Question 3.11: Estimate the power required to compress 5000 kmol/h of HCl a...

For HCl, P_{c} = 82 bar, T_{c} = 324.6 K
C^{°}_{p} = 30.30 – 0.72 × 10^{-2} T + 12.5 × 10^{-6} T^{2} – 3.9 × 10^{-9} T^{3} kJ/kmol K

Estimate T_{2} from equations 3.35  and 3.36a

T_{2}=T_{1} \left(\frac{p_{2} }{p_{1} } \right) ^{m}      (3.35)

m=\frac{(\gamma-1)}{\gamma E_{p}}      (3.36a)
For diatomic gases γ\backsimeq 1.4.
Note: γ could be estimated from the relationship γ = \frac{C_{p} }{C_{v} } = \frac{C_{p} }{C_{p}-R }

At the inlet conditions, the flow rate in 10^{3}/s
= \frac{5000}{3600}\times 22.4\times \frac{288}{273}\times \frac{1}{5}=6.56

From Figure 3.6  E_{p} = 0.73
From equations 3.36a and 3.35

m=\frac{(\gamma-1)}{\gamma E_{p}}      (3.36a)

T_{2}=T_{1} \left(\frac{p_{2} }{p_{1} } \right) ^{m}      (3.35)

m= \frac{1.4-1}{1.4\times 0.73} =0.39

T_{2}=288\left(\frac{15}{5} \right) ^{0.39} =442 K
T_{ r \left(mean\right) } =\frac{442+228}{2\times 324.6}=1.03
P_{ r \left(mean\right) } =\frac{5+15}{2\times 82}=0.12

AtT_{\left(mean\right) }C^{°}_{p} = 29.14 kJ/kmol K
Correction for pressure from Figure 3.2, 2 kJ/kmol K
C_{p} =29.14 + 2 \simeq 31 kJ/kmol K

From Figures 3.8, 3.9 and 3.10 at mean conditions:
X = 0.18, Y = 1.04, Z = 0.97
Z at inlet conditions = 0.98
From equations 3.36  and 3.38

m=\frac{ZR}{C_{p} }\left(\frac{1}{E_{p} } +X\right)       (3.36)

n\frac{1}{Y-m\left(1+X\right) }      (3.38)

m=\frac{0.97×8.314}{31}\left(\frac{1}{0.73 } +0.18\right) = \underline{\underline{0.40} }
n=\frac{1}{1.04 – 0.4\left(1+0.18\right) }=\underline{\underline{1.76} }
From equation 3.31

W polytropic = 0.98 × 288 × 8.314 × \frac{1.76}{1.76 – 1} \left(\left(\frac{15}{5}\right) ^{\left(1.76-1\right)/1.76 } -1\right)
=\underline{\underline{3299 kJ/kmol} }
Actual work required =\frac{polytropic  work}{E_{p} }
=\frac{3299}{0.73} = \underline{\underline{4520 kJ/kmol} }
Power = \frac{4520}{3600} \times 5000=6275 KW
Say,\underline{\underline{6.3 MW} }
T_{2} =288\left(\frac{15}{5} \right) ^{0.4} =\underline{\underline{447 K} }