A rigid tank whose volume is 25 m3 is initially containing a

A rigid tank whose volume is 2.5 m^3 is, initially containing air at 1.5 bar, 325 K, is connected by a valve to a large vessel holding air at 8 bar, 295 K. The valve is opened only as long as required to fill the tank with air to a pressure of 4 bar and a temperature of 300 K. As this occurs, a paddle wheel adds 150 kJ per kg of work to the tank (where the kg mass for this work represents the mass amount at the end of the process). Assume that the air behaves as an ideal gas with constant specific heats. Determine the following. (a) The heat transfer between the tank contents and the surroundings in kJ. (b) The amount of entropy production during the process in kJ/K, assuming that the temperature for heat transfer is 295 K.

Solution

a)

Initial mass in tank m1 = P1*V1 / (R*T1)

= 1.5*10⁵ * 2.5 / (287*325)

m1 = 4.02 kg

Final mass in tank m2 = P2 * V2 / (R*T2)

= 4*10⁵ * 2.5 / (287 * 300)

m2 = 11.614 kg

Energy balance: Q - W + h_in*(m2 - m1) = U2 - U1

Q - m2*(-150) + h_in*(m2 - m1) = m2*u2 - m1*u1

Q = -m2*150 - m2*(h_in - u2) - m1*(u1 - h_in)

From ideal gas properties of air, at T = 295 K, we get h_in = 295.17 kJ/kg, s_in = 1.68515 kJ/kg-K

At T1 = 325 K, we get u1 = 232.02 kJ/kg, s1 = 1.78249 kJ/kg-K

At T2 = 300 K, we get u2 = 214.07 kJ/kg, s2 = 1.70203 kJ/kg-K

Q = -11.614*150 - 11.614*(295.17- 214.07) - 4.02*(232.02 - 295.17)

Q = -2430.33 kJ

b)

m2*s2 - m1*s1 + (m2 - m1)*s_in

= 11.614*1.70203 - 4.02*1.78249 + (11.614 - 4.02)*1.68515

= 25.3988 kJ/K

Entropy lost to environment = Q/T = -2430.33 / 295 = -8.238 kJ/K

Entropy production = 25.3988 - 8.238 = 17.16 kJ/K