NivaarExam PrepOfficial exam papers ↗

23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2017

Question 6 of 6: Air-Cycle Air-Conditioner — Coefficient of Performance

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2017 — 16-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: two parts — Part A (Q1–Q3) Process Balances, Part B (Q4–Q6) Chemical Thermodynamics; the candidate answers two questions from each part (four constitute a complete paper, equal value). All six questions are solved below for completeness. Property data not printed on the paper (molar volumes at STP, gas constant, Rankine conversion) are stated explicitly in each Given block as open-book look-ups.

Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — recycle balances, humidity and adiabatic-flame (energy) balances; Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (8th ed.) — multi-unit separation balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — cubic equations of state, residual properties, reaction/phase equilibrium and gas-cycle work; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 6: Air-Cycle Air-Conditioner — Coefficient of Performance (Part B — 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. Ideal-gas air. States: (1) compressor inlet $T_1=75\,{}^{\circ}\text{F}=534.67$ R at 1 atm; (2) compressor outlet; (3) after cooling $T_3=100\,{}^{\circ}\text{F}=559.67$ R at high $P$; (4) turbine outlet $T_4=55\,{}^{\circ}\text{F}=514.67$ R at 1 atm, into the house. Turbine work drives the compressor; net external work is the shortfall.

Find. COP $=\dfrac{\text{cooling delivered}}{\text{net work in}}$ for (a) reversible and (b) 70%-efficient compressor and turbine.

CompressorCooler(to 100 F)TurbineAir from house75 F, 1 atmhigh PQ to outside air100 F, high PTo ductwork55 F, 1 atm
Figure 3 — House air is compressed, cooled against outside air at high pressure, then expanded through a turbine that helps drive the compressor. The cold turbine exhaust (55 F) enters the house and absorbs heat as it warms back to 75 F.

Approach. Compressor and turbine share the same pressure ratio, so for isentropic operation the temperature ratio $r=(P_2/P_1)^{(\gamma-1)/\gamma}=T_3/T_4$; the cooling effect is $C_p(T_1-T_4)$ and the net work is $W_{\text{comp}}-W_{\text{turb}}$. $C_p$ and $\gamma$ cancel in the COP.

  1. Cooling effect (both cases). The 55 °F supply air warms to the 75 °F house temperature, absorbing $$Q_C=C_p(T_1-T_4)=C_p(534.67-514.67)=20\,C_p\ (\text{per mole}).$$
  2. Pressure ratio from the reversible turbine (case a). Isentropic turbine: $T_4=T_3/r$, so $$r=\frac{T_3}{T_4}=\frac{559.67}{514.67}=1.0874,$$ and the isentropic compressor gives $T_2=T_1\,r=534.67(1.0874)=581.4$ R.
  3. Net work, reversible (case a). $W_{\text{comp}}=C_p(T_2-T_1)=46.75\,C_p$ and $W_{\text{turb}}=C_p(T_3-T_4)=45.0\,C_p$, so $$W_{\text{net}}=C_p(46.75-45.0)=1.749\,C_p,\qquad \boxed{\text{COP}_{\text{rev}}=\frac{20}{1.749}=11.4.}$$ Near-reversible, the turbine almost pays for the compressor, so the tiny net work gives a very high COP.
  4. Pressure ratio from the 70% turbine (case b). The actual drop is fixed ($T_3-T_4=45$ R) but now equals $\eta_s$ times the isentropic drop: $45=0.70\,(T_3-T_{4s})\Rightarrow T_{4s}=495.4$ R, hence $$r=\frac{T_3}{T_{4s}}=\frac{559.67}{495.4}=1.1298.$$
  5. Net work and COP at 70% (case b). Isentropic compressor outlet $T_{2s}=T_1r=604.1$ R; the actual rise is larger by $1/\eta_s$: $W_{\text{comp}}=C_p(T_{2s}-T_1)/\eta_s=99.12\,C_p$. With $W_{\text{turb}}=45\,C_p$, $$W_{\text{net}}=C_p(99.12-45)=54.12\,C_p,\qquad \boxed{\text{COP}_{70\%}=\frac{20}{54.12}=0.370.}$$

The dramatic collapse from 11.4 to 0.37 is the key lesson: because the reversible cycle’s net work is a small difference of two large, nearly equal work terms, modest component inefficiencies inflate the net work enormously — air-cycle cooling is only attractive when the machinery is very efficient.

CaseNet workCOP
(a) Reversible1.75 $C_p$11.4
(b) 70% efficient54.1 $C_p$0.370
Back to the paper →