Question 3 of 7: Two-Stage Water Recovery with Interstage Compression

Question 3: Two-Stage Water Recovery with Interstage Compression (Part B — 25 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. Feed 100 mol/s (75 N₂ + 25 H₂O) at 300 °C, 2 bar. First contactor at 2 bar; gas leaves saturated at 40 °C. Interstage compressor to 6 bar, 200 °C. Second contactor at 6 bar; gas leaves saturated at 25 °C. Liquid water fed to each contactor enters at 20 °C. $M_{\text{H}_2\text{O}}=0.01802$ kg/mol. N₂ enthalpies (kJ/mol, ref 25 °C): $\hat H(300)=8.12$, $\hat H(200)=5.13$, $\hat H(40)=0.438$, $\hat H(25)=0$.

Find. (a) $\hat H$ of H₂O(vap) and H₂O(liq) at the tabulated temperatures; (b)–(d) contactor water feeds, gas/liquid flows and compositions, and the compressor power.

Approach. The nitrogen is inert and passes through unabsorbed, so at each saturated outlet $y_{\text{H}_2\text{O}}=P^{sat}(T)/P$ fixes the water carried by the gas. An adiabatic energy balance on each contactor (with the cold liquid feed as the only free stream) sets the required water rate; the compressor duty is the enthalpy rise of the gas across it.

(a) Enthalpy table. Steam-table values converted to kJ/mol ($\times0.01802$); the vapour figures at 200–300 °C are low-pressure superheated steam:

1. (b) Water carried out of contactor 1. Saturation at 40 °C, 2 bar gives $y_1=P^{sat}(40)/P=0.0738/2=0.0369$. With 75 mol/s N₂ inert, $$\dot n_{G,1}=\frac{75}{1-y_1}=77.9\ \text{mol/s},\qquad \dot n_{w,\text{gas}}=77.9(0.0369)=2.87\ \text{mol/s}.$$
2. Energy balance on contactor 1 (adiabatic ⇒ $\dot H_{in}=\dot H_{out}$). The only unknown is the liquid feed $\dot n_{\text{H}_2\text{O},1}$ (20 °C in, 40 °C out): $$75(8.12)+25(55.41)+\dot n_1(1.511)=75(0.438)+2.87(46.38)+\dot n_2(3.018),$$ with water balance $\dot n_2=25+\dot n_1-2.87$. Solving, $$\dot n_{\text{H}_2\text{O},1}=\boxed{1170\ \text{mol/s}\;(=21.1\ \text{kg/s})},\qquad \dot n_2=1192\ \text{mol/s liquid out}.$$ Leaving gas: 77.9 mol/s at $y_{\text{H}_2\text{O}}=0.0369$ (75 N₂ + 2.87 H₂O).
3. (c) Compressor power. The compressor raises the 40 °C gas to 200 °C; adiabatic shaft work equals the enthalpy rise (N₂ + its 2.87 mol/s water): $$\dot W=75(5.13-0.438)+2.87(51.88-46.38)=351.9+15.8=\boxed{368\ \text{kW}}.$$
4. (d) Second contactor. Saturation at 25 °C, 6 bar gives $y_2=P^{sat}(25)/6=0.03169/6=0.00528$, so $$\dot n_{G,2}=\frac{75}{1-y_2}=75.4\ \text{mol/s},\qquad \dot n_{w,\text{gas,2}}=0.398\ \text{mol/s}.$$ Energy balance (gas in 200 °C, out 25 °C; liquid in 20 °C, out 25 °C) with $\dot n_4=2.87+\dot n_3-0.398$: $$75(5.13)+2.87(51.88)+\dot n_3(1.511)=75(0)+0.398(45.88)+\dot n_4(1.889)\;\Rightarrow\;\dot n_{\text{H}_2\text{O},3}=\boxed{1357\ \text{mol/s}},\ \dot n_4=1359\ \text{mol/s}.$$ The vented gas carries only 0.398 mol/s water — 98.4 % of the original 25 mol/s has been recovered.