Question

A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air enters the compressor at 520 R and the turbine at 2000 R. Accounting for the variation of specific heats with temperature, determine (a) the air temperature at the compressor exit, (b) the back work ratio, and (c) the thermal efficiency.

Answer 1

Answer:

The answers to the question are as follows

(a) The air temperature at the compressor exit is 557.56 K

(b) The back work ratio is 0.502

(c) The thermal efficiency is 0.482 or 48.2%

Explanation:

First we list out the given variables

Air temperature compressor inlet = 520 R

Air temperature turbine inlet = 2000 R

Pressure ratio of working fluid 10

The heat added to the air t the compressor is given by

Process 1 - 2      $\frac{T_{2} }{T_{1} } = (\frac{P_{2} }{P_{1} } )^{\frac{\gamma - 1}{\gamma} }$ = $(\frac{10}{1} )^{\frac{1.4 - 1}{1.4} }$ = 1.93

Therefore T₂ = T₁ × 1.93 = 288.89 K × 1.93 = 557.56 K

(a) the air temperature at the compressor exit 557.56 K

(b) The back work ratio is given by $\frac{T_{2} - T_{1} }{T_{3} - T_{4} }$

For the turbine we have $\frac{T_{3} }{T_{4} } = (r_{p} )^{\frac{\gamma - 1}{\gamma} } = (10)^{\frac{1.4 - 1}{1.4} }$ = 1.93

which gives T₄ = $\frac{T_{3} }{1.93}$ = $\frac{1111.11}{1.93}$ = 575.705 K

Therefore the back work ratio = $\frac{557.56 K - 288.89 K}{1111.11 K - 575.705 K}$ = 0.502

(c) The thermal efficiency of a Brayton Cycle is given by

$\eta_{cycle} = 1-\frac{1}{r_{p} ^{\frac{\gamma -1}{\gamma} } }$ = $1-\frac{1}{10^{\frac{1.4 - 1}{1.4} } } =$ 0.482

The cycle efficiency = 0.482 or 48.2%