Question

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 1

Solution:

For isobaric process 1-2,

W12 = / PAV

W12 = P(V2 - Ví) = 2 x 10 (0.3-0.1) = 40 kJ

Using the ideal gas equation,

PV = mRT

.. (200) (0.1) = m(0.287)(300)

tim = 0.2323 kg

Therefore,

W1240 0.2323 == 172.2 kJ/kg m

For isothermal process 2-3,

W23 = PxVà ln () = (200)(0.3) In () = 41.589 kJ/kg

Therefore,

W23 41.589 – 179.031 kJ/kg m 0.2323

Now,

Total work,

W = W12 + W23 = 172.2 +179.031 = 351.231 kJ/kg

To =T; x 2 = 300 x 3 = 900 K

For isobaric process 1-2,

Q12 = (uz - u) + W12

Using the ideal gas properties of air table,

For,

Ti = 300 K, uz = 214.07 kJ/kg

T2 = 900 K, uz = 674.58 kJ/kg

Therefore,

Q12 = (674.58 - 214.07) + 172.2 = 632.71 kJ/kg

For isothermal process 2-3,

Q23 = (uz – uz) + W 23

Q23 = 0 + 179.031 = 179.031 kJ/kg

Therefore,

Total heat transfer,

Q = Q12+Q23 = 632.71 +179.031 = 811.741 kJ/kg