A polytropic process of air from 150 psia, 300°F, and 1 ft3 occurs to P = 20 psia in accordance with PV1.3 = 2 C. Determine (a) t and V , (b) ∆U, ∆H and ∆S, (c) 22 ∫ pdV and - ∫ Vdp, (d) Compute the heat from the polytropic specific heat and check by the equation Q = ∆U + ∫ pdV, (e) Find the nonflow work and (f) the steady flow work for ∆K = 0. Continue solving for: (d) Compute the heat from the polytropic specific heat and check by the equation Q = ∆U + ∫ pdV, (e) Find the nonflow work and (f) the steady flow work for ∆K = 0. Answer with complete solution and with cancellation of units.
A polytropic process of air from 150 psia, 300°F, and 1 ft3 occurs to P = 20 psia in accordance with PV1.3 = 2 C. Determine (a) t and V , (b) ∆U, ∆H and ∆S, (c) 22 ∫ pdV and - ∫ Vdp, (d) Compute the heat from the polytropic specific heat and check by the equation Q = ∆U + ∫ pdV, (e) Find the nonflow work and (f) the steady flow work for ∆K = 0. Continue solving for: (d) Compute the heat from the polytropic specific heat and check by the equation Q = ∆U + ∫ pdV, (e) Find the nonflow work and (f) the steady flow work for ∆K = 0. Answer with complete solution and with cancellation of units.
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A polytropic process of air from 150 psia, 300°F, and
1 ft3 occurs to P = 20 psia in accordance with PV1.3 = 2
C. Determine (a) t and V , (b) ∆U, ∆H and ∆S, (c) 22
∫ pdV and - ∫ Vdp, (d) Compute the heat from the polytropic specific heat and check by the equation Q = ∆U + ∫ pdV, (e) Find the nonflow work and (f) the steady flow work for ∆K = 0.
Continue solving for:
(d) Compute the heat from the polytropic specific heat and check by the equation Q = ∆U + ∫ pdV,
(e) Find the nonflow work and
(f) the steady flow work for ∆K = 0.
Answer with complete solution and with cancellation of units.
Transcribed Image Text:(k-n
here cn = cv()
Cn= 0子14*
1.4-1-3
= -0.05713 BTUllbm°R
AS = 0.5332 *(-0.05713) ln (44:39)
760
AS = 0.014164 BTU /OR
(c)
SPdv = MACT,-T,)
non flow work
= Wne
|-n
O•5332* 0•0686 * (477.39-760)
|-|: 3
Wnf
Wnf = 34·46 BTU
f R = Cp -Cv = 0•0686 Btu/lbm°R
Flow Work =
n Wn f
(rdP = 1:3 (34-46)
44.80
- JvdP =- 44.80 BTU
Transcribed Image Text:Solution: From ideal gas equation.
PV = MRT → m=
RT,
I psia = 144 ebl ff?
R = 53.3 fH ./(6-aix) °R
(150* 144) *I
53.3 * 760
m = 0.5332
(a) from poly tropic process
1•3-1
20
Tz
P,
760
150
Te = 477.39° R
Te = 17.39 ° F
Va
150
4.71 H
» * = (.
20
Vz = 4.71 f#³
(6) Change in Internal energy = oU = m cu (Ts -Ti)
0·1714 Btu/lbmoR
AU = 0•5332*0•1714 (477.39- 760)
AU = - 25·83 BTU
Change in En thalpey = oH= mcp(T, - T.)
AH = 0.5332 *0-24 (47-39 - 760)
AH = - 36.I65 BTU
{co =
= 0-24 Btu/lbm R}
Change in Entropy = AS = mCn en ( TalTi)
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