PHE Design

parameter paremetr value from to multiplying factorfactor

length 1 m ft 0.305pressure 1 kg/cm2 psi 14.223Temperature 1 C F 33.800

Sp. Heat 1 Btu/lb0F 1.000693Density 1 kg/m3 lb/ft3 0.062Viscosity 1 cp lb/(ft)(hr) 2.420Thermal cond. 1 Kcal/h-m-C Btu/h-ft-F 0.672

heat t. coeff 1 kcal/(hr)(m2)C 0.2051 m2 in2 1549.9651 m2 ft2 10.76

ft2 m2 0.09291 (hr)(m2)(0C)/Kcal (hr)(ft2)(0F)/Btu 4.882391

Kcal/kg 0C

Btu/hr(ft2)(0F)

multiplying factorfactor

Refer appendix table 1. (Kern)

1.002084 1 kj/kgC=0.239 Btu/lbF

SI unitsDATA INPUT Hot Fluid: Styrene

T1, Inlet temperature 43 CT2, outlet temperature 20 CW, mass flow rate 21240 kg/hr

c, sp. Heat at inlet 0.43

c, sp. Heat at outlet 0.41

Average c 0.42s, specific gravity, inlet 0.88511s, specific gravity, outlet 0.90418Average s 0.894645Density, inlet 885.11 kg/m3Density, outlet 904.18 kg/m3Average density 894.645 kg/m3

µ, viscosity inlet 0.55 cp µ, viscosity outlet 0.75 cp

0.65 cp k, Thermal cond. Inlet 0.11 Kcal/h-m-Ck, Thermal cond. Outlet 0.12 Kcal/h-m-CAverage k 0.115 Kcal/h-m-CRd, dirt factor 0.000407 h-m2-C/kcal

Tube side

BWG thicknessID, bwgOD, bwg 16

flow area per tubesurface area per feattriangular pitchsquare pitchPasses 2.0length of tube

Step 1 Heat Heat load from hot fld. 205178.40 Kcal/hr

Step 2 LMTD 20.713 C

Kcal/kg 0C

Kcal/kg 0C

Kcal/kg 0C

Average µ

Assume 1-2 Exchanger Number of tube passesR 4.600S 0.139

from graph Ft 0.965

true temp diff. 19.987622012 C

Step 3 Caloric temperature del t cold terminal 13

del t hot terminal 31del tc/del th 0.4194Kc 0.2000Fc 0.4000Tc 29.2tc 9

trial U: Assume U from table 8 U

Area 0Number of tubes

from table 9square pitch nearest count tubes

pitchShell IDOD tubeC', clearance between tubesDe

triangular pitch nearest count tubespitchShell IDOD tubeC', clearance between tubesDe

Corrected Ud Corrected surface of tubes, A

Hot fluid : Tube side (styrene) and

Step 4 flow area per tube a't

Total flow area at

UD

Step 5 Mass velocity Gtvelocity vt 1.30 m/s

Step 6 at tc, consider props. NreL/D

Step 7 evaluate jh from fig 24. jhfor 25% cut segmental baffles

Step 8

Step 9 tube side h.t.coeff.calculate viscosity correcti φt

Step 10 hio

Pressure drop

step 11 Reynolds number Nrefriction factor fspecific gravity s

v2/2g0.053 Kg/cm2

0.136 Kg/cm2

Total ΔPt 0.189 Kg/cm2ΔPt allowable 0.350 Kg/cm2

step 12 Clean overall coefficient

step 13 Dirt Factor Rd given 0.00041 (hr)(m2)(0C)/Kcal

0.00043 (hr)(m2)(0C)/Kcal

k(cµ/k)1/3

hi/φt

ΔPr, return loss

ΔPt, calculated

Uc

Rd calculated

FPS units Formulae

109.4 F68 F

46728 lb/hr

0.43029799

0.41028413

0.4202910628.367135158 API24.995388086 API

26.66329382 API55.141764539 lb/ft3

56.32981286 lb/ft355.735788699 lb/ft3

1.331 lb/(ft)(hr)1.815 lb/(ft)(hr)1.573 lb/(ft)(hr)

0.0739162427 Btu/h-ft-F0.0806359012 Btu/h-ft-F

0.077276072 Btu/h-ft-F

0.065 in from table 100.62 in from table 100.75 in from table 10

0.302 from table 100.1963 from table 10

2.016 ft

813069.53 Btu/hr

37.283

Btu/lb0F

Btu/lb0F

Btu/lb0F

in2

ft2

WcΔT

0F

2.04.600 refer fig 180.139 refer fig 180.965 refer fig 18

35.9777

23.4

55.80.41940.20000.4000 refer fig 17

84.56 Hot F. outlet + Fc*diff. in temp48.2 Cold F. inlet + Fc*diff. in temp

120

188.3270951159.961505065

52 refer table 9.1 in refer table 9.

