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Sagot :
Air pressure at the end of the Turbine exit is : 730.57 K
Given data :
Pressure ratio ( p₂ / p₁ )= 10.9
minimum temperature ( T₁ ) = 280 K
maximum temperature ( T₃ ) = 1410 K
Assuming :
constant specific heat and efficiency of 100%
Determine the Air pressure at the end of the turbine exit
For air :
Cp = 1.005 kJ/kg.k, Cv = 0.718 kJ/kg.k, v = 1.4
Given that efficiency for compressor and turbine is 100% the process ( 1-2 , 3 - 4 ) will all be isentropic
We will Apply the formula below
[tex]\frac{T_{2} }{T_{1} } = ( \frac{p_{2} }{p_{1} } )^{\frac{v-1}{v} } = ( \frac{V_{1} }{V_{2} } )^{v-1}[/tex]
Insert values into equation ( 1 )
T₂ = 551.147 K ( temperature at compressor exit )
Next : Determine the value of the temperature at Turbine exit ( T₄ )
T₃ / T₄ = 10.7^[tex]^{\frac{1.4-1}{1.4} }[/tex]
Therefore : T₄ = 1410 / 10.7^0.286
= 1410 / 1.93
= 730.57 K
Hence we can conclude that the Air pressure at the end of the Turbine exit is : 730.57 K .
Learn more about Turbine : https://brainly.com/question/894340
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