Air expands through a turbine from 9 bar, 920 K to 2 bar, 510 K. The inlet velocity is small compared to the exit velocity of 110 m/s. The turbine operates at steady state and develops a power output of 3220 kW. Heat transfer between the turbine and its surroundings and potential energy effects are negligible. Calculate the mass flow rate of air, in kg/s, and the exit area, in m².

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Netta00 Netta00 · Answer 1 · 2019-10-04T09:59:20+02:00

Answer:

m =7.93 kg/s

[tex]A_2=0.054\ m^2[/tex]

Explanation:

Given that

[tex]T_1=920\ K[/tex]

[tex]T_2=510\ K[/tex]

Inlet velocity is small so

[tex]V_1=0[/tex]

[tex]V_2=110\ m/s[/tex]

Power develops ,W= 3220 KW

Now from first law of thermodynamics for open system at steady state

[tex]m.h_1+\dfrac{m.V_1^2}{2000}+Q=m.h_2+\dfrac{m.V_2^2}{2000}+W[/tex]

Here given Q=0

m is the mass flow rate in kg/s.

For air

,[tex]C_p=1.005 KJ/kg.k[/tex]

h = m. Cp. T

[tex]m.h_1=m.h_2+\dfrac{m.V_2^2}{2000}+W[/tex]

[tex]m\times 1.005\times 920=m\times 1.005\times 510+\dfrac{m\times 110^2}{2000}+3220[/tex]

[tex]m\times (1.005\times 920- 1.005\times 510-\dfrac{ 110^2}{2000})=3220[/tex]

So

m =7.93 kg/s

Mass flow rate of air = 7.93 Kg/s

[tex]m=\rho_2A_2V_2[/tex]

For ideal gas air

[tex]\rho_2=\dfrac{P_2}{RT_2}[/tex]

[tex]\rho_2=\dfrac{200}{0.287\times 510}\ kg/m^3[/tex]

[tex]\rho_2=1.33\ kg/m^3[/tex]

[tex]m=\rho_2A_2V_2[/tex]

[tex]7.93=1.33\times A_2\times 110[/tex]

[tex]A_2=0.054\ m^2[/tex]

This is the exit area.