5 DPE Plate HE Design (Syrup Water)

Thermal design of plate HE syrup – cooling water (counter flow)

Calculation procedure

Material for courses DPE and PRO III (remember part 3).

Prepared by: Pavel Hoffman

1. Given data

Cooled solution (syrup):

Amount of incoming solution MS (kg/s)

Temperature of incoming solution tS0 (°C)

Required (or calculated) temperature of outgoing syrup tS1 (°C)

Incoming solution concentration xS0 (%)

Cooling liquid (cooling water):

Amount of incoming water MW (kg/s)

Temperature of incoming water tW0 (°C)

Given (or calculated) temperature of outgoing water tW1 (°C)

Next parameters needed for calculations:

Gap between plates s (mm)

Thickness of plates sp (mm)

Length of interplate channel L (mm)

With of interplate channel b (mm)

Equivalent diameter of channel de (m)

Flow clear area of channel fp (m2)

Heat transfer area of plate AP (m2)

Thermal conductivity of plates p (W/mK)

document.doc 1 / 59 Print date: 8 / 2007 P. Hoffman

Fouling thickness (incrustation) on side of cooling water sWi (mm)

Thermal conductivity of fouling on side of cooling water Wi (W/mK)

Fouling thickness (incrustation) on side of cooled syrup sSi (mm)

Thermal conductivity of fouling on side of cooled syrup Si (W/mK)

Max. allowable pressure loss on side of cooling water pWmax (kPa)

Max. allowable pressure loss on side of cooled syrup pSmax (kPa)

Max. allowable pressure in plates pmax (kPa)

Minimal economical temperature difference in HE tHEmin (°C)

2. Specification of physical parameters needed for calculations

Density of cooling water W (kg/m3)

Dynamic viscosity of cooling water W (Pa*s)

Kinematic viscosity of cooling water W (m2/s)

Specific heat of cooling water cW (J/kgK)

Thermal conductivity of cooling water W (W/mK)

Prandtl number of cooling water PrW (-)

Density of cooled syrup S (kg/m3)

Dynamic viscosity of cooled syrup S (Pa*s)

Kinematic viscosity of cooled syrup S (m2/s)

Specific heat of cooled syrup cS (J/kgK)

Thermal conductivity of cooled syrup S (W/mK)

Prandtl number of cooled syrup PrS (-)

(all parameters are found from tables etc. for average temperatures)


Fig. 1.: Example of chart of plate HE with 2 sections

Fig.2.: Temperature profile on heat transfer surface in cooler

Fig.3.: Temperature profile in 1 section of cooler


tHEmin

MS - syrup

MW - water

tS1 = ?

tS0

tW1

tW0

syrup syrup

cooling water

1.section 2.section

cooling water

cooling water

sSi; Si

sw; w

sWi; Wi

twW

q

twS

tW

tS

tW

plate fouling

twfW

W

heating of cooling waterS

cooling of syrup

cooling water(boundary layer)

fouling

twfS

tS

cooled syrup(boundary layer)

Fig. 4. Examples of plate HE design


a) Inter plate channels arrangement– flow pattern

b) Plate HE with connecting plate (with necks) between two sections


c) Plate HE (cooler) with press plate closed and open

d) Two types of plates with various profiles

Lb

3. Basic thermal balance of cooler

Heat transferred from syrup to cooling water (it is heated)

QHW = MW * cW * (tW1 – tW0) = k * ATreal * tL

(W; kg/s, J/kg°C, °C; W/m2K, m2, K)

Heat that is taken from cooled syrup – without heat loss it is (principle of energy

conservation in HE)

QHW = QCS

QCS = MS * cS * (tS0 – tS1) (W)

If we take into account heat loss in HE QL (W)

where QL is determined from a calculation or estimated on the basis of our pre-

vious experiments or knowledge as a part of total transferred heat (in % trans-

ferred heat – easier calculation)

QL = QCS * z / 100

QHW = QCS – QL = MS * cS * (tS0 – tS1) – QL (W)

or QHW = MS * cS * (tS0 – tS1) * (1 – z / 100) (W)

Specification of outgoing temperature of heated cooling water (if we have given

the outgoing temperature of cooled syrup)

For heat loss QL= 0


tW1 = tW0 + (MS*cS*(tS0 – tS1)) / (MW*cW) (°C)

If we take into account heat loss in HE QL (W) or z (in % of transferred heat)

– owing to heat loss we have to transfer less heat from the hot syrup to the cool-

ing water (it is that tW1 will be lower than for the case without heat loss)

tW1 = tW0 + (MS*cS*(tS0 – tS1) – QL) / (MW*cW) (°C)

or tW1 = tW0 + (MS*cS*(tS0 – tS1)*(1-z / 100)) / (MW*cW) (°C)

Specification of outgoing temperature of cooled syrup (if we have given outgoing

temperature of cooling water)

For heat loss Qz = 0

tS1 = tS0 - (MW*cW*(tW1 – tW0)) / (MS*cS) (°C)

If we take into account heat loss in HE QL (W) or z (in % of transferred heat)

– owing to heat loss the hot syrup will be cooler → for needed outgoing tempera-

ture we will transfer less heat to the cooling water

tS1 = tS0 - (MW*cW*(tW1 – tW0) + QL) / (MS*cS) (°C)

or tS1 = tS0 - (MW*cW*(tW1 – tW0)*(1+z / 100)) / (MS*cS) (°C)

These outgoing temperatures we have to compare with a reality of heat transfer

(depending on given data). It is that for example we compare the calculated

temperature of cooled syrup with the given temperature of cooling water. If it is

tS1 – tW0 < tHEmin (°C)


given data are unreal (see fig. 2 and 3) and it is necessary to design the cooler for

this new outgoing temperature of cooled syrup

tS1 = tW0 + tHEmin (°C)

Otherwise, with regard of temperature drops on the cooling water side, in the

fouling layer, in plate material and on the syrup side (again with fouling layer),

it would not be economical or real to transfer the needed amount of heat (or the

heat transfer area would be too large).

Then we can specify the mean logarithmic temperature difference in the HE

(counter flow) – see fig.3

(K)

4. Thermal calculation of cooler

As values of heat transfer coefficients for heating of the cooling water and

cooled syrup in HE channels depend on (among others) speeds in these channels

(an as we do not know a total heat transfer area of the HE we do not know these

speeds too) it is impossible to calculate these coefficients directly. Therefore we

must use an iterative method.

