Thermodynamics Assignment - Desklib

This article provides solutions to various thermodynamics problems and assignments. It covers topics such as COP, net power, efficiency, and heat transfer. The article also references two textbooks on thermodynamics. Subject: Thermodynamics

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
3

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.
8