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parameter paremetr value from to multiplying factorfactor
length 1 m ft 0.305pressure 1 kg/cm2 psi 14.223Temperature 1 C F 33.800
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
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
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
0.07 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
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