Wednesday, October 22, 2008

TE 303 - HW #5, P7 - Waste Heat Steam Generator - 15 pts

At steady-state, water enters the waste heat recovery steam generator shown in the diagram at 42 psia and 220oF and exits at 40 psia and 320oF. The steam is then fed into a turbine from which it exits at 1 psia and a quality of 90%.



Air from an oven exhaust enters the steam generator at 360oF and 1 atm with a volumetric flow rate of 3000 ft3/min and exits at 280oF and 1 atm. Ignore all heat exchange with the surroundings and any changes in potential and kinetic energies. Determine the power developed by the turbine in horsepower.

CP,air = 7.05 Btu/lbmole-oF.

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1 comment:

Anonymous said...

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HINTS:
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* Remember to look at the comments from Dr. B's class

* Apply the first law to the heat exchanger to determine the mass flow rate of the water in streams 1,2, and 3. Then, apply the first law to the turbine to determine its power output in hP.

* The analysis of the heat exchanger can be done in multiple ways but should yield the same result.

* You can assume that the air behaves as an ideal gas with a constant Cp (given in problem). You can write the ideal gas EOS to include the rates: P Vdot = ndot R T = mdot R T / MW.

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ANSWERS:
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* mass flow rate at 1 = 2.803 lbm/min
* Wsh = 12.9 hP