Container A of volume 0.6 m3 with nitrogen at 3 MPa and 400 C is connected throu

Question

Container A of volume 0.6 m3 with nitrogen at 3 MPa and 400 C is connected through a pipe with a valve to container B of volume 1 m3, which contains nitrogen at 0.2 MPa and 150 C. The containers and the piping are well insulated. The valve is opened, and nitrogen flows slowly from A to B until the pressures and temperatures equalize, at which time the valve is closed. Find (a) the final temperature in each container (K), (b) the final mass in each container (kg), and (c) the total change in entropy (kJ/K).

Explanation / Answer

For nitrogen, molecular mass M = 28 Hence, Gas constant R = Ru/M = 8314/28 = 297 J/kg-K Pa*Va = m_a*R*Ta 3*10^6 * 0.6 = m_a*297*(400+273) m_a = 9 kg Similarly, Pb*Vb = m_b*R*Tb 0.2*10^6 *1 = m_b*297*(150+273) m_b = 1.6 kg After mixing, V = V_a + V_b = 1 + 0.6 = 1.6 m3 and m = m_a + m_b = 9 + 1.6 = 10.6 kg a) Energy balance: m_a*u_a + m_b*u_b = (m_a+m_b)*u or, m_a(u - u_a) + m_b(u - u_b) = 0 or, m_a*Cv*(T - Ta) + m_b*Cv*(T - Tb) = 0 Thus, T(m_a + m_b) = m_a*Ta + m_b*Tb T = (m_a*Ta + m_b*Tb)/(m_a + m_b) T = [9*(400+273) + (1.6*(150+273)]/(9+1.6) T = 631 K (= 358 deg C) b) After mixing, P*V = mRT P*1.6 = 10.6*297*631 P = 12.42 bar After mixing, mass in tank A is, m'_a = PV_a/RT = 12.42*10^5 *0.6/(297*631) = 4 kg After mixing, mass in tank B is m'_b = PV_b/RT = 12.42*10^5 *1/(297*631) = 6.6 kg c) Change in entropy = m_a*R ln(V/Va) + m_b*R ln(V/Vb) = 9*297*ln(1.6/0.6) + 1.6*297*ln(1.6/1) = 2845 J/K = 2.845 kJ/K

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