10 in refer table 9.0.75 in refer table 9.0.25 in0.95 ft figure 28.

560.9375 in

10 in0.75 in

0.18750.55 ft figure 28.

163.3216 calculate from tria. Pitch

138.37

0.302 in2 where a't, n are area of tube and

0.0545 Area=Nt*a't/(144*n)

0F

0F0F

Btu/(hr)(ft2)(oF)

ft2 Q=UA Δt

ft2

Q=UA Δt

ft2

856957.724.27 ft/sec

28147.58309.68

90 from fig 24

0.1580482351 Btu/(hr)(ft2)(0F/ft)

275.30982883 Btu/(hr)(ft2)(0F)1 assumed

227.5894585 Btu/(hr)(ft2)(0F)

28147.580.000205 figure 260.894645

0.087 psi figure 270.7779622085 (4*n/s)*(v2/2g)

1.997 psi

2.775 psi return loss + ΔPt4.978 psi given in problem statement

194.92 Uc=hio*ho/(hio+ho)

0.00199 (hr)(ft2)(0F)/Btu given in problem statement

0.00210 (hr)(ft2)(0F)/Btu

lb/(hr)(ft2)

Nre=DGt/µ

ht=jh*(k/D)*(cµ/k)^(1/3)*φt

hio/φt = (hi/φt)*(ID/OD)

ΔPt = f*Gt2*L*n/(5.22*1010D*s*φt)

Btu/(hr)(ft2)(oF)

Rd = (Uc-UD)/(Uc*UD)

SI unitsCold Fluid : Chilled water

t1, Inlet temperature 7 Ct2, outlet temperature 12 Cw, mass flow rate 41036 kg/hr

c, sp. Heat at inlet 1.01 Kcal/kg 0C

c, sp. Heat at outlet 1 Kcal/kg 0C

Average c 1.005s, specific gravity, inlet 0.9996s, specific gravity, outlet 0.9991Average s 0.99935Density, inlet 999.6 kg/m3Density, outlet 999.1 kg/m3Average density 999.35 kg/m3


1.34k, Thermal cond. Inlet 0.5 Kcal/h-m-Ck, Thermal cond. Outlet 0.5 Kcal/h-m-CAverage k 0.5Rd, dirt factor 0.000205 h-m2-C/kcal

Shell side

IDBaffle Space, BPasses

Average µ

Cold fluid: shell side (Chilled water)

Step 4 Flow area calculation

Step 5 Mass vel, Gs

as

vsStep 6 at Tc, consider props. De

NreStep 7 evaluate jh from fig 28. jh

for 25% cut segmental baffles

Step 8

Step 9 shell side h.t.coeff.calculate viscosity correction factor φs

Step 10 ho

Pressure drop

step 11 Reynolds number Nrefriction factor fspecific gravity slength of tube LNo. of crosses N+1

dia of shell Ds

0.241 Kg/cm2ΔPt allowable 0.350 Kg/cm2

k(cµ/k)1/3

ho/φs

ΔPs calculated

FPS units Formulae

44.653.6

90279.2

1.0106999

1.000693

1.005696510.05662310.12746510.09203562.274415 lb/ft362.243266 lb/ft362.258841 lb/ft3

3.43643.04923.2428

0.33598290.33598290.3359829

10 in3.33 in

1

0.0578704 using triangula pitch

1560024.6 Gs=W/as

ft2

lb/(hr)(ft3)

6.9602992 ft/s0.95 ft considering triangular pitch from fig. 28

38084.97150 from fig 28

0.7167051 Btu/(hr)(ft2)(0F/ft)

1357.96761 assumed

1357.9676

38084.970.0015 figure 29

0.9993516 ft

57.60

0.833 ft

3.536 psi4.978 psi given in problem statement

Nre=DGs/µ

ho=jh*(k/D)*(cµ/k)^(1/3)*φs

ΔPs = f*Gs2*Ds*(N+1)/(5.22*1010Ds*s*φs)

