Multistage Compression for Air Liquefaction: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2014, Q5

Question 5: Multistage Compression for Air Liquefaction (Part C, 30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Air (ideal gas, $\gamma = 1.4$, $C_p = \tfrac72 R = 29.1$ J/mol·K) from 1.0 bar, 25 °C to 180 bar. Each stage: inlet 25 °C, actual outlet $\le 200$ °C, isentropic efficiency $\eta = 0.80$; intercooling to 25 °C between stages.

Find. the number of stages and the total compression work.

Figure 4 — Four compressor stages with intercooling back to 25 °C; equal pressure ratio per stage keeps each actual outlet below the 200 °C limit.

Approach. The 200 °C limit caps the per-stage pressure ratio; the smallest integer number of equal-ratio stages that reaches 180 fixes the design, and each stage’s actual work is the isentropic work divided by $\eta$.

Maximum pressure ratio per stage. The actual outlet is $T_2 = T_1 + (T_{2s}-T_1)/\eta$. Setting $T_2 = 473.15$ K with $T_1 = 298.15$ K gives the largest allowable isentropic outlet $T_{2s} = T_1 + \eta(T_2-T_1) = 438.2$ K, hence $$r_{\max} = \left(\frac{T_{2s}}{T_1}\right)^{\gamma/(\gamma-1)} = (1.470)^{3.5} = 3.85.$$
Number of stages. Equal ratios give $r = (180)^{1/N}$. Three stages need $r=5.65>3.85$ (outlet 263 °C — too hot); four stages give $r = 180^{1/4}=3.66 \le 3.85$. Therefore $$\boxed{N = 4\ \text{stages}}\quad(r = 3.66\ \text{each}).$$
Actual outlet check. $T_{2s} = 298.15(3.66)^{0.2857}=432.0$ K, so $T_2 = 298.15 + (432.0-298.15)/0.80 = 465.5$ K $=192\,{}^{\circ}$C $<200\,{}^{\circ}$C. ✓
Work per stage and total. Each identical stage (inlet 25 °C) does isentropic work $C_p(T_{2s}-T_1)=29.1(133.9)=3896$ J/mol, so the actual work is $3896/0.80 = 4870$ J/mol. For four stages, $$W_{\text{total}} = 4(4870) = \boxed{1.95\times10^{4}\ \text{J/mol} = 19.5\ \text{kJ/mol}}\;(\approx 673\ \text{kJ/kg air}).$$

Quantity	Result
Pressure ratio per stage	3.66 (max allowed 3.85)
Number of stages	4
Actual stage outlet temperature	465.5 K (192 °C)
Total compression work	19.5 kJ/mol (≈ 673 kJ/kg)

Concept & Theory

Topic. Staged adiabatic compression with intercooling; isentropic efficiency; ideal-gas compressor work.

Key relations. Isentropic path $T_{2s}/T_1 = (P_2/P_1)^{(\gamma-1)/\gamma}$; efficiency $\eta = (T_{2s}-T_1)/(T_2-T_1)$; adiabatic work $W = C_p(T_2-T_1)=C_p(T_{2s}-T_1)/\eta$.

Why this works. Intercooling returns every stage to the same inlet temperature, so all stages are identical and the temperature-limit constraint becomes a single cap on the pressure ratio. Equal ratios minimise total work for a fixed overall ratio and stage count.

Common pitfall. Applying the 200 °C limit to the isentropic outlet rather than the actual (larger) outlet — that would wrongly allow a bigger ratio and too few stages. Also, dividing by $\eta$ once for the whole train instead of per stage.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §7.1–7.3 (compression processes, isentropic efficiency).

Question 5 of 7: Multistage Compression for Air Liquefaction

Question 5: Multistage Compression for Air Liquefaction (Part C, 30 marks)