Big iteration loop

We estimate a value of coefficient of heat passage k (overall heat transfer co-

efficient) in the HE and on the base we can calculate the heat transfer area,

number of plates and liquids flow rates in one channel. If is the liquid speed too

high we design a HE with several channels connected in parallel (their number


we can estimate according our practical knowledge with optimal speeds for vari-

ous liquids – from points of views of heat passage coefficient → heat transfer

area → capital cost and pressure loss → pump consumption → running costs).

Then we calculate new values of heat transfer coefficients and a new value of

the coefficient of heat passage. If there is a difference we repeat the procedure.

Simultaneously we must check pressure losses in the HE. If they are too high we

must use a solution with lower losses (it is to increase a number of parallel con-

nected channels for a liquid in question). On the contrary if are the pressure

losses too low it is good to reduce number of parallel connected channels and so

increase speed. Thus increase coefficient of heat passage and decrease needed

heat transfer area (cheaper HE). More see example about effect of speeds on a

HE size.

4.1. Estimation of coefficient of heat passage and preliminary calculation of cooler (coolers)

Estimation of coefficient of heat passage

kest = ? (W/m2K) x compare with calculated and event. change estimation

Heat transfer area of cooler (coolers) needed for syrup cooling

(m2)

Number of plates in cooler (coolers) – theoretical estimation

nPT = ATC / AP choose whole number (-)

Note 1:


Exact equation for calculation valid for all types of HE is

Q = k * A * tL * F

where F is a factor with values usually in range from 0.8 to 1.0. A HE with factor

F < 0.8 is not acceptable as it is not stable in operation. For our case of a pure

counter flow is the F = 1.0. Similarly it is for pure parallel flow, evaporators and

condensers.

Fig. 5.: Examples of temperature profiles and their F factors:


F = 1

Pure parallel flow in 1 or 2 passes Pure counter flow in 1 or 2 passes

Steam condensation and liquid heating in 2 passes (= as in 1)

Coolant boiling and liquid cooling in 2 passes (= as in 1)

HE with pure counter flow and 2 passes for both fluids

F factor values dependence on parameters P and R

F

F < 1

Combination of counter and parallel flows or cross flow

Evaporator

P = (tC1 – tC0) / (tH0 – tC0) = heating of cool liquid / max. possible heating

R = (tH0 – tH1) / (tC1 – tC0) = cooling of hot liquid / heating of cool liquid

Note 2:

Another variant very often used for plate HE design.

According experiences speeds of both liquids in interplate channels are in the 1st

iteration estimated. On the basis of these speeds we can specify a flow pattern. It

is numbers of parallel channels in 1 pass. From it we can specify exact speeds of

both liquids wi and for this iteration Rei, Nui and ki. Then we can specify the

heat transfer area A and from it number of passes connected in series. Next pro-

cedure is analogous to the previous case. This procedure is used in the examples

prepared in Excel.

Choice of HE arrangement

Number of parallel channels in 1 pass and number of passes in HE we

choose from a point of view of optimal speeds of liquid in channels it is that coef-

ficient of heat passage and pressure losses must be optimal. Criterion of a

proper choice (design) is a HE size (mass, total heat transfer area) of the cooler

and total pressure losses on both liquids → economic appraisal.

Choice of a number of parallel interplate channels in the cooler – syrup side

(with regard to the recommended speed)


npcScalc = MS / (S * fP *wSrec) (-; kg/s, kg/m3, m2, m/s)

npcS = ? choose a whole number (-)

Choice of a number of parallel interplate channels in the cooler – cooling water

side (with regard to the recommended speed)

npcWcalc = MW / (W * fP *wWrec) (-; kg/s, kg/m3, m2, m/s)

npcW = ? choose a whole number (-)

Choice of a number of passes connected in series in HE (section) – syrup side

(one half of channels is for cooled syrup, the second one is for cooling water)

npsS = nPT / (2 * npcS) (-)

npsS = ? choose a whole number (-)

Choice of a number of passes connected in series in HE (section) – cooling water

side

npsW = nPT / (2 * npvW) (-)

npsW = ? choose a whole number (-)

A maximal difference between products npcS*npsS and npcW*npsW can be

+/- 1 (ev. max. 2) else it is impossible to set plates together according calcu-

lated values of numbers. (for = 2 in 1 pass is not the same number of parallel

channels as in others – for one liquid) – see fig. 6b.


Total number of plates in the cooler (section) – again it must be a whole number

nPTreal = npsS * npcS + npsW * npcw + 1 (-)

(one end plate must close a last channel – see fig. 6a)

Real heat transfer area of one cooler (section) is

AHEreal = Ap * nPHT = Ap * (nPTreal – 2) (m2)

Both end plates close interplate channels and they do not participate on heat

transfer (see fig. 6.).

Fig.6.: Example of a flow pattern in a plate HE (plates arrangement)

a) Flow pattern: syrup 1 x 4; cooling water 2 x 2

Flow pattern on this example:

cooling water npcW = 2; npsW = 2 syrup npcS = 4; npsS = 1

= npcS*npsS – npcW*npsW

=4*1 – 2*2 = 0 O.K.

nPTreal = 1 * 4 + 2 * 2 + 1 = 9

nPTHEreal = nPTreal – 2 = 9 – 2 = 7

Cross section clear areas in 1 pass:cooling water f1pW = 2 * fp (m2)syrup f1pS = 4 * fp (m2)

b) Flow pattern: syrup 1 x 5; cooling water 2 x 2:

Flow pattern on this example:


syrup

1 2 3 4 5 6 7 8

cooling water

9

f1pS = 4 * fPf1pW = 2 * fP

syrup

1 2 3 4 5 6 7 8

cooling water

9

f1pS = 5 * fpf1pW = 2 * fp

10

cooling water npcW = 2; npsW = 2 syrup npcS = 5; npsS = 1

= npcS*npsS – npcW*npsW

=5*1 – 2*2 = 1 O.K.

nPTreal = 1 * 5 + 2 * 2 + 1 = 10

nPTHEreal = nPTreal – 2 = 10 – 2 = 8

Specific heat flux in the HE (cooler) – through plates = heat transfer area

qHE = QHE / AHEreal QHE = QCW (W/m2)

4.2. Heat transfer for forced convection in interplate channel

Cross section clear area in one pass of the HE (section) – cooling water side

f1pW = npcW * fp (m2)

Cooling water speed in cannels

wW = MW / (W * f1pW) (m/s)

The speed we compare with a recommended one and ev. change a number of

parallel channels or parallel HE.

