Question

Hot combustion gases enter a gas turbine at 0.8 MPa and 0.538 m³/kg at a rate of 2.1 kg/s, and exit at 0.1 MPa and 2.29 m³/kg. Heat is lost from the turbine to the surroundings at a rate of 150 kW and the internal energy decreases 904 kW in the process. What is the power output of the gas turbine? (Assume no change in velocity or elevation across the turbine.)

Answer 1

In above problem all the values are given in SI units.

i.e. ;

- massflow rate in kg/s (m)

- Internal energy in J/s (u)

- Pressure in Pa (p)

- Specific volume m³/kg (v)

- Heat lost Q in J/s

So our answer will be in W. Now calculate the Workdone by turbine.

Now according to First law of thermodynamic:

dQ = dU + dW

Here heat is rejected to surrounding so it will be negative.

Now (internal energy + pressure energy+ velocity energy + elevation energy) Q = (internal energy + pressure energy + velocity energy + elevation energy) + W.

But in the given data change in velocity and elevation is zero so these both energies become zero.

(u1 + mp1v1) Q = ( u2 + mp2v2) + W    (Where the W is the workdone)

So W= {(u1 - u2) + (mp1v1 - mp2v2)} - Q.

Now putting all the values

So after calculating above equation

W = 1176.94 kW.

So Final answer is:- Power output of turbine is 1176.94 kW