UofL student designs a steady-state gas turbine which operates at 627degree C an

Question

UofL student designs a steady-state gas turbine which operates at 627degree C and 1.2 MPa (state 1) at a rate of 2.5 kg/s and leave at 527degree C and 500 kPa (state 2). It is estimated that heat is lost from the turbine at a rate of 20-kW for an actual process. using ideal gas air properties for the combustion gases and assuming the surrounding to be at 27degree C and 100-kpa and ignore kinetic and potential energy change, determine a. The actual power outputs [KW] b. The reversible power output (KW] in case of adiabatic process, c. The energy destroyed within the turbine [KW], d. The second-law efficiency of the turbine. (Note: Please set your assumptions clearly and fill out below table for partial credits).

Explanation / Answer

The actual power out put Wt = m x cp x ( T2 -T1 )

= 2.5 x 1.05 (900 - 800)

= 2.5 x 1.05 x 100

= 262.5 Kw

reversible power output during adiabatic process = mx cp x (t4 - t1 )

= 2.5 x 1.05 x (900-300)

= 1575 kw

The exergy destroyed the turbine = 262.5 - 20 = 242.5 kw

second law of efficiency = net work done / heat supplied

heat supplied = cp (T2 - T1 ) = 1.05 (800-500) =315 kj/s

work done =(T2-T3) - (T1 - T4 )/315

=( 600 - 500 )- (300 -500 )

EFFICIENCY = 300 /315

= 0.952

CP ASSUMED AS 1.05 KJ/KG K

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