Question

Air enters an adiabatic turbine at 800 kPa and 870 K with a velocity of 60 m/s, and leaves at 120 kPa and 520 K with a velocity of 100 m / s. The inlet area of the turbine is 90 cm2. What is the power output?

Answer 1

The power output of the turbine is 603 KW.

Step-by-step explanation:

Turbine is the thermodynamic open system in which fluid looses thermal energy into kinetic energy. Kinetic energy then converted into electric energy.

Here, fluid is air which passes through turbine at 800 Kpa and 870 K with a velocity of 60 m/s.

The turbine is an adiabatic turbine that means there is no heat transfer from the surrounding. Finally the air leaves the turbine at 120 Kpa and 520 K with a velocity of 100 m/s. The turbine inlet area is 90 cm2

Given:

Inlet pressure is
`P_(1)=800`kpa.

Inlet temperature is
`T_(1)=870`K.

Inlet velocity is
`V_(1)=60` m/s.

Outlet pressure is
`P_(2)=120`Kpa.

Outlet temperature is
`T_(2)=520`K.

Outlet velocity is
`V_(2)=100` m/s.

Inlet area of turbine is A=90 cm2.

Step1

Convert the area into SI unit as follows:

`A=90 cm^(2)((1 m^(2))/(10^(4)cm^(2)))`

`A=0.009 m^(2)`

Step 2

Consider air as an ideal gas. So, ideal gas equation is applicable. For air, gas constant is 287 j/kgK.

Ideal gas equation is expressed as follows:

`P=\rho RT`

Here, P is pressure, T is temperature and \rho is density.

Density of air is calculated by ideal gas equation as follows:

`\rho =(P)/(RT)`

`\rho =(800* 10^(3))/(287*870)`

`\rho =3.2039 kg/m^(3)`

Step 3

Mass flow rate is calculated as follows:

`\dot{m}=\rho  AV_(1)`

`\dot{m}=3.2039* 0.009*60`

`\dot{m}=1.73 Kg/s`

Step 4

Steady state equation is the equation of first law of thermodynamics for the open system

Steady state equation for the turbine as follows:

`h_(1)+(v^(2)_(1))/(2000)+Z_(1)+Q=h_(2)+(v^(2)_(2))/(2000)+Z_(2)+W`

Heat transfer is zero as the process is adiabatic. So value of Q is zero.

Turbine is taken as at the same level. So the value of
`Z_(1)` is equal to
`Z_(2)`.

Substitute the value of Q as zero and tex]Z_{1}[/tex] is equal to
`Z_(2)` in steady state equation as follows:

`h_(1)+(v^(2)_(1))/(2)+Z_(1)+0=h_(2)+(v^(2)_(2))/(2)+Z_(1)+W`

`h_(1)+(v^(2)_(1))/(2)+0=h_(2)+(v^(2)_(2))/(2)+W`

`W=(h_(1)-h_(2))+(v^(2)_(1)-v^(2)_(2))/(2000)`

`W=c_(p)(T_(1)-T_(2))+(60^(2)-100^(2))/(2000)`

Specific heat at constant pressure is 1.005 kj/kgK for air.

Substitute the values of temperature and specific heat at constant temperature in the above simplified steady state equation as follows:

W=1.005(870-520)-3.2

W=351.75-3.2

W=348.55 Kj/kg.

Step 5

Power of the turbine is calculated as follows:

`P=\dot{m}W`

`P=1.73*348.55`

P=603 KW

Thus, the power output of the turbine is 603 KW.