Page 362 - Analysis, Synthesis and Design of Chemical Processes, Third Edition

For the moment, ignore this and return to the heuristic analysis:
ΔT = [(105 – 40) – (38 – 30)]/ln[(105 – 40)/(38 – 30)] = 27.2°C
lm
Q = 1085 MJ/h = 301 kW (from Table 1.7)

A = Q/UΔT F = (301,000)/(850)/(27.2)/(0.90) = 14.46 m 2
lm
From Rule 9, Table 11.11, this heat exchanger should be a double-pipe or multiple-pipe design.
Comparing our analysis with the information in Table 1.7 we get
Heuristic: Double-pipe design, Area = 14.5 m 2

Table 1.7: Multiple-pipe design, Area = 12 m 2

Again, the heuristic analysis is close to the actual design. The fact that the minimum approach temperature
of 10°C has been violated should not cause too much concern, because the actual minimum approach is
only 8°C and the heat exchanger is quite small, suggesting that a little extra area (due to a smaller overall
temperature driving force) is not very costly.
c. P-101
From Table 11.9, we use the following heuristics:
3
Rule 1: Power(kW) = (1.67)[Flow(m /min)]ΔP(bar)/ε
Rules 4–7: Type of pump based on head
From Figure 1.5 and Tables 1.5 and 1.7, we have
Flowrate (Stream 2) = 13,300 kg/h

Density of fluid = 870 kg/m 3
ΔP = 25.8 – 1.2 = 24.6 bar = 288 m of liquid (head = ΔP/ρg)
3
Volumetric flowrate = (13,300)/(60)/(870) = 0.255 m /min
Fluid pumping power = (1.67)(0.255)(24.6) = 10.5 kW

From Rules 4–7, pump choices are multistage centrifugal, rotary, and reciprocating. Choose reciprocating
to be consistent with Table 1.7. Typical ε = 0.75.
Power (shaft power) = 10.5/0.75 = 14.0 kW → compares with 14.2 kW from Table 1.7.
d. C-101
From Table 11.10, we use the following heuristics:
a
Rule 2: W rev adiab = mz RT [(P /P ) – 1]/a
2
1
1
1
From Table 1.7, we have flow = 6770 kg/h, T = 38°C = 311 K, mw = 8.45, P = 23.9 bar, P = 25.5
1 1 2
k = 1.41 (assume) and a = 0.2908
m = (6770)/(3600)/(8.45) = 0.223 kmol/s
W = (223)(1.0)(8.314)(311){ (25.5/23.9) 0.2908 – 1)/0.2908 = 37.7 kW using a compressor
rev adiab
efficiency of 75%
W = (37.7)/(0.75) = 50.3 kW → This checks with the shaft power requirement given in Table
actual
1.7.
e. T-101
From Table 11.13, we use the following heuristics:
Rule 5: Optimum reflux in the range of 1.2–1.5 R min
Rule 6: Optimum number of stages approximately 2N min
Rule 7: N = ln{ [x/(1 – x)] /[x/(1 – x)] }/ln α
min oυhd bot
Rule 8: R min = {F/D}/(α – 1)
Rule 9: Use a safety factor of 10% on number of trays.

Rule 14: L max = 53 m and L/D < 30