Question

refrigerator working on Bell-Coleman cycle operates between pressure limits of 1.05 bar and 8.5 bar. Air is drawn from the cold chamber at 10°C. Air coming out of compressor is cooled to 30°C before entering the expansion cylinder. The expansion and compression follow the law of = constant. Heat extracted from the cold chamber per kg of air

Answer 1

Answer:

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Answer 2

Answer:

Heat extracted from cold chamber per kg of air (283 – 178) = 105 kJ/kg.  

Explanation:

shows the working cycle of the refrigerator.  

Given : p2 = 1.05 bar ;  

p1 = 8.5 bar ;  

T3 = 10 + 273 = 283 K ;  T4 = 30 + 273 = 303K

Considering polytropic compression 3-4, we have

$\frac{T_{4}}{T_{3}}=\frac{P_{1}}{P_{2}} ^\frac{n-1}{n}$

$\frac{T_{4}}{T_{3}}=\frac{8.05}{1.05} ^\frac{1.35-1}{1.35}\\\frac{T_{4}}{T_{3}}=\frac{8.05}{1.05} ^\frac{1.35-1}{1.35}\\\frac{T_{4}}{T_{3}}=(7.666)^{0.259}\\T_{4}=T_{3}*(7.666)^{0.259}\\T_{4}=283*1.6947409485564\\T_{4}=479$

Again, considering polytropic expansion 1-2, we have

$\frac{T_{1}}{T_{2}}=\frac{P_{1}}{P_{2}} ^\frac{n-1}{n}$

$\frac{T_{1}}{T_{2}}=\frac{8.05}{1.05} ^\frac{1.35-1}{1.35}\\\frac{T_{1}}{T_{2}}=\frac{8.05}{1.05} ^\frac{1.35-1}{1.35}\\\frac{T_{1}}{T_{2}}=(7.666)^{0.259}\\=1.694740\\T_{2}=\frac{T_{1}}{1.694740} \\T_{2}=\frac{303}{1.694740} \\T_{2}=178$

Heat extracted from cold chamber per kg of air = cp (T3 – T2) = 1.003 (283 – 178) = 105 kJ/kg.  

Heat rejected in the cooling chamber per kg of air = cp (T4 – T1) = 1.003 (479 – 303) = 176 kJ/kg.  

Since the compression and expansion are not isentropic, difference between heat rejected and heat absorbed is not equal to the work done because there are heat transfers to the surroundings and from the surroundings during compression and expansion.