When you are getting ready to turn you must signal at least
2 answers:
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Answer:
a) specific work output of the actual turbine is 73.14 Btu/lbm
b) the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R
c) Isentropic efficiency of the turbine is 70.76%
Explanation:
Given the data in the question;
For an adiabatic turbine; heat loss Q = 0
For Initial State;
p₁ = 120 psia
T₁ = 500°F = 959.67°R
from table; { Gas Properties of Air }
At T₁ = 959.67°R
= 0.74102 Btu/lbm°R
= 230.98 Btu/lbm
For Finial state;
p₂ = 15 psia
T₂ = 200°F = 659.67°R
= 0.64889 Btu/lbm°R
= 157.84 Btu/lbm
we know that R for air is 0.06855 Btu/lbm.R
a)
The specific work output of the actual turbine Wₐ is;
W =
-
we substitute
W = 230.98 - 157.84
W = 73.14 Btu/lbm
Therefore, specific work output of the actual turbine is 73.14 Btu/lbm
b)
amount of specific entropy generation during the irreversible process.
To determine the entropy generation ;
= ΔS =
-
=
-
- R ln(
)
we substitute in our values
= 0.64889 - 0.74102 - 0.06855 ln(
)
= 0.64889 - 0.74102 + 0.1425457
= 0.050416 Btu/lbm°R
Therefore, the amount of specific entropy generation during the irreversible process is 0.050416 Btu/lbm°R
c)
Isentropic efficiency of turbine η
η = {actual work output] / [ ideal work output ] = (
-
) / (
-
)
Now, for an ideal turbine;
ΔS = 0 = -
so, - s₁ =
-
- R ln(
)
0 = -
- R ln(
)
=
+ R ln(
)
we substitute
= 0.74102 + 0.06855 ln(
)
= 0.74102 - 0.1425457
= 0.59847 Btu/lbm°R
Now, from table; { Gas Properties of Air }
At = 0.59847 Btu/lbm°R;
= 127.614 Btu/lbm
η = [(
-
) / (
-
)] × 100%
we substitute
η = [( 230.98 - 157.84 ) / ( 230.98 - 127.614 )] × 100%
η = [ 73.14 / 103.366] × 100%
η = 0.70758 × 100%
η = 70.76%
Therefore, Isentropic efficiency of the turbine is 70.76%