Question 6 of 9: Moist Air Heated at Constant Volume in a Sealed Rigid Vessel

Question 6: Moist Air Heated at Constant Volume in a Sealed Rigid Vessel (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. Sealed rigid vessel (fixed volume, fixed mass, no moisture added or removed). Initial: $m_{total}=1$ kg moist air, $P_1=101.325$ kPa, $T_{db,1}=20\ ^\circ$C, $\phi_1=60\%$. Final: $T_{db,2}=50\ ^\circ$C.

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

Approach

Because both the dry-air mass and the water-vapor mass are individually fixed in a sealed rigid vessel, the humidity ratio $W$ never changes (no condensation occurs on heating, since heating only drives the mixture further from saturation). With volume and both component masses fixed, each ideal-gas partial pressure scales directly with absolute temperature, so the total pressure simply scales the same way. The heat transfer is the change in the mixture's total internal energy at constant $W$.

1. Initial humidity ratio and split of the 1 kg total mass. At $T_{db,1}=20\ ^\circ$C, $\phi_1=60\%$, $P_1=101.325$ kPa: $$W_1=0.008773\ \text{kg water/kg dry air}\quad(\text{partial vapor pressure }p_{v,1}=1.409\ \text{kPa}).$$ Dry-air and vapor masses in the 1 kg total: $m_a=1/(1+W_1)=0.99130$ kg, $m_v=1-m_a=0.008697$ kg.
2. Final pressure. With $V$, $m_a$, and $m_v$ all fixed, both the dry-air and vapor partial pressures scale with absolute temperature ($p_{a,2}=p_{a,1}T_2/T_1$, $p_{v,2}=p_{v,1}T_2/T_1$), so the total pressure scales the same way: $$P_2=P_1\frac{T_2}{T_1}=101.325\times\frac{323.15}{293.15}=\boxed{111.69\ \text{kPa}}.$$
3. Final relative humidity. The humidity ratio is unchanged ($W_2=W_1=0.008773$, since no water was added, removed, or condensed); evaluating the saturation state at $T_{db,2}=50\ ^\circ$C and $P_2=111.69$ kPa with that fixed $W$: $$\phi_2=\boxed{12.51\%}.$$ This drop from 60% to 12.5% makes physical sense: the same absolute amount of water vapor is now spread through the same volume at a much higher temperature, so it is far below its (much higher) saturation capacity at 50°C.
4. Heat transfer. Using the mixture's total internal energy per unit mass of dry air (accounts for both the dry-air and water-vapor internal-energy changes at their own partial pressures): $$Q=m_a\left(u_2-u_1\right)=0.99130\times(u_2-u_1)=\boxed{21.73\ \text{kJ}}.$$ (Positive: heat must be added, consistent with raising the temperature of a fixed-volume, fixed-mass gas mixture.)