Hot combustion gases (assumed to have the properties of air at room temperature)

Question

Hot combustion gases (assumed to have the properties of air at room temperature) enter a gas turbine at 0. 8 MPa and 1500 K at a rate of 2. 1 kg / s, and exit at 0. 1 MPa and 800 K If heat is lost from the turbine to the surroundings at a rate of 150 kJ / s, find the power output of the gas turbine State the system assumptions and write the energy balance equation for this system, If heat is lost from the turbine to the surroundings at a rate of 150 kJ / s find the power output of the gas turbine.

Explanation / Answer

(a)assumptions 1 Steady operating conditions exist. 2 The working fluid is that can be assumed to be an idea gas . 3 Kinetic and potential energy changes are negligible. 4 The variation of specific heats with temperature is to be considered. energy equation for any control volume system energy(in) - energy(out) + Q(heat) + w(work done on the system)= change in the energy of control volume as this is steady state the change in energy of control volume will be zero and work done on the system is also zero and heat is being disputed to the surrounding so Q will be negative. energy terms can be replaced by the enthalpy because all the other energies are not changing enthalpy(in) - enthlpy(out) -Q= w(output) (b) applying the values of the enthalpy 2.1*1635.97- 2.1*821.95 - 150 = w w(output)=1559.442Kj/s

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