Cross section clear area in one pass of the HE (section) – cooled syrup side

f1pS = npcS * fp (m2)


Cooled syrup speed in channels

wS = MS / (S * f1pS) (m/s)

The speed we compare with a recommended one and ev. change a number of

parallel channels or parallel HE.

For next calculations we will use equations from literature, e.g. criteria equa-

tions that are valid for the type of plates.

Cooling liquid side

Determination of Reynolds criterion

ReW = wW * de / W (-)

Determination of Nusselt criterion for turbulent flow

Analogous to tubular HE is for the Nusselt criterion following equation used

Nu = a * Reb * Prc * (μw / μ)0,14 = * de /

Sieder-Tate correction (μw / μ)0,14 is in practice usually neglected as it has very

small effect – especially with regard to vague effect of fouling. Value of heat

transfer coefficient or Nu depends on a character of corrugation of plates (see

fig. 4d). It follows from it that for various types of plates are in the criteria equa-

tion various constants. For this example we will use following equation that was

derived from our experiments that were done in VÚPCHT Praha on designed

plates (producer CHS Chotěboř, now Tenez Chotěboř).

NuW = 0,0303 * ReW0,809 * PrW

0,43 (-)


Determination of heat transfer coefficient

W = NuW * W / de (W/m2K)

Cooled syrup side

Determination of Reynolds criterion

ReS = wS * de / S (-)

Again as in previous we use the criteria equation designed for the used type of

plates

NuS = 0,0303 * ReS0,809 * PrS

0,43 (-)

Determination of heat transfer coefficient

R = NuR * R / de (W/m2K)

4.3. Heat passage coefficient including fouling

(W/m2K)

This value we compare with the estimated one (see chap. 4.1.) and if there is a

difference we repeat these calculations till

kcalc kest

4.4. Determination of pressure loss in the HE (cooler)


Pressure loss in plate HE depends among others on speed, character of corruga-

tion of plates and their arrangement. Analogous to the heat transfer we use cri-

teria equation derived from our experiments for used type of plates.

Cooling water side

EuW = 460 * ReW-0,264 (-)

Note: This equation is valid for common working conditions, it is with common

fouling layer and directions changes, inlet and outlet necks etc.

Pressure loss in one interplate channel

pW1 = EuW * W *wW2 (Pa; -, kg/m3, m/s)

Total pressure loss in the HE (or one section)

pWT = pW1 * npsW (Pa; Pa, -)

Cooled syrup side

EuS = 460 * ReS-0,264 (-)

Note: Like for previous the equation is valid for working conditions, it is with

common fouling on plates.

Pressure loss in one interplate channel

pS1 = EuS * S *wS2 (Pa; -, kg/m3, m/s)

Total pressure loss in the HE (or one section)


pST = pS1 * npsS (Pa; Pa, -)

These pressure losses we compare with given maximal allowable and according

situation (for corresponding liquid) repeat this calculation with a new number of

parallel channels npci and from it calculated number of passes connected in se-

ries npsi.

Too low pressure loss in HE results in low value of heat passage coefficient and

consequently too large heat transfer area = too expensive HE. Therefore it is im-

portant to optimize capital and running costs of a designed HE.

Note:

Calculation for tubular HE

Specification of roughness of tubes

kT = ? (-)

Coefficient of friction losses for turbulent flow in tubes

(-)

Friction loss in tubes (syrup side)

(Pa; -, m, -, m, m/s, kg/m3)

Where nPS = number of passes on the syrup side in a HE

nHES = number of HE connected in series

LT = tubes lenght

dTi = internal tubes diameter

wST = syrup speed in tubes


Pressure loss owing to sudden enlargement (from tubes to chamber)

(Pa; m/s, kg/m3, -)

where n is number of sudden enlargements.

Pressure loss owing to sudden contraction (from chamber to tubes)

For the case it is possible to use following equation

pLSC 0,6 * pLSE

Pressure loss in valves (approximately)

Coefficient of pressure loss in valves

(-; m, m)

where Dvalve is diameter of seat of valve and Zvalve is valve stroke above the seat.

(Pa; -, m/s, kg/m3)

where speed in valve is

Note: For valves, flap valves (butterfly valves), spherical valves, slide valves etc. it is better to calculate pressure losses according characteristics given by a manufacturer.

Total pressure loss on the syrup side

pLS = pLTS + pLSES +pLSCS + pLvalvesS (Pa)

4.5. Specification of necks diameters

For the neck diameters specification we can use following equations


→

where is M (kg/s) mass flow of a fluid

(steam, vapour, inerts, solution, condensate ...)

n (-) number of necks for the fluid

wrec (m/s) reccomended speed for the fluid

(kg/m3) density of the fluid

Reccomended speeds (according my practical experiences)

These speeds are valid for common cases of evaporators and heat exchangers in

food and chemical industries.

Steam (vapour) in inlet necks 10 - 25 m/s

Condensate in outlet necks (tubes) 0,2 - 0,5 m/s

(condensate is on boundary line vapour/water) if pressure decreases owing

to pressure losses condensate starts superheated and vapour forms (in valves,

steam traps, pipelines etc.,) two-phases flow in pipeline >> volume)

Solution inlet or outlet for HEs 1 - 3 m/s

Solution outlet for evaporators 1 - 2 m/s

(boiling liquid is on boundary line vapour/liquid)

Inerts outlet 10 - 15 m/s

5. Specification of maximal performance of designed HE – syrup cooler

As the real number of plates and their configuration are little different from cal-

culated values we want to know what is the real maximal performance of this

HE. Analogical is a calculation if we want to use an existing HE for other appli-

cation etc. In our case we want to know what the lowest attainable temperature

of cooled syrup is or what the maximal attainable temperature of cooling water

is.


Select number of parallel channels on side of cooled syrup

npcS = ? (-)

Select number of parallel channels on side of cooling water

npcW = ? (-)

Speed in channels – side of syrup

wS = MS / (S * fP * npcS) (m/s)

Speed in channels – side of water

wW = MW / (W * fP * npcW) (m/s)

Select number of passes connected in series – side of syrup

npsS = ? (-)

Select number of passes connected in series – side of water

npsW = ? (-)

Total number of plates in the cooler (section)

nPTreal = npsS * npcS + npsW * npcw + 1 (-)

Total real heat transfer area of one cooler (section) is

AHE = Ap * nPHT = Ap * (nPTreal – 2) (m2)

Heat transfer for forced convention in channels


On the basis of our previous results we estimate outlet temperatures of syrup

and cooling water - (i –iteration) and from them physical properties. The next

procedure is similar to the previous.

tS1i = ? and tW1i = ?

