P P= = =V V 2

2

2 1 2P exitVLh z z f K K

D g⎛ ⎞= − + + +⎜ ⎟⎝ ⎠

ent

where and K1exitK = ent = 0.5. The friction factor is a function of Reynolds number and roughness. For galvanize iron pipe =0.15 mmε , and for water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 .


3

2 2

4 0.1m s4 m=5.66sπ 0.15m

V VA Dπ

= = =V

( )( )( )3

4 2

999.6kg m 5.66m s 0.15m657,700

12.9 10 Ns mDRe ρµ −= = =

×V


1.11 1.111 6.9 0.15 150 6.91.8log 1.8log 0.0201

3.7 3.7 657,700D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Therefore, ( ) ( )( )

2

2

5.66m s10010m+ 0.0201 +0.5+1 14.3m0.15 2 9.81m sPh ⎡ ⎤⎛ ⎞= − =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Because this is positive, the device is a pump. Answer b) Power is

( )3 2

3 2

kg m m N s999.6 0.1 9.81 14.3m =14,020W=14.0kWs kg mm sP P PW mgh Vghρ

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= = = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

Answer

9- 40

9-37 Ski resorts pump water to make snow when the weather does not cooperate. Consider a resort that uses 100 gal/min of 35 °F water. It is pumped from the water holding pond through a 4-in. diameter, 3000-ft steel pipe to the top of the mountain. The elevation difference is 950 ft. The gage pressure required at the nozzle at the end of the pipe is 150 lbf/in.2. Determine the required pumping power (in hp).

Approach:

By applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the required pump power.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no turbine, so . At point 1 the area of the pond is large, and the gage pressure is atmospheric, so . Ignoring entrance loss, the losses are line and exit. Therefore,

0Th =

1 10 and 0P=V =

( )2

22 1 2P e

P VLh z z f K xitg D gρ⎛ ⎞= + − + +⎜ ⎟⎝ ⎠

where . 1exitK = The friction factor is a function of Reynolds number and roughness. For commercial steel pipe

=0.00015 ftε , and for water from Appendix B-6 at 35 ºF, 5 3114 10 lbm ft s , 62.4 lbm ftµ ρ−= × = .

The velocity is: ( )( )( )

( )

3

2 2

4 100gal min 0.1337gal ft 1min 60s4 f=2.61sπ 0.33ft

V VA Dπ

= = =V t

( )( )( )3

5

62.4 lbm ft 2.61ft s 0.33ft47,060

114 10 lbm ft sDRe ρµ −= = =

×V


1.11 1.111 6.9 0.00015 0.33 6.91.8log 1.8log 0.0223

3.7 3.7 47,060D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Therefore, ( )( ) ( )

( )( ) ( ) ( )( )

2 22 2

3 2 2

150lbf in. 12in. ft 32.2ft lbm lbf s 2.61ft s3000+950ft+ 0.0223 +10.3362.4 lbm ft 32.2ft s 2 32.2ft s

=346+950+21.5=1318ft

Ph ⎡ ⎤⎛ ⎞= ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Power is

( )2

3 3 2

lbm gal gal 1min ft lbf s 1hp s62.4 100 0.1337 32.2 1318ftmin 60s 32.2ft lbm 550 ft lbfft ft sP P PW mgh Vghρ

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎛= = = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎝⎝ ⎠

⎞⎟⎠

=33.3 hp Answer

9- 41

9-38 In the western United States, many crops are irrigated, and water must be pumped long distances. Consider a system that consists of a 1-m diameter, 2-km long steel pipe, which connects a river to an irrigation canal. The canal’s elevation is 50 m higher than that of the river. For water at 15 °C, a pump with a mechanical efficiency of 80%, and neglecting minor losses, determine the power required to pump 2.5 m3/s of water (in kW).

Approach:

The steady, incompressible flow energy equation is used to calculate the required pump head. Once that is known, the required pumping power can be calculated.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The areas at points 1 and 2 the areas are large and the pressure is atmospheric, so 1 2 10 and P P2= = =V V .

There is no turbine, so , and minor losses are neglected. Therefore, 0Th =2

2 1 2PVLh z z f

D g= − +


3

2 2

4 2.5m s4 m=3.18sπ 1m

V VA Dπ

= = =V

The friction factor is a function of Reynolds number and roughness. For the pipe =0.045mmε , and for water from Appendix A-6 at 15 ºC, 4 211.2 10 Ns m , 999kg mµ ρ−= × = 3 . The Reynolds numbers are:

( )( )( )3

64 2

999kg m 3.18m s 1m2.81 10

11.2 10 Ns mDRe ρµ −= = = ×

×V


1.11

6

1 0.045 1000 6.91.8log 0.01133.7 2.81 10

ff

⎡ ⎤⎛ ⎞= − + → =⎢ ⎥⎜ ⎟ ×⎝ ⎠⎢ ⎥⎣ ⎦

Therefore,

( ) ( )( )

2

2

3.18m s200050m 0.0113 61.7m1 2 9.81m sPh ⎛ ⎞= + =⎜ ⎟

⎝ ⎠ Answer

Pumping power, taking into account pump efficiency, is:

( )( )( )( )( )

( )( )

3 3 2 2999kg m 2.5m s 61.7m 9.81m s Ns kgm=1,890kW

0.80 1000W kWP

P

mghW

η= =

9- 42

9-39 Air at 105 kPa and 25 °C flows from a 7.5-cm circular duct into a 22.5-cm circular duct. The downstream pressure is 6.5-mm of water higher than the upstream pressure. Determine:

a. the average air velocity approaching the expansion (in m/s) b. the volumetric flow rate (in m3/s) c. the mass flow rate (in kg/s).

Approach:

Applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the flow rate through the expansion. The loss coefficient across the expansion must be evaluated.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . The pipe is horizontal, so 0T ph h= = 1z z2= . The only loss is due to the sudden expansion. Solving for the pressure rise:

( )2

2 2 12 1 1 22 2expP P Kρ ρ− = − −

VV V

From conservation of mass, ( ) ( )21 2 2 1 1 2 1 1 2m m A A D D= → = =V V V

Substituting this into the pressure drop equation and solving for velocity:

( )

( )

0.5

2 11 4

1 2

2

1 exp

P P

D D Kρ

⎧ ⎫−⎪ ⎪= ⎨ ⎬⎡ ⎤− −⎪ ⎪⎣ ⎦⎩ ⎭

V

Assuming an ideal gas, the air density is:

( )( )( )( )

2

3

105kN m 28.97 kg kmol kg=1.2288.314kJ kgK 25+273 K m

PMRT

ρ = =

Using Figure 9-15b for a sudden expansion, with ( ) ( )2 2

1 2 1 2 7.5 22.5 0.111 0.76expA A D D K= = = → ≈ The pressure rise is obtained from the given information and water density: ( )( )( )( )3 2 2

2 1 1000kg m 9.81m s 0.0065m kNs 1000kg m =0.0638kN mP P ghρ− = = 2

( )( )( ) ( )

0.52

1 43

2 0.0638kN m 1000kg m kNs m=21.4s1.228kg m 1- 7.5 22.5 -0.76

⎧ ⎫⎪ ⎪= ⎨ ⎬

⎡ ⎤⎪ ⎪⎣ ⎦⎩ ⎭

V Answer

b) The volume flow rate is: ( )( )( )2 321.4m s π 4 0.075m =0.0944m sV A= =V Answer c) Mass flow rate is: ( )( )3 3m= V= 1.228kg m 0.0944m s 0.116kg sρ = Answer

9- 43

9-40 In some high rise buildings, water is stored in an elevated tank on the roof to minimize pressure fluctuations in the system. Consider water that is pumped through a 10-cm steel pipe to the roof of a 200-m tall building; the pump is on the ground floor. For a water temperature of 10 °C and a flow of 0.02 m3/s, what is the pressure at the pump discharge (in kPa)?

Approach:

The steady, incompressible flow energy equation is used directly to calculate the pressure at point 1.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The area at point 2 is large and the pressure is atmospheric, so 2 20 and 101.3kPaP= =V . There is no pump or turbine, so , and all minor losses except the exit are neglected. Therefore,

0T Ph h= =

( )2 2

1 11 2 2 12 2exit

LP P g z z K fD

ρ ρ ρ⎛ ⎞= − + − + +⎜ ⎟⎝ ⎠

V V

Note that , so 1exitK =

( )2

11 2 2 1 2

LP P g z z fD

ρ ρ= + − +V


3

2 2

4 0.02m s4 m=2.55sπ 0.10m

V VA Dπ

= = =V

The friction factor is a function of Reynolds number and roughness. For the pipe =0.045mmε , and for water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 . The Reynolds number is:

( ) ( )( )3

4 2

999.6kg m 2.55m s 1m198,000

12.9 10 Ns mDRe ρµ −= = =

×V


1.111 0.045 100 6.91.8log 0.0184

3.7 198,000f

f

⎡ ⎤⎛ ⎞= − + → =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

Therefore,

( )( ) ( ) ( )22 22

1 3

2.55m s200101.3 999.6 9.81m s 200 0.01841000 0.10 2 1000

kg kN s kN sP kPa mkg m kg mm

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

( )101.3 1960 120 2181kPa absolute =2080kPa (gage)= + + = Answer

9- 44

9-41 A fluid flows by gravity down an 8-cm galvanized iron pipe. The pressures at the higher and lower locations are 120 kPa and 140 kPa, respectively. The horizontal distance between the two locations is 30 m, and the pipe has a slope of 1m rise per 10 m of run (horizontal distance). For a fluid with a kinematic viscosity of 10-6 m2/s and a density of 900 kg/m3, determine the flow rate (in m3/s).

