A piston-cylinder device initially contains 2.8 lbm of water, 1.2 lbm of which i

Question

A piston-cylinder device initially contains 2.8 lbm of water, 1.2 lbm of which is in the liquid form, at a pressure of 140 psia. Heat is now added to the water at constant pressure until a final temperature of 450 oF is reached.
a) Determine the initial values of temperature, specific volume and specific internal energy
b) Determine the final values of specific volume and specific internal energy
c) Calculate the overall change in volume.
d) Calculate the overall change in total internal energy.
e) Show the process (State 1 - State 2) on a T-v diagram (show the initial and final values of T & v).

Explanation / Answer

The cylinder is a control volume of air. It is insulated and therfore adiabatic, so Q = 0. By the continuity equation, m2 = m1 = m (26) From the energy equation (5.11), m(u2-u1) = 1Q2 -1W2 = -1W2 (27) From the entropy equation (8.37), m(s2-s1) = Z dQ T + 1S2gen = 0+1S2gen (28) Pressure and temperature are known at State 1, but only pressure is known at State 2, so T2 must be obtained. Assume a reversible process. 1S2gen = 0)s2 -s1 = 0 (29) State 1: Using Table A.7, u1 = 1095.2 kJ/kg and s±T1 = 8.5115 kJ kg·K m = P1V1 RT1 = 15000 · 9 · 10-6 0.287 · 1380 = 0.000341 kg (30) State 2: From the entropy equation, s2 = s1 so from Eq. 8.19, s±T2 = s±T1+Rln P2 P1 = 8.5115+0.287ln µ 200 15000 ¶ = 7.2724 kJ kg ·K (31) Then, use Table A.7 to interpolate and find T2 = 447.2 K and u2 = 320.92 kJ/kg. V2 =V1 T2P1 T1P2 = 9 · 447.2 · 15000 1380 · 200 = 218.7 cm3 Then, use Table A.7 to interpolate and find T2 = 447.2 K and u2 = 320.92 kJ/kg. V2 =V1 T2P1 T1P2 = 9 · 447.2 · 15000 1380 · 200 = 218.7 cm3 (32) )L2 = V2 Acyl = 218.7 5 = 43.74 cm (33) 1w2 = u1-u2 = 774.3 kJ/kg (34) 1W2 = m1w2 = 0.264 kJ = 264 J (

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