- Side of cooled syrup

Remember that physical parameters depends (among others) on temperature → iteration

Specification of Reynolds and Nusselt critera

ReSi = wSi * de / Si (-)

NuSi = 0,0303 * ReSi0,809 * PrSi

0,43 (-)

Specification of heat transfer coefficient

Si = NuSi * Si / de (W/m2K)

- Side of cooling water

Specification of Reynolds and Nusselt critera

ReWi = wWi * de / Wi (-)

NuWi = 0,0303 * ReWi0,809 * PrWi

0,43 (-)

Specification of heat transfer coefficient

Wi = NuWi * Wi / de (W/m2K)

Heat passage coefficient including fouling

(W/m2K)


Mean logarithmic temperature difference in the HE (counter flow)

(K)

Maximal value of heat transfered from cooled syrup in cooling water

(without heat loss)

QHWi = kcalci * ATreal * tLi (W)

New temperature of outgoing cooled syrup is (i + 1 iteration)

tS1i+1 = tS0 – QHWi / (MW * cW) (°C)

New temperature of outgoing heated cooling water is (i + 1 iteration)

tW1i+1 = tW0 + QHWi / (MW * cW) (°C)

Now we compare these temperatures with previous i –iteration. If there is a dif-

ference we repeat these calculations till are temperatures the same.

In supplement are results of several variants calculated in Excel file. Here you

can see for example an effect of a variant with 2 sections or higher pressure loss

on heat transfer area of HE.

Generalization of the procedure on other types of HE

In a case of HE where is a phase change (condensation, boiling) is the procedure

similar, it is that we firstly estimate heat passage coefficient = big iteration loop


(1st iteration level). Because for condensation heat transfer coefficient depends

on a temperature difference between condensing steam and wall (that is un-

known), we must estimate it too = small iteration loop in the big one (2nd itera-

tion level). When is the calculated wall temperature on the condensing steam

side equal to the estimated one we can continue in calculations in the big itera-

tion loop. Therefore is this procedure more complicated.

Our calculations (temperatures specification) go from this standpoint: Specific

heat flux through all interfaces must be the same (from condensing steam to in-

crustation, through incrustation, through wall, through incrustation and from

incrustation to boiling liquid). In this way we can specify temperatures on single

interfaces (see fig. 2 – there are other indices).

q = s * (ts“ – tis) = is / sis *(tis – tws) =w / sw * (tws – twb) =

= ib / sib * (twb – tib) = b * (tib – tb)

In the equation is:

s – steam; is – incrustation on steam side; ws – wall on steam side;

wb – wall on boiling side; ib – incrustation on boiling side; b – boiling liquid



M2 - syrup

M1 - water

t22

t21

t12

t11

Temperatures in 1 section

t112

M2 - syrup

M1 - water

t221

t21

t122

t121

t222

t111

Temperatures in 2 sections

syrup

water

1st section

Connection of 2 HE sections(cooling water in parallel, syrup in series)

water (50%)

water (50 %)

syrup

2nd section

water syrup

1.section

Syrup and cooling water flow in 1 section

watersyrup

Two variants of the plate cooler design

Comparison of results of 3 variants of the syrup cooler design

Msyrob = 6705 kg/h Mcooling water = 15000 kg/h

Variant

Parameter 1 section

< p

1 section

> p

2 sections

>p

Q (kW) 191,0 190,3 205,0

k (W/m2K) 257 385 485

302

AHET (m2) 47,4 30,9 42,7

number of plates 111 73 102

mass of cooler (kg) 599 394 551

temperatures (°C)

- syrup

- cooling water

80,0 → 40,6

35,0 → 46,0

80,0 → 40,9

35,0 → 45,9

80,0 → 38,4

35,0 → 56,0

35,0 → 37,6

pressure loss (kPa)

- syrup

- cooling water

77

15

203

50

184,6

12,3

14,1


Approximate economical appreciation of 3 variants

with 1 and 2 sections (both for higher p L)

Cooler with 2 sections

tSyrOut = 38,4 °C AHE = 42,7 m2 Q = 205,0 kW

Cooler with 1 section

tSyrOut = 40,9 °C AHE = 30,9 m2 Q = 190,3 kW

Data for economical appreciation

Cost of stainless steel heat transfer area (plates) CA = 8000 Kč/m2

Cost of “cool” CCOOL = 250 Kč/GJ

Costs for salaries will be the same for both variants. Costs of maintaining, deprecia-

tion etc. will be supposed as some percent of capital costs. Therefore we can neglect

them and we can calculate only with these two costs.

Syrup pre-cooled in the HE must be after-cooled in another cooler with refrigerant to

a needed temperature 10 °C with ice water from a cooling circuit – e.g. with an am-

monia compressor.

Additional costs for heat transfer area for variant with 2 sections

CCA = A * CA = (42,7 – 30,9) * 8000 = 94 400,- Kč


This value represents higher capital cost.

Additional costs for maintaining and depreciation

We assume a depreciation time 12 years. Corresponding depreciation rate is c. 8

%. Cost for maintaining we assume as c. 4 % of acquisition cost. Then these ad-

ditional costs for maintaining and depreciation are

CMD = CCA * (D + M) = 94 000 * (8 + 4) / 100 = 11 330,- Kč/year

Amount of energy saved for 2 nd step of syrup cooling

QCOOL = Q2 – Q1 = 205,0 – 190,3 = 14,7 kW

Amount of saved energy (cool) per year

Line working time = 200 days / year = 200 * 24 = 4 800 h/y

QCOOLY = QCOOL *

QCOOLY = 14,7 * 4800 * 3600 / 1 000 000 = 254,0 GJ / year

Cost of saved cool per year

CCCOOLY = QCOOLY * CCOOL = 254,0 * 250 = 63 500,- Kč/year


Annual profit from the variant with 2 sections

AP = CCCOOLY - CMD = 63 500 – 11 330 = 52 170,- Kč/year

Simple rate of return of the investment

SRR = CCA / AP = 94 400 / 52 170 = 1,81 years


Example of centrifugal pump flow control by throttling

Design of plate HE water / syrup - example DPE - 1

Given data: Water and syrup flowrates, cooling water inlet and outlet temperatures , syrup inlet temperature

Specify: Available syrup outlet temperature, number andconfiguration of plates1st variant = t11,t12,M1,t21,M2 t22 = ?

