An 0.5 m^3 rigid tank contains equal volumes of Freon-12 vapor and Freon-12 liqu
ID: 1862594 • Letter: A
Question
An 0.5 m^3 rigid tank contains equal volumes of Freon-12 vapor and Freon-12 liquid at 312K. Additional Freon-12 is added to the tank until the total mass of Freon-12 (liquid + vapor) is 400 kg. (Some vapor is led off to maintain the original temperature and pressure with a final mass of 400 kg.) At 312K, P sat for Freon-12 is 0.9334 MPa. Under these saturation conditions,
vf = 0.000795 m^3/kg and vg = 0.01872 m^3/kg.
What is the final mass of Freon-12 vapor?
a. 10 kg
b. 39 kg
c. 100 kg
d. 300 kg
Explanation / Answer
Initially .25 m^3 Liq. and .25 m^3 Vap. is present.
Initial mass of liquid = .25/.000795 = 314.4665 Kg
Initial mass of vapour = .25/.01872 = 13.3547 Kg
Reaminig volume = 400 - 314.4665 - 13.3547 = 72.1788 Kg
It is equal to 72.1788*.000795 = .05738 m^3 volume of liquid.
So to maintain constant volume , temperature and pressure, 0.05738 m^3 of vapour be displaced ...
Mass of displaced vapour 0.05738/.01872 = 3.065
So Remaining volume of Freon-12 vapour = 13.3547-3.065 = 10.28 which is approximately equal to 10 as given in option.
b) Now from 400 Kg of mixture, mass of liquid Freon-12 = 397-10.28=389.72 Kg
(because we have learnt in previous problem that total mass has dropped by 3.065 Kg)
It's volume = 386.655*0.000795=0.307 which is approximately equal to 0.31
c) Freon-12 added = 72.1788 Kg as shown earlier
d) The nominal values used for air at 300 K are CP = 1.00 kJ/kg.K, Cv = 0.718 kJ/kg.K
SO,
Change in Internal Energy per unit mass = Cv*delta(T)
=0.718*(550-100)= 323 KJ/Kg
which is approximately equal to 320KJ/Kg
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