Ideal-Gas Polytropic Process; Steady-Flow Air Compression: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2013, Q1

Question 1: Ideal-Gas Polytropic Process; Steady-Flow Air Compression

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 closed body of ideal gas with fixed specific heats, no mass flow — the pressure change alone (with $n$) fixes the temperature change. Part (b): a compressor with a stated heat loss, a known isentropic (reversible) work requirement, and a compression efficiency relating actual to ideal work.

Given data
Quantity	Symbol	Part (a)	Part (b)
Inlet pressure	$P_1$	1380 kPa	100 kPa
Exit pressure	$P_2$	345 kPa	500 kPa
Inlet temperature	$T_1$	$60\,{}^{\circ}\text{C}$	$27\,{}^{\circ}\text{C}$
Polytropic exponent	$n$	1.20	—
Specific heat	$C_p$	1.047 kJ/kg·K	1.005 kJ/kg·K (standard air)
Gas constant	$R$	0.258 kJ/kg·K	0.287 kJ/kg·K (standard air)
Heat loss	$q_{\text{loss}}$	—	11.2 kJ/kg
Isentropic (reversible) work	$W_s$	—	307.5 kJ/kg
Compression efficiency	$\eta_c$	—	75%

Find. (a) The change in specific enthalpy $\Delta h$ and specific internal energy $\Delta u$. (b) The change in specific entropy $\Delta s$ of the air during the actual (irreversible, heat-losing) compression.

Approach. Part (a) uses the polytropic temperature–pressure relation to get $T_2$, then the ideal-gas relations $\Delta h = C_p\Delta T$ and $\Delta u = C_v\Delta T$ (state functions — the closed-system polytropic path does not change this). Part (b) first converts the given isentropic work to the actual work via the compression efficiency, uses the steady-flow energy equation (with the stated heat loss) to get the actual exit enthalpy and hence $T_2$, then applies the ideal-gas entropy-change relation between the two pressure/temperature states.

Part (a) — specific heat at constant volume. For an ideal gas, $C_v = C_p - R$: $$C_v = 1.047 - 0.258 = 0.789\text{ kJ/kg}\cdot\text{K}$$
Part (a) — exit temperature from the polytropic relation. For a polytropic process $Pv^n = \text{const}$ on an ideal gas, $$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{n-1}{n}}$$ With $T_1 = 333.15\text{ K}$, $P_2/P_1 = 345/1380 = 0.2500$ and $(n-1)/n = 0.20/1.20 = 1/6$: $$T_2 = 333.15\times(0.2500)^{1/6} = 333.15\times0.7937 = 264.42\text{ K} = -8.73\,{}^{\circ}\text{C}$$ The pressure drops by a factor of 4, so the gas expands and cools.
Part (a) — enthalpy and internal-energy change. Both are state functions of temperature alone for an ideal gas: $$\Delta h = C_p(T_2-T_1) = 1.047\times(264.42-333.15) = 1.047\times(-68.73)$$ $$\Delta u = C_v(T_2-T_1) = 0.789\times(-68.73)$$ $$\boxed{\Delta h = -71.96\text{ kJ/kg}, \qquad \Delta u = -54.23\text{ kJ/kg}}$$
Part (b) — actual compressor work from the efficiency. Compression efficiency compares the ideal (isentropic) work needed to the larger actual work a real, irreversible compressor must supply for the same pressure rise, $\eta_c = W_s/W_a$: $$W_a = \frac{W_s}{\eta_c} = \frac{307.5}{0.75} = 410.00\text{ kJ/kg}$$
Part (b) — actual enthalpy rise from the steady-flow energy equation. With heat LEAVING the gas ($q = -q_{\text{loss}}$) and work done ON the gas ($w = -W_a$ using the "work done by the system" sign convention), the SFEE $q - w = \Delta h$ gives $$\Delta h_{\text{actual}} = W_a - q_{\text{loss}} = 410.00 - 11.2 = 398.80\text{ kJ/kg}$$ so the actual exit temperature is $$T_2 = T_1 + \frac{\Delta h_{\text{actual}}}{C_p} = 300.15 + \frac{398.80}{1.005} = 696.97\text{ K} = 423.82\,{}^{\circ}\text{C}$$ Irreversibility drives the actual enthalpy rise well above the ideal 307.5 kJ/kg even though some heat is lost — friction and turbulence inside a real compressor generate far more internal heating than the jacket loses to the surroundings.
Part (b) — entropy change. For an ideal gas between two $(T,P)$ states, $$\Delta s = C_p\ln\frac{T_2}{T_1} - R\ln\frac{P_2}{P_1} = 1.005\ln\frac{696.97}{300.15} - 0.287\ln\frac{500}{100}$$ $$\Delta s = 1.005(0.8425) - 0.287(1.6094) = 0.8467 - 0.4619$$ $$\boxed{\Delta s = +0.3848\text{ kJ/kg}\cdot\text{K}}$$ The positive sign confirms the process is irreversible overall: the entropy generated by friction/turbulence inside the machine outweighs the entropy carried away with the lost heat.

