THERMODYNAMICS CHAPTER 4 SOLUTION MANUAL.

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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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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) T2, 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 T1 = 90 F + 460 = 550 °R m = 5 lb Solution: (a) Q = mcv(T2 – T1) i = R = 58.8 x = 0.0756 ii. cv = R/(k-1) = 0.0756/(1.26-1) cv = 0.29 iii. Q = mcv(T2 – T1) 20 = (5)(0.29)( T2- 550) T2 = 563.8 °R (b) i. cp = (kR)/(k-1) = (1.26)(0.0756)/(1.26-1) cp = 0.366 ii. ΔH = mcp(T2 – T1) = (5)(0.366)(563.8-550) ΔH = 25.25 BTU (c) ΔS = mcvln( ) = (5)(0.29)ln ) ΔS = 0.036 (d) ΔU = mcv(T2 – T1)

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= (5)(0.29)(563.8-550) = 20.01 BTU

2. A reversible, non flow, constant volume process decreases the internal energy by 316.5 KJ

for 2.268KG of a gas R=430 J/KG-K and k=1.35. for process determine: a.)the work ; b.) the heat and c.) the change in entropy if the initial temperature is 204.4 °C?

Given:

U = -316.5 kJ k = 1.35

m = 2.268 kg T1 = 204.4 +273 = 477.4 K

R = 430 J/kg.K

Solution:

(a) Wn = __ pdv ; constant volume

Wn = 0 (b) Q = U + Wn = -316.5 + 0 Q = -316.5 kJ (c) i. cv = R/(k-1) = 430/(1.35-1) cv = 1228.57 J/kg.K = 1.22857 kJ/kg.K ii. finding for T2

Q = mcv(T2-T1) -316.5 = (2.268)(1.22857)(T2-477.4) T2 = 363.81 K iii. ΔS = mcvln( ) = (2.268)(1.22857)ln ) ΔS = -0.757 kJ/K

3.10ft^3 vessel of hydrogen at a pressure of 305 psia is vigorously stirred by paddles until the pressure becomes 400 psia. determine ∆U and W. no heat is transferred, Cv = 2.434 btu / lb.R.

Given:

V1 = 10 ft3

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P1 = 305 psia Q = 0 = 43920 lb/ft2 P2 = 400 psia = 57600 lb/ft2 Solution: (a) i. R(hydrogen) = 765.9 lb.ft/lb.R cv = 2.434 BTU/lb.R ii. ∆U = mcv(T2-T1) = mcv( ) ∆U = (p2-p1) = (57600-43920) ∆U = 434.75 BTU

(b) Irreversible nonflow constant volume Q = U + Wn ;Q = 0

Wn = -434.75 BTU

4. 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 T1 = 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(T2-T1) 300 = (3)(0.1221)(T2 - 560) T2 = 1379 R or 919 °F (b) Q= mcp(T2-T1) = H ∆H = 300 BTU (c) Wn = p(V2-V1)

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= p( - ) ; p1= p2 = pmR( - ) Wn = mR(T2-T1) = (3)(38)(1379-560) 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(T2-T1) = (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 R = 0.287 kJ/kg.K V1 = 0.567 m3 cv = 0.7816 V2 = 0.283 m3 cp = 1.00625 Solution: (a) i. ∆U = mcv(T2-T1) = mcv( ) ∆U = (V2-V1) = x (0.283-0.567) ∆U = 490 kJ (b) ∆H = mcp(T2-T1) = mcp( ) ∆H = (V2-V1)

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= (0.283-0.567) ∆H = -686.56 kJ (c) Q = mcp(T2-T1)= ∆H Q = -686.39 kJ (d) ∆S = mcpln( ) = mcpln( ) ∆S = mcpln( ) = cpln( ) = (1.00625)ln(0.283/0.567)

=

-0.699 kJ/kgK (e) Wn = p(V2-V1) = (689.5)(0.283-0.567) Wn = -195.82 kJ

6. Four pounds of air gain 0.491 Btu/°R of entropy during a non-flow isothermal process. If

P1 = 120 psia and V2 = 42.5 ft³, find a. V1 and T1 b. Wnf c. Q and d. Change in U.

