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:
The correct answer will be "18 degrees AOB".
Explanation:
The given value is:
KTAS = 120
For determining bank angles through standard rate switches, as we understand
⇒
On substituting the given value of KTAS, we get
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Answer:
L= 312.75 mm
Explanation:
given data
elastic modulus E = 106 GPa
cross-sectional diameter d = 3.9 mm
tensile load F = 1660 N
maximum allowable elongation ΔL = 0.41 mm
to find out
maximum length of the specimen before deformation
solution
we will apply here allowable elongation equation that is express as
ΔL = ....................1
put here value and we get L
L =
solve it we get
L = 0.312752 m
L= 312.75 mm