Answer :
Answer:
a ) [tex]\dot m = 351.49 kg/s[/tex]
b) [tex]\dot m_{actual} = 1046.15 kg/s[/tex]
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
[tex]\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}[/tex]
[tex]\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}[/tex]
[tex]T_2' = 610.181 K[/tex]
AT the end of isentropic process (expansion) temperature is
[tex]\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}[/tex]
[tex]\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}[/tex]
[tex]T_4' = 491.66 K[/tex]
isentropic work is given as
[tex]w(compressor) = CP (T_2' -T_1)[/tex]
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
[tex]\dot m = \frac{70000}{510.88 - 311.73}[/tex]
[tex]\dot m = 351.49 kg/s[/tex]
b) actual mass flow rate uis given as
[tex]\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}[/tex]
[tex]\dot m_{actual} = 1046.15 kg/s[/tex]