Check: Part (b) treats the air with standard properties $C_p=1.005$, $R=0.287$ kJ/kg·K (the exam names the fluid as "air" but does not reprint these values in this sub-part, unlike Part (a)'s generic ideal gas). Note that the printed isentropic work is NOT consistent with standard air: $C_pT_1\big[(P_2/P_1)^{R/C_p}-1\big]=1.005\times300.15\times(5^{0.2856}-1)=176.0$ kJ/kg, not 307.5 kJ/kg. The printed 307.5 kJ/kg is a given datum of the question, so it is used as stated; the entropy change follows from the actual exit state it implies, and the method is unchanged if the examiner's figure was intended for a different gas model.

Question 1 — results
Quantity	Value
(a) Exit temperature $T_2$	264.42 K ($-8.73\,{}^{\circ}\text{C}$)
(a) $\Delta h$	$-71.96$ kJ/kg
(a) $\Delta u$	$-54.23$ kJ/kg
(b) Actual work input $W_a$	410.00 kJ/kg
(b) Actual exit temperature $T_2$	696.97 K ($423.82\,{}^{\circ}\text{C}$)
(b) $\Delta s$	$+0.3848$ kJ/kg·K

Topic. Ideal-gas polytropic state changes, and steady-flow compressor efficiency combined with the entropy balance.

Key relations. Polytropic relation $T_2/T_1=(P_2/P_1)^{(n-1)/n}$; ideal-gas $\Delta h=C_p\Delta T$, $\Delta u=C_v\Delta T$, $C_v=C_p-R$; steady-flow energy equation $q-w=\Delta h$; compressor efficiency $\eta_c=W_s/W_a$ (ideal over actual — the opposite ratio from a turbine's efficiency); ideal-gas entropy change $\Delta s = C_p\ln(T_2/T_1)-R\ln(P_2/P_1)$.

Why this works. Enthalpy and internal energy are state functions of temperature alone for an ideal gas, so Part (a) needs only the endpoint temperatures, not the polytropic path details, once $T_2$ is found. In Part (b), a compressor's irreversibilities always make the actual work exceed the ideal (isentropic) work for the same pressure ratio, which is why $\eta_c$ is defined as ideal/actual (inverted from a turbine, where irreversibility reduces the actual output below ideal). The entropy relation follows directly from combining the first law and $Tds=dh-vdP$ for an ideal gas.

Common pitfall. Using the turbine-style efficiency ratio ($W_a/W_s$) for a compressor inflates the actual work rather than correctly increasing it beyond the ideal value; and forgetting the heat-loss sign in the SFEE (heat leaving must be subtracted, not added) understates $T_2$ and $\Delta s$.

Source. Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 3–4 (ideal-gas properties, polytropic processes) and Ch. 7 (entropy change, isentropic/adiabatic efficiencies).

Question 1 of 8: Ideal-Gas Polytropic Process; Steady-Flow Air Compression

Question 1: Ideal-Gas Polytropic Process; Steady-Flow Air Compression