MECH 310 Problems 6.129 Solution, Fall 2011
American University of Beirut · Thermodynamics I · 1 page · with solution
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Problem 6.129
C.V. Cylinder volume.
Continuity Eq.6.15: Mp - My = My
Energy Eq.6.16: MpUp - MyUVy = Miphjine + Qcy - 1W2
Process: Py is constant to stops, then constant V to state 2 at P>
PiV__ 300 x 0.25
Statel: P}.T; my =RT, 0.287 = 290.2 = 0.90 kg
State 2:
Open to P> = 400 kPa. T, = 350K
400 = 1
My = 9.987 = 350 7 3-982 kg
m, = 3.982 - 0.90 = 3.082 kg
Only work while constant P
1W2 = Py (V2 - Vy) = 300(1 - 0.25) = 225 kJ
Energy Eq.: Qey + mjhj = mgu - myuy + {Wo
Qcy = 3.982 0.717 x 350 - 0.90 x 0.717 x 290.2 + 225
- 3.082 « 1.004 = 600 = -819.2 kJ
We could also have used the air tables A.7.1 for the u’s and hj.