SI unitsDATA INPUT Hot Fluid: Hotwater

T1, Inlet temperature 80 CT2, outlet temperature 74 CW, mass flow rate 6854.25 kg/hr

c, sp. Heat at inlet 1

c, sp. Heat at outlet 1

Average c 1s, specific gravity, inlet 0.97s, specific gravity, outlet 0.97Average s 0.97Density, inlet 970 kg/m3Density, outlet 970 kg/m3Average density 970 kg/m3


0.60 cp k, Thermal cond. Inlet 0.10 Kcal/h-m-Ck, Thermal cond. Outlet 0.10 Kcal/h-m-CAverage k 0.10 Kcal/h-m-CRd, dirt factor

Tube side

BWG thicknessID, bwgOD, bwg 16

flow area per tubesurface area per feattriangular pitchsquare pitchPasses 4.00length of tube

Step 1 Heat Heat load from cold fld. 41125.50 Kcal/hr

Step 2 LMTD 19.496 C

Kcal/kg 0C

Kcal/kg 0C

Kcal/kg 0C

Average µ

Assume 1-4 Exchanger Number of tube passesR 1.200S 0.200

from graph Ft 0.965

true temp diff. 18.813 C

Step 3 Caloric temperature del t cold terminal 19

del t hot terminal 20del tc/del th 0.9500Kc 0.2000Fc 0.4000

Caloric temperature, hot Tc 76.4Caloric temperature, cold tc 57

trial U: Assume U from table 8 U

Area 0Number of tubes

from table 9square pitch nearest count tubes

pitchShell IDOD tubeC', clearance between tubesDe

triangular pitch nearest count tubespitchShell IDOD tubeC', clearance between tubesDe

Corrected Ud Corrected surface of tubes, A

Hot fluid : Tube side (Hot water) and

Step 4 flow area per tube a't

Total flow area at

UD

Step 5 Mass velocity Gtvelocity vt 1.01 m/s

Step 6 at tc, consider props. NreL/D


Step 8

Step 9 tube side h.t.coeff.calculate viscosity correcti φt

Step 10 hiohio

Pressure drop

step 11 Reynolds number Nrefriction factor fspecific gravity s

v2/2g0.077 Kg/cm2

0.136 Kg/cm2

Total ΔPt 0.213 Kg/cm2ΔPt allowable 0.250 Kg/cm2

step 12 Clean overall coefficient

step 13 Dirt Factor 0.00010 (hr)(m2)(0C)/Kcal

0.00209 (hr)(m2)(0C)/Kcal

k(cµ/k)1/3

hi/φt

ΔPr, return loss

ΔPt, calculated

Uc

Rd given

Rd calculated

FPS units Formulae

176 F165.2 F

15079.35 lb/hr

1.00

1.00

1.0014.38 API14.38 API14.38 API60.43 lb/ft360.43 lb/ft360.43 lb/ft3

1.45 lb/(ft)(hr)1.45 lb/(ft)(hr)1.45 lb/(ft)(hr)0.07 Btu/h-ft-F0.07 Btu/h-ft-F0.07 Btu/h-ft-F

0.07 in from table 100.62 in from table 100.75 in from table 10

0.30 from table 100.20 from table 10

412 ft

162969.84 Btu/hr

35.09

Btu/lb0F

Btu/lb0F

Btu/lb0F

in2

ft2

WcΔT

0F

41.20 refer fig 180.20 refer fig 18

0.965 refer fig 18

33.86

34.20

36.000.950.200.40

169.52 Hot F. outlet + Fc*diff. in temp134.60 Cold F. inlet + Fc*diff. in temp

50

96.24940.860

40 refer table 9.1 in refer table 9.

10 in refer table 9.0.75 in refer table 9.0.250.95 ft figure 28. also can be calculated

320.9375 in

10 in0.75 in

0.18750.55 ft figure 28.

