Here is the question: A closed rigid tank with .02lb water at 120F and 50% quali
ID: 1861167 • Letter: H
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