Given: m = 4lb Rair = 53.34 lb.ft/lb.R ∆S = 0.491 BTU/R P1 = 120 psia V1 = 42.5 ft3 Solution:

(a) i. ∆S = (0.491 BTU/R)(778 lb.ft/BTU) = 382 lb.ft/R Rair = 53.34 lb.ft/lb.R P1 = (120 lb/in2 )(144 in2/ft2)= 17280 lb/ft2 ii. ∆S= mRln(p1/p2) 382 = (4)(53.34)[ln(17280)]-ln(p2) - ln(17280)]-= -ln(p2) e7.9669 = eln(p2) p2 = 2883.91 lb/ft2

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iii. p1V1 = p2V2 V1 = p2V2/p1 = (2883.91)(42.5)/(17280) V1 = 7.093 ft3 iv. p1V1 = mRT T = p1V1/mR = (17280)(7.093)/(4)(53.34) T = 574.46 R (b) W = p1V1ln(V2/ V1) = (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 p1 = 96 kPa/kJ V1 = 7.65 m3min P2 = 620kPa/kJ Solution: (a) i. p1V1 = p2V2 V2 = p1V1/p2 = (96)(7.65)/620 V2 = 1.185 m3/min ii. Wn = p1V1ln(V2/ V1) = (96)(7.65)ln( ) Wn =-1369.63 kJ/min

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iii. ∆S = mRln(p1/p2) ; Rair = 0.287 kJ/kg.K

= (10)(0.287)ln(

20)

∆S = -5.35 kJ/mink

8. one pound of an ideal gas undergoes an isentropic process from 93.5 psig and a volume of 0.6ft3 to a final volume of 3.6 ft3. Ifcp=0.124 and cv=0.093 BTU/lb.R, what are (a)T2(b)p2(c) ∆H and (d)W

Given:

P1 = (95.3 +14.7) psia = 110 x = 15480 lb/ft 2 V1 = 0.6 ft3 V2 = 3.6 ft3 Cp = 0.124 Cv = 0.093 BTU/lb.R Solution: (a) i. R= cp-cv = 0.124-0.093 R = 0.031 BTU/lb.R ii. T1= T1= 394.063 R iii. ../.. k= cp/cv = 1.3333 ../.. T2= T1[ = (394.063)[ T2= 216.87 R or -243.13 F (b) p1 = p2 p2 = = 10.09 psia (c) ΔH = mcp() = (1)(0.124)(216.87-394.063) ΔH = -21.97 BTU

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(d) W =

=

W = 16.48 BTU

9. A certain ideal gas whose R=278.6 j/kg.K and cp=1.015 kJ/kg.K expands isentropically from 1517kpa,288 C to 965 kPa. For 454g/s of this gas determine (a)Wn(b)V2(c) ∆ U (d) ∆H

Given:

R = 278.6 J/kg.K or 0.2786 kJ/kg Cp = 1.015 kJ/kg.K P1 = 1517 kJ T1 = 288 C +273 = 561 P2 = 965 kJ m = 454 g/s or 0.454 kg/s

Solution:

(a) i. cv=cp-R =1.015-0.2786 cv=0.7364 ii. k=cp/cv =(1.015)/( 0.7364) k= 1.378 iii. T2= T1[ =(561)[ T2 = 495.53 K ../.. Wn = = Wn = 21.9 kJ/s (b) p2V2 =mRT2 V2 =

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= 0.6495 m3/s (c) ΔU = mcv(T2-T1) = (0.454)(0.7364)(495.53-561) ΔU = -21.888 kJ/s (d) ΔH = mcp(T2-T1) = (0.454)(1.015)( 495.53-561) ΔH = -30.169 kJ/s

10.A polytropic process of air from . . . ../..

Given:

P1=150 psia T1=300 F+460 =760 R V1=1 ft3 P2=20 psia n=1.3 Rair=(53.34 lb.ft/lb.R)( ) = 0.06856 BTU/lb.R cv=0.1714 cp= 0.24

Solution:

(a) i. = T2=(760)[ = 477.39 R -460 =17.40 F ii. p1V1n=p2V2n V2 = [p1V1n/p2]1/n =[(150)(1)1.3/(20)]1/1.3 V2 = 4.711 ft3 (b.) i. p1V1=mRT1 m= = 0.5328 lb ii. ΔU = mcv(T2-T1) = (0.5328)(0.1714)(477.39-760) = -25.81 BTU

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iii. ΔH=mcp(T2-T1) = (0.5328)(0.24)(477.39-760) = -36.14 BTU iv. ../.. k=cp/cv =0.24/0.1714 k=1.40 ../.. cn=cv[(k-n)/1-n] =(0.1714)[(1.40-1.3)/(1-1.3)] cn=-0.571 ΔS= mcnln(T2/T1) =(0.5328)(-0.0571)ln(477.39/760) ΔS=0.0141 BTU/R (c) i.pdV= = = 34.41 BTU ii.- Vdp =n(pdV) = (1.3)(34.41) = 44.73 BTU (d)Q= mcn(T2-T1) =(0.5328)(-0.0571)(477.39-760) Q=8.60 BTU

Check:

Q=ΔU+ pdV = -25.81 + 34.41 Q=8.60 BTU (e) Wn = = Wn = 34.41 BTU (f) Ws = Q- ΔU = 8.60-(-36.14) Ws = 44.7 BTU

Figure

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