Approach:

Flow is set by a balance among elevation difference, pressure difference, and line losses. Line losses depend on velocity, so this maybe an iterative solution depending on whether this is a laminar or a turbulent flow. The steady, incompressible flow energy equation can be used to determine the flow.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + + +∑V V

There is no pump or turbine, so , the pipe diameter is constant, 0P Th h= = 1 2=V V , there are no minor losses, and with ( ) 2 2Lh f L D= V g :

( )2

1 22 1 2

P P Lz z fg D gρ−

= − +V

Note that ( ) . From the problem geometry: 2 1 3mz z− ≠ −

( ) ( )1 o o2 1tan 1 10 5.71 30sin 5.71 2.985mz zθ −= = → − = − = −

The friction factor depend on Reynolds number, so

( )( )

6 2

m s 0.08m80,000

10 m sD DRe ρµ ν −= = = =

VV V V (1)

For any reasonable velocity the flow will be turbulent, so with =0.15mm, D 0.15 80 0.00188ε ε = =

1.111 1.8log

3.7D

Refε⎡ ⎤⎛ ⎞= − +⎢⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

6.9⎥ (2)


( )

( )( )( )( )

22 2

3 2 2

kN 1000kgm120-140m sm kNs 30m=-2.985m+f

0.08m900kg m 9.81m s 2 9.81m s

⎛ ⎞⎜ ⎟

⎛ ⎞⎝ ⎠⎜ ⎟⎝ ⎠

V

( 2-2.265m=-2.985m+19.11 m sf V ) (3) The above three equations can be solved iteratively to determine velocity and, hence, flow rate. Performing the iteration: V (m/s) = 1.24 m/s Re=98,900 f = 0.0247 For volume flow rate

( )3

2m m1.24 0.08m =0.00623s 4 s

V A π⎛ ⎞= = ⎜ ⎟⎝ ⎠

V Answer

9- 45

9-42 A new factory is to be built that requires 0.03 m3/s of water. The water main from which the water will be obtained is 150-m from the factory. The water main pressure is 400 kPa (gage), and the factor needs 100 kPa (gage) water at a location 10-m above the water main. Assuming that galvanized steel pipe will be used, determine the minimum pipe diameter needed (in m).

Approach:

Friction pressure loss depends on the friction factor and velocity, both of which depend on the pipe diameter. An iterative solution is required. The steady, incompressible flow energy equation is used to find the required diameter.

Assumptions:


Solution:

From point 1 to point 2, the steady, incompressible flow energy equation is:

2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + + +∑V V

There is no pump or turbine, so . The pipe diameter is constant, so 0P Th h= = 1 2=V V . Only line losses are present Therefore,

( )2

1 22 1 2

P P Lz z fg D gρ−

= − +V


3

2 2 2

4 0.3m s4 0.0382 m=sπ m

V VA D DDπ

= = =V (1)

The friction factor is a function of Reynolds number and roughness. For water from Appendix A-6 at 10 ºC (assumed), 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 . The Reynolds numbers are:

( )( )( )3 2

4 2

999.6kg m 0.0382 m s m 29,60012.9 10 Ns m

D DDReD

ρµ −= = =

×V (2)

For any reasonable size pipe, the flow will be turbulent, and from Table 9-2, with =0.15mmε , so

1.111 1.8log

3.7D

Refε⎡ ⎤⎛ ⎞= − +⎢⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

6.9⎥ (3)

Substituting known values into the pressure drop equation

( ) ( )

( )( )( )( )

22 2

3 2

400-100 kN m 1000kgm kNs m s150=10m+2 9.81m s999.6kg m 9.81m s

fD

⎛ ⎞⎜ ⎟⎝ ⎠

V

2 2

2.6942.694

f DD

= → =V f V (4)

The above four equations are solved iteratively: D = 0.104 m V = 3.54 m/s Re = 285,000 f = 0.0223 The next standard pipe size greater than 0.104 would be used. Answer

9- 46

9-43 For the system shown below, a water flow rate of 3 m3/s is to be pumped from the lower to the upper reservoir through a 1-m diameter commercial steel pipe. The pump has a mechanical efficiency of 80%. Neglecting minor losses, determine the power required (in kW).

Approach:

The required pump head can be obtained directly from the steady, incompressible flow energy equation. Once the head is determined, the pumping power can be calculated.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + + +∑V V

There is no turbine, so . The areas at 1 and 2 are large, 0Th = 1 2 0= =V V , and pressure is atmospheric, so . The losses include inlet, exit, and line losses. Therefore, 1P P= 2

( )2

2 1 2P ent exitLh z z K K fD g

⎛ ⎞= − + + +⎜ ⎟⎝ ⎠

V


3

2 2

4 3m s4 =3.82sπ 1m

V VA Dπ

= = =V m

The friction factor is a function of Reynolds number and roughness. For water from Appendix A-6 at 10 ºC (assumed), 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 . The Reynolds numbers are:

( )( )( )3

64 2

999.6kg m 3.82m s 1m2.96 10

12.9 10 Ns mDRe ρµ −= = = ×

×V

Assuming a sharp edged entrance and exit, from Figures 9-13 and 9-14, 0.5 and 1.0ent exitK K= = . The flow is turbulent and from Table 9-2, =0.045mmε , so

1.11 1.11

6

1 6.9 0.045 1000 6.91.8log 1.8log 0.01133.7 3.7 2.96 10

D fRef

ε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟ ×⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Substituting known values into the pressure drop equation

( ) ( ) ( )( )

23.82m s245= 730-673 m+ 0.5 1 0.01131 2 9.81m sPh ⎡ ⎤⎛ ⎞+ + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

57.0m+3.2m=60.2m= Answer

( )( )( ) ( )( )3 3 2 2999.6kg m 3m s 9.81m s 60.2m 1kNs 1000kgm

=2,213kW0.80

P

P

mghW

η= = Answer

9- 47

9-44 Fresh air is distributed in a factory through a 250-ft long rectangular galvanized duct, which is 36-in. by 6-in. For a flow rate of 5000 ft3/min of 60 °F air at 14.2 lbf/in.2, determine the fan pumping power required if the fan has a mechanical efficiency of 65% (in hp).

Approach:

The required fan head can be obtained directly from the steady, incompressible flow energy equation. Once we determine the required head, the fan power can be calculated.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no turbine, so . The duct has constant area and is horizontal 0Th = 1 2 1and z z2= =V V . Point 1 is just upstream of the fan and point 2 is outside of the duct, so both are at atmospheric pressure and . Therefore, with only an exit minor loss and line loss:

1P P= 2

2

2P exitVLh f K

D g⎛ ⎞= +⎜ ⎟⎝ ⎠

where . Assuming air is an ideal gas: 1exitK =

( )( )( )

( )( )

2 2 2

3

14.2 lbf in. 28.97 lbm lbmol 144in. ft lbm=0.07371545ft lbf lbmR 60+460 R ft

PMRT

ρ = =

The friction factor is a function of Reynolds number and roughness. For the duct (from Table 9-2), =0.0005 ftε , and for air from Appendix B-7 at 60 ºF, 51.214 10 lbm ft sµ −= × .


( )( )

35000ft min 1min 60s ft=55.60.5ft 3ft s

VA

= =V

Because this is a non-circular duct, we need the hydraulic diameter:

( )

( )( )( )

4 0.5ft 3ft4 4 =0.857ft2 2 0.5+3 ft

xh

wetted

A HWDp H W

= = =+

( )( )( )3

5

0.0737 lbm ft 55.6ft s 0.857ft289,000

1.214 10 lbm ft shD

Reρµ −= = =

×V


1.11 1.111 6.9 0.0005 0.857 6.91.8log 1.8log 0.0186

3.7 3.7 289,000D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Therefore,

( ) ( )( )

2

2

55.6ft s2500.0186 +1 =308ft0.857 2 32.2ft sPh ⎡ ⎤⎛ ⎞= ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Power is

( )

( )

3

3 2

2

lbm ft 1min ft 1hp s0.0737 5000 32.2 308ftmin 60s 550 ft lbfft s

32.2ft lbm0.65lbf s

P PP

P P

mgh VghW

ρη η

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝⎝ ⎠= = =⎛ ⎞⎜ ⎟⎝ ⎠

⎞⎟⎠

= 5.3hp Answer

9- 48

9-45 Water at 20 °C is to be siphoned from a large tank, as shown below. The siphon is a 2.5-cm diameter smooth tube and has a reentrant inlet. Determine:

a. the volume flow rate if only the minor losses are taken into account (in m3/s) b. the volume flow rate if both the minor and line losses are taken into account (in m3/s).

