Common Data Question:
In a simple Bryaton cycle, the pressure ratio is $8$ and temperature at the entrance of compressor and turbine are $300 K$ and $1400 K$ respectively. Both compressor and gas turbine have isentropic efficiencies equal to $80$ %. or the gas, assume a constant value of $c_{p}$ (specific heat at constant pressure) equal to $1 k J / k g K$ and ratio of specific heats as $14$ . Neglect changes in kinetic and potential energies

The thermal eficiency of the cycle in percentage % is

24.8

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38.6

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44.8

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53.1

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Solution

The correct option is A $24.8$
For adiabatic process (3-4s\)
$\frac{T_{3}}{T_{4 s}} = {(\frac{p_{2}}{p_{1}})}^{\frac{γ - 1}{γ}} = (8)^{\frac{1.4 - 1}{1.4}}$
$= (8)^{0.2857} = 1.811$
or $T_{4 s} = \frac{T_{3}}{1.811} = \frac{1400}{1.811}$
$= 773.05 K$
$η - T = \frac{A c t u a l f a l l i n t e m p e r a t u r e}{I s e n t r o p i c f a l l i n t e m p e r a t u r e}$
$η = \frac{T_{3} - T_{4}}{T_{3} - T_{4 s}}$
$0.8 = \frac{1400 - T_{4}}{1400 - 773.05}$
$0.8 = \frac{1400 - T_{4}}{626.95}$
or $1400 - T_{4} = 0.8 \times 626.95$
$1400 - T_{4} = 501.56$
or $T_{4} = 898.44 K$
Power develoed by the turbine,
$W_{T} = c_{p} (T_{3} - T_{4})$
$= 1 (1400 - 898.44) = 501.56 k J / k g$
Heat supplied,
$q_{2 - 3} = c_{p} (T_{3} - T_{2})$
$= 1 (1400 - 604.12)$
$= 795.88 k J / k g$
Thermal efficiency of the cycle,
$η_{t h} = \frac{n e t p o w e r o u t p o u t : W_{n e t}}{H e a t s u p p l i e d : (q_{2 - 3})}$
$= \frac{197.44}{795.88} = 0.2480$
$= 24.80$ %

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