Thermodynamics Assignment - Engineering Thermodynamics Solutions

This document presents a comprehensive solution to a thermodynamics assignment, addressing various problems related to heat transfer, heat pumps, and thermodynamic cycles. The assignment covers calculations of the Coefficient of Performance (COP) for different scenarios, including refrigerators and heat pumps, and assesses the feasibility of claims based on thermodynamic principles. It involves determining heat transfer rates, net power output, and efficiency calculations for different systems, such as power plants and heating systems. The solutions include detailed calculations, formulas, and explanations to help students understand the concepts and solve similar problems. The assignment also explores the impact of heat losses and provides answers to questions about the validity of given claims based on calculated COPs. References from textbooks on thermodynamics are included to support the solutions.

Running Head: THERMODYNAMICS ASSIGNMENT 1
Thermodynamics Assignment
Student’s Name
Institutional Affiliation

Thermodynamics Assignment 2
Question 1
Temperature of source , T s=95o C
Temperature of refrigerator ,T L=0o C=0+273=273 k
Temperature of the environment , T E =19o C
COP=3.1
But COP= 1
T H
T L
−1
3.1= 1
T H
273 −1
T H =361.06=88.06o C
The claim is possible but not probable.
Question 2
˙QH =70000 kJ
h
T H =26o C=26+273=299 K
T L=−4o C=−4+273=269 K
Heat Pump COP= 1
1− T L
T H
2

Thermodynamics Assignment 3
COP= 1
1−269
299
=9.97
Recall COP of heat pump is also= 1
1− ˙QL
˙QH
9.97= 1
1− ˙QL
70000
˙QL=62976.58 kJ /h
Net Power , wnet = ˙˙QH −QL
wnet =70000−62976.58=7023.42kJ /h
wnet =7023.42
3600 =1.95 kW
Question 3
wnet =760 MW
Efficiency , η=54 %=0.54
QH = wnet
η = 760
0.54 =1407.41 MW
Rate of heat transfer QL=QH −wnet=14 O7.41−760=647.41 MW
The amount of heat transfer will be lower as some heat is lost through the pipes and other
components.
Question 4
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Thermodynamics Assignment 4
wnet =180 kW
Efficiency , η=3 %=0.03
˙QH = wnet
η = 180
0.03 =6000 kW
But 1kW =1 kJ
s ∧Ihr =3600 s
˙QH =6000 ×3600=21.6× 106 kJ /h
Question 5
QH =750 kJ
wnet output =250 kJ
QL(Rejected heat)¿ QH −wnet output=700−250=450 kJ
Question 6
wnet output =155 MW
Heating volume , V =155 kJ / Kg
Mass of coalused =56 tons
h =56 × 1000
3600 =15.56 Kg/ s
winput=30000× 15.56=466,666.67 kJ
s =466.67 MW
Efficiency , η= wnet output
winput
×100 %= 155
466.67 ×100=33.21 %
Question 7
4

Thermodynamics Assignment 5
Energy released fromthe house=70000 kJ /h
House generated energy=8000 kJ / h
COP=2.5
˙QH =70000−8000=62000
wnet input = ˙QH
COP × 1
3600 = 62000
2.5 × 1
3600 =6.89 kW
Question 8
˙QH =110000 kJ
h
T H =25o C=25+ 273=298 K
T L=4o C=4 +273=277 K
Heat Pump COP= 1
1− T L
T H
COP= 1
1−277
298
=¿
Recall COP of heat pump is also= 1
1− ˙QL
˙QH
14.19= 1
1− ˙QL
70000
5

Thermodynamics Assignment 6
˙QL=102248.06 kJ /h
Net Power , wnet = ˙QL− ˙QH
wnet =110000−102248.06=7751.94 kJ /h
wnet =7751.94
3600 =2.15 kW
A. Heat pump powerful enough
Question 9
No
Question 10
T L=−12o C=−12+273=261 K
T H =25o C=25+ 273=298 K
But COP= 1
T H
T L
−1
= 1
298
261 −1
=7.05
The claim is valid because the COP calculated above is more than the claimed COP of 6.5.
Question 11
wnet =2.15 kW
Heat Pump COP= 1
1− T L
T H
6

Thermodynamics Assignment 7
COP=8.7
T H =24o C=24+273=297 K
Subsituting∧solving , 8.7= 1
1− T L
297
T L=262.86 K
Recall COP of heat pump is also= 1
1− ˙QL
˙QH
wnet =2.15 kW =2.15× 3600=7740 kJ /h
wnet = ˙QH − ˙QL
˙QL= ˙QH −wnet = ˙QH −7740
8.7= 1
1− ˙QH−7740
˙QH
8.7= ˙QH
7740
˙QH =67338.77 kJ /h
7

Thermodynamics Assignment 8
References
Cengel , Y. A., & Boles, M. A. (2018). Thermodynamics: An Engineering Approach. New York:
McGraw-Hill Education.
Cengel, Y. A., Cimbala, J. M., & Turner, R. H. (2016). Fundamentals of Thermal-Fluid
Sciences. New York: McGraw-Hill Higher Education.
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