Consider 2 kg of steam initially at 0.150 MPa and 400 K in a well-insulated, iso

Question

Consider 2 kg of steam initially at 0.150 MPa and 400 K in a well-insulated, isolated, closed, rigid container. This steam is heated to 700 K. a. What is v2? b. What is the change in internal energy of the steam in J? c. What is the amount of work done by the system? d. Determine the amount of heat added to the system in kJ.

Explanation / Answer

just please change values........ P1 = 90 kPa T1 = 22 deg. C = 295.15 K P2 = 900 kPa 1) Reversible adiabatic process, k = 1.4 PV^k = constant P1V1^k = P2V2^k P2/P1 = (V1/V2)^k PV = RT PV/T = R = constant P1V1/T1 = P2V2/T2 V1/V2 = (P2/P1)*(T1/T2) P2/P1 = [(P2/P1)*(T1/T2)]^k P2/P1 = (P2/P1)^k * (T1/T2)]^k P2/P1 / (P2/P1)^k = (T1/T2)]^k T2 = T1*(P2/P1)^k-1/k a) Constant specific heat, Energy balance, du = ?Q - ?W, dQ = 0(adiabatic) ?CvdT = -??W, Cv = constant Cv*(T2 - T1) = -W W = Cv*(T1-T2) b) variable specific heat du = ?Q - ?W, dQ = 0(adiabatic) ?du = - ??W W = U(T1) - U(T2) --> internal energy as function of temperature can be found in table of property of thermodynamic or any other media. 2) m = 5 kg T1 = 427 C = 700 K P1 = 600 kPa P2 = 100 kPa W = 600 kJ (a)?S Energy balance ?U = Q - W, Q = 0 (adiabatic) ?U = - W Cv(T2-T1) = - W Cv(T1-T2) = W T2 = T1 - W/Cv Tds = dh - vdp ds = dh/T - (v/T)dp, dh/dT = Cp and v/T = R/p ?s = Cp*ln(T2/T1) - R ln(P2/P1) ?S = m*[Cp*ln(T2/T1) - R ln(P2/P1)] b) If ?S > 0, then the process is realistic. 3) Pi = 4 Mpa Ti = 350 C Pe_sat = 120 kPa (saturated vapor) Isentropic efficiency, ?s = hi - he(s) / hi - he hi = h(Ti,Pi) he = h(Pe_sat) he(s) = h(si, Pe_sat)

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