Question

A simple Brayton cycle using air as the working fluid has a pressure ratio of 10.9. The minimum and maximum temperatures in the cycle are 280 K and 1410 K. Assuming constant specific heats, an efficiency of 100% for compressor and turbine, Determine:

Answer 1

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

$\frac{T_{2} }{T_{1} } = ( \frac{p_{2} }{p_{1} } )^{\frac{v-1}{v} } = ( \frac{V_{1} }{V_{2} } )^{v-1}$

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^$^{\frac{1.4-1}{1.4} }$

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 .

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