Here is the question: A closed rigid tank with .02lb water at 120F and 50% quali

Question

Here is the question:

A closed rigid tank with .02lb water at 120F and 50% quality.recieves 8btu from heat transfer.Determine the new Temperature, pressure and quality of the mixture.

Here is what I know: STATE1 : .01lb vapor, .01lb liquid

p1 - 1.695 lbf/in2

vf - .01621

vg - 203

work should be 0

ke and pe should be 0 so q = m(u2-u1)

by calculation v1 = 101.5    u1 should = 568.945 and h1 should equal 600.75

With nothing constant except volume I dont understand how to get any further. Please help, any advice would be appreciated!

Thank you

Explanation / Answer

Here the Heat added to the system is used only to increase the enthalpy of the system because the added heat will convert the remaining 50 % of water to steam.

Initial enthalpy of the system= 87.99 + 0.5*(1114-87.99) = 601 btu/lbm (steam tables)

Amount of heat added = 8 btu

Amount of heat added per unit of lbu = 8/ .02 =400 btu/lbm

therefore resulating enthalpy of mixture = 601+400 = 1001 btu/lbm < (hg = 1114)

Therefore suprisingly the final temperature of the gas is same but the percentage of steam in it changes.(i.e quality changes)

So final temperature is 120F

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