Given dataType of HE: CHX 1000Sections: Plate cooler of syrup with cooling water from cooling towers

1 section given revise data given data

Medium 1 - heated (cooling): water from water fromcol. tow. col. tow.

- inlet temperature t11 (oC) = 35 35 - outlet temperature-ev. change acc. results t12 (oC) = 48 46,0 temp. calculated from HE balance t12calc (oC) = 45,8 45,9

- flowrate (in sum for 2 sections 15 t/h) M1 (kg/h) = 15000 15000 - max. allowable pressure loss p1max (kPa) = 150 150 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2

Medium 2 - cooled: syrup syrup - inlet temperature t21 (oC) = 80 80 - outlet temperature - calculated t22 (oC) = 42,3 40,9 (estimation for 1.iteration - change till be the same) 41,0 40,9 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure loss p2max (kPa) = 200 200 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2

Connection (parallel or counter flow): counter fl. counter fl.

M2 - syrup

M1 - water

t22

t21

t12

t11

Temperature course in 1 section

tHEmin

Connection of 1 section of HE

syrup

water syrup

water

1.section


Note: In HE inflow all cooling water and cooled hot syrup

Order: Example of DPEVarious variants of design - 1 sections

Made by: P. HoffmanDate april 2003

CONSTANTS FOR CALCULATIONS

- gap between plates s (mm) = 4 4 - channel width b (mm) = 418 418 - channel length L (mm) = 1220 1220 - ekvivalent diameter of channel de (m) = 0,0079 0,0079

- flow clear area of channel f (m2) = 0,00167 0,00167 - heat transfer area of 1 plate A (m2) = 0,435 0,435

- recommended speed - medium 1: water w1rec (m/s) = 0,25 0,25 - medium 2: syrup w2rec (m/s) = 0,25 0,25

- relations for calculation of physical parameters are from my sources (for syrup find in handbooks - for water program calculates automatically)

- minimal economical temperature difference in HE 3 3tminHE = t22-t11 tminHE (°C) =

- maximal allowable pressure in HE pmax (kPa) = 300 300

- criteria equations for calculations of heat transfer coefficient and pressure loss must be choosen according type of plates

- plate thickness sp (mm) = 1,3 1,3

- thermal conductivity of plate p (W/mK) = 15 15

- mass of 1 plate Mp (kg) = 5,4 5,4 - mass of frame Mfr (kg) = 200 200 (choose from producer data)

- heat loss in HE z (%) = 0 0 (in % transfered heat)


HE CALCULATION FOR GIVEN DATA

1. Specification of physical parameters for both media

- Medium 1: water from water from - mean temperature t1 (oC) = 41,5 40,5 - specific heat c1 (kJ/kg.K) = 4,177 4,177 - density 1 (kg/m3) = 993,5 993,8 - thermal conductivity 1 (W/m.K) = 0,6292069 0,62793753 - dynamic viscosity 1 (Pa.s) = 0,000637 0,000649 - kinematic viscosity 1 (m2/s) = 6,41E-07 6,53E-07 - Prandtl number Pr1 (-) = 4,17 4,25

- Medium 2 (for 1. iteration): syrup syrup - mean temperature t2 (oC) = 60,5 60,4 - specific heat c2 (kJ/kg.K) = 2,670 2,610

(ev. change according results of iter.)

2. Heat input of the HE

Q1 = M1*c1*(t12-t11)/3600 Q1 (kW) = 226,3 191,5Q2 = 100*Q1/(100-z) Q2 (kW) = 226,3 191,5t22 = t21-(t12-t11)*M1*c1/(M2*c2)1.iteration t22 (oC) = 34,50 40,62

unreal given data!! 40,6Comparison with water inlet temperature t11 (oC) = 35,0 35,0Minim. possible temp. of cooled syrup t22minp (°C) = 38,0 38,0Max. possible temp. of heated water t12maxp (°C) = 47,0 46,7 It is necessary to change given data !!!!next calculations in the column are not good - new calc. are in the right column !!

- Medium 2 (for next calculations): syrup syrup - mean temperature t2 (oC) = 57,2 60,3 - specific heat c2 (kJ/kg.K) = 2,670 2,610 - density 2 (kg/m3) = 1353,0 1360,0 - thermal conductivity 2 (W/m.K) = 0,3910 0,3850 - dynamic viscosity 2 (Pa.s) = 0,025000 0,063000 - kinematic viscosity 2 (m2/s) = 1,85E-05 4,63E-05 - Prandtl number Pr2 (-) = 170,72 427,09


3. Liquids (media) speeds

calculated speed in a case of flow only through 1 interplate channel w1c = M1/(3600*1*f) w1c (m/s) = 2,51 2,51w2c = M2/(3600*2*f) w2c (m/s) = 0,82 0,82

calculated number of parallel channels in 1 pass npc1c = w1c/w1rec npc1c (-) = 10,03 10,03npc2c = w2c/w2rec npc2c (-) = 3,29 3,28

choosen number of parallel channel in 1 pass npc1 water from npc1 (-) = 6 6npc2 syrup npc2 (-) = 3 3

speeds in interplate channelsw1 = M1/(3600*1*f*npc1) w1 (m/s) = 0,418 0,418w2 = M2/(3600*2*f*npc2) w2 (m/s) = 0,274 0,273

4. Specification of heat passage coefficient

Re1 = w1*de/1 water Re1 (-) = 5165,603 5071,93755

Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 56,53 56,18

1 = Nu1*1/de 1 (W/m2*K) = 4489 4452

Re2 = w2*de/2 syrup Re2 (-) = 117,69 46,70

Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 13,087 9,191

2 = Nu2*2/de 2 (W/m2*K) = 646 447

k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 524 385

5. Specification of number of plates in HE (section)

tlog = abs((t22-t11)-(t21-t12))/ln(abs((t22-t11)/(t21-t12))) tlog (oC) = 7,82 15,76


calculated total heat transfer area of HEATC = Q1*1000/(k*dtlog) ATC (m2) = 55,18 31,59

calculated number of plates in HE (section)nPT = ATC/A1 nPT (-) = 126,8 72,6

calculated number of passes connected in series every with npc parallel channels nps1c = nPT/(2*npc1) nps1c (-) = 10,57 6,05

nps2c = nPT/(2*npc2) nps2c (-) = 21,14 12,10

choosen number of passes connected in series every with npc parallel channels nps1c = nPT/(2*npc1) nps1c (-) = 5 6

nps2c = nPT/(2*npc2) nps2c (-) = 10 12

Note: Maximal difference between products (nps1*npc1) and (nps2*npc2) can be at most +/- 1 plate

nps1*npc1 = 30 36nps2*npc2 = 30 36

total number of plates in HE (section)np = nps1*npc1+nps2*npc2+1 np (-) = 61,0 73,0

Total heat transfer area oh HE (section)AHEreal = (np-2)*A1 AHEreal (m2) = 25,7 30,9

6. Specification of pressure losses

Medium 1 water from water fromcol. tow. col. tow.

