In a reverse Brayton cycle working at a pressure ratio of 5- the temperatures at the inlet of compressor and expander are 10-C and 25-C respectively- If the compression and expansion takes place according to the law pv1-2 - constant- the COP of the plant is -Use Cp-1-005kJ-kgK and R - 0-287 kJ-kgK-

T_1=283K So, T_2=T_1(r_p)^n 1/n nbsp;=283×(5)^1.2 1/1.2 T_2=370.06K So, nbsp; T_4=T_3/(r_p)^n 1/n nbsp;=298/5^1.2 1/1.2 T_4=227.88K Net refrigeratio ...

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Byju's Answer

Standard XII

Chemistry

Heat Capacity at Constant Volume

In a reverse ...

Question

In a reverse Brayton cycle working at a pressure ratio of 5, the temperatures at the inlet of compressor and expander are $10^{\circ} C$ and $25^{\circ} C$ respectively. If the compression and expansion takes place according to the law $p v^{1.2}$ = constant, the COP of the plant is _____.

Use $C_{p} = 1.005 k J / k g K$ and R = 0.287 kJ/kgK.

1.899

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Solution

The correct option is A 1.899

$T_{1} = 283 K$

So, $T_{2} = T_{1} {(r_{p})}^{n - 1 / n} = 283 \times {(5)}^{1.2 - 1 / 1.2}$

$T_{2} = 370.06 K$

So, $T_{4} = \frac{T_{3}}{{(r_{p})}^{n - 1 / n}} = \frac{298}{5^{1.2 - 1 / 1.2}}$

$T_{4} = 227.88 K$

Net refrigeration effect = $c_{p} (T_{1} - T_{4}) = 1.005 (283 - 227.88)$

RE = 55.3956 kJ/kg

For polytropic process 1 – 2,

$W_{c} = \frac{n}{n - 1} R T_{1} (r_{p}^{n - 1 / n} - 1) = \frac{n}{n - 1} R (T_{2} - T_{1})$

$W_{c} = \frac{1.2}{1.2 - 1} \times 0.287 (370.06 - 283)$

$W_{c} = 149.92 k J / k g$

For polytropic process 3 – 4,

$W_{T} = \frac{n}{n - 1} R T_{4} (r_{p}^{\frac{n - 1}{n}} - 1) = \frac{n}{n - 1} R (T_{3} - T_{4})$

$W_{T} = \frac{1.2}{1.2 - 1} \times 0.287 (298 - 227.88)$

$W_{T} = 120.75 k J / k g$

So, $C O P = \frac{R E}{W_{c} - W_{T}} = \frac{55.3956}{149.92 - 120.75}$

COP = 1.899

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