Approach:

The flow can be determined by applying the steady, incompressible flow energy equation between points 1 and 2. For part (b) an iterative solution is required.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . We assume the area at 1 large, so 0T ph h= = 1 0≈V , and the pressure at 1 and 2 are atmospheric, so P1 = P2. The losses, in general, include one entrance, one bend, and line loss:

2

2L ent bendLh f K KD g

⎛ ⎞= + +⎜ ⎟⎝ ⎠

∑ V

Combining and solving for the velocity:

( ) 0.5

2 121 ent bend

g z zK K f L D

−⎡ ⎤= ⎢ ⎥+ + +⎣ ⎦

V

From Figure 9-13a, for a reentrant entrance Kent = 0.8. From Table 9-3, for a 180º bend, Kbend = 1.5. a) Ignoring line loss, f = 0

( )( ) 0.5

2 9.81m s 1.5m m=2.991+0.8+1.5+0 s

⎡ ⎤= ⎢ ⎥⎣ ⎦

V

( )( )( )2 32.99m s π 4 0.025m =0.00147 m sV A= =V b) The line loss will decrease the flow, so the procedure we will follow is: assume a velocity, calculate Reynolds number, evaluate friction factor, and then calculate velocity from the complete velocity equation. For smooth tubing =0ε , and for water from Appendix A-6 at 20 ºC,

4 29.85 10 Ns m , 998.2kg mµ ρ−= × = 3 . The friction factor is a function of Reynolds number and roughness.

( )( )( )3

4 2

998.2kg m m s 0.025m25,340

9.85 10 Ns mDRe ρµ −= = =

×

VV V

For a velocity even one-fourth the velocity obtained without line loss, the flow still would be turbulent, so

21.11 1.111 6.9 0 6.91.8log 1.8log 1.8log

3.7 3.7D f

Re Re Refε

−⎡ ⎤ ⎡ ⎤ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎡ ⎤= − + = − + → = +⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

6.9

Guess V = 2 m/s → Re = 50,700 → f = 0.0207 →

( )( )

( )( )

0.52 9.81m s 1.5m m=2.47

1+0.8+1.5+ 0.0207 0.1+0.1+1.5+0.1 /2 0.025 sπ⎡ ⎤

= ⎢ ⎥⎢ ⎥⎣ ⎦

V

For a second iteration: V = 2.47 m/s → Re = 62,500 → f = 0.0197 → V = 2.49 m/s This is close enough so ( )( )( )2 32.49m s π 4 0.025m =0.00122m sV A= =V Answer

9- 49

9-46 A pump draws 40 °F water from a lake through 20 ft of commercial steel pipe; the line has a reentrant inlet and a 90° regular flanged elbow. The pump elevation is 12 ft above the lake surface. For a design flow rate of 100 gal/min, the head at the suction side of the pump must not be less than –20 ft of water. Determine the minimum pipe diameter (in in.).

Approach:

The pressure drop in the inlet piping must be calculated, and the pressure at point 2 expressed in terms of head. The steady, incompressible flow energy equation must be used. Because friction factor depends on velocity and, hence, diameter, an iterative solution is required.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump, so . We assume the area at 1 large, so 0ph = 1 0≈V , and the pressure at 1 is atmospheric, so using gage pressure P1 = 0. Let . The losses, in general, include one entrance, one bend, and line loss: 1 0z =

2 2

2 2 220

2 2ent bendP Lz f K Kg g Dρ

⎛ ⎞= + + + + +⎜ ⎟⎝ ⎠

V Vg

Solving for 2P gρ : 2

2 22 2 1

2ent bendP Lh z f K Kg D gρ

⎛ ⎞= = − − + + +⎜ ⎟⎝ ⎠

V


( )( )( )

( )

3

2 2 2

4 100gal min 0.1337ft gal 1min 60s4 0.284 ft=sft

V VA D DDπ π

= = =V 2

For commercial steel pipe =0.00015ftε , and for water from Appendix B-6 at 40 ºF, 5104 10 lbm ft s , 62.4 lbm ftµ ρ−= × = 3 . From Figure 9-13a, for a reentrant entrance Kent = 0.8. From Table 9-3,

for a 90º bend, Kbend = 0.3. The friction factor is a function of Reynolds number and roughness.

( )( )( )3 2

5

62.4 lbm ft 0.284 ft s ft 17,000104 10 lbm ft s

D DDReD

ρµ −= = =

×V

For any pipe with a diameter of less than about 8 ft, the flow will be turbulent, so

1.111 61.8log

3.7D

Refε⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

.9

We know that . The procedure is: guess diameter, calculate velocity, Reynolds number, and friction factor, and then calculate h

2 20fth ≤ −

2 from the energy equation and compare to the limit.

D (ft) V (ft/s) Re f h2 (ft) 0.2 7.10 85,000 0.0214 -15.4

0.17 9.83 100,000 0.0215 -19.0 0.16 11.1 106,300 0.0216 -21.2

Based on these calculations, the standard pipe size about 0.17 ft = 2.04 in. in diameter or larger would be chosen for the suction pipe. Answer

9- 50

9-47 Large farm implements and road construction equipment use hydraulically actuated cylinders to position scoops, cutting blades, and other tools. High-pressure pumps are used to circulate the hydraulic fluid (ρ = 880 kg/m3 and µ = 0.033 N·s/m2). Consider a hydraulic system that has a pump outlet pressure of 20 MPa and which requires a minimum pressure at the hydraulic cylinder of 18 MPa at a flow rate of 0.0005 m3/s. If the hydraulic fluid flows through 25 m of smooth, drawn steel tubing, determine the minimum tubing diameter required (in cm).

Approach:

The steady, incompressible flow energy equation must be used. For line friction losses, friction factor is a function of velocity, which depends on pipe diameter. Depending on the flow, an iterative solution may be required.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . The pipe diameter and elevation are constant, so and . Ignoring minor losses:

0p Th h= = 1 =V V2

21z z=

2

1 2

2P P Lf

g D gρ−

=V

The friction factor is a function of Reynolds number and roughness. However, because of the low flow rate and large viscosity, we can speculate that the flow might be laminar. Therefore, we will assume (initially) that the flow is fully developed laminar flow. For this situation, 64f R= e Substituting into the pressure drop relation, using 24V A V Dπ= =V , and solving for diameter:

( )( )( )

( )( )

1 41 4 2 3

2

128 0.033Ns m 25m 0.0005m s128 =0.00957m=9.57mm2000kN m 1000N 1kN

LVDPµ

π π

⎡ ⎤⎡ ⎤⎢ ⎥= =⎢ ⎥∆ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Checking Reynolds number

2

4 4D m VRe Re VA DD

ρ ρ ρµ ρ ππ

= → = = → =V V

µ

( )( )( )( )

3 3

2

4 880kg m 0.0005 s=1770

0.033Ns m 0.00957mRe

π=

This is laminar, so our assumption was valid. The diameter would be the next standard diameter greater than 9.57 mm. Answer

Comments:

If the flow had been turbulent, we would have needed to solve the pressure drop equation for diameter, and then iterated to find a solution. We would have assumed a friction factor, calculated a diameter, calculated Reynolds number, and then checked the friction factor for convergence.

9- 51

9-48 Water at 70 °F with a flow rate of 30 gal/min flows from a 1-in. diameter tube into a 2-in. diameter tube through a sudden expansion. Determine the pressure rise across the expansion (in lbf/in.2).

Approach:

Applying the steady, incompressible flow energy equation between points 1 and 2, we can determine the pressure rise across the expansion.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . The pipe is horizontal, so 0T ph h= = 1z z2= . The only loss is due to the sudden expansion. Solving for the pressure rise:

( )2

2 2 12 1 1 22 2expP P Kρ ρ− = − −

VV V

From conservation of mass, ( ) ( )21 2 2 1 1 2 1 1 2m m A A D D= → = =V V V

Substituting this into the pressure drop equation and simplifying:

42

1 12 1

2

12 exp

DP P K

Dρ

⎡ ⎤⎛ ⎞⎢ ⎥− = − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

V


( )

3

1 2 2

4 30gal min 0.1337gal ft 1min 60s4 f=12.2sπ 1 12 ft

V VA Dπ

= = =1

V t

Using Figure 9-15b for a sudden expansion, with ( ) ( )2 21 2 1 2 1 2 0.25 0.56expA A D D K= = = → ≈

Therefore, ( )( )23 4 22

2 1

62.2 lbm ft 12.2ft s 1 lbf s 11+ -0.562 2 32.2ft lbm 12in.

P P⎡ ⎤ ⎛ ⎞⎛ ⎞ ⎛ ⎞− = ⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎝ ⎠⎣ ⎦

ft

2

lbf0.380in.

= Answer

9- 52

9-49 A Class 100 clean room is to be supplied with 15 m3/min of air, which enters the duct (shown below) at 100 kPa, 25 °C. All entrances and exits are all sharp edged. Determine:

a. the pressure in the clean room (in kPa) b. the fan power required (in W).

Approach:

The pressure in the clean room can be determined by applying the steady, incompressible flow energy equation between points 2 and 3. The fan power can be obtained by applying the energy equation between points 1 and 3.

Assumptions:


Solution:

a) The steady, incompressible flow energy equation is: 22

3 32 22 32 2P T L

PPz h z h h

g g g gρ ρ+ + + = + + + +∑VV

There is no pump or turbine, so . The pipe is horizontal, so 0T ph h= = 1 2z z z3= = . The pressure loss is due to line loss, plus one expansion and one contraction, so

2 2

2 2L entLh f K KD g g g

= + +V V 2

2exitV

Assuming the room at 2 and the outside at 3 are large, 2 3 0≈ ≈V V . Solving the energy equation for P2 (note that ): 3P P= 1

2

2 3 2ent exitLP P f K KD

ρ⎛ ⎞= + + +⎜ ⎟⎝ ⎠

V


( ) ( )

( )

3

2 2

4 15m min 1min 60s4 m=5.09s0.25m

V VA Dπ π

= = =V

The friction factor is a function of Reynolds number and roughness. For smooth pipe =0ε , and for air from Appendix A-7 at 100 kPa, 25 ºC, 5 21.832 10 Ns m , 1.169kg mµ ρ−= × = 3 .