Eu1 = 418,6*(Re1^-0.264)*x x (-) = 1,1 1,1(x=correction according experiments Eu1 (-) = 48,19 48,42 incl. fouling)pressure loss in 1 channelp11 = Eu1*1*w1^2/1000 p11 (kPa) = 8,37 8,40

total pressure loss in HE (section)p1T = nps1*p11 p1T (kPa) = 41,8 50,4

maximal allowable pressure loss p1max = 150 150


Medium 2 syrup syrup

Eu2 = 418,6*(Re2^-0.264)*x x (-) = 1,1 1,1(equation is valid for water and milk Eu2 (-) = 130,77 166,91 ( x=correction)pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) = 13,33 16,92



maximal allowable pressure in HE (kPa) pmax = 300

If there is a high difference between calculated and allowable pressure lossesit is good to change HE arrangement (numbers of parallel channels) andrepeat the calculation.

NEW CALCULATION - HE maximal output - specification of max. heating of 2.medium t22 for choosen arrangement

- specification of max. cooling of 1.medium t12 for choosen arrangement

newly choosen number of parallel channels in 1 pass (for next calcullations if pressure loss is unsatisfactory)npc1 water from npc1 (-) = 6 6npc2 syrup npc2 (-) = 3 3

speeds in passes (channels)w1 = M1/(3600*1*f*npc1) w1 (m/s) = 0,42 0,42w2 = M2/(3600*2*f*npc2) w2 (m/s) = 0,27 0,27

newly choosen number of passes in series nps1 = nPT/(2*npc1) nps1*npc1 = 5 6

nps2 = nPT/(2*npc2) nps2*npc2 = 10 12



nps1*npc1 = 30 36nps2*npc2 = 30 36

total number of plates in HEnp = nps1*npc1+nps2*npc2+1 np (-) = 61 73

Total heat transfer area oh HE (section)

AHEreal = (np-2)*A1 AHEreal (m2) = 25,7 30,9

Specification of heat transfer passage

Re1 = w1*de/1 Re1 (-) = 5165,6 5071,9

Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 56,53 56,18

1 = Nu1*1/de 1 (W/m2*K) = 4489 4452

Re2 = w2*de/2 Re2 (-) = 117,69 46,70

Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 13,09 9,19

2 = Nu2*2/de 2 (W/m2*K) = 646 447

k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 524 385

Specification of max. cooling of mediun 2 (t22) in HE (section) and heating of medium 1 (t12) don´t iterate

unrealistic given data

estimation of t22 for i- iteration t22i (°C) = 42,50 40,86calculated value of t22 t22i+1 (°C) = 42,26 40,86estimation of t12 for i- iteration t12i (°C) = 56,66 45,94calculated value of t12 t12i+1 (°C) = 45,78 45,93


Max. amount of transfered heat in HEQ=k*AHE*tlog Q (kW) = 187,7 190,3

Temperature of cooled medium 2 t22i+1 (°C) = 42,3 40,9Temperature of heated medium 1 t12i+1 (°C) = 45,8 45,9


Pressure losses checking


Eu1 = 418,6*(Re1^-0.264)*x Eu1 (-) = 48,19 48,42

pressure loss in 1 channelp11 = Eu1*1*w1^2/1000 p11 (kPa) = 8,37 8,40



Eu2 = 418,6*(Re2^-0.264)*x Eu2 (-) = 130,77 166,91pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) =

13,33 16,92total pressure loss in HE (section)p2T = nps2*p21 p2T (kPa) = 133,3 203,0

comparison of calculated and maximal allowed pressure losses (kPa)calculated max. allowedp1T = 41,8 p1max = 150 150p2T = 133,3 p2max = 200 200

maximal allowable pressure in HE (kPa)pmax = 300

If pressure losses are higher than maximal allowable we must increase numberof parallel channels npc1 or npc2 and repeat the procedure.It they are too lower we can reduce the number of parallel channels and repeat the calculations.

If results are OK it is possible to print results.

Print of results


GIVEN DATA

Type of HE: CHX 1000Section: Plate cooler of syrup with cooling water from cooling towers


- inlet temperature t11 (oC) = 35 35 - outlet temperature t12 (oC) = 45,8 45,9 - flowrate M1 (kg/h) = 15000 15000 - max. allowable pressure lost p1max (kPa) = 150 150 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2

Medium 2 - cooled: syrup syrup

- inlet temperature t21 (oC) = 80 80 - outlet temperature t22 (oC) = 42,3 40,9 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure lost p2max (kPa) = 200 200 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2

- heat loss in HE (in % transfered heat) 0 0


Note: In HE inflow all cooling water and cooled hot syrup0

Customer:

Action: Example of DPEOrder No.: Various variants of design - 1 sections

Made by: P. HoffmanDate: april 2003


CALCULATED VALUES


Number of parallel channels in 1 pass npc1 (-) = 6 6Number of passes in series nps1 (-) = 5 6Liquid speed w1 (m/s) = 0,42 0,42Heat transfer coefficient 1 (W/m2*K) = 4489 4452Total pressure loss in HE (section) p1T (kPa) = 42 50



Heat passage coefficient k(W/m2*K) = 524 385Mean log. temperature difference tlog (oC) = 13,95 16,02Total number of plates in HE (section) np (-) = 61 73Total heat transfer area of HE (section) AHE (m2) = 25,67 30,89

Max. allowable pressure in HE (section) pmax (kPa) = 300 300

Plates (channels) arrangement npc nps medium 1 - water from cooling towers 6 * 5

------------------------------------------------------------------------------------------------ medium 2 - syrup 3 * 10

