A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage
at 12 MPa, 5608C and expands to 1 MPa, where some of the steam is extracted and diverted to the open feedwater heater operating at 1 MPa. The remaining steam expands through the second turbine stage to the condenser pressure of 6 kPa. Saturated liquid exits the open feedwater heater at 1 MPa. The net power output for the cycle is 330 MW. For isentropic processes in the turbines and pumps.
Determine:
a. the cycle thermal efficiency.
b. the mass flow rate into the first turbine stage, in kg/s.
c. the rate of entropy production in the open feedwater heater, in kW/K.
Determine:
a. the cycle thermal efficiency.
b. the mass flow rate into the first turbine stage, in kg/s.
c. the rate of entropy production in the open feedwater heater, in kW/K.
1 answer:
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Answer:
Maximum number of vehicle = 308
Explanation:
See the attached file for the calculation.
Answer:
(a) 0 kJ
(b) 9.81 kJ
(c) 31.32 m/s
Explanation:
(a)
From the law of conservation of energy, energy can only be transformed from one state to another. At a height of 50 m, all the kinetic energy is converted to potential energy hence KE=0
(b)
Potential energy, PE=mgh where m is the mass, g is acceleration due to gravity and h is the height
Substituting 50 m for h and 20 Kg for m, taking g as 9.81 then
PE=20*9.81*50=9810 J=9.81 kJ
(c)
Relating the equation of potential energy to the equation of kinetic energy, which is
where v is the velocity of the mass
Substituting 50 m for h and taking g as 9.81 then