94.22 calculate from tria. Pitch

51.07

0.302 in2 where a't, n are area of tube and nu

0.021 Area=Nt*a't/(144*n)

0F

0F0F

Btu/(hr)(ft2)(oF)

ft2 Q=UA Δt

ft2

ft2

719015.363.31 ft/sec

25584.80232.26

85 from fig 24

0.193 Btu/(hr)(ft2)(0F/ft)

316.7511 assumed

261.85 450 from fig 25 w.r.t. Tc

25584.800.00021 figure 26

0.970.069 psi figure 27

1.1381443299 (4*n/s)*(v2/2g)

1.992 psi

3.130 psi return loss + ΔPt3.556 psi given in problem statement

106.60 Uc=hio*ho/(hio+ho)

0.00049 (hr)(ft2)(0F)/Btu

0.01020 (hr)(ft2)(0F)/Btu

lb/(hr)(ft2)

Nre=DGt/µ

ht=jh*(k/D)*(cµ/k)^(1/3)*φt

hio/φi = (hi/φt)*(ID/OD)

ΔPt = f*Gt2*L*n/(5.22*1010D*s*φt)

Btu/(hr)(ft2)(oF)

Rd = (Uc-UD)/(Uc*UD)

SI unitsCold Fluid : Monochlorobenzene

t1, Inlet temperature 55 Ct2, outlet temperature 60 Cw, mass flow rate 24700 kg/hr

c, sp. Heat at inlet 0.33 Kcal/kg 0C

c, sp. Heat at outlet 0.33 Kcal/kg 0C

Average c 0.33 Kcal/kg 0Cs, specific gravity, inlet 1.08s, specific gravity, outlet 1.08Average s 1.08Density, inlet 1080.00 kg/m3Density, outlet 1080.00 kg/m3Average density 1080.00 kg/m3


0.55 cp k, Thermal cond. Inlet 0.10 Kcal/h-m-Ck, Thermal cond. Outlet 0.10 Kcal/h-m-CAverage k 0.10 Kcal/h-m-CRd, dirt factor

Shell side

IDBaffle Space, BPasses

Average µ

Cold fluid: shell side (Monochlorobenzene)

Step 4 Flow area calculation

Step 5 Mass vel, Gs

as

Step 6 at Tc, consider props. DeNre


Step 8

Step 9 shell side h.t.coeff.calculate viscosity correction factor φs

Step 10 ho

Pressure drop

step 11 Reynolds number Nrefriction factor fspecific gravity slength of tube LNo. of crosses N+1

dia of shell Ds

0.018 Kg/cm2ΔPt allowable 0.100 Kg/cm2

k(cµ/k)1/3

ho/φs

ΔPs calculated

FPS units Formulae

131 F140 F

54340 lb/hr

0.33

0.33

0.33-0.48 API-0.48 API-0.48 API67.28 lb/ft367.28 lb/ft367.28 lb/ft3

1.32 lb/(ft)(hr)1.32 lb/(ft)(hr)1.32 lb/(ft)(hr)0.07 Btu/h-ft-F0.07 Btu/h-ft-F0.07 Btu/h-ft-F

10 in5.00 in

1

Btu/lb0F

Btu/lb0F

Btu/lb0F

0.0868056 using triangula pitch

625996.8 Gs=W/as

ft2

lb/(hr)(ft3)

0.95 ft considering triangular pitch from fig. 2837437.93

110 from fig 28

0.129 Btu/(hr)(ft2)(0F/ft)

179.801 assumed

179.80

37437.930.0015 figure 29

1.0812 ft

28.80

0.833 ft

0.263 psi1.422 psi given in problem statement

Nre=DGs/µ

ho=jh*(k/D)*(cµ/k)^(1/3)*φs

ΔPs = f*Gs2*Ds*(N+1)/(5.22*1010Ds*s*φs)

SI DATA INPUT Hot Fluid: Ethanol

T1, Inlet temperature 78T2, outlet temperature 40W, mass flow rate 25



Average c 2.93s, specific gravity, inlets, specific gravity, outletAverage sDensity, inlet 0Density, outlet 0Average density at 59 C 775

µ, viscosity inlet 0.60µ, viscosity outlet 0.60

0.60k, Thermal cond. Inlet 0.15k, Thermal cond. Outlet 0.15Average k 0.15Rd, dirt factor

