Centrifugal air compressor with kinetic-energy terms: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2018, Q4

Question 4 — Centrifugal air compressor with kinetic-energy terms (Part A, 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. Adiabatic centrifugal air compressor, air as an ideal gas ($k=1.4$, $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$).

Quantity	Value
Mass flow, $\dot m$	9.1 kg/s
Inlet, state 1	100 kPa, 15 °C (288.15 K)
Inlet / outlet velocity	110 m/s / 90 m/s
Pressure ratio	4 : 1
Isentropic efficiency, $\eta_c$	0.80

Find. (a) outlet pressure, (b) outlet temperature, (c) shaft power.

Figure 4 — Adiabatic compressor control volume. The steady-flow energy balance includes the kinetic-energy change between the 110 m/s inlet and the 90 m/s discharge.

Approach. Get the outlet pressure from the ratio, the ideal temperature rise from the isentropic relation scaled by $\eta_c$, then the power from the steady-flow energy balance including the kinetic-energy change.

Outlet pressure. Directly from the 4:1 ratio: $$P_2=4P_1=4(100)=\boxed{400\ \text{kPa}}$$
Isentropic outlet temperature. $$T_{2s}=T_1\left(\frac{P_2}{P_1}\right)^{(k-1)/k}=288.15(4)^{0.2857}=428.2\ \text{K}$$
Actual outlet temperature from $\eta_c$. The isentropic efficiency scales the enthalpy (temperature) rise: $$\Delta T_\text{act}=\frac{T_{2s}-T_1}{\eta_c}=\frac{428.2-288.15}{0.80}=175.1\ \text{K}$$ $$T_2=T_1+\Delta T_\text{act}=288.15+175.1=463.2\ \text{K}=\boxed{190\ ^\circ\text{C}}$$
Shaft power (steady-flow energy balance with kinetic energy). For an adiabatic compressor, $$\dot W=\dot m\left[(h_2-h_1)+\tfrac{1}{2}(V_2^{2}-V_1^{2})\right]=\dot m\left[c_p\Delta T_\text{act}+\tfrac{1}{2}(V_2^{2}-V_1^{2})\right]$$ The kinetic term is $\tfrac12(90^2-110^2)=-2.0\times10^{3}\ \text{J/kg}=-2.0\ \text{kJ/kg}$ (the flow decelerates, returning a little energy), and $c_p\Delta T_\text{act}=1.005(175.1)=176.0\ \text{kJ/kg}$: $$\dot W=9.1\,(176.0-2.0)=9.1(174.0)=1583\ \text{kW}$$ $\dot W\approx 1.58$ MW

Check — property assumption.

Constant $c_p=1.005\ \text{kJ/kg}\cdot\text{K}$ is used over a 288–463 K span; an air-table calculation with variable specific heats lowers the power by roughly 2–3 %. The isentropic efficiency is applied to the static-enthalpy rise; treating it on stagnation enthalpies changes the result by well under 1 % because the kinetic terms are small (≈1 % of the work).

Quantity	Result
(a) Outlet pressure	400 kPa
Isentropic outlet temp $T_{2s}$	428 K (155 °C)
(b) Actual outlet temperature	463 K ≈ 190 °C
Specific work (incl. KE)	174.0 kJ/kg
(c) Shaft power	≈ 1.58 MW (1583 kW)

Concept & Theory

Topic. Adiabatic compressor with isentropic efficiency and non-negligible flow velocities.

Key relations. Isentropic ideal-gas relation $T_{2s}/T_1=(P_2/P_1)^{(k-1)/k}$; compressor isentropic efficiency $\eta_c=(h_{2s}-h_1)/(h_2-h_1)$; steady-flow energy balance $\dot W=\dot m[\Delta h+\tfrac12\Delta V^2]$.

Why this works. The pressure ratio alone fixes the ideal exit temperature; real irreversibilities make the actual temperature rise larger by the factor $1/\eta_c$, since the same pressure rise costs more enthalpy. The kinetic-energy change is retained because the question gives distinct inlet and outlet velocities — although here it only trims the power by about 1 %, omitting it would be an incomplete answer to a question that explicitly supplies both speeds.

Common pitfall. Dividing rather than multiplying by $\eta_c$ (a compressor needs more work than ideal), or dropping the kinetic-energy term the question deliberately provides.

Source. Çengel & Boles, Thermodynamics §7 (isentropic efficiencies of compressors) and §5 (steady-flow energy balance with kinetic energy).

Question 4 of 8: Centrifugal air compressor with kinetic-energy terms

Question 4 — Centrifugal air compressor with kinetic-energy terms (Part A, equal value)