Question 6 of 9: Moist Air in a Closed Rigid Vessel — Heating at Constant Humidity Ratio & Volume

Question 6: Moist Air in a Closed Rigid Vessel — Heating at Constant Humidity Ratio & Volume (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. $m_{total}=1$ kg (moist air), $P_1=101.325$ kPa, $T_{db,1}=20\ ^\circ$C, $\phi_1=60\%$. Closed rigid vessel: volume and dry-air mass are both fixed, so the humidity ratio $W$ is unchanged (no water enters or leaves) and the specific volume per kg dry air is unchanged (rigid vessel, fixed dry-air mass). Final state: $T_{db,2}=50\ ^\circ$C.

Find. $Q$ [kJ]; $P_2$ [kPa]; $\phi_2$.

Approach

Fix state 1 fully (humidity ratio $W_1$, specific volume per kg dry air $v_{da,1}$, enthalpy per kg dry air $h_{da,1}$) from the given $T,P,\phi$. Since the vessel is rigid and closed, $W_2=W_1$ and $v_{da,2}=v_{da,1}$; combined with the given $T_2$, this pins down state 2 completely and $P_2$ falls out as the value that makes the humid-air equation of state consistent with the unchanged specific volume. Heat transfer follows from the first law at constant volume, $Q=\Delta U$ (no boundary work in a rigid vessel).

1. State 1 (fully defined). From $T_1=20\ ^\circ$C, $P_1=101.325$ kPa, $\phi_1=60\%$: $$W_1=0.008773\ \text{kg}_{w}/\text{kg}_{da},\quad v_{da,1}=0.8418\ \text{m}^3/\text{kg}_{da},\quad h_{da,1}=42.38\ \text{kJ/kg}_{da}.$$ (Every specific property in this solution is carried per kilogram of dry air — the same basis the supplied psychrometric chart uses, which reads $v\approx0.84\ \text{m}^3/\text{kg}_{da}$ at 20°C and 60% RH.)
2. Constraints carried to state 2. Rigid, closed vessel $\Rightarrow$ $W_2=W_1=0.008773$ and $v_{da,2}=v_{da,1}=0.8418\ \text{m}^3/\text{kg}_{da}$ (both dry-air mass and total volume are fixed). With $T_2=50\ ^\circ$C also given, these three constraints (T, W, v) over-determine the usual two-property humid-air state and instead solve directly for the final total pressure $P_2$ that makes them consistent: $$P_2=\boxed{111.72\ \text{kPa}}.$$
3. Final relative humidity. At $T_2=50\ ^\circ$C, $W_2=0.008773$, $P_2=111.72$ kPa: $$\phi_2=\boxed{12.51\%}.$$ (Physically sensible: heating at fixed absolute moisture content always drives $\phi$ down, since the saturation vapor pressure rises steeply with $T$ while the actual vapor partial pressure changes only through the small total-pressure rise.)
4. Heat transfer (first law, rigid vessel). No boundary work since $V$ is fixed, so $Q=\Delta U$. Using $u_{da}=h_{da}-Pv_{da}$ (general definition of enthalpy, applied per kg dry air with the total mixture pressure) and dry-air mass $m_{da}=m_{total}/(1+W_1)=0.9913$ kg: $$u_{da,1}=h_{da,1}-P_1v_{da,1}=42.38-101.325\times0.8418=-42.92\ \text{kJ/kg}_{da}.$$ $$u_{da,2}=h_{da,2}-P_2v_{da,2}=73.05-111.72\times0.8418=-21.00\ \text{kJ/kg}_{da}.$$ $$Q=m_{da}(u_{da,2}-u_{da,1})=0.9913\times(-21.00-(-42.92))=\boxed{21.73\ \text{kJ}}.$$ (Both $h$ and $v$ must be on the same basis in $u=h-Pv$; pairing a per-kg-dry-air enthalpy with a per-kg-moist-air specific volume, $0.8345\ \text{m}^3/\text{kg}_{ha}$, shifts $Q$ by about 0.4%.)