Step 1 Heat Heat load from cold fld. 2783.50

Step 2 LMTD 19.254Assume one thermal plateHot side Number of passes 1Cold side Number of passes 1

for 1:1pass arrangementfrom graph Ft 0.965

true temp diff. 18.580

Step 3

trial U: Assume U from U 2000

Area 72.285

Average µ

Step 4 Dimension of plate width, w 0.5length of plate L 1.5hole dia port dia 120hole area Ahole 0.01131plate thickness tp 0.001Area of plate Ap 0.75

No. of plates 96.38Adjusted plates 97

Area provided Apro 72.75No. of channels per pass 48

Assume gap between plates y 3Equivalent dia De 0.006Flow area Af 0.0015

Hot fluid : Ethanol

Step 5 mass velocity Gpe 347.22velocity ue 0.45Reynolds num Nre 3472.22

Pr 11.959

Step 6 heat transfer coefficient c 0.26a 0.65b 0.4(µ/µw) 1hpe 3440.447

Fouling coefficient of ethanol hfe 10000.000Fouling coefficient of water hfw 10000.000thermal cond of plat mat. kp 21.000

Step 7 Overall heat transfer coeff U 1697.57

Area provided should greater than Area required Area required from new U A req 85.16

TRIAL 2 providing 85.16 m2 areaStep 8 No of plates Np 113.55

Adjusted plates Np adjusted 113.00Hot side Number of passes 2

Cold side Number of passes 1Hot side No. of channels per pass 28Cold side No. of channels per pass 56

Step 9 mass velocity Gpe 595.24velocity through channel upe 0.77Reynolds num Nre 5952.38

hpe 4883.93

Step 10 Overall heat transfer coeff U 1966.48Area Required A req 73.52

Excess area provided 15.84

Step 11 channel pressure drop, ΔPpJf 0.0442327596

path length Lp 3ΔPp 40443.97

velocity thru hole uh 2.851ΔPpo 8189.7

Total pressure drop ΔP 48633.66

Pressure drop ΔP = channel pressure drop + port pressure drop = ΔPp + ΔPpo

Port pressure drop, Δppo

units FPS units Formulae

C 172.4 FC 104 F

kg/s 55 lb/hr

2.93

2.93

2.93#DIV/0! API#DIV/0! API#DIV/0! API

kg/m3 0.00 lb/ft3kg/m3 0.00 lb/ft3kg/m3 48.28 lb/ft3

cp 1.45 lb/(ft)(hr)cp 1.45 lb/(ft)(hr)cp 1.45 lb/(ft)(hr)

W/m C 0.10 Btu/h-ft-FW/m C 0.10 Btu/h-ft-FW/m C 0.10 Btu/h-ft-F

kW 11030.30 Btu/hr

C 34.66

11

0.965 refer fig 18

C 33.44

W/m2 C

m2

KJ/kg 0C Btu/lb0F

KJ/kg 0C Btu/lb0F

KJ/kg 0C Btu/lb0F

WcΔT

0F

0F

Btu/(hr)(ft2)(oF)

ft2 Q=UA Δt

mm

mmm2m

m2

m2

mmm

m2

kg/m2 s mE/total gap aream/s

W/m2 C table 6.37W/m2 C table 6.37W/m C titanium

W/m2 C 1/U = 1/hpe+1/hfe + tp/kp + 1/hpw + 1/hfw

provide this area for next trial

hp*de/Kp = cRea Prb (µ/µw)

DensG )/(

Re

kg/m2 sm/s

W/m2 Cm2

mN/m2

N/m2

N/m2

ΔPp = 8Jf*(Lp/de)ρu2/2

ΔPpo = 1.3*(ρu2/2)*Np

mass velocityvelocityReynolds num

Step 5 heat transfer coefficient

1/U = 1/hpe+1/hfe + tp/kp + 1/hpw + 1/hfw

No of platesAdjusted platesHot side

Cold side

Cold side

mass velocityvelocity through channelReynolds num

Pressure drop

channel pressure drop, ΔPp

path length

velocity thru hole

Total pressure drop


SI units FPS units FormulaeCold Fluid : Cooling water

t1, Inlet temperature 32 C 89.6 Ft2, outlet temperature 40 C 104 Fw, mass flow rate 83.12 kg/s 182.86252 lb/hr

c, sp. Heat at inlet 4.19 4.19

c, sp. Heat at outlet 4.19 4.19

Average c 4.19 4.19s, specific gravity, inlet 1.00 10.00 APIs, specific gravity, outlet 1.00 10.00 APIAverage s 1.00 10.00 APIDensity, inlet 1000.00 kg/m3 62.30 lb/ft3Density, outlet 1000.00 kg/m3 62.30 lb/ft3Average density 1000.00 kg/m3 62.30 lb/ft3

