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

Answer:

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

b)  $\dot m_{actual} = 1046.15 kg/s$

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

$\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}$

$\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}$

$T_2' = 610.181 K$

AT the end of isentropic process (expansion) temperature is

$\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}$

$\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}$

$T_4' = 491.66 K$

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

$\dot m = \frac{70000}{510.88 - 311.73}$

$\dot m = 351.49 kg/s$

b) actual mass flow rate uis given as

$\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}$

$\dot m_{actual} = 1046.15 kg/s$