SSC JE Mechanical Engineering 11 Oct 2020 Shift 1 Papers

Question : 172 of 200

Marks: +1, -0

An air refrigeration system works on Bell – Coleman cycle. It draws air at the rate of 1 kg/s from the cold chamber at 1 bar and 5° C. Air is compressed to 7 bar and then is cooled to 25°C before sending it to expansion cylinder. Given γ = 1.4

c_p

= 1.005 kJ/kg-K and

7^{0.286} = 1.745

. Refrigeration effect for this system in tonne per hour, is:

15.8
30.6
3.06
1.58

Validate

Solution:

P_1 = P_4 = 1

bar ,

P_2 = P_3 = 7

bar,

T_1 = 5 °C = 5 + 273 = 278 K,

T_3 = 25 °C = 25 + 273 = 298 K

m˙=1 Kg/s

, \γ = 1.4,

c_p

= 1.005 kJ/kg-K,

7^{0.286 }= 1.745

(R E)=C_{p}(T_{1}-T_{4})

Apply isentropic relation between 3 & 4

{{T}_{4}}/{{T}_{3}}=({{P}_{4}}/{{P}_{3}})^{{\γ-1}/{\γ}}

{{T}_{4}}/{298}=({1}/{7})^{{1.4-1}/{1.4}} \

⇒{T}_{4}=298 \×({1}/{7})^{0.286}

T_4 = 170.77

RC = 1 ×1.005 (278 - 170.77) = 107.76 KW

Since 1TR = 3.5 kW

RC=107.76/3.5=30.79 TR

Hence the nearest answer is an option (2) 30.6 TR

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