Question 1 of 8: Sealed Freon-12 Tube and Polytropic Nitrogen Compression

Question 1: Sealed Freon-12 Tube and Polytropic Nitrogen Compression (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)

Given. Sealed (rigid, closed) glass tube of Freon-12 vapour; State 1 at $30\ ^\circ\text{C}$; cooled at constant total volume until dew just forms at State 2, $10\ ^\circ\text{C}$.

Find. The pressure at State 1.

Approach. The tube is sealed and rigid, so the specific volume is constant throughout the cooling. Condensation first appears when the state reaches the saturated-vapour line; that fixes $v$, and reading the 30 °C superheated table at that $v$ gives $P_1$.

1. Fix the specific volume. The first liquid at State 2 means State 2 is saturated vapour at $10\ ^\circ\text{C}$: $v_1=v_2=v_g(10\ ^\circ\text{C})=0.040914\ \text{m}^3/\text{kg}$.
2. Locate State 1 on the superheated table. At $30\ ^\circ\text{C}$ the specific volume $0.040914$ falls between $0.40\ \text{MPa}$ ($v=0.047971$) and $0.50\ \text{MPa}$ ($v=0.037464$).
3. Interpolate for pressure. $P_1=0.40+\dfrac{0.047971-0.040914}{0.047971-0.037464}(0.10)=\boxed{0.467\ \text{MPa}}$ (≈ 467 kPa).

Because the tube is sealed, the process is a vertical line on the $T$–$v$ plane; the pressure at State 1 could never be measured directly, but the phase-change observation at 10 °C pins the specific volume and lets the superheated table recover it.

Part (b)

Given. Nitrogen ($R=0.2968\ \text{kJ/kg}\cdot\text{K}$, $c_v=0.743$, $\gamma=1.4$) compressed reversibly, $P_1=100\ \text{kPa}$, $T_1=20\ ^\circ\text{C}=293.15\ \text{K}$, to $P_2=500\ \text{kPa}$ with $pv^{1.3}=\text{const}$ ($n=1.3$).

Find. Work and heat transfer per unit mass.

Approach. Find the end temperature from the polytropic relation, use the closed-system polytropic work formula, and close the first law for the heat transfer (cross-checked with the polytropic specific heat).

1. End temperature. $T_2=T_1\left(\dfrac{P_2}{P_1}\right)^{\frac{n-1}{n}}=293.15\,(5)^{0.2308}=\boxed{425.0\ \text{K}}$.
2. Boundary work. Per unit mass $w=\dfrac{R(T_1-T_2)}{n-1}=\dfrac{0.2968(293.15-425.0)}{0.3}=-130.5\ \text{kJ/kg}$, i.e. $\boxed{130.5\ \text{kJ/kg of work input}}$.
3. Heat transfer (first law). $\Delta u=c_v(T_2-T_1)=0.743(131.9)=98.0\ \text{kJ/kg}$, so $q=\Delta u+w=98.0-130.5=\boxed{-32.5\ \text{kJ/kg}}$ (rejected).
4. Cross-check. Polytropic specific heat $c_n=c_v\dfrac{n-\gamma}{n-1}=0.743\dfrac{-0.1}{0.3}=-0.248$; $q=c_n(T_2-T_1)=-32.6\ \text{kJ/kg}$. ✓

Because the polytropic index lies between unity (isothermal) and $\gamma$ (adiabatic), the compression sheds heat: the gas warms less than in an adiabatic compression, and the rejected 32.5 kJ/kg is exactly what keeps the entropy falling in the $T$–$s$ view.