( )( )( )3

5 2

1.169kg m 5.09m s 0.25m81,200

1.832 10 Ns mDRe ρµ −= = =

×V


1.11 1.111 6.9 0 6.91.8log 1.8log 0.0186

3.7 3.7 81, 200D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Using Figure 9-15a for the entrance and Figure 9-15b for a sudden expansion, with 1 2 0 0.5 and 1.0ent exitA A K K= → ≈ =

Therefore:

( ) ( )2 2

2

5.09m s15 kg kNs100kPa+ 0.0186 +0.5+1 1.1690.25 2 1000kg mm

P⎛ ⎞⎡ ⎤ ⎛ ⎞= ⎜ ⎟⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠ ⎝ ⎠

=100+0.0397=100.04kPa Answer b) For fan power, we recognize that the pressure drop from just after the fan to the clean room is the same as from the clean room to the outside (calculated above). Using the energy equation, again assuming incompressible flow, and taking into account the two identical lengths of pipe:

3

23 2

m kN 1min 1000W s2 -2 15 0.0397 =-19.9Wmin 60s kN mmtotW V P V P

⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= − ∆ = − ∆ = ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

Answer

9- 53

9-50 In Problem P 9-49, the sharp-edged entrances and exits are replaced with well-rounded entrances and exits. For the same fan power as in the original installation, determine the new volumetric flow rate (in m3/min).

Approach:

We will use the information from the solution to Problem P 9-49. The main difference is that Kent and Kexit will change because of the well-rounded entrances and exits.

Assumptions:


Solution:

We have two equations to combine: ( )23 2 32 2W V P A P P= − ∆ = − −V

2

2 3 2ent exitLP P f K KD

ρ⎛ ⎞− = + +⎜ ⎟⎝ ⎠

V

The friction factor is a function of velocity, but the new flow probably will not cause a significant change in the friction factor. Therefore, we will assume it is approximately constant. Combining the two equations and solving for velocity:

( )

1 3

2

4

ent exit

WD fL D K Kπ ρ

⎡ ⎤−= ⎢ ⎥

+ +⎢ ⎥⎣ ⎦V

For well-rounded entrances and exits, from Figure 9-13d and Figure 9-14d, 0.04 and 1.0ent exitK K≈ = .

( )( )( )( )

( ) ( )

1 3

2

2 3

-4 -19.9W 1J Ws 1Nm 1J kgm Ns m=5.4315 sπ 0.25m 1.169kg m 0.0186 +0.04+1

0.25

⎡ ⎤⎢ ⎥⎢ ⎥=

⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

The volume flow rate is: ( )( )( ) ( )2 35.43m s π 4 0.25m 60s 1min =16.0m sV A= =V

16.0 15.0%increase 100 6.7%15.0−

= × = Answer

Comments:

Note that the new Reynolds number is 86,600 and the friction factor is 0.0184, which would not change our answer very much.

9- 54

9-51 Frictional pressure loss in fluid flow is converted to unwanted thermal energy. Consider an 18 gal/min flow of 70 °F water through a 1.25-in. diameter smooth tube. The tube is sloped so that the pressure remains constant throughout the tube. Determine: a) the slope (in ft/100 ft), b) the heat transfer per 100 ft of tube if the temperature remains constant (Btu/hr), and c) the temperature rise if the tube is perfectly insulated (in °F).

Approach:

For constant pressure flow, the steady, incompressible flow energy equation is used. Frictional losses can be calculated because the flow is given. With constant temperature, Eq. 9-29 can be used to calculate the heat transfer; the same equation can be used to calculate temperature rise if the pipe is insulated.

Assumptions:


Solution:

a) The slope (in ft/100 ft of pipe) requires that the elevation difference, 2z z1− , be determined. Using the steady,

incompressible flow energy equation: 2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V


neglected. For constant pressure flow 1P P2= . Therefore, 2

2 1 2LLh z z fD g

− = − = −V

Velocity is :( )( )( )

( )

3

2 2

4 15gal min 0.1337ft gal 1min 60s4 f=3.93sπ 0.104ft

V VA Dπ

= = =V t

The friction factor is a function of Reynolds number and roughness. Assuming a smooth pipe = 0ε , and for water from Appendix B-6 at 70 ºF, 5 365.8 10 lbm fts , 62.2 lbm ftµ ρ−= × = :

( )( )( )3

5

62.2 lbm ft 3.93ft s 0.104ft38,700

65.8 10 lbm ftsDRe ρµ −= = =

×V

The flow is turbulent, so 1.111 0 6.91.8log 0.022

3.7 38,700D f

f

⎡ ⎤⎛ ⎞= − + → =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

( ) ( )( )

22

2 1 2

3.93ft s1000.022 5.06ft2 0.104 2 32.2ft sL

Lh z z fD g

⎛ ⎞− = − = − = − = −⎜ ⎟⎝ ⎠

V

So the slope is -5.06 ft/100 ft or ( )1 osin 5.06 100 2.9θ −= − = − . Answer b) For the heat transfer , using Eq. 9-29 ( )2 1Lh u u q= − − g⎡ ⎤⎣ ⎦ . For isothermal flow, , so that 2 1 0u u− =

L Lq h g Q mq mh g Ah gρ= − → = = − = − V L

( ) ( )2

22 2

lbm ft π ft 1Btu lbf s Btu62.2 3.93 0.104ft 5.06ft 32.2 =-0.0135s 4 778ft lbf 32.2ft lbm sft s

Q⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= − ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠Answer

c) For an insulated pipe, q = 0, so again using Eq. 9-28, and assuming an ideal liquid with constant specific heat: ( ) ( )2 1 2 1 2 1L Lh u u q g u u h g c T T h= − − → − = → − =⎡ ⎤⎣ ⎦ L g

( ) ( )( )( )( )

2o

2 1 2

5.06ft 32.2ft s 1Btu 778ft lbf=0.0065 F

1.00Btu lbmR 32.2ft lbm lbf sLT T h g c− = =

9- 55

9-52 A gas turbine power plant consists of a compressor, a combustor in which the fuel and air are mixed and combusted, and a turbine that drives an electrical generator. The air compression process takes from 40 to 80% of the turbine output power, leaving only 20-60% to drive the electric generator. Some gas turbine plants store compressed air in salt domes or caverns for use during times when additional electric power is needed; the compressed air to be supplied to the power plant is taken from the stored air instead of just using the air compressor. Consider the system shown in the figure below. The air reservoir, which fills with 10 °C water when the air has been used, is connected to the outside by a 30-cm cast iron pipe. During charging of the reservoir with air, the air pressure, P, increases. Determine the gage pressure P required to produce a water flow rate of 0.15 m3/s (in kPa).

Approach:

The required pressure can be determined by applying the steady, incompressible flow energy equation between points 1 and 2. The frictional losses must be calculated.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . We assume the surface of the reservoir is large, so , and the pressure at 2 is atmospheric. Expressing P

0T ph h= = 1 0≈V

1 and P2 in terms of gage pressures, 2, 0gageP = Therefore:

( )2

21, 1 2 2gage LP g z z gρ ρ ρ= − + + ∑V

h

The losses include line loss, one entrance, and two bends:

2

22L ent exit bend

Lh f K K KD g

⎛ ⎞= + + +⎜ ⎟⎝ ⎠

V

The entrance has an area ratio of zero, so from Figure 9-15a, Kent = 0.5. Assuming the bends are regular, 90º, flanged, from Table 9-3, Kbend = 0.3. The velocity is determined from conservation of mass for incompressible flow:

( )( )

3

2 2 2

4 0.15m s4 m=2.12s0.30m

V VA Dπ π

= = =V

The friction factor is a function of Reynolds number and roughness. For cast iron pipe =0.26mmε , and for water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 .

( )( )( )3

4 2

999.6kg m 2.12m s 0.30m493,000

12.9 10 Ns mDRe ρµ −= = =

×V


1.11 1.111 6.9 0.26 300 6.91.8log 1.8log 0.0196

3.7 3.7 493,000D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( ) ( ) ( ) ( )2 2

1, 2

2.12m skg m 850 kg kNs999.6 9.81 50m + 0.0196 +0.5+1+2 0.3 999.60.30 2 1000kg mm s mgageP

⎧ ⎫⎛ ⎞⎪ ⎪⎛ ⎞⎛ ⎞ ⎡ ⎤ ⎛ ⎞= ⎨ ⎬⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭

=620kPa (gage) Answer

9- 56

9-53 Water at 20 °C is pumped from a reservoir through a 20-cm commercial steel pipe for 5 km from the pump outlet to a reservoir whose surface is 150-m above the pump. The flow rate is 0.10 m3/s. Determine:

a. the pressure at the pump outlet (in kPa) b. the pumping power required (in W).

Approach:

We can determine the pressure at 2 by using the steady, incompressible flow energy equation between points 2 and 3. The pump power can be calculated with the same equation but applied between points 1 and 2.

