a.) The mass flow rate of the steam at each of the two exits.
b.) The diameter in meters of the duct through which the 0.5 MPa steam is extracted if the velocity there is 20 m/s.
Sunday, April 15, 2007
HW #5, P4 - Analysis of a Two-Stage, Adiabatic Turbine - 6 pts
A well-insulated two-stage turbine operating at steady-state is shown in the diagram. Steam enters at 3 MPa and 400oC with a volumetric flow rate of 85 m3/min. Some steam is extracted from the turbine at a pressure of 0.5 MPa and a temperature of 180oC. The rest expands to a pressure of 6 kPa and a quality of 90%. The total power developed by the turbine is 11,400 kW. Changes in kinetic and potential energies are negligible. Determine:
a.) The mass flow rate of the steam at each of the two exits.
b.) The diameter in meters of the duct through which the 0.5 MPa steam is extracted if the velocity there is 20 m/s.
a.) The mass flow rate of the steam at each of the two exits.
b.) The diameter in meters of the duct through which the 0.5 MPa steam is extracted if the velocity there is 20 m/s.
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6 comments:
Dr b,
How do we determine the values of H2, H3 and H4 if they are different? Or may we assume that they all are the same? Has the pressure changed to .5 MPa before or after the splitter? Has it changed to 6 kPa before or after the second turbine?
confused,
Use the picture to sort out where stuff is coming in and going out. It helped me to think of the system as both turbines and the splitter. Then you don't have to consider states 2 and 3 at all. The pressure has changed to .5 MPa after the splitter and 6 kPa after the second turbine.
The problem statement seems kind of vague... is P3 0.5 MPa or 6kPa?
Confused:
H2 = H3 = H4 because the stream splitter is just a tee in a pipe. P2 = P3 = P4 = 0.5 MPa. P5 = 6 kPa.
Turbines drop P and turn that energy (enthalpy) into shaft work.
Sparticus:
Thank you again.
Questioning:
P2 = P3 = P4 = 0.5 MPa. P5 = 6 kPa.
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