3.93 w Air contained in a piston-cylinder assembly undergoes two processes in series, as shown in…
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Question “3.93 w Air contained in a piston-cylinder assembly undergoes two processes in series, as shown in…”
3.93 w Air contained in a piston-cylinder assembly undergoes two processes in series, as shown in Fig. P3:93. Assuming ideal gas behavior for the air, determine the work and heat transfer for the overall process, each in kJ/kg. Isothermal process Ti = 300 K (bar) 1 °C 0.1 0.2 0.3 0.4 V (m) 0.5 0.6 FIGURE P3.93
Answer
Solution:
1-2 isobaric process
The ideal gas equation is
Therefore,
For isothermal processes 2-3
Therefore,
Now,
1-2 isobaric process
Use the ideal gas properties and air table
For,
Therefore,
For isothermal processes 2-3
Therefore,
W12 =/ PAV
W12 = P(V2-Vi) = 2×10 (0.3-0.1), = 40 kJ
PV = mRT
.. (200) (0.1) = m(0.287)(300)
tim = 0.2323 kg
W1240 0.2323 == 172.2kJ/kg
W23 = PxVa () = (200.(0.3) In ) = 41.589 KJ/kg
W23 41.589-179.031 KJ/kg m 0.2323
W = W12 +W23 = 172.2 +179.031 = 351 kJ/kg
To =T; 300 x 3 = 9000 K
Q12 = (uz – u) + W12
Ti = 300 K, uz = 214.07 kJ/kg
T2 = 900 K, uz = 674.58 kJ/kg
Q12 = (674.58 – 214.07) + 172.2 = 632.71 kJ/kg
Q23 = (uz – uz) + W 23
Q23 = 0 + 179.031 = 179.031 kJ/kg
Q = Q12+Q23 = 632.71+179.031 = 8111.741 kJ/kg
Conclusion
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