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Question 9.7: Actual Gas-Turbine Cycle with Regeneration Determine the the...

Actual Gas-Turbine Cycle with Regeneration

Determine the thermal efficiency of the gas-turbine described in Example 9-6 if a regenerator having an effectiveness of 80 percent is installed.

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The gas-turbine discussed in Example 9-6 is equipped with a regenerator. For a specified effectiveness, the thermal efficiency is to be determined.

Analysis The T-s diagram of the cycle is shown in Fig. 9-41 . We first determine the enthalpy of the air at the exit of the regenerator, using the definition of effectiveness:

 

\epsilon =\frac{h_{5}-h_{2 a}}{h_{4 a}-h_{2 a}}

 

0.80 =\frac{\left(h_{5}-605.39\right) \mathrm{kJ} / \mathrm{kg}}{(880.36-605.39) \mathrm{kJ} / \mathrm{kg}} \rightarrow h_{5}=825.37 \mathrm{~kJ} / \mathrm{kg}

 

Thus,

q_{\mathrm{in}}=h_{3}-h_{5}=(1395.97-825.37) \mathrm{kJ} / \mathrm{kg}=570.60 \mathrm{~kJ} / \mathrm{kg}

 

This represents a savings of 220.0 kJ/kg from the heat input requirements. The addition of a regenerator (assumed to be frictionless) does not affect the net work output. Thus,

 

\eta_{\mathrm{th}}=\frac{w_{\mathrm{net}}}{q_{\mathrm{in}}}=\frac{210.41 \mathrm{~kJ} / \mathrm{kg}}{570.60 \mathrm{~kJ} / \mathrm{kg}}=0.369 \text { or } 36.9 \%

 

 

This represents a savings of 220.0 kj/kg from the heat input requirements.
The addition of a regenerator (assumed to be frictionless) does not affect the net work output. Thus,

 

\eta_{\mathrm{th}}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{210.41 \mathrm{~kJ} / \mathrm{kg}}{570.60 \mathrm{~kJ} / \mathrm{kg}}=0.369 \text { or } 36.9 \%

 

 

Discussion Note that the thermal efficiency of the gas turbine has gone up from 26.6 to 36.9 percent as a result of installing a regenerator that helps to recuperate some of the thermal energy of the exhaust gases.

9.41

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