Assumptions:


Solution:

a) The steady, incompressible flow energy equation between points 2 and 3 is:

22

3 32 22 32 2P T L

PPz h z h h

g g g gρ ρ+ + + = + + + +∑VV

The reservoir is large, so . There is no pump or turbine, so 3 0≈V 0P Th h= = . Expressing P2 and P3 in terms of gage pressure, so . The losses include line loss and exit loss (3, 0gP = 1exitK = ). Solving for 2,gP

( ) ( )2 2 2

2 2 22, 3 2 3 22 2 2g exit

L LP g z z K f g z z fg g D g D

ρ ρ⎡ ⎤ ⎡

= − − + + = − +⎢ ⎥ ⎢⎣ ⎦ ⎣

V V V V 22

2⎤⎥⎦

The friction factor is a function of Reynolds number and roughness. For water from Appendix A-6 at 20 ºC, 4 29.85 10 Ns m , 998.2kg mµ ρ−= × = 3 . For commercial steel pipe,

0.045mm, 0.045 200 0.000225Dε ε= = = .


3

2 2

4 0.10m s4 m=3.18sπ 0.20m

V VA Dπ

= = =V

( )( )( )3

-4 2

998.2kg m 3.18m s 0.20m645,000

9.85×10 Ns mDRe ρµ

= = =V

This is turbulent flow so

1.11 1.111 6.9 0.000225 6.91.8log 1.8log 0.0152

3.7 3.7 645,000D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


( ) ( ) ( )2

2, 3 2

3.18m skg m 5000 kN998.2 9.81 150m + 0.0152 =3390kPa (gage)0.2 2 1000Nm sgP

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

Answer

b) Applying the same energy equation between points 1 and 2. Assume pressure 1 is atmospheric, and the pipe diameters at 1 and 2 are the same. Therefore:

2, 1, 2,g g gP

P P Ph

g gρ ρ−

= =

The pumping power is

( )( )( )2, 3 22, 0.10m s 3390kN m 1kWs 1kNm =339kWg

P g

PW mgh Vg VP

gρ

ρ⎛ ⎞

= = = =⎜ ⎟⎝ ⎠

Answer

9- 57

9-54 Two reservoirs are connected by three galvanized iron pipes in series. The first pipe is 600-m long, 20-cm diameter; the second pipe is 800-m long, 30-cm diameter; and the third pipe is 1200-m long, 40-cm in diameter. For a flow of 0.15 m3/s of water at 10 °C, determine the elevation difference between the reservoirs (in m).

Approach:

The elevation difference drives the flow. That pressure head is consumed by the frictional losses in the line. The steady, incompressible flow energy equation is used to calculate the elevation difference.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

At points 1 and 2 the areas are large and the pressure is atmospheric, so 1 2 10 and P P2= =V V = . There is no pump or turbine, so , and minor losses are neglected. Therefore,

0P Th h= =22 2

1 2 2 2C CA A B B

A B CA B C

L VL V L Vz z f f f

D g D g D g− = + +

2

The velocities are: ( )( )

3

2 2

4 0.15m s4 m=4.77sπ 0.20m

AA A

V VA Dπ

= = =V

Using conservation of mass: ( ) ( ) ( )( ) ( )( )

2 2

2 2

4.77 m s 0.20 0.30 =2.12 m s

4.77 m s 0.20 0.40 =1.19m sB A A B

C A A C

V V D D

V V D D

= =

= =

The friction factor is a function of Reynolds number and roughness. For galvanized iron pipe =0.15mmε , and for water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 . The Reynolds numbers are:

( )( )( )

( )( )

3

4 2

999.6 kg m 4.77 m s 0.20m739,000

12.9 10 Ns m493,000

369,000

A

B A A B

C A A C

DRe

Re Re D D

Re Re D D

ρµ −= = =

×

= =

= =

V

T the flow is turbulent, so

1.111 0.15 200 6.91.8log 0.0188

3.7 739,000 AA

ff

⎡ ⎤⎛ ⎞= − + → =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

Similarly, and . 0.0176Bf = 0.0170Cf =Therefore,

( ) ( )( ) ( ) ( )

( ) ( ) ( )( )

2 2

1 2 2

4.77 m s 2.12m s 1.19m s600 800 12000.0188 + 0.0176 + 0.01760.20 0.30 0.402 9.81m s 2 9.81m s 2 9.81m s

z z ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2

2

=65.4m+10.7m+3.8m=79.9m Answer Comments:

Note that pipe A losses dominate the total pressure loss. For a given elevation difference, if more flow were desired through this system, increasing the diameter of pipe A would be the first option to try.

9- 58

9-55 In a water system, a reservoir is connected to a canal with an 8-in. cast iron pipe. The system has three regular 90° threaded elbows, a half closed gate valve, and the exit from the reservoir is sharp edged. With an elevation difference of 55 ft between the reservoir surface and the pipe outlet, 3 ft3/s of water at 50 °F flows through the pipe. Determine the total length of straight pipe in the system (in ft).

Approach:

The length can be determined by applying the steady, incompressible flow energy equation between points 1 and 2. The frictional losses must be calculated.

Assumptions:


Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . We assume the surface of the reservoir and canal are large, so , and the pressure at 1 and 2 are atmospheric, so P

0T ph h= =

1 2 0≈ ≈V V 1 = P2. Therefore, taking into account all the minor losses:

( )2

2 10 32valve exit bend

Lz z f K K KD g

⎛ ⎞= − + + + +⎜ ⎟⎝ ⎠

V

Solving for the length:

( ) ( )2 1

2

23valve exit bend

g z zDL K Kf

−⎡ ⎤= − − + +⎢ ⎥

⎣ ⎦VK

From Figure 9-15b, Kexit = 1.0. From Table 9-3, Kbend = 1.5 and Kvalve = 2.1. For cast iron pipe =0.00085ftε , and for water from Appendix B-6 at 50 ºF, 5 388 10 lbm ft s , 62.4 lbm ftµ ρ−= × = . The velocity is determined from conservation of mass for incompressible flow:

( )( )

3

2 2

4 3ft s4 f=8.59s0.667ft

V VA Dπ π

= = =V t


( )( )( )3

5

62.4 lbm ft 8.59ft s 0.667ft406,300


×V


1.11 1.111 6.9 0.00085 0.667 6.91.8log 1.8log 0.0214

3.7 3.7 406,300D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( )( )

( )[ ]( )

2

2

2 32.2ft s -55ft0.667ft - - 2.1+1+3 1.5 =1260ft0.0214 8.59ft s

L⎡ ⎤⎢ ⎥=⎢ ⎥⎣ ⎦

Answer

9- 59

9-56 On your land high in the Rocky Mountains, you decide to produce your own electric power for your vacation home using a hydroturbine. The surface of the small lake from where you will get the 50 °F water is 500 ft above where you will locate the turbine. You connect the lake and turbine with 1000 ft of 6-in. cast iron pipe. The turbine discharge is the same diameter as the inlet and is open to the atmosphere. Determine the maximum power that can be produced (in W).

Approach:

The elevation difference drives the flow. Pressure head is consumed by the frictional losses in the line. Turbine power can be calculated with the steady, incompressible flow energy equation.

Assumptions:

1. The system is steady with constant properties. 2. The flow is fully developed. 3. Neglect minor losses.

Solution:


1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

At points 1 and 2 the areas are large and the pressure is atmospheric, so 1 2 10 and P P2= =V V = . There is no pump, ( ), and the losses are line and exit losses. Therefore, 0Ph =

2 2

1 2 2 2T exitV VLh z z f K

D g g⎛ ⎞

= − − +⎜⎝ ⎠

⎟

T

(1)

where and . The friction factor is a function of Reynolds number and roughness, so to find the maximum possible power, an iterative solution is required.. For cast iron pipe

1exitK = TW mgh==0.00085 ftε , and for water

from Appendix B-6 at 50 ºF, 5 388 10 lbm ft s , 62.4 lbm ftµ ρ−= × = .


3

2 2

4 ft s4 f=5.09sπ 0.5ft

VV V VA Dπ

= = =V t

( )( )( )3

5

62.4 lbm ft 5.09V ft s 0.5ft180,600

88 10 lbm ft sDRe Vρµ −= = =

×V (2)

For any reasonable flowrate, the flow is turbulent, so 1.111 61.8log

3.7D

Refε .9⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦ (3)

Therefore, ( )( )

2

2

5.09 ft s1000500ft +10.5 2 32.2ft sT

Vh f⎡ ⎤⎛ ⎞= − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

(4)

( ) ( )2

223 2

lbm ft ft lbfs 1hps W62.4 5.09 0.5 32.2 ft 7464 s 4 32.2ft lbm 550ft lbf hpft sT T TW D gh V ft hπ πρ

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

V

84.6 WT TW h V= (5) The procedure to follow is to use the above five equations to calculate the maximum power: guess a volume flow rate, calculate velocity, Reynolds number, friction factor, turbine head, and turbine power. Increase the volume flow and recalculate until a maximum power is obtained. Proceeding with the calculations: ( )3ft sV ( )ft sV ( )WTW 2.6 13.24 76,590 2.7 13.75 76,830 Answer 2.8 14.26 76,760 At maximum power, f = 0.0274 and Re = 487,500.

Comments: For zero power output from the turbine, the volume flow rate would be 4.73 ft3/s and V = 24.1 ft/s.

9- 60

9-57 Fire trucks have pumps to boost the pressure of the water supplied by a fire hydrant. Consider a fire truck that has a 250-ft long, 2-in. diameter smooth fire hose. Water must reach the nozzle at the hose exit at 100 lbf/in.2 (gage). Water from the hydrant reaches the pump inlet at 60 lbf/in.2 (gage). If the design pressure drop specification for the hose is 25 psi/100 ft of length, determine:

a. the design flow rate (in gal/min) b. the nozzle exit velocity (in ft/s) c. the pump power required if the pump has a mechanical efficiency of 75% (in hp).