µ, viscosity inlet cp 0.00 lb/(ft)(hr)µ, viscosity outlet cp 0.00 lb/(ft)(hr)

0.72 cp 1.74 lb/(ft)(hr)k, Thermal cond. Inlet W/m C 0.00 Btu/h-ft-Fk, Thermal cond. Outlet W/m C 0.00 Btu/h-ft-FAverage k 0.62 W/m C 0.42 Btu/h-ft-FRd, dirt factor

KJ/kg 0C Btu/lb0F

KJ/kg 0C Btu/lb0F

KJ/kg 0C Btu/lb0F

Average µ, 36 C

Cold fluid: Cooling water

Gpw 1154.44 kg/m2 suw 1.15 m/sNre 9620.29Pr 4.839

c 0.26a 0.65b 0.4(µ/µw) 1hpw 19685.95

Np 113.55Np adjusted 113.00Number of passes 2

Number of passes 1

No. of channels per pass 56

Gpw 989.52upw 0.99 m/sNre 8245.96hpw 17809.06

Jf 0.0401124Lp 1.5ΔPp 39275.73

uh 7.346ΔPpo 35080.3

ΔP 74356.01

SI DATA INPUT Hot Fluid: Ethanol

T1, Inlet temperature 78T2, outlet temperature 40W, mass flow rate 25



Average c 2.93s, specific gravity, inlets, specific gravity, outletAverage sDensity, inlet 0Density, outlet 0Average density at 59 C 775

µ, viscosity inlet 0.60µ, viscosity outlet 0.60

0.60k, Thermal cond. Inlet 0.15k, Thermal cond. Outlet 0.15Average k 0.15Rd, dirt factor

Step 1 Heat Heat load from cold fld. 2783.50

Step 2Assume one thermal plateHot side Number of passes 1Cold side Number of passes 1

for 1:1 pass arrangementfrom graph Ft 0.965

true temp diff. 18.580

Step 2 Calculation of bulk mean temperatureLMTD 19.254

Heat capacity ratio C* 0.211Step 3 Heat transfer effectiveness

Average µ

Є 0.826

Step 4 Assume infinite number of channelsNTU 1.974

Step 5 Dimension of plate width, w 0.5length of plate L 1.5hole dia port dia 120hole area Ahole 0.01131plate thickness tp 0.001Area of plate Ap 0.75Assume one plate No. of plates 1

Assume gap between plates y 3Equivalent dia De 0.006Flow area Af 0.0015

Hot fluid : EthanolStep 5b mass velocity Gpe 16666.67

velocity ue 21.51Reynolds num Nre 166666.67

Pr 11.959

Step 6 heat transfer coefficient c 0.26a 0.65b 0.4(µ/µw) 1hpe 42601.190

Fouling coefficient of ethanol hfe 10000.000Fouling coefficient of water hfw 10000.000thermal cond of plat mat. kp 21.000

Step 7 Overall heat transfer coeff U 3633.79

Step 8 Calculation of thermal platesmcp hot 73250.00mcp cold 347937.50

thermal plates N 2.76

Trial 2

No of channels N+1 3.76

TRIAL 2 providing 85.16 m2 areaStep 8 No of plates Np 0.00

Adjusted plates Np adjusted 113.00Hot side Number of passes 2Cold side Number of passes 1Hot side No. of channels per pass 28Cold side No. of channels per pass 56

Assume N+1 channel & calculate NTU from Є-NTU relation. Re do step 8 until matches with assumed value

Step 9 mass velocity Gpe #REF!velocity through channel upe #REF!Reynolds num Nre #REF!

hpe #REF!

Step 10 Overall heat transfer coeff U #REF!Area Required A req #REF!

Excess area provided #REF!

Step 11 channel pressure drop, ΔPpJf #REF!

path length Lp 3ΔPp #REF!

velocity thru hole uh 2.851ΔPpo 8189.7

Total pressure drop ΔP #REF!

