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In a Bell Coleman cycle, air enters the compressor at a pressure of 2.5 bar at a temperature of 100C. It is then compressed to a pressure of 6.25 bar and then cooled in a cooler to a temperature of 250C in the cooler before being expanded in the expansion cylinder upto a pressure of 2.5 bar. The COP will be ______ .
    Correct answer is between '3.1,3.5'. Can you explain this answer?
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    In a Bell Coleman cycle, air enters the compressor at a pressure of 2....
    Bell Coleman cycle is also known as Reversed Brayton cycle
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    In a Bell