Question #215007

Steam enters with negligible velocity a well-insulated turbine operating at 

steady state at 4 MPa, 320°C. The steam expands to an exit pressure of 0.07 

MPa, a quality (dryness fraction) of 0.95 and a velocity of 90 m/s. The diameter 

of the exit is 0.6 m. Neglecting potential energy effects, find the power 

developed by the turbine in kW



1
Expert's answer
2021-07-08T13:17:24-0400

Gives


P1=4MPa,T1=320°c,P2=0.07Mpa,v=90m/secP_1=4MPa,T_1=320°c,P_2=0.07Mpa, v=90m/sec

r=0.3m

T2=27°cT_2=27°c

Quality=0.95

m˙=Av\dot{m}=Av


A=πr2=3.14×0.32=0.2826m2A=\pi r^2=3.14\times0.3^2=0.2826m^2

m˙=0.2826×90=25.434kg/sec\dot{m}=0.2826\times90=25.434kg/sec

Power

P=dQdtP=\frac{dQ}{dt}

We know that

Q˙=m˙cpT\dot{Q}=\dot{m}c_p∆T


dQdt=25.43×4.18(32027)=31145.13J/sec\frac{dQ}{dt}=25.43\times4.18(320-27)=31145.13J/sec

Now

P=31.145KJ/secP=31.145KJ/sec


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