Pressure drop ΔP = channel pressure drop + port pressure drop = ΔPp + ΔPpo


units FPS units Formulae

C 172.4 FC 104 F

kg/s 55 lb/hr

2.93

2.93

2.93#DIV/0! API#DIV/0! API#DIV/0! API

kg/m3 0.00 lb/ft3kg/m3 0.00 lb/ft3kg/m3 48.28 lb/ft3

cp 1.45 lb/(ft)(hr)cp 1.45 lb/(ft)(hr)cp 1.45 lb/(ft)(hr)

W/m C 0.10 Btu/h-ft-FW/m C 0.10 Btu/h-ft-FW/m C 0.10 Btu/h-ft-F

kW 11030.30 Btu/hr

11

0.965 refer fig 18

C 33.44

C 34.66

KJ/kg 0C Btu/lb0F

KJ/kg 0C Btu/lb0F

KJ/kg 0C Btu/lb0F

WcΔT

0F

0F

for Ch<Cc, Ch= Cmin

mm

mmm2m

m2

mmm

m2

kg/m2 s mE/total gap aream/s

W/m2 C table 6.37W/m2 C table 6.37W/m C titanium

W/m2 C 1/U = 1/hpe+1/hfe + tp/kp + 1/hpw + 1/hfw

J/s CJ/s C

hp*de/Kp = cRea Prb (µ/µw)

mp

p

tUA

mcNTUN

min)(

)(

)(

)(

)(

,,min

,,

,,min

,,

icih

icocc

icih

ohihh

ttC

ttC

ttC

ttC

-NTU relation. Re do step 8 until matches with assumed value

mp

p

tUA

mcNTUN

min)(

kg/m2 sm/s

W/m2 Cm2

mN/m2

N/m2

N/m2

ΔPp = 8Jf*(Lp/de)ρu2/2

ΔPpo = 1.3*(ρu2/2)*Np

Fundamentals of Heat Exchanger Design, 0471321710

mass velocityvelocityReynolds num

Step 5 heat transfer coefficient

1/U = 1/hpe+1/hfe + tp/kp + 1/hpw + 1/hfw

)(

)(

)(

)(

,,min

,,

,,min

,,

icih

icocc

icih

ohihh

ttC

ttC

ttC

ttC

No of platesAdjusted platesHot sideCold side

Cold side

mass velocityvelocity through channelReynolds num

Pressure drop

channel pressure drop, ΔPp

path length

velocity thru hole

Total pressure drop


SI units FPS units FormulaeCold Fluid : Cooling water

t1, Inlet temperature 32 C 89.6 Ft2, outlet temperature 40 C 104 Fw, mass flow rate 83.12 kg/s 182.86252 lb/hr

c, sp. Heat at inlet 4.19 4.19

c, sp. Heat at outlet 4.19 4.19

Average c 4.19 4.19s, specific gravity, inlet 1.00 10.00 APIs, specific gravity, outlet 1.00 10.00 APIAverage s 1.00 10.00 APIDensity, inlet 1000.00 kg/m3 62.30 lb/ft3Density, outlet 1000.00 kg/m3 62.30 lb/ft3Average density 1000.00 kg/m3 62.30 lb/ft3

µ, viscosity inlet cp 0.00 lb/(ft)(hr)µ, viscosity outlet cp 0.00 lb/(ft)(hr)

0.72 cp 1.74 lb/(ft)(hr)k, Thermal cond. Inlet W/m C 0.00 Btu/h-ft-Fk, Thermal cond. Outlet W/m C 0.00 Btu/h-ft-FAverage k 0.62 W/m C 0.42 Btu/h-ft-FRd, dirt factor

KJ/kg 0C Btu/lb0F

KJ/kg 0C Btu/lb0F

KJ/kg 0C Btu/lb0F

Average µ, 36 C

Cold fluid: Cooling waterGpw 55412.88 kg/m2 suw 55.41 m/sNre 461774.04

c 0.26a 0.65b 0.4(µ/µw) 1hpw 243760.39

Pr 4.839

Np 0.00Np adjusted 113.00Number of passes 2Number of passes 1

No. of channels per pass 56

Gpw #REF!upw #REF! m/sNre #REF!hpw #REF!

Jf #REF!Lp 1.5ΔPp #REF!

uh 7.346ΔPpo 35080.3

ΔP #REF!

Date post:	18-Nov-2014
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PHE Design

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