Question 4 of 9: Isothermal Compression of Wet Steam in a Piston-Cylinder

Question 4: Isothermal Compression of Wet Steam in a Piston-Cylinder (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. $m=2$ kg H₂O. $V_1=0.3$ m$^3$ ($v_1=0.15$ m$^3$/kg). $V_2=0.1\,V_1=0.03$ m$^3$ ($v_2=0.015$ m$^3$/kg). $T=150\ ^\circ$C constant throughout.

Find. Magnitude and direction of $W$ and $Q$.

Approach

At 150°C, $v_g\approx0.3928$ m$^3$/kg and $v_f\approx0.00109$ m$^3$/kg, so both $v_1=0.15$ and $v_2=0.015$ m$^3$/kg fall strictly inside the two-phase dome — the entire process therefore stays at the constant saturation pressure $P_{sat}(150\ ^\circ\text{C})$, even though the problem never states "constant pressure" directly. Recognizing this collapses the boundary-work integral to $W=\int P\,dV=P(V_2-V_1)$, a simple algebraic evaluation rather than a path-dependent integral.

1. Confirm the two-phase region and get the (constant) pressure. $$P=P_{sat}(150\ ^\circ\text{C})=\boxed{476.16\ \text{kPa}}.$$ Qualities: $x_1=(v_1-v_f)/(v_g-v_f)=0.3805$, $x_2=(v_2-v_f)/(v_g-v_f)=0.03554$ (both between 0 and 1, confirming the two-phase assumption at both endpoints).
2. Boundary work. Since $P$ is constant across the whole process: $$W=m\,P\,(v_2-v_1)=2\times476.16\times(0.015-0.15)=\boxed{-128.56\ \text{kJ}}.$$ Negative sign: work is done on the system (compression), magnitude 128.56 kJ.
3. Internal energy change. $u_1=1365.01$ kJ/kg ($x_1=0.3805$), $u_2=700.16$ kJ/kg ($x_2=0.03554$): $$\Delta U=m(u_2-u_1)=2\times(700.16-1365.01)=-1329.71\ \text{kJ}.$$
4. Heat transfer (first law). $$Q=\Delta U+W=-1329.71+(-128.56)=\boxed{-1458.27\ \text{kJ}}.$$ Negative sign: heat leaves the system, magnitude 1458.27 kJ — consistent with the problem statement that heat transfer occurs from the H₂O.