Engineering Thermodynamics Solutions Manual – Prof. T.T. Al-Shemmeri | Complete Worked Solutions to Chapter 4 Tutorial Problems | Bookboon 2012 Edition (ISBN 978-87-403-0267-7)

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Engineering Thermodynamics 2

,Solutions Manual

Prof. T.T. Al-Shemmeri




Prof. T.T. Al-Shemmeri

Engineering Thermodynamics Solutions
Manual




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3

,Engineering Thermodynamics Solutions Manual First Law of Thermodynamics N.F.E.E Applications




Engineering Thermodynamics Solutions Manual Contents


4.1




4.1 First Law of Thermodynamics
N.F.E.E Applications
1. In a non-flow process there is heat transfer loss of 1055 kJ and an internal energy increase of 210
kJ. Determine the work transfer and state whether the process is an expansion or compression.


[Ans: -1265 kJ, compression]


Solution:

Closed system for which the first law of Thermodynamics applies,


Q - W = U


1055 – W = 210


Hence the work done can be found as:


W = -1265 kJ


Since negative, it must be work input, ie compression.


2. In a non-flow process carried out on 5.4 kg of a substance, there was a specific internal energy

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, Engineering Thermodynamics Solutions Manual First Law of Thermodynamics N.F.E.E Applications
decrease of 50 kJ/kg and a work transfer from the substance of 85 kJ/kg. Determine the
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0




k0 heat transfer and state whether it is gain or loss.
k0 k0 k0 k0 k0 k0 k0 k0 k0




[Ans: 189 kJ, gain]
k0 k0 k0




Solution:

Closed system for which the first law of Thermodynamics
k0 k0 k0 k0 k0 k0 k0 k0




k0 applies, Q - W = U
k0 k0 k0 k0 k0




Q = U + W
k0 k0 k0




= 5.4x (-50) +5.4 x 85
k0 k0 k0 k0 k0




= + 189 kJ,
k0 k0 k0




Since Q is positive, it implies heat is entering the control volume, ie Gain.
k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0 k0




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