Question 1 of 8: Cooling Saturated Steam and Compressor Power

Question 1: Cooling Saturated Steam and Compressor Power (equal value)

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.

Part (a) — heat removed from the rigid vessel

Given. A rigid, closed vessel of volume $V=0.0011\ \text{m}^3$ contains saturated (dry) steam at $P_1=95\ \text{kPa}$. It is cooled at constant volume to $T_2=25\ ^\circ\text{C}$.

Find. The heat $Q$ that must be removed so the contents reach 25 °C.

Approach. The vessel is rigid, so the boundary work is zero and the closed-system first law reduces to $Q=\Delta U$; the constant specific volume fixes the quality (hence $u_2$) at 25 °C.

1. Mass in the vessel. At 95 kPa the saturated-vapour specific volume (interpolated between 90 and 100 kPa) is $v_g=1.7817\ \text{m}^3/\text{kg}$, so$$m=\frac{V}{v_1}=\frac{0.0011}{1.7817}=6.17\times10^{-4}\ \text{kg}.$$
2. Final state at constant volume. Cooling in a rigid vessel keeps $v_2=v_1$. At 25 °C, $v_f=0.001003$ and $v_g=43.34\ \text{m}^3/\text{kg}$, so the mixture is wet with quality$$x_2=\frac{v_1-v_f}{v_g-v_f}=\frac{1.7817-0.001003}{43.339}=\boxed{0.0411}.$$
3. Internal energies. $u_1=u_g(95\ \text{kPa})=2504.4$ kJ/kg; at 25 °C ($u_f=104.9$, $u_{fg}=2304.9$):$$u_2=u_f+x_2u_{fg}=104.9+0.0411(2304.9)=199.6\ \text{kJ/kg}.$$
4. First law at constant volume. With $W=0$,$$Q=\Delta U=m(u_2-u_1)=6.17\times10^{-4}(199.6-2504.4)=\boxed{-1.42\ \text{kJ}}.$$ The negative sign confirms heat is removed: about 1.42 kJ must be extracted.

Part (b) — power to drive the compressor

Given. Air (ideal gas, $R=0.287\ \text{kJ/kg}\cdot\text{K}$, $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$) enters at $P_1=95\ \text{kPa}$, $T_1=25\ ^\circ\text{C}$ with negligible velocity and volume flow $\dot V_1=300\ \text{m}^3/\text{min}=5\ \text{m}^3/\text{s}$; it leaves at $P_2=200\ \text{kPa}$, $T_2=120\ ^\circ\text{C}$ through area $A_2=0.028\ \text{m}^2$.

Find. The compressor power input $\dot W$ (adiabatic assumed).

Approach. Apply the steady-flow energy equation between inlet and outlet, keeping the kinetic-energy term because the discharge velocity is appreciable; the inlet velocity is negligible.

1. Mass flow from inlet density. $\rho_1=P_1/RT_1=95/(0.287\cdot298.15)=1.110\ \text{kg/m}^3$, so$$\dot m=\rho_1\dot V_1=1.110(5.0)=\boxed{5.55\ \text{kg/s}}.$$
2. Discharge velocity. $\rho_2=P_2/RT_2=200/(0.287\cdot393.15)=1.773\ \text{kg/m}^3$; from continuity$$V_2=\frac{\dot m}{\rho_2A_2}=\frac{5.55}{1.773(0.028)}=111.8\ \text{m/s}.$$
3. Energy balance (adiabatic). With $V_1\approx0$,$$\dot W=\dot m\!\left[c_p(T_2-T_1)+\tfrac12V_2^2\right]=5.55\!\left[95.5+\frac{111.8^2}{2000}\right]=5.55(95.5+6.3)=\boxed{565\ \text{kW}}.$$