Polytropic Expansion and Non-Adiabatic Compression: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2013, Q1

Question 1: Polytropic Expansion and Non-Adiabatic 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.

Given. Part (a): a fixed mass of ideal gas expands polytropically ($PV^{n}=\text{const}$, $n=1.20$) from $P_1=1380$ kPa, $T_1=60\ ^\circ\text{C}=333.15$ K to $P_2=345$ kPa, with $c_v=1.047$ and $R=0.258\ \text{kJ/kg}\cdot\text{K}$ (so $c_p=c_v+R=1.305$). Part (b): air ($c_p=1.005$, $R=0.287\ \text{kJ/kg}\cdot\text{K}$) is compressed in steady flow from $P_1=100$ kPa, $T_1=27\ ^\circ\text{C}=300.15$ K to $P_2=500$ kPa; isentropic work $w_s=307.5$ kJ/kg, compression efficiency $\eta_c=0.75$, heat loss $q=11.2$ kJ/kg.

Find. (a) $\Delta h$ and $\Delta u$ per unit mass for the polytropic expansion; (b) the specific-entropy change $\Delta s$ of the air across the actual compressor.

Approach. Part (a): a polytropic process fixes $T_2$ from the pressure ratio, then the ideal-gas caloric relations give $\Delta u=c_v\Delta T$ and $\Delta h=c_p\Delta T$. Part (b): scale the isentropic work by the efficiency to get the actual work, close the steady-flow energy balance for the real exit temperature, then evaluate the ideal-gas entropy change.

(a) Final temperature of the polytropic expansion. Along $PV^n=\text{const}$ the temperature follows $T_2=T_1\left(P_2/P_1\right)^{(n-1)/n}$:$$T_2=333.15\left(\frac{345}{1380}\right)^{0.20/1.20}=333.15(0.25)^{0.1667}=\boxed{264.4\ \text{K}}\;(-8.7\ ^\circ\text{C}),\quad \Delta T=-68.73\ \text{K}.$$
(a) Internal-energy and enthalpy change. For an ideal gas these depend only on temperature, so $\Delta u=c_v\Delta T$ and $\Delta h=c_p\Delta T$:$$\Delta u=1.047(-68.73)=\boxed{-71.96\ \text{kJ/kg}},\qquad \Delta h=1.305(-68.73)=\boxed{-89.69\ \text{kJ/kg}}.$$Both are negative because the gas cools as it expands.
(b) Actual compressor work. The compression efficiency relates the ideal to the actual work, $\eta_c=w_s/w_\text{act}$, so$$w_\text{act}=\frac{w_s}{\eta_c}=\frac{307.5}{0.75}=410.0\ \text{kJ/kg (work input)}.$$
(b) Actual exit temperature from the energy balance. The steady-flow first law with work input $w_\text{act}$ and heat loss $q=-11.2$ kJ/kg gives $\Delta h=w_\text{act}+q=410.0-11.2=398.8$ kJ/kg, hence$$T_2=T_1+\frac{\Delta h}{c_p}=300.15+\frac{398.8}{1.005}=\boxed{697.0\ \text{K}}\;(423.8\ ^\circ\text{C}).$$
(b) Entropy change of the air. For an ideal gas with constant specific heats, $\Delta s=c_p\ln(T_2/T_1)-R\ln(P_2/P_1)$:$$\Delta s=1.005\ln\!\frac{697.0}{300.15}-0.287\ln\!\frac{500}{100}=0.8466-0.4619=\boxed{+0.385\ \text{kJ/kg}\cdot\text{K}}.$$

Check

In (b) the entropy rise is positive even though the machine loses heat: the internal irreversibility of the 75 %-efficient compression (dissipation) more than offsets the entropy carried out with the rejected heat. Had the compression been reversible and adiabatic, $\Delta s$ would be exactly zero. Note that the printed data over-determine part (b): for air with $c_p=1.005$ the isentropic work between 100 and 500 kPa is only $c_p T_1\!\left[(P_2/P_1)^{0.2857}-1\right]=176$ kJ/kg, not the 307.5 kJ/kg the paper states (307.5 kJ/kg would correspond to a pressure ratio near 12). The printed 307.5 kJ/kg is used as given — it is supplied precisely so no air table is needed — which is why the actual exit temperature comes out as high as 697 K; the inconsistency is stated here as an assumption under exam Note 1.

Quantity	Result
(a) Polytropic exit temperature	264.4 K (−8.7 °C)
(a) Change in internal energy	Δu = −71.96 kJ/kg
(a) Change in enthalpy	Δh = −89.69 kJ/kg
(b) Actual exit temperature	697.0 K (423.8 °C)
(b) Entropy change of air	Δs = +0.385 kJ/kg·K

Concept & Theory

Topic. Ideal-gas polytropic process and non-adiabatic steady-flow compression.

Key relations. Polytropic $T_2/T_1=(P_2/P_1)^{(n-1)/n}$; caloric $\Delta u=c_v\Delta T$, $\Delta h=c_p\Delta T$; SFEE $\Delta h=w_\text{in}+q$; ideal-gas entropy $\Delta s=c_p\ln(T_2/T_1)-R\ln(P_2/P_1)$.

Why this works. Because $u$ and $h$ of an ideal gas are functions of temperature alone, the caloric changes need only $T_1$ and $T_2$; the polytropic relation supplies $T_2$ without integrating the path. In (b) the compression efficiency converts the tabulated isentropic work into the real work, and the energy balance then fixes the real end state so the entropy function can be evaluated.

Common pitfall. Forgetting the heat loss in the (b) energy balance (which would give $T_2\approx708$ K), or applying $\Delta s=0$ because the process is “a compression” — it is neither reversible nor adiabatic here.

Source. Çengel & Boles, Thermodynamics §4 (polytropic work, closed systems), §7 (entropy change of ideal gases) and §7–12 (isentropic/compression efficiency).

Question 1 of 8: Polytropic Expansion and Non-Adiabatic Compression

Question 1: Polytropic Expansion and Non-Adiabatic Compression (equal value)