Question

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:

1. 15.8
2. 30.6
3. 3.06
4. 1.58

Answer 1

Correct Answer - Option 2 : 30.6

Concept:

Bell Coleman or Reversed Brayton cycle: there are two isobaric and two isentropic process in this cycle which are as follows:

• 1-2 Isentropic compression
• 2-3 Constant pressure heat rejection
• 3-4 isentropic expansion
• 4-1 Constant pressure heat additionThe pressure at point 1 and 2 are equal ⇒ P₁ = P₄

The pressure at point 3 and 4 are equal ⇒ P₂ = P₃ 

In isentropic process : \(PV^{\gamma} = Constant\), \(TV^{\gamma\ -\ 1} = Constant\),  \(T\propto P^{\frac{\gamma\ -\ 1}{\gamma}} \)

Refrigeration effect (RE) = C_p (T1 - T4) kJ/Kg

Refrigeration capacity \(RC = \dot{m}\times RE\) kW

1 Tonne Refrigeration per hour  = 3.5 kW

Calculation:

Given:

P₁ = P₄ = 1 bar , P₂ = P3 = 7 bar, T₁ = 5 °C = 5 + 273  = 278 K, T₃ = 25 °C = 25 + 273 = 298 K

\(\dot{m} = 1\ Kg/s\) , γ = 1.4 cp = 1.005 kJ/kg-K,  70.286 = 1.745

(RE) = Cp (T1 - T₄)

Apply isentropic relation between 3 & 4

\({\tfrac{T_4}{T_3}} = \left ({\frac{P_4}{P_3}}\right )^{\frac{\gamma\ -\ 1}{\gamma}}\)

\({\tfrac{T_4}{298}} = \left ({\frac{1}{7}}\right )^{\frac{1.4\ -\ 1}{1.4}}\) ⇒ \({T_4} = 298\times\left ({\frac{1}{7}}\right )^{0.286}\) 

\(\)T₄ = 170.77 K

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

Since 1TR = 3.5 kW

 \(RC = \frac{107.76}{3.5} = 30.79\ TR\)

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