The rigid tank in the figure below has a volume of 60 L and initially contains a

Question

The rigid tank in the figure below has a volume of 60 L and initially contains a two-phase liquid vapor mixture of H20 at a pressure of 15 bars and a quality of 20%. As the tank contents are heated, a pressure-regulating valve keeps the pressure constant in the tank by allowing saturated vapor to escape. Heating continues until no liquid remains in the tank. Neglecting kinetic and potential energy effects, determine
a) The amount of heat transfer in kJ
b) The mass of the vapor that escapes in kg

Explanation / Answer

At 15 bar and quality = 20 %

specific internal energy =

u1 = uf@15 + 0.20 x u fg@15bar

   = 843.075 + 0.20 x 1751.6

= 1193.395 KJ/Kg

when all liquid is converted into vapor:

u2 = u g@15bar

u2 = 2594.675 KJ/Kg

V initial = 60 L = 0.06 m^3

V final = 0

At P =15 bar . T saturation = 198.30 C = 471.3 K

P.V =m.R water x T

15 x 105 x 0.06 = m x 461.89 x 471.3

mass total = 0.413 Kg

Out of this 80 % is liquid water.

mass evaporated = 0.80 x 0.413 = 0.33 Kg

Now Q = U + W

Q = (U2 - U1) + P.dV

   = 0.413 x (2594.675 - 1193.395) x 103 + 15 x 105 x (0.06 - 0)

   = 668.73 KJ

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