Mass of plates in HE (section) (kg) 394,2Mass of HE (section) with frame (kg) 594,2

Total mass of full HE (section) (kg) 767,1Mass of liquids in HE (section) (kg) 172,9





Given dataType of HE: CHX 1000Sections: Plate cooler of syrup with cooling water from cooling towers

2 sections 1.section 2.section


- inlet temperature t11 (oC) = 35,0 35,0 - outlet temperature-ev. change acc. results t12 (oC) = 56,1 37,5 temp. calculated from HE balance t12calc (oC) = 56,0 37,6


Medium 2 - cooled: syrup syrup - inlet temperature t21 (oC) = 80,0 43,0 - outlet temperature - calculated t22 (oC) = 43,1 38,5 (estimation for 1.iteration - change till be the same) 43,0 38,5 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure loss p2max (kPa) = 150 150 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2


syrupsyrup

water water

voda

1.section 2.section

Connection of 2 sections in HE

t122

M2 - syrup

M1 - water

t221

t21

t121

t11

Temperature course in 2 sections

tHEmin

t222

t11


Note: Both coolers sections are connected in series (syrup outlet from 1.section is inlet to 2.); both sections are cooled withthe same water - 50 % goes in 1. and 50 % in 2.section)

Order: Example of DPEVarious variants of design - 2 sections - syrup in series, cooling water in parallel







- minimal economical temperature difference in HEtminHE = t22-t11 tminHE (°C) = 3 3










- Medium 1: water from water fromcol. tow. col. tow.

- mean temperature t1 (oC) = 45,6 36,3 - specific heat c1 (kJ/kg.K) = 4,179 4,176 - density 1 (kg/m3) = 992,0 995,3 - thermal conductivity 1 (W/m.K) = 0,6342123 0,6223785 - dynamic viscosity 1 (Pa.s) = 0,000593 0,000703 - kinematic viscosity 1 (m2/s) = 5,98E-07 7,07E-07 - Prandtl number Pr1 (-) = 3,86 4,64





43,06 38,53Comparison with water inlet temperature t11 (oC) = 35,0 35,0Minim. possible temp. of cooled syrup t22minp (°C) = 38,0 38,0Max. possible temp. of heated water t12maxp (°C) = 59,0 37,8

- Medium 2 (for next calculations): syrup syrup - mean temperature t2 (oC) = 61,5 40,8 - specific heat c2 (kJ/kg.K) = 2,670 2,610 - density 2 (kg/m3) = 1353,0 1360,0 - thermal conductivity 2 (W/m.K) = 0,3910 0,3850 - dynamic viskosity 2 (Pa.s) = 0,025000 0,063000 - kinematic viskosita 2 (m2/s) = 1,85E-05 4,63E-05 - Prandtl number Pr2 (-) = 170,72 427,09








Re1 = w1*de/1 water Re1 (-) = 2775,4137 2807,9556

Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 33,07 36,16

1 = Nu1*1/de 1 (W/m2*K) = 2647 2840

Re2 = w2*de/2 syrup Re2 (-) = 117,69 35,03

Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 13,087 7,283

2 = Nu2*2/de 2 (W/m2*K) = 646 354

k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2)

k (W/m2*K) = 485 302







nps2c = nPT/(2*npc2) nps2c (-) = 9,96 4,67


nps2c = nPT/(2*npc2) nps2c (-) = 10 5


nps1*npc1 = 30 20nps2*npc2 = 30 20



in the 1.section is syrup cooled of c. 37 °C in the 2.section is syrup cooled of c. 4,6 °C S = 42,6







Medium 2 syrup syrupEu2 = 418,6*(Re2^-0.264)*x x (-) = 1,1 1,1(equation is valid for water and milk Eu2 (-) = 130,77 180,08 ( x=correction)pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) = 13,33 10,27



maximal allowable pressure in HE pmax = 300




newly choosen number of parallel channels in 1 pass (for next calcullations if pressure loss is unsatisfactory)npc1 water from npc1 (-) = 6 5npc2 syrup npc2 (-) = 3 4


newly choosen number of passes in series nps1 = nPT/(2*npc1) nps1 (-) = 5 4

nps2 = nPT/(2*npc2) nps2 (-) = 10 5



nps1*npc1 = 30 20nps2*npc2 = 30 20


Total heat transfer area of HE (section)AHEreal = (np-2)*A1 AHEreal (m2) = 25,7 17,0


Re1 = w1*de/1 Re1 (-) = 2775,4 2808,0

Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 33,07 36,16

1 = Nu1*1/de 1 (W/m2*K) = 2647 2840

Re2 = w2*de/2 Re2 (-) = 117,69 35,03

Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 13,09 7,28

2 = Nu2*2/de 2 (W/m2*K) = 646 354

k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 485 302

Specification of max. cooling of mediun 2 (t22) in HE (section) and heating of medium 1 (t12)








Eu1 = 418,6*(Re1^-0.264)*x Eu1 (-) = 56,77 56,60




Eu2 = 418,6*(Re2^-0.264)*x Eu2 (-) = 130,77 180,08pressure loss in 1 channelp21 = Eu2*2*w2^2/1000 p21 (kPa) = 13,33 10,27







Print of results

GIVEN DATA t22 = ?



- inlet temperature t11 (oC) = 35 35 - outlet temperature t12 (oC) = 56,0 37,6 - flowrate (in sum for 2 sections 15 t/h) M1 (kg/h) = 7500 7500 - max. allowable pressure lost p1max (kPa) = 150 150 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2



- heat loss in HE (in % transfered heat) 0 0 Connection (parallel or counter flow): counter fl. counter fl.

