THERMODYNAMICS CHAPTER 4(1-10) by Hipolito Sta. Maria 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)
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.
= (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 m = 2.268 kg R = 430 J/kg.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 cv= 2.434 BTU/lb.R
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 V1 = 0.567 m3 V2 = 0.283 m3
= -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 ∆S = 0.491 BTU/R P1 = 120 psia V1 = 42.5 ft3
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.
∆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
(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)[
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