Approach:

Flow rate, nozzle exit velocity, and pump power are calculated with the steady, incompressible flow energy equation. However, the equation must be applied across three different segments of the system.

Assumptions:


Solution:

a) To find the design flow rate, apply the steady, incompressible flow energy equation between points 2 and 3:

22

3 32 22 32 2P T L

PPz h z h h

g g g gρ ρ+ + + = + + + +∑VV

The duct is constant area and horizontal, so 2 3=V V and 2z z3= There is no pump or turbine, so , and there are no minor losses:

0P Th h= =

( ) 0.52

2 32 3 22

22

P P DP P Lfg D g fLρ ρ

−⎡ ⎤−= → = ⎢ ⎥

⎣ ⎦

V V

From the design pressure drop specification:

( )

( )2

22 3

25lbf in.250ft =62.5lbf in.

100ftP P

⎡ ⎤⎢ ⎥− =⎢ ⎥⎣ ⎦

The friction factor is a function of Reynolds number and roughness, so an iterative solution is required. For water from Appendix B-6 at 70 ºF, 5 365.8 10 lbm fts , 62.2 lbm ftµ ρ−= × = .


3

2

4 ft s ft=45.84s2 12ft

VV VA π

= =V

( )( )( )3

5

62.2 lbm ft 45.84 ft s 2 12ft722,200

65.8 10 lbm fts

VDRe Vρµ −= = =

×V

For any reasonable flow rate, this is turbulent flow, so assuming a smooth duct:

1.111 61.8log

3.7D

Refε⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

.9

( ) ( )( )( )

( ) ( )( )

0.50.5 2 22 3

2 2 03

2 62.5lbf in. 2 12ft 32.2ft lbm lbf s2 2.492 ft=s62.2 lbm ft 250ft 1ft 12in.

P P DfL ffρ

⎡ ⎤−⎡ ⎤⎢ ⎥= =⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

V .5

3ft45.84

sV = V

Now perform the iteration: ( )3ft sV V (ft/s) Re f V (ft/s) ( )3ft sV

0.5 22.9 361,000 0.0139 21.2 0.462 0.462 21.2 333,500 0.0141 21.0 0.458

This is satisfactory convergence, so the volume flow rate is:

9- 61

( )( )( )2 30.458ft s 60s min 1gal 0.1337ft =206gal minV = Answer b) To find the nozzle exit velocity, we apply the energy equation between points 3 and 4. Assuming no losses:

2 2

3 3 4 4

2 2P Pg g gρ ρ+ = +

V Vg

( ) ( )( )( )( )

0.50.5 2 2 23 4 2

4 3 23

2 100lbf in. 32.2ft lbm lbf s2 ft ft21.0 124s s62.2 lbm ft 1ft 12in.

P Pρ

⎡ ⎤−⎡ ⎤ ⎛ ⎞⎢ ⎥= + = + =⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ ⎣ ⎦V V Answer

c) To find the pumping power, applying the energy equation from point 1 to2:

2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The duct is constant area and horizontal, so 1 2=V V and 1z z2= There is no turbine ( ) and no minor losses, and incorporating other losses in terms of pump efficiency:

0Th =

1 2p

P Ph

gρ−

= and ( )1 21 2P p

P PW mgh Vg V P P

gρ

ρ−

= = = −

This is the power into the fluid. So taking into account pump efficiency:

( ) ( ) ( ) ( )( )3 2

1 20.458ft s 60- 100+62.5 lbf in. 1hps 550ft lbf 144in. ft

0.75PP

V P PW

η

⎡ ⎤− ⎣ ⎦= =2 2

= -16.4 hp Answer

9- 62

9-58 Water is pumped from a lake to a pond that is 50 m above the lake. A suction pipe runs from the lake to a pump, and a connecting pipe runs from the pump to the pond. The suction pipe is constructed of 10-cm diameter cast-iron pipe (assume no minor losses). The connecting pipe also is 10-cm diameter cast iron and has five long radius 90° threaded elbows. The pump can be located in one of three places: 1) level with the lake surface, and the suction pipe would be 6-m long and the connecting pipe would total 150-m long, 2) 10-m below the lake surface, and the suction pipe would be 11-m long and the connecting pipe would total 160-m long, and 3) 5-m above the lake surface, and the suction pipe would be 8-m long and the connecting pipe would total 145-m long. For a flow of 0.025 m3/s, determine which installation requires the smallest required pumping power (in W).

Approach:

The pumping power is calculated with the steady, incompressible flow energy equation applied between points 1 and 2. The preferred pump location can be determined by comparing losses for each system.

Assumptions:


Solution:

The steady, incompressible flow energy equation between points 1 and 2 is:

2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The lake and pond surface areas are large, so 1 2 0≈ ≈V V and 1P P2= There is no turbine, so . Therefore:

0Th =

2 1P Lh z z h= − +∑ All piping is the same diameter. Minor losses in all pump locations include an entrance loss, an exit loss, and 5 elbow losses. The only difference in these locations is the length of straight pipe. The location level at the lake and 5 m above the lake have total pipe lengths of 156 m; the location 10 m below the surface has a length of 163 m. So based solely on the pumping power, the preferred location is either level or above the lake. Pumping power is:

2

2 1 52P p ent exit bend

LW mgh mg z z f K K KD g

⎡ ⎤⎛ ⎞= = − + + + +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

V

The friction factor is a function of Reynolds number and roughness. For water from Appendix B-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 .


3

2 2

4 0.025m s4 m=3.18s0.10m

V VA Dπ π

= = =V

( )( )( )3

4 2

999.6kg m 3.18m s 0.10m246, 400

12.9 10 Ns mDRe ρµ −= = =

×V

This is turbulent flow so for cast iron pipe, 0.26mm, 0.26 100 0.0026Dε ε= = = :

1.11 1.111 6.9 0.0026 6.91.8log 1.8log 0.0257

3.7 3.7 246,400D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

From Table 9-3, Kbend = 0.7, and from Figure 9-15, Kent = 0.5 and Kexit = 1.0 The mass flow rate is: ( )( )( )( )23999.6kg m 3.18m s π 4 0.10m =25.0kg sm Aρ= =V The power is:

[ ] [ ] ( )( )

2 2

2 2

3.18m skg m 156 Ns W s25.0 9.81 50m+ 0.0257 +0.5+1.0+5 0.7s 0.10 kg m N ms 2 9.81m sPW

⎡ ⎤ ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎢ ⎥= ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎢ ⎝ ⎠ ⎥ ⎝ ⎠⎝ ⎠⎣ ⎦

= 17,960 W Answer

9- 63

9-59 Many universities have a central facility that produces chilled water for use in cooling all the buildings on campus. The water at 10 °C is continuously circulated through a closed flow loop and used as needed. Consider a system that consists of 5 km of 30-cm commercial steel pipe with a flow rate of 0.15 m3/s. The pump has a mechanical efficiency of 75% and is driven by a motor that has an efficiency of 92%. Determine:

a. the pressure drop (in kPa) b. the pumping power required (in kW) c. the annual cost if electricity costs $0.10/kWhr and the system runs 7,500 hr/yr.

Approach:

Pressure drop and pumping power both can be calculated with the steady, incompressible flow energy equation. The equation must be applied across different sections of the system.

Assumptions:


Solution:

a) The steady, incompressible flow energy equation between points 2 and 1 is:

2 2

2 2 1 12 12 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The pipe diameter is constant, so and horizontal 1 =V V2 21z z= There is no pump or turbine, so . Ignoring minor losses,

0P Th h= =

2 2

2 1 2 2LL LP P g h g f fD g D

ρ ρ ρ⎛ ⎞

− = = =⎜ ⎟⎝ ⎠

∑ V V

The friction factor is a function of Reynolds number and roughness. For water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 .

Velocity is: ( )( )

3

2 2

4 0.15m s4 m=2.12s0.30m

V VA Dπ π

= = =V

( )( )( )3

4 2

999.6kg m 2.12m s 0.30m493,300

12.9 10 Ns mDRe ρµ −= = =

×V

This is turbulent flow so for commercial steel pipe, 0.045mm, 0.045 300 0.00015Dε ε= = = :

1.11 1.111 6.9 0.00015 6.91.8log 1.8log 0.0148

3.7 3.7 493,300D f

Refε⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − + = − + → =⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

( ) ( )2

2 1 3

2.12m s5000 kg 1kN0.0148 999.6 =555kPa0.30 2 1000Nm

P P ⎛ ⎞⎛ ⎞ ⎛ ⎞− = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

Answer

b) To find the pumping power, applying the energy equation from point 1 to2:

1 2p

P Ph

gρ−

= and ( )1 21 2P p

P PW mgh Vg V P P

gρ

ρ−

= = = −

( )32

kN 1kW s0.15m s -555 =-83.2kW1kN mmPW⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ Answer

The minus sign indicates power is input to the pump. c) For the cost of running the pump:

( )( )( )

( )( )$0.10 kWhr 83.2kW 7500hr yr

=$90,500 yr0.75 0.92P m

CWtCostη η

= = Answer

9- 64

9-60 The Alaskan oil pipeline is 48-in. in diameter, with a wall roughness of approximately 0.0005 ft. The design flow rate is 1.6 × 106 barrels per day (1 barrel = 42 gal). To limit the required pipe wall thickness, the maximum allowable oil pressure is 1200 psig. To keep dissolved gases in solution in the crude oil, the minimum oil pressure is 50 psig. The oil has ρ = 58 lbm/ft3 and µ = 0.0113 lbm/ft·s. Determine:

a. the maximum spacing between pumping stations (in km) b. the pumping power at each station if the pump mechanical efficiency is 85% (in kW).

