The gas turbine power plant (shown in the figure below) operates in steady-state

Question

The gas turbine power plant (shown in the figure below) operates in steady-state regime with the constant mass flow rate m.dot=15 kg/s. The atmospheric air (P=101 kPa) is first fed into the compressor, then, it flows through the heat exchanger to be heated up and, eventuallym it flows through the gas turbine to produce work. The shaft of the turbine is connected to the shaft of the compressor, so that 60% of the turbine power output is utilized as a power input to the compressor (i.e. (Wc)=0.6*(Wt) ). The remaining 40% of the turbine power output are used to drive the electric generator connected to the turbine shaft. By assuming that (1) both a compressor and a turbine are adiabatic, (2) kinetic and potential energy changes through each component of the power plant are negligible, and (3) negligible pressure drop through heat exchanger, and by using information from the given table below, determine the following information:

a) Could you use the Carnot Principles to determine the maximum possible effieciency of this power plant? Why or why not?

b) List your assumptions/observations

c) Power input to the compressor [kJ/s]?

d) Power output of the turbine [kJ/s]?

e) Net power output of the power plant [kJ/s]?

f) Magnitude and direction of heat transfer in the heat exchanger [kJ/s]?

g) Fill in the missing information in the table:

Explanation / Answer

mass flow rate = 15 kg/s

P1 = 101 kPa

Wc = 0.6 Wt

Egen = 0.4 Wt

a) we cannot use carnot's efficieny because this is not a constant temperature-constant entropy cycle.

c) input to compressor = m*R*(T1-T2)/n-1 = 15*0.287*(400-20) / 0.4 = 4089.75 kJ/s

d) output by turbime = Wc/0.6 = 6816.25 kJ/s

e)Net power output = o.4*Wt = 2726.5 kJ/s

f) Heat transferred by Exchanger = m*Cv*(T3-T2) = 15 * 0.287 * 2.5 * (1333.33 - 700 ) = 6186.25 kJ

from location 3 to location 2

g) 6816.25 = 15*0.287*(T-700 )/ 0.4

=> T = 1333.33 C

And as negligible pressure drop across heat exchanger, P = 400 kPa

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