Adiabatic Compression of Moist Air: 04-BS-10, December 2018, Q5

Question 5: Adiabatic Compression of Moist Air (15 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. Inlet: $T_1=25\ ^\circ$C, $P_1=105$ kPa, $\phi_1=85\%$, $\dot V_1=0.3$ m³/s. Exit: $T_2=97\ ^\circ$C, $P_2=200$ kPa. Insulated, steady-flow, negligible KE/PE. Humidity ratio unchanged through the compressor ($\omega_2=\omega_1$, no moisture added or removed).

Find. (a) $\phi_2$; (b) $\dot W_{in}$ [kW].

Approach

Fix the humidity ratio and dry-air specific volume at the inlet state, convert the given volumetric flow rate to a dry-air mass flow rate, then apply the stated hint ($\omega_2=\omega_1$) at the exit state to read off the exit relative humidity. The compressor power follows from an energy balance on the moist-air stream (dry air plus the water vapor it carries), with $Q=0$.

Inlet humidity ratio and dry-air mass flow. $$\omega_1=0.622\,\frac{\phi_1 P_{sat}(T_1)}{P_1-\phi_1 P_{sat}(T_1)}=\boxed{0.016453\ \text{kg}_{v}/\text{kg}_{da}}$$ using $P_{sat}(25\ ^\circ\text{C})=3.169$ kPa. The dry-air specific volume from the inlet partial pressure of dry air, $P_{a,1}=P_1-\phi_1 P_{sat}=102.31$ kPa, gives $v_{a,1}=0.8363$ m³/kg, so $$\dot m_a=\frac{\dot V_1}{v_{a,1}}=\frac{0.3}{0.8363}=0.3587\ \text{kg/s}.$$
Exit relative humidity (part a). With $\omega_2=\omega_1=0.016453$ fixed and $P_2=200$ kPa, the exit vapor partial pressure is $$P_{v,2}=\frac{\omega_2 P_2}{0.622+\omega_2}=\frac{0.016453\times200}{0.63845}=5.155\ \text{kPa}.$$ At $T_2=97\ ^\circ$C, $P_{sat}(97\ ^\circ\text{C})=91.87$ kPa (interpolated), so $$\phi_2=\frac{P_{v,2}}{P_{sat}(T_2)}=\boxed{0.0561\ (5.61\%)}.$$
Compressor power (part b). Energy balance per kg of dry air (moist-air specific enthalpy $H$ carries both the dry-air and the $\omega$-weighted water-vapor contributions together), $Q=0$: $$\dot W_{in}=\dot m_a\,(H_2-H_1)$$ $$H_1=67.03\ \text{kJ/kg}_{da},\qquad H_2=141.72\ \text{kJ/kg}_{da}$$ $$\dot W_{in}=0.3587\times(141.72-67.03)=0.3587\times74.69=\boxed{26.79\ \text{kW}}.$$

Quantity	Result
(a) $\phi_2$	0.0561 (5.61%)
(b) $\dot W_{in}$	26.79 kW

Concept & Theory

Topic. Steady-flow adiabatic compression of moist air, using conservation of the humidity ratio (no condensation, no added moisture) to close the exit state.

Key relations. $\omega=0.622\,P_v/(P-P_v)$ links vapor partial pressure and humidity ratio at any state; $\omega$ is conserved whenever no liquid water crosses the control volume boundary. Energy balance on the combined dry-air + vapor stream: $\dot W=\dot m_a[\Delta h_a+\omega\Delta h_v]$.

Why this works. A compressor has no mechanism to add or remove water — the mass of vapor per kg of dry air is fixed by mass conservation alone, independent of the energy balance that determines the temperature rise. That decoupling is exactly what the given hint states, and it is what lets part (a) be solved before part (b) needs any energy balance at all.

Common pitfall. Assuming the exit state is still at 85% relative humidity or re-computing $\omega$ from some assumed exit humidity — $\phi$ is not conserved through a temperature change (only $\omega$ is); $\phi$ must instead be recovered afterward from the (fixed) $\omega$ and the new saturation pressure at the higher exit temperature.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 14 (Gas–Vapor Mixtures and Air-Conditioning).

Question 5 of 9: Adiabatic Compression of Moist Air

Question 5: Adiabatic Compression of Moist Air (15 marks)

Approach