Question

(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet temperature of 300 K, and a turbine inlet temperature of 1000 K. Determine the required mass flow rate of air for a net power output of 70 MW, assuming both the compressor and the turbine have an isentropic efficiency of (a) 100 percent and (b) 85 percent. Answers: (a) 352 kg/s, (b) 1037 kg/s

Answer 1

a )
`\dot m = 351.49 kg/s`

b)
`\dot m_(actual) = 1046.15 kg/s`

Step-by-step explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature = 1000 K

AT the end of isentropic process (compression) temperature is

`(T_2')/(T_1) = rp ^{(\gamma -1)/(\gamma)}`

`(T_2')/(300) = 12^{(1.4 -1)/(1.4)}`

`T_2' = 610.181 K`

AT the end of isentropic process (expansion) temperature is

`(T_3)/(T_4') = rp ^{(\gamma -1)/(\gamma)}`

`(1000')/(T_4') = 12^{(1.4 -1)/(1.4)}`

`T_4' = 491.66 K`

isentropic work is given as

`w(compressor) = CP (T_2' -T_1)`

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

`\dot m = (70000)/(510.88 - 311.73)`

`\dot m = 351.49 kg/s`

b) actual mass flow rate uis given as

`\dot m_(actual) = (70000)/(51.088* 0.85 - (311.73)/(0.85))`

`\dot m_(actual) = 1046.15 kg/s`