SOLUTION MANUAL
OF THERMODYNAMICS
By Hipolito Sta. Maria
Answered by: ENGR. NASER A. FERNANDEZ
Published by: ‘I Think, Therefore I’m An Atheist’ Enterprises
and Priority Development Fund (PDF)
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THERMODYNAMICS CHAPTER 4 SOLUTION by HIPOLITO
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Download THERMODYNAMICS CHAPTER 4 SOLUTION by HIPOLITO and more Exercises Thermodynamics in PDF only on Docsity!
SOLUTION MANUAL
OF THERMODYNAMICS
By Hipolito Sta. Maria
Answered by: ENGR. NASER A. FERNANDEZ
Published by: ‘I Think, Therefore I’m An Atheist’ Enterprises
and Priority Development Fund (PDF)
CHAPTER 4
1.A perfect gas has a value of R = 58.8 ft.lbf/lbm - °R and k = 1.26. If 20 Btu are added to 5 lbm of this gas at constantvolume when the initial temperature is 90 °F, find (a) T 2 , Change in H, Change in S, Change in U and (b) Work for a non flowprocess.
Given:
R = 58.8 Q = 20 BTU
k = 1.26 T 1 = 90 F + 460 = 550 °R
m = 5 lb
Solution:
(a) Q = mcv(T 2 – T 1 ) i = R = 58.8 x = 0.
ii. cv = R/(k-1) = 0.0756/(1.26-1) cv = 0.
iii. Q = mcv(T 2 – T 1 ) 20 = (5)(0.29)( T 2 - 550) T 2 = 563.8 °R
(b) i. cp = (kR)/(k-1) = (1.26)(0.0756)/(1.26-1) cp = 0.
ii. ΔH = mcp(T 2 – T 1 ) = (5)(0.366)(563.8-550) ΔH = 25.25 BTU
(c) ΔS = mcvln( )
= (5)(0.29)ln )
ΔS = 0.
(d) ΔU = mcv(T 2 – T 1 ) = (5)(0.29)(563.8-550) = 20.01 BTU
Solution:
(a) i. R(hydrogen) = 765.9 lb.ft/lb.R cv = 2.434 BTU/lb.R ii. ∆U = mcv(T 2 - T 1 ) = mcv( )
∆U = (p 2 - p 1 )
= (57600-43920) ∆U = 434.75 BTU
(b) Irreversible nonflow constant volume Q = U + Wn ;Q = 0 Wn = -434.75 BTU
- Three pounds of a perfect gas with R = 38 ft.lb/lb.R and k = 1.667 have 300 Btu of heat added
during the reversible nonflow constant pressure change of state. The initial temperature is 100.
Determine (a) final temperature, (b) ∆H, © W, (d) ∆U and (e) ∆S.
Given: R = 38 lb.ft/lb.R Q = 300 BTU k = 1.667 T 1 = 100 F + 460 = 560 °R m = 3 lb
Solution:
(a) i. cp = (kR)/(k-1) = (1.667)(38)/(1.667-1) = 94.97 x = 0.1221 BTU/lb.R ii. Q = mcp(T 2 -T 1 ) 300 = (3)(0.1221)(T 2 - 560) T 2 = 1379 R or 919 °F (b) Q= mcp(T 2 -T 1 ) = H ∆H = 300 BTU
(c) Wn = p(V 2 -V 1 ) = p( - ) ; p 1 = p 2
= pmR( - )
Wn = mR(T 2 -T 1 )
Wn = 120.008 BTU
(d) i. cv = R/(k-1) = 38/(1.667-1) = 56.97 x cv = 0.0732 BTU/lb.R ii. ∆U = mcv(T 2 - T 1 ) = (3)(0.0732)(1379-560) ∆U = 179.85 BTU (e) ∆S = mcpln( ) = (3)(0.1221)ln(1379/560) ∆S = 0.3301 BTU/R
5. While the pressure remains constant at 689.5 kPa, the volume of a system of air changes from
0.567 m³ to 0.283 m³, what are a. Change in U b. Change in H c. Q d. Change in S e. if
the process is non-flow and internally reversible, what is the work?
Given:
P = 689.5 kPa
V 1 = 0.567 m^3 V 2 = 0.283 m^3
Solution:
(a) i. ∆U = mcv(T 2 - T 1 ) = mcv( )
∆U = (V 2 - V 1 )
= x (0.283-0.567)
∆U = 490 kJ
(b) ∆H = mcp(T 2 - T 1 ) = mcp( )
∆H = (V 2 - V 1 )
= (0.283-0.567) ∆H = - 686.39 kJ
(c) Q = mcp(T 2 - T 1 )= ∆H Q = -686.39 kJ
(b) W = p 1 V 1 ln(V 2 / V 1 ) = (17280)(7.093)ln(42.5/7.093) = (219443.50 lb.ft)( ) W = 282.06 BTU
(c) Q = U + W; U= 0 Q = 282.06 BTU
(d) U= 0
7. If 10 kg/min of air are compressed isothermally from P1=96kPa and V1=7.65m^3/min to
P3=620kPa, find the work change of entropy and the heat for: a) nonflow process and b) steady
flow process with v1=15m/s and v2=60m/s.
Given:
m = 10 kg/min
p 1 = 96 kPa/kJ
V 1 = 7.65 m^3 min
P 2 = 620 kPa/kJ
Solution:
(a) i. p 1 V 1 = p 2 V 2 V 2 = p 1 V 1 /p 2 = (96)(7.65)/ V 2 = 1.185 m^3 /min
ii. Wn = p 1 V 1 ln(V 2 / V 1 ) = (96)(7.65)ln( ) Wn = -1369.63 kJ/min
iii. ∆S = mRln(p 1 /p 2 ) ; Rair = 0.287 kJ/kg.K = (10)(0.287)ln( 20 ) ∆S = - 5.35 kJ/minK