University of Glasgow ENG3053 Thermodynamics Exam

Degrees of MEng, BEng, MSc and BSc in Engineering
ENG3053
Thermodynamics of Energy Systems
Wednesday, 10 December 2025, 09:30 – 11:00
Attempt ALL questions.
TOTAL MARKS AVAILABLE
75
The numbers in square brackets in the right-hand margin indicate the marks allotted to the
part of the question against which the mark is shown. These marks are for guidance only.
Data sheets are included at the end of the exam paper.
An electronic calculator may be used provided that it does not have a facility for either
textual storage or display, or for graphical display.
PLEASE SEE BELOW FOR INSTRUCTIONS ON HOW TO ANSWER:
• Script books will NOT be used in this exam.
• You will be issued with THREE A3 double sided answer sheets.
• Answer only one question on each A3 answer sheet.
• Ensure you CLEARLY add your GUID and date of birth on each answer sheet. GUID
goes on front and back.
• Continuation answer sheets (pink) are available on request.
• Please ensure you CLEARLY add your GUID, date of birth and question number to
all continuation sheets. GUID goes on front and back.
OVER
Page 1 of 6
Q1
As schematically shown in Figure Q1 overleaf, a vapour compression cycle refrigeration
system uses R134a as refrigerant and the condensing temperature is 35 ◦C. It has two
evaporators for different cooling loads. The intermediate-temperature evaporator provides
cooling at 0 ◦C, and its cooling capacity is 400 kW. The low-temperature evaporator provides
cooling at -20 ◦C, and its cooling capacity is 500 kW. Assume the states of refrigerant at the
exit of the evaporator, the condenser and the flash tank are saturated, and the compression
processes are isentropic. The enthalpies of some states are shown as Table Q1.
Table Q1 Enthalpy of several state points.
i.
State
1
2
3
4
5
7
h (kJ/kg)
386.4
402.46
398.52
421.57
248.79
227.26
Sketch a Pressure-Enthalpy diagram for the system as shown in Figure Q1 overleaf
and indicate the state points 1-8 on the Pressure-Enthalpy diagram.
[4]
ii.
Describe the function of the flash tank in this system and explain its benefits.
[4]
iii.
Determine the mass flow rates of refrigerant passing through the two compressors.
[kg/s]
[9]
iv.
Determine the COP of this refrigeration system.
[6]
v.
Determine the heat rejection by refrigerant within the condenser.
[2]
Continued overleaf
Page 2 of 6
4
5
Condenser
Compressor II
Expansion valve III
𝑚ሶ3
Evaporator II
6
3
2
Expansion valve II
𝑚ሶ2
Flash tank
6
Compressor I
𝑚ሶ1
7
Evaporator I
Expansion valve I
8
1
Figure Q1
Continued overleaf
Page 3 of 6
Q2
A 4-liter, CI, V8 automotive engine operates on a four-stroke cycle at 3300 RPM. The AirFuel ratio is AF=14, a fuel heating value is 44,000 kJ / kg, and the combustion efficiency is
97%. Density of air at atmospheric pressure is 1.181kg/m3. At this speed, the measured brake
output torque is 210 N∙m. Air enters the engine at 77 kPa and 55 ◦C, and the mechanical
efficiency is 87%.
Determine:
i.
The rate of fuel flow into the engine [kg/s]
[6]
ii.
Brake power [kW]
[3]
iii.
Brake thermal efficiency [%]
[6]
iv.
Brake specific fuel consumption [kg/kW∙𝑠]
[5]
v.
Indicated thermal efficiency [%]
[5]
Continued overleaf
Page 4 of 6
Q3
A 4-cylinder, 4 L, SI automobile engine operates on a 4-stroke air-standard Otto cycle at 3500
RPM. The engine has compression ratio of 8.5:1. Fuel is isooctane with AF=14, a heating value
of 44,300 kJ/kg, and the combustion efficiency 100%. A mechanical efficiency is 86%, and
stroke-to-bore ratio is set at S/B =1.025. At the start of the compression stroke, conditions in
the cylinder combustion chamber are at 90 kPa and 70 °C. There is 4% exhaust residual left
over from previous cycle, which needs to be taken into account.
Determine:
i.
sketch a P-V and T-S diagram of the 4-stroke air-standard Otto cycle
ii.
the temperature [K], volume [𝐿] and pressure [kPa] at each state of the cycle for one
iii.
[6]
cylinder
[15]
Net indicated work [kJ] based on the cycle
[4]
Continued overleaf
Page 5 of 6
Data Sheet
Formula
𝑚ሶ𝑎 = 𝜂𝑣 𝜌𝑎 𝑉𝑑 𝑁⁄𝑛
𝑄𝑖𝑛 = 𝑚𝑓 𝑄𝐻𝑉 𝜂𝑐
𝑃𝑣 = 𝑅𝑇
𝑊ሶ𝑏 = 2𝜋𝑁𝜏
𝑄 = 𝑚𝑐𝑣,𝑝 (∆𝑇)
̅𝑝 = 2𝑆𝑁
𝑈
sfc = 𝑚ሶ𝑓 ⁄𝑊ሶ
𝑁𝑊
𝑛
𝑊
mep =
𝑉𝑑
𝑊ሶ =
For isentropic process:
𝑇𝑃(𝑘−1)⁄𝑘 = constant
𝑇𝑣 𝑘−1 = constant
𝑃𝑣 𝑘 = constant
𝑤1−2 = (𝑃2 𝑣2 − 𝑃1 𝑣1 )⁄(1 − 𝑘) = 𝑅(𝑇2 − 𝑇1 )⁄(1 − 𝑘)
Air in the surrounding
𝑐𝑝 = 1.005 kJ⁄kg ∙ K
𝑐𝑣 = 0.718 kJ⁄kg ∙ K
𝑘 = 1.40
𝑅 = 0.287 kJ⁄kg ∙ K
𝜌 = 1.181 kg/m3 at standard conditions
Mixture in the engine and exhaust gas
𝑐𝑝 = 1.108 kJ⁄kg ∙ K
𝑐𝑣 = 0.821 kJ⁄kg ∙ K
𝑘 = 1.35
𝑅 = 0.287 kJ⁄kg ∙ K
End of question paper
Page 6 of 6