phe design
TRANSCRIPT
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
µ, viscosity inlet 1.42 cp µ, viscosity outlet 1.26 cp
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
µ, viscosity inlet 0.60 cp µ, viscosity outlet 0.60 cp
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 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 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
µ, viscosity inlet 0.55 cp µ, viscosity outlet 0.55 cp
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 7 evaluate jh from fig 28. jhfor 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
0.018 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
c, sp. Heat at inlet 2.93
c, sp. Heat at outlet 2.93
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
Port pressure drop, Δppo
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
c, sp. Heat at inlet 2.93
c, sp. Heat at outlet 2.93
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
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
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
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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
Port pressure drop, Δppo
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!