Answer:
The mass flow rate of the air through the compressor
m'=0.6857
Explanation:
The law of gases help to find the mass flow in the gas-turbine
Ein=Eout
[tex]E_{in} =E_{out}\\W_{in}+m'*(h_{i}+\frac{v_{i}^{2}}{2})=Q_{out}+m'*(h_{o}+\frac{v_{o}^{2}}{2})\\W_{in}-Q_{out}=m'*(h_{o}-h_{i}+\frac{v_{o}^{2}-v_{i}^{2}}{2})[/tex]
Pressure determinate the 'h' the gas turbine move so
[tex]h_{i}=T1\\T1=25 C= 298K=h_{i}=298.2 \frac{kj}{kg}\\h_{o}=T2\\T2=347 C= 620K=h_{o}=628.07 \frac{kj}{kg}[/tex]
revolving for m'
[tex]m'=\frac{W_{in}-Q_{out}}{h_{o}-h_{i}+(\frac{vi^{2} -0}{2})(\frac{1\frac{kj}{kg} }{1000} \frac{m^{2} }{s^{2} } )} \\m'=\frac{254\frac{ki}{s} -1500\frac{ki}{60s}}{628.07-298.2+(\frac{90^{2} -0}{2s})(\frac{1\frac{kj}{kg} }{1000} \frac{m^{2} }{s^{2} } )} \\m'=\frac{229}{333.92}\\m'=0.6857[/tex]
