## Thermodynamics: Addition of saturated steam to closed tank

1. The problem statement, all variables and given/known data
A well-insulated closed tank is of volume 70 m^3. Initially, it contains 25000 kg of water distributed between liquid and vapor phases at 30 deg C. Saturated steam at 11 bar is admitted to the tank until pressure reaches 7 bar. What mass of steam is added?

2. Relevant equations
m' = m2 - m1 (where m' is mass of steam, m2 is final mass in tank and m1 is initial mass in tank, m1 = 25000 kg)
m2H2 - (m1H1 + m'H') = 0
m2/m1 = (U1 - H')/(U2 - H') = (H' -U1)/(H' - U2)

3. The attempt at a solution
Q = 0 if well insulated
Assume W = 0
Kinetic and potential energies can be ignored
V(tank) = constant and is taken to be the control volume
H' = 2779.7 kJ/kg (from steam tables)

I know m1 and H'. I thought if volume of tank is constant than i can calculate P1 from P1V1=P2V2 but then that just gives me that the pressure is constant too. I need to figure out the enthalpies of the contents in the tank before and after the addition of the sat. steam and then I should be okay. Can anyone help? Thanks!
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 I have since worked further on this problem and ended up with 5 equations, 5 unknowns but it just seems too complicated for this problem. Can anybody help? m2 = m' + m1 = m' + 25000 = m(Liquid)2 + m(Vapor)2 (at 7 bar) m(L)2*V(L)2 + m(V)2*V(V)2 = 70 m^3, where V(L)2=0.1108 & V(V)2=0.2727 H2 = m(V)2*H(V)2 + m(L)2*H(L)2, where H(V)2=2762 & H(L)2=697.061 m1H1 +m'H' = m2H2, where m1=25000, H'=2779.7 H1=m(V)1*H(V)1 + m(L)1*H(L)1 = 3.146 x 10^6 kJ
 I think you went down the wrong road with this problem. I would approach this problem by simply adding up the enthalpy H. You know the enthalpy of the original contents and the incoming enthalpy. You also know the enthalpy of the final state. Since the system is closed, it should be a simple relation of; Hf = Hi + Hadded H = h*m Since you know the condition of enthalpy added, you just have to reverse calculate to get your mass.

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