Sunday, April 15, 2007
HW #5, P3 - Effluent Pressure in a Non-Adiabatic Steam Diffuser - 6 pts
Steam enters a diffuser at a pressure of 14.7 psia, a temperature of 300oF and a velocity of 500 ft/s. Steam exits the diffuser as a saturated vapor with negligible kinetic energy. Heat transfer occurs from the steam to the surroundings at a rate of 19.59 Btu/lbm of flowing steam. Neglecting potential energy effects, determine the exit pressure in psia. Assume the diffuser operates at steady-state.
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10 comments:
can this problem be solved without knowing the cross sectional area?
Are you sure that 19.59 Btu per lbm is a rate? Doesn't a rate always include some sort of per unit time?
I would have to agree with anonymous with 19.59 Btu/lbm not being a rate.
When I solved this problem I got 79 psia, not 60 as you posted on the website. I used H[hat]2 = Q[hat] (which is negative) plus H[hat]1 + (velocity2)^2/gc
Anon 6:45, 4:11, 4:58:
Yes, you can solve this problem without knowing the cross-sectional area for flow.
Qdot / mdot = 19.59 Btu / lbm
Heat is transferred at a rate equal to 19.59 Btu for each pound-mass of fluid flowing through the process. You can think of this as :
Qhat = Qdot / mdot
Questioning:
Yes, Qhat is a negative number. But I think you slipped a sign when you moved the effluent kinetic energy term from one side of the eqn to the other.
Qhat = H2hat - H1hat + v2^2/(2gc)
becomes:
H2hat = Qhat + H1hat - v2^2/(2gc)
Dr. B,
I messed up on typing. i would assume that because the problem statement says that Ek final is negligible that we are only concerned with Ek initial.
If so, then H2hat = Qhat + H1hat + v1^2/(2gc), correct?
I did find my mistake though, which was that I used gc, not 2*gc. In this case, I get 60.78 psia, which I assume is close enough to your answer that there are just some rounding errors...
I got 60.78 psia as well.
Questioning 9:15:
Yes, you are correct. And yes, 60.78 psia is close enough ! Good work.
Anonymous Coward:
Great ! You are all set on this one.
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