In an aircraft refrigeration working on reverse Brayton cycle, the pressure ratio was 5. The temperature of air after the condenser and evaporator is $15^{0} C$ and $6^{0} C$ respectively. If the refrigeration capacity is 6 tonnes, the mass of air circulated per minute is
(Use, $γ = 1.5 a n d C_{p} = 1.005 k J / k g K)$

10.52

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11.39

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13.26

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16.57

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Solution

The correct option is B 11.39
(b)

$T_{1} = 273 + 6 = 279 K$

$T_{3} = 273 + 15 = 288 K$

So, $T_{4} = \frac{T_{3}}{r_{p}^{γ - 1 / γ}}$

$T_{4} = \frac{288}{5^{0.5 / 1.5}}$

$T_{4} = 168.423$

Now, for process 4-1

$\cdot m_{a} \times c_{p} \times (T_{1} - T_{4}) = 3.5167$

$\cdot m_{a} \times 1.005 \times (279 - 168.423) = 3.5167 \times 6$

${\cdot m}_{a} = 0.18986 k g / s$

${\cdot m}_{n} = 11.3922 k g / m i n$

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