Note: Both coolers sections are connected in series (syrup outlet the same water - 50 % goes in 1. and 50 % in 2.section)

Customer:Action: Example of DPEOrder No.: Various variants of design - 2 sections - syrup in series,

cooling water in parallel



CALCULATED VALUES



Medium 2 syrup syrupNumber of parallel channels in 1 pass npc2 (-) = 3 4Number of passes in series nps2 (-) = 10 5Liquid speed w2 (m/s) = 0,27 0,20Heat transfer coefficient 2 (W/m2*K) = 646 354Total pressure loss in HE (section) p2T (kPa) = 133 51



Plates (channels) arrangement npc nps medium 1 - water from cooling towers 6 * 5

----------------------------------------------------------------------------------------------- medium 2 - syrup 3 * 10

Mass of plates in HE (section) (kg) 329,4 221,4Mass of HE (section) with frame (kg) 529,4 421,4

Total mass of full HE (section) (kg) 672,9 517,5Mass of liquids in HE (section) (kg) 143,5 96,1





Given data

Type of HE: CHX 1000Sections: Plate cooler of syrup with cooling water from cooling towers

effect of pressure losses < pL > pL


- inlet temperature t11 (oC) = 35 35 - outlet temperature-ev. change acc. results t12 (oC) = 46,0 46,0 temp. calculated from HE balance t12calc (oC) = 46,0 46,0



- inlet temperature t21 (oC) = 80 80 - outlet temperature - calculated t22 (oC) = 40,6 40,6 (estimation for 1.iteration - change till be the same) 40,7 40,7 - flowrate M2 (kg/h) = 6705 6705 - max. allowable pressure loss p2max (kPa) = 100 250 - fouling thickness s2i (mm) = 0,1 0,1 - fouling thermal conductivity 2i (W/mK) = 2 2

Connection of 1 section of HE

syrup

water syrup

water

1.section

M2 - syrup

M1 - water

t22

t21

t12

t11

Temperature course in 1 section

tHEmin



Note: In HE inflow all cooling water and cooled hot syrup

Order: Example of DPEVarious variants of design - 1 sections







- minimal economical temperature difference in HEtminHE = t22-t11 tminHE (°C) = 3 3










- Medium 1: water from water from - mean temperature t1 (oC) = col. tow. col. tow. - specific heat c1 (kJ/kg.K) = 40,5 40,5 - density 1 (kg/m3) = 4,177 4,177 - thermal conductivity 1 (W/m.K) = 993,8 993,8 - dynamic viscosity 1 (Pa.s) = 0,6279375 0,62793753 - kinematic viscosity 1 (m2/s) = 0,000649 0,000649 - Prandtl number Pr1 (-) = 6,53E-07 6,53E-07 - Prandtlovo číslo Pr1 (-) = 4,25 4,25





40,6 40,6Comparison with water inlet temperature t11 (oC) = 35,0 35,0Minim. possible temp. of cooled syrup t22minp (°C) = 38,0 38,0Max. possible temp. of heated water t12maxp (°C) = 46,7 46,7

- Medium 2 (for next calculations): syrup syrup - mean temperature t2 (oC) = 60,3 60,3 - specific heat c2 (kJ/kg.K) = 2,610 2,610 - density 2 (kg/m3) = 1360,0 1360,0 - thermal conductivity 2 (W/m.K) = 0,3850 0,3850 - dynamic viskosity 2 (Pa.s) = 0,063000 0,063000 - kinematic viskosita 2 (m2/s) = 4,63E-05 4,63E-05 - Prandtl number Pr2 (-) = 427,09 427,09







4. Určení součinitele prostupu tepla


Re1 = w1*de/1 water Re1 (-) = 2766,5114 7607,90633

Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 34,40 77,99

1 = Nu1*1/de 1 (W/m2*K) = 2726 6180

Re2 = w2*de/2 syrup Re2 (-) = 28,02 46,70

Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 6,080 9,191

2 = Nu2*2/de 2 (W/m2*K) = 295 447

k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 257 394







nps2c = nPT/(2*npc2) nps2c (-) = 10,86 11,81


nps2c = nPT/(2*npc2) nps2c (-) = 11 12


nps1*npc1 = 55 36nps2*npc2 = 55 36




Médium 1 water from water fromcol. tow. col. tow.






Eu2 = 418,6*(Re2^-0.264)*x x (-) = 1,1 1,1(equation is valid for water and milk Eu2 (-) = 191,01 166,91 ( x=correction)




maximal allowable pressure in HE (kPa) pmax = 300




newly choosen number of parallel channels in 1 pass (for next calcullations if pressure loss is unsatisfactory)ppk1 water from ppk1 (-) = 11 4ppk2 syrup ppk2 (-) = 5 3


newly choosen number of passes in series nps1 = nPT/(2*npc1) nps1*npc1 = 5 9

nps2 = nPT/(2*npc2) nps2*npc2 = 11 12


Note: Maximal difference between products (nps1*npc1) and (nps2*npc2) can be at most +/- 1 plate nps1*npc1 = 55 36

nps2*npc2 = 55 36




Re1 = w1*de/1 Re1 (-) = 2766,5 7607,9

Nu1 = 0,0303*(Re1^0.809)*(Pr1^0.43) Nu1 (-) = 34,40 77,99

1 = Nu1*1/de 1 (W/m2*K) = 2726 6180

Re2 = w2*de/2 Re2 (-) = 28,02 46,70

Nu2 = 0,0303*(Re2^0.809)*(Pr2^0.43) Nu2 (-) = 6,08 9,19

2 = Nu2*2/de 2 (W/m2*K) = 295 447

k = 1/(1/1+sp/d+si1/i1+si2/i2+1/2) k (W/m2*K) = 257 394

Specification of max. cooling of mediun 2 (t22) in HE (section) and heating of medium 1 (t12)








Eu1 = 418,6*(Re1^-0.264)*x Eu1 (-) = 56,82 43,50




Eu2 = 418,6*(Re2^-0.264)*x Eu2 (-) = 191,01 166,91








Print of results

GIVEN DATA



- inlet temperature t11 (oC) = 35 35 - outlet temperature t12 (oC) = 46,0 46,0 - flowrate M1 (kg/h) = 15000 15000 - max. allowable pressure lost p1max (kPa) = 100 200 - fouling thickness s1i (mm) = 0 0 - fouling thermal conductivity 1i (W/mK) = 2 2



- heat loss in HE (in % transfered heat) 0 0


Note: In HE inflow all cooling water and cooled hot syrup0

Customer:Action: Example of DPEOrder No.: Various variants of design - 1 sections



CALCULATED VALUES







Plates (channels) arrangement npc nps medium 1 - water from cooling towers=G33411 * 5

---------------------------------------------------------------------- medium 2 - syrup 5 * 11

Mass of plates in HE (section) (kg) 599,4 394,2Mass of HE (section) with frame (kg) 799,4 594,2

Total mass of full HE (section) (kg) 1063,5 767,1Mass of liquids in HE (section) (kg) 264,1 172,9


5 DPE Plate HE Design (Syrup Water)

Documents

cooling water ts1

cooling liquid cooling

heat loss tw1

heat loss ql

hot syrup

w i w heating of cooling

mw water tw

account heat loss