Real Freon-12 Vapour-Compression Cycle from Test Data: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2016, Q4

Question 4: Real Freon-12 Vapour-Compression Cycle from Test Data

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. This is REAL test data, not an idealized four-state cycle: the compressor inlet is superheated above the evaporator-outlet state (suction-line pick-up), the condenser inlet is desuperheated below the compressor-outlet state (discharge-line loss), and the expansion-valve inlet is subcooled further below the condenser-outlet state (liquid-line heat gain). Compression is NOT adiabatic — 3.35 kJ/kg is rejected from the refrigerant while it is being compressed.

Given data (R-12, all states along the loop)
Station	Pressure	Temperature
Compressor inlet	150 kPa	$0\,{}^{\circ}\text{C}$
Compressor outlet	700 kPa	$70\,{}^{\circ}\text{C}$
Condenser inlet	700 kPa	$60\,{}^{\circ}\text{C}$
Condenser outlet	700 kPa	$15\,{}^{\circ}\text{C}$
Expansion-valve inlet	700 kPa	$20\,{}^{\circ}\text{C}$
Evaporator outlet	150 kPa	$-10\,{}^{\circ}\text{C}$

Fig. 4 — T-s diagram of the actual (non-idealized) R-12 cycle: suction superheat, discharge desuperheat and liquid subcooling are all visible as departures from the idealized four-corner cycle.

Find. The cycle COP and the "thermal efficiency of the compression process" (read as the compressor's isentropic efficiency, since compression here is explicitly non-adiabatic and a comparison against the ideal adiabatic work is the only quantity this phrase can sensibly mean).

Approach. Evaluate R-12 enthalpy at each printed state directly; the refrigeration effect follows from the isenthalpic throttle (valve-inlet enthalpy = evaporator-inlet enthalpy); the actual compressor work follows from a steady-flow energy balance that includes the stated heat loss; the isentropic work uses the compressor-inlet entropy carried to the condenser pressure.

Key enthalpies (R-12). $$h_{\text{comp,in}}(0\,{}^{\circ}\text{C},150\text{ kPa})=355.50\text{ kJ/kg},\quad s_{\text{comp,in}}=1.6159\text{ kJ/kg}\cdot\text{K}$$ $$h_{\text{comp,out}}(70\,{}^{\circ}\text{C},700\text{ kPa})=393.81\text{ kJ/kg}$$ $$h_{\text{valve,in}}(20\,{}^{\circ}\text{C},700\text{ kPa})=219.16\text{ kJ/kg},\qquad h_{\text{evap,out}}(-10\,{}^{\circ}\text{C},150\text{ kPa})=349.55\text{ kJ/kg}$$
Refrigeration effect. The expansion valve is isenthalpic, so $h_{\text{evap,in}}=h_{\text{valve,in}}$: $$q_{\text{evap}}=h_{\text{evap,out}}-h_{\text{valve,in}}=349.55-219.16=\boxed{130.4\text{ kJ/kg}}$$
Actual compressor work. Steady-flow energy balance with heat REJECTED during compression, $q_{\text{out}}=3.35$ kJ/kg: $$w_{\text{actual}}=(h_{\text{comp,out}}-h_{\text{comp,in}})+q_{\text{out}}=(393.81-355.50)+3.35=\boxed{41.66\text{ kJ/kg}}$$
Coefficient of performance. $$\text{COP}=\frac{q_{\text{evap}}}{w_{\text{actual}}}=\frac{130.4}{41.66}=\boxed{3.13}$$
Isentropic compressor work (for the compression-efficiency comparison). Carrying $s_{\text{comp,in}}=1.6159\text{ kJ/kg}\cdot\text{K}$ to 700 kPa isentropically gives $T_{\text{out},s}=57.7\,{}^{\circ}\text{C}$ and $h_{\text{out},s}=385.30$ kJ/kg, so $$w_{\text{isentropic}}=h_{\text{out},s}-h_{\text{comp,in}}=385.30-355.50=29.80\text{ kJ/kg}$$
"Thermal efficiency" of the compression process. $$\eta_{\text{compression}}=\frac{w_{\text{isentropic}}}{w_{\text{actual}}}=\frac{29.80}{41.66}=\boxed{71.5\%}$$

Question 4 — results
Quantity	Value
Refrigeration effect, $q_{\text{evap}}$	130.4 kJ/kg
Actual compressor work, $w_{\text{actual}}$	41.66 kJ/kg
Coefficient of performance, COP	3.13
Isentropic compressor work, $w_{\text{isentropic}}$	29.80 kJ/kg
Compression-process efficiency, $\eta_{\text{compression}}$	71.5%

Topic. Vapour-compression refrigeration analysed from raw test data rather than idealized saturation states, including non-adiabatic compression.

Key relations. Throttle: $h$ constant; refrigeration effect $q_{\text{evap}}=h_{\text{out,evap}}-h_{\text{in,evap}}$; steady-flow energy balance with heat loss $w_{\text{actual}}=(h_{\text{out}}-h_{\text{in}})+q_{\text{out}}$; COP$=q_{\text{evap}}/w_{\text{actual}}$; isentropic efficiency $\eta=w_s/w_{\text{actual}}$.

Why this works. Because the problem supplies six distinct temperature readings around the loop rather than four idealized saturation states, every property lookup must use the ACTUAL printed state (own $T$, own $P$) rather than assuming the compressor inlet equals the evaporator outlet or the condenser inlet equals the compressor outlet — those apparent "duplicates" are deliberately different states representing real suction-line and discharge-line heat transfer.

Common pitfall. Using $h_{\text{evap,out}}$ (the evaporator's own exit state) as if it equalled $h_{\text{comp,in}}$ in the actual-work balance, or forgetting the sign of $q_{\text{out}}$ (heat LEAVES the refrigerant during compression here, so it ADDS to the work the compressor must do to reach the same exit enthalpy, rather than subtracting).

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 11 (vapour-compression refrigeration, actual vs. ideal cycles).

Question 4 of 8: Real Freon-12 Vapour-Compression Cycle from Test Data

Question 4: Real Freon-12 Vapour-Compression Cycle from Test Data