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25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2017

Question 4 of 8: Ammonia Refrigeration Cycle

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

Notes on this paper

National Examinations, May 2017 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

It is solved as the thermodynamics/heat-transfer exam it actually is.

Question 4: Ammonia Refrigeration Cycle (20 marks)

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. Liquid ammonia at 1200 kPa, 27°C before the throttle (state 3); evaporator at a uniform 250 kPa, exit at -10°C (state 1, compressor inlet); $\dot W_{comp}=2.12\text{ kW}$; $\dot m=0.007\text{ kg/s}$.

Entropy s (kJ/kg·K)Temperature Tsaturation dome1234
Vapour-compression cycle on the T–s plane (schematic, saturation dome shown).

Find. (a) quality leaving the throttle valve, (b) evaporator heat-absorption rate, (c) COP.

Approach. Read $h_f$ at 27°C for the sub-cooled/near-saturated liquid entering the throttle (standard approximation $h\approx h_f@T$); throttling is isenthalpic, so this fixes $h$ at the evaporator inlet; evaluate quality against saturation properties at 250 kPa; find the evaporator-exit enthalpy from the (superheated) state at -10°C; close the cycle with an energy balance on the evaporator and the definition of COP.

  1. Enthalpy before the throttle (interpolated saturated-liquid table, 26°C/28°C). $$h_3=h_f@27\,{}^{\circ}\text{C}\approx\tfrac12(303.6+313.2)=308.4\text{ kJ/kg}$$
  2. Saturation state at 250 kPa (interpolated between -14°C/246.51 kPa and -12°C/267.95 kPa). $T_{sat}\approx-13.7\,{}^{\circ}\text{C}$, $h_f\approx118.4\text{ kJ/kg}$, $h_g\approx1427.4\text{ kJ/kg}$, $h_{fg}\approx1309.0\text{ kJ/kg}$.
  3. Quality after throttling ($h_4=h_3$, isenthalpic). $$x_4=\dfrac{h_4-h_f}{h_{fg}}=\dfrac{308.4-118.4}{1309.0}=\boxed{0.145\ (14.5\%)}$$
  4. Evaporator-exit enthalpy. Exit at -10°C is $\approx3.7\,{}^{\circ}\text{C}$ of superheat above $T_{sat}(250\text{ kPa})=-13.7\,{}^{\circ}\text{C}$. Using an ammonia-vapour $c_p\approx2.6\text{ kJ/kg}\!\cdot\!\text{K}$ near this state, $h_1\approx h_g+c_p\Delta T_{sh}=1427.4+2.6(3.7)=1437.0\text{ kJ/kg}$.
  5. Evaporator heat-absorption rate. $$\dot Q_{evap}=\dot m(h_1-h_4)=0.007(1437.0-308.4)=\boxed{7.90\text{ kW}}$$
  6. Coefficient of performance. $$\text{COP}=\dfrac{\dot Q_{evap}}{\dot W_{comp}}=\dfrac{7.90}{2.12}=\boxed{3.73}$$
Question 4 — final results
QuantityValue
Quality leaving throttle valve$x_4 = 0.145\ (14.5\%)$
Evaporator heat-absorption rate$\dot Q_{evap} = 7.90\text{ kW}$
Coefficient of performance$\text{COP} = 3.73$
This affects only steps 4–6 (evaporator duty and COP); the throttling quality in step 3 is unaffected and uses only the saturated-liquid/saturated-vapour tables transcribed in full.