Approach:

The maximum length and pumping power are calculated with the steady, incompressible flow energy equation. However, the equation must be applied across different segments of the system.

Assumptions:


Solution:

a) To find the maximum length, apply the steady, incompressible flow energy equation between points 2 and 1:

2 2

2 2 1 12 12 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

The pipe is constant area and horizontal, so 2 1=V V and 2z z1= There is no pump or turbine, so ,

and there are no minor losses:

0P Th h= =2

2 3 2

2P P Lf

g D gρ−

=V

Solving for length ( )2 3

2

2 P P DL

fρ−

=V

Velocity is :

( )( )( )( )( )

( )

6 3

2 2

4 1.6×10 barrel day 42gal barrel 0.1337ft gal 1day 24hr 1hr 3600s4 ft=8.28sπ 4ft

V VA Dπ

= = =V

The friction factor is a function of Reynolds number and roughness. With the given properties

( )( )( )358lbm ft 8.28ft s 4ft

170,0000.0113lbm fts

DRe ρµ

= = =V

This is turbulent flow so with 0.0005ft, 0.0005 4 0.000125Dε ε= = =

1.11 1.111 6.9 0.000125 6.91.8log 1.8log

3.7 3.7 170,000D

Refε⎡ ⎤ ⎡ ⎤

⎥⎥⎦

⎛ ⎞ ⎛ ⎞= − + = − +⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢⎣ ⎦ ⎣

0.0168f→ =

( ) ( )( )

( )( )( ) ( )

2 2

2 23

2 1200-50 lbf in. 4 ft 32.2ft lbm lbf s=638,600ft =121mi

58lbm ft 0.0168 8.28ft s 1ft 12in.L = Answer

b) To find the pumping power, applying the energy equation from point 1 to 2:

1 2p

P Ph

gρ−

= and ( )1 21 2P p

P PW mgh Vg V P P

gρ

ρ−

= = = −

This is the power into the fluid. So taking into account pump efficiency: ( )( )( )( ) ( )( )22 28.28ft s -1150lbf in. 4 4ft 1hps 550ft lbf 144in. ft =-31,300hpPW π= 2 This does not take into account pump efficiency, so

-31,300hp -36,800hp0.85PW = = Answer

Comments: The minus sign indicates that power must be supplied to the pump.

9- 65

9-61 For air at 300 K and one atmosphere, a fan performance curve can be approximated with , where h is the pressure rise across the fan in cm of water and V is the air flow rate in

m

4 270 3 10h −= − × V3/min. The fan discharges into a smooth rectangular duct 20-cm by 40-cm. Determine:

a. the flow rate if the duct is 30-m long (in m3/min) b. the flow rate if the duct is 75-m long (in m3/min).

Approach:

We must balance the pressure loss in the duct against the pressure head produced by the fan. Pressure drop is calculated directly with the steady, incompressible flow energy equation. Because the friction factor depends on flow rate, an iterative solution is required.

Assumptions:

1. The system is steady. 2. The flow is fully developed. 3. Properties are constant. 4. The duct is smooth.

Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

Let location 1 be just upstream from the fan and station 2 at the end of the duct, so 1P P2= . The duct is constant area and horizontal, so and There is no turbine, so 1 =V V2 21z z= 0Th = , and there are no minor losses:

2

2PLh fD g

=V

(1)

where hP is produced by the fan (expressed in consistent units). The friction factor is a function of Reynolds number and roughness. For air from Appendix A-7at 100 kPa, 300 K, 5 21.846 10 Ns m , 1.1774kg mµ ρ−= × = 3 .

Velocity is : ( ) ( )

( )( )

3m min 1min 60s m=0.2080.20m 0.40m s

VV VA

= =V (2)

Because this is a non-circular duct, the hydraulic diameter must be used:

( )( )( )

4 0.20m 0.40m4=0.2667m

2 0.20m+0.40mx

hwetted

AD

p= =

( )( )( )3

5 2

1.1774kg m 0.208 m s 0.267m3540

1.846 10 Ns mh

VDRe V

ρµ −= = =

×V

(3)

For any volume flow greater than 1 m3/min, this is turbulent flow so with a smooth duct:

1.11 1.111 6.9 0 6.91.8log 1.8log

3.7 3.7D

Re Refε⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞= − + = − +⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢⎣ ⎦ ⎣

⎤⎥⎥⎦

(4)

The solution procedure is to solve the above four equations simultaneously using an iterative procedure: guess a volume flow rate V ; calculate velocity V, Reynolds number Re, and friction factor f. Then calculate the pressure drop across the duct and the pressure rise across the fan. When these two quantities are equal, then the flow has been determined. Performing the iterations, we obtain For L = 30 m

3285m min 59.3m s 0.0116 387m of air = 45.5cm of waterfanV f h= = = =V Answer For L = 75 m

3198m min 41.1m s 0.0123 495m of air = 58.3cm of waterfanV f h= = = =V Answer

9- 66

9-62 A town water system is constructed to supply water at a flow rate of 0.04 m3/s as shown below. Available cast iron pipe is to be used and the gate valve is fully open. The water is at 20 °C. Determine the height to which the upper reservoir dam (reservoir surface elevation) must be built (in m).

Approach:

The elevation difference drives the flow. That pressure head is consumed by the frictional losses in the line. The steady, incompressible flow energy equation is used to calculate the elevation of the upper reservoir.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

At points 1 and 2 the areas are large and the pressure is atmospheric, so 1 2 10 and P P2= =V V = . There is no pump or turbine, so . Therefore, 0P Th h= =

2 2

1 2 2 2A A B B

ent valve contract A exit BA B

L Lz z K K K f K f

D g D g⎡ ⎤ ⎡

= + + + + + +⎢ ⎥ ⎢⎣ ⎦ ⎣

V V⎤⎥⎦

The velocities are: ( )( )

3

2 2

4 0.04m s4 m=1.27sπ 0.20m

AA A

V VA Dπ

= = =V

Using conservation of mass: ( ) ( )( )2 21.27 m s 0.20 0.15 =2.26m sB A A BD D= =V V The friction factor is a function of Reynolds number and roughness. For cast iron pipe =0.26mmε , and for water from Appendix A-6 at 10 ºC, 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 . The Reynolds numbers are:

( )( )( )

( )

3

4 2

999.6kg m 1.27 m s 0.20m197,000

12.9 10 Ns m263,000

A

B A A B

DRe

Re Re D D

ρµ −= = =

×

= =

V


1.111 0.26 200 6.91.8log 0.0221

3.7 197,000 AA

ff

⎡ ⎤⎛ ⎞= − + → =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

Similarly, 0.0233Bf = From Table 9-3, Figures 9-13, 9-14, and 9-15, Kent = 0.5, Kvalve = 0.15 (assumed fully open), Kexit = 1.0 For ( ) ( )2 20.15 0.20 0.56B A B AA A D D= = = , and Kcontract ≈ 0.15. Therefore,

( ) ( )( ) ( ) ( )

( )2 2

1 2 2

1.27 m s 2.26m s200 150175m 0.5 0.15 0.15 0.0221 + 1 0.02330.20 0.152 9.81m s 2 9.81m s

z ⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞= + + + + +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣

⎤⎥⎦

=175m+1.88m+6.32m=183.2m Answer

9- 67

9-63 The pump in an existing water system (shown below) fails and must be replaced. A duplicate is not available, so a manufacturer proposes a pump with a pump curve: 6 24 10 0.0038 86Ph V V−= − × + + , where

is in gal/min and hV p is in ft. The gate valve is fully open and a friction factor of 0.018 can be used for both pipes. Determine the flow rate in the system (in gal/min).

Approach:

The pump must supply enough head to overcome frictional losses in the suction and discharge pipes, as well as the elevation head. The steady, incompressible flow energy equation is used to calculate the required pump head. The given pump head curve is equated to the head loss calculation to determine the operating point.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + + +∑V V

There is no turbine, so . The areas at 1 and 2 are large, so 0Th = 1 2 0= =V V , and pressure is atmospheric, so . The losses include entrance, exit, valve, and line losses. Therefore, 1P P= 2

( )2 2

2 1 2 2A A B B

P ent A valve exit BA B

L Lh z z K f K K f

D g D g⎛ ⎞ ⎛

= − + + + + +⎜ ⎟ ⎜⎝ ⎠ ⎝

V V⎞⎟⎠

From conservation of mass: ( )2A B B A A Bm m D D= → =V V

Substituting this into the above pump head equation:

( )4 2

2 1 2A A B

P ent A valve exit BA B B

L D Lh z z K f K K f

D D D g

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − + + + + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

VA


( )

3

2 2

4 gal min 0.1337 ft gal 1min 60s4 f=0.00409sπ 10 12ft

AA A

VV V VA Dπ

= = =V t (1)

The head loss must be balance against the head added by the pump: (2) 6 24 10 0.0038 86Ph V V−= − × + +

where hp is in ft and V is in gal/min. From Table 9-3 and Figures 9-13 and 9-14, 0.5, 1.0, and 0.15ent exit valveK K K= = = , assuming the valve is fully open. Substituting the known information into the head loss equation:

( ) ( ) ( ) ( )4 2

A2

40 0.833 8001204-1152 ft+ 0.5+ 0.018 + 0.15+1+ 0.0180.833 0.667 0.667 2 32.2ft sPh

⎡ ⎤⎛ ⎞ ⎛ ⎞= ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

V

Simplifying: (3) 2A52.0ft+0.882Ph = V

The above three equations are solved simultaneously to find the operating point of the pump. Solving them, we obtain: 1450gal minV = 5.93ft sA =V Answers 83.1ftPh =

9- 68

9-64 In drier regions, large central pivot sprinkler systems are used to irrigate large areas. Consider the simplified schematic of a portion of such a sprinkler (shown below). Water at 10 °C is pumped through the spray arm that is constructed of 2.5-cm diameter galvanized iron. The flow area of each nozzle is 1.5 cm2. The pressure at the first nozzle is 250 kPa (gage). Ignoring friction in each nozzle but not in the connecting lengths of pipe, determine the flow rate through the sprinkler (in m3/s).

Approach:

The total flow is such that all the pressure head at location 1 is used to balance line losses in the galvanized pipe and the creation of kinetic energy at the nozzle exits. The flow through each nozzle is different and can be estimated using the Bernoulli equation. Because the line losses are a function of flow rate and friction factor, an iterative solution is required. The steady, incompressible flow energy equation must be used.

Assumptions:


Solution:

The frictional losses between each nozzle can be calculated with the steady, incompressible flow energy equation.

For flow between location 1 and 2: 2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

There is no pump or turbine, so . The pipe diameter and elevation are constant, so and . Ignoring minor losses:

0p Th h= = 1 =V V2

21z z=

2

122 1 12 2

LP P fDρ= −

V

Note that velocity and friction factor are different for each segment of pipe. The velocity is: 2

12 12 124V V A V Dπ= = The flow between points 1 and 2 is the difference between what enters the system and what is leaving through the nozzle: 12 1 1jV V V= − Assuming no losses in the nozzle, we use the Bernoulli equation to calculate the flow through the nozzle:

2211 1

1 12 2j j

j

PPz z

g g g gρ ρ+ + = + +

VV

Neglecting the inlet velocity and the elevation difference, 1 0≈V and 1 jz z= , and expressing the pressures in gage pressure, , we obtain: 0jP =

( )0.51 12jV P ρ=

and the volume flow rate through a jet is 1 1j j jV A= V .

Assuming turbulent flow in each pipe segment, with 4 212.9 10 Ns m , 999.6kg mµ ρ−= × = 3 from Appendix A-6 and from Table 9-2 for galvanized pipe, 0.15mmε = so that 0.15 25 0.006Dε = = . The flow is turbulent, so

1.111 61.8log

3.7D

Refε⎡ ⎤⎛ ⎞= − +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

.9 where Re Dρµ

=V

To solve for the total flow, we generalize the above equations for each pipe segment and nozzle: 1 1 2 3 4j j jV V V V V= + + + j (1)

2

122 1 12 2

LP P fDρ= −

V (2)

9- 69

2

233 2 23 2

LP P fDρ= −

V (3)

2

344 3 34 2

LP P fDρ= −

V (4)

12 1 1jV V V= − (5)

23 12 2jV V V= − (6)

34 23 3jV V V= − (7)

( )0.51 12jV P ρ= (8)

( )0.52 22jV P ρ= (9)

( )0.53 32jV P ρ= (10)

( )0.54 42jV P ρ= (11)

The eleven unknowns in this system of equations are: . Solving the system of equations with appropriate software:

1 2 3 4 2 3 4 12 23 34 1, , , , , , , , , ,j j j jV V V V P P P V V V V

31

31 1

32 2

33 3

34 4

312

323

334

0.00699m s

0.00336 m s 250kPa (gage)

0.00179m s 70.9kPa (gage)

0.00104m s 23.9kPa (gage)

0.00081m s 14.6kPa (gage)

0.00363m s

0.00185m s

0.00081m s

j

j

j

j

V

V P

V P

V P

V P

V

V

V

=

= =

= =

= =

= =

=

=

=

Answers

Comments:

Note that with this design different flows exit at each nozzle. This would not be a good design, since a farmer would want uniform coverage on a field. Hence, different size nozzles or some other modification would be used to ensure equal flow through all nozzles.

9- 70

9-65 Two pipes are connected in parallel. The first pipe is 2-cm in diameter and 100-m long with a friction factor of 0.012. The second pipe is 5-cm in diameter 50-m long with a friction factor of 0.010. Determine the ratio of the flow rates in the two pipes.

Approach:

We assume the same pressure drop across these two pipes and ignore minor losses. A ratio of the steady, incompressible flow energy equations written for both pipes will give the ratio of the velocities, which can be used to obtain a ratio of the flow rates.

Assumptions:


Solution:


2 2

1 1 2 21 22 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + + +∑V V

There is no pump or turbine, so . The pipe is constant area, so 0P Th h= = 1 2=V V , and the pipe is horizontal, so . Therefore, 1z z= 2

2

1 2

2P P Lf

g D gρ−

=V

Applying this equation to pipe A and pipe B, and taking the ratio

( )( )

2

0.51 2

21 2

21

2

A AA

A A A B B

B A A BB BBB

B

LfP P g D g f L Df L DLP P g f

D g

ρ

ρ

−⎡ ⎤ ⎛ ⎞⎣ ⎦ = = → = ⎜ ⎟−⎡ ⎤ ⎝ ⎠⎣ ⎦

VVVV

A

Substituting in known information:

( )( )( )( )( )( )

0.50.010 50 0.02

0.4080.012 100 0.05

A

B

⎡ ⎤= =⎢ ⎥⎢ ⎥⎣ ⎦

VV

Because 2 4V A Dπ= =V V , from conservation of mass:

( )2 20.020.408 0.065

0.05A A A

B BB

V DDV

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

VV

Answer

Comments:

This answer is reasonable using the given friction factors. In actual practice, with a known total flow, the friction factors would need to be evaluated, and the solution would be iterative.

9- 71

9-66 For a storm sewer modification project, a 24-in. pipe and a 30-in. pipe both open at their ends to the atmosphere are to be joined using three existing (but under-utilized pipes) as shown on the figure. All the pipes are concrete, and the friction factors are shown on the figure. The branches are horizontal For a total flow rate of 20 ft3/s of 60 °F water, determine the flow rate in each of the three connecting pipes (in ft3/s) and the elevation difference from the entrance to the exit.

Approach:

We know the total flow through the system and that the head loss across the middle three branches (B, C, and D) is the same. There are four unknowns in the problem: the head loss from 2 to 3, and the volume flows through the three branches. We use conservation of mass and the steady, incompressible flow energy equations to obtain the required four equations.

Assumptions:


Solution:

Conservation of mass with constant density water for the whole system is: where tot B C Dm m m m= + + m Vρ= . Therefore, (1) tot B C DV V V V= + + From point 2 to point 3, the steady, incompressible flow energy equation is:

22

3 32 22 32 2P T L

PPz h z h h

g g g gρ ρ+ + + = + + + + +∑VV

There is no pump or turbine, so , and we ignore minor losses. Each of the pipes is constant area, so the velocity is constant. The pipe is horizontal, so

0P Th h= =

2z z3= . Therefore,

2

2 3

2P P Lf

g D gρ−

=V

Let ( )2 3brh P P gρ= − , which is the same for all three branches. Applying the pressure drop equation to each branch:

2

2B B

br BB

Lh f

D g=

V (2)

2

2C C

br CC

Lh f

D g=

V (3)

2

2D D

br DD

Lh f

D g=

V (4)

For each of the velocities: 2

4V VA Dπ

= =V

We have four equations and four unknowns: , , ,B C D bV V V h r . With the given information, the equations can be solved simultaneously (iteratively) for the unknowns. Doing so, we obtain: 3 3 39.23ft s 4.20ft s 6.57 ft s 26.7ftB C D brV V V h= = = = Answers Now we need to evaluate the elevation difference between points 1and 4. Using the energy equation

2 2

1 1 4 41 42 2P T L

P Pz h z h h

g g g gρ ρ+ + + = + + + +∑V V

With no pump or turbine, so , no minor losses, and 0P Th h= = 1P P4= , and noting that there are line losses in pipe 1-2 and 3-4 and we know the line loss between 2-3, since we calculated that above:

9- 72

2 22 2 2

34 34 34 34 3412 12 12 12 121 4 12 34 12 34

12 34 12 34

1 12 2 2 2 2 2br br

L LL Lz z h f f f f h

g g D g D g D g D g⎛ ⎞⎛ ⎞

− = − + + + + = − + + + +⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

V VV V V 2V

For the velocities

( )( )

( )( )

3 3

12 342 2 212

4 20ft s 4 20ft s4 ft=2.83 =4.07s sπ 3ft π 2.5ft

V VA Dπ

= = = =V V ft

( ) ( )( ) ( ) ( )

( )2 2

1 4 2 2

2.83ft s 4.07 ft s2300 4000-1+ 0.022 + 1+ 0.020 +26.7ft3 2.52 32.2ft s 2 32.2ft s

z z ⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞− = ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

1.97ft+8.48ft+26.7ft=37.2ft= Answer

9- 73

ch09

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