Friday, April 14, 2006
HW #5 - P #9 - Operation of a Pressure Cooker - 8 pts
A 4 L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of water for one hour, determine the highest rate of heat transfer allowed.
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7 comments:
graham 4:38 PM
Mass does leave the system. So, yes, you cancelled too many things.
Initially, the pot contains half and half liq and vap , by volume. At the end it contains all vap. Where did all the water go ? It went out through the check valve at the top of the lid which keeps the P constant throughout this process.
Yes, Wsh = 0.
Getting the quality and U^ intial and final for the system is a good idea.
H^ out is H^ of sat vap at 175 kPa.
Solve the transient mass and energy balance eqns for mout and Q.
This may be really obvious, but how do you solve for x at State 1?
I'd have backsolved and figured it out on my own, but the U1^ is dependent on x...
to find the x of the first state you use the specific volumes of sat liq and vapor. since half the initial volume is liq you can divide .002 by the specific volume to get a mass. x is then equal to the mass of vapor/mass total.
So I cancel till i get the equation deltaEsys=Q-mout(H^). How do I solve for delta Esys? Is it U^2(m2)-U^1(m1)? or am i really confused
Steve 2:37 PM
The vapor and liquid initially in the pot are saturated. Look up the Vhat of each. Use these to determine the mass of liquid and the mass of vapor in the pot. x = mvap/(mvap+mliq).
Anon 5:01 PM
Yes. Thank you for the explanation.
Anon 5:04 PM
It is U2^(m2)- U1^(m1).
You are not really confused !
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