Air expands through an ideal turbine from 1 MPa, 900 K to 0.1 MPa, 500K. The inlet velocity is small compared to the exit veloci
ty of 100 m/s. The turbine operates at steady state and develops a power output of 3,200 kW. Heat transfer between the turbine and its surroundings and any potential energy effects are negligible.
a. Calculate the mass flow rate of air, in kg/s, and the exit area, in m^2.
a. Calculate the mass flow rate of air, in kg/s, and the exit area, in m^2.
1 answer:
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Answer:
4m/s
Explanation:
We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted
Mathematically
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We also know that Power = force x velocity ..................(i)
The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load
=> power = mg x Velocity........(ii)
While hoisting the load at at constant speed only the potential energy of the mass increases
Thus Potential energy = Mass x g x H...................(iii)
where
g = accleration due to gravity (9.81m/s2)
H = Height to which the load is hoisted
Equating equations (ii) and (iii) we get
m x g x v =
thus we get v = H/t
Applying values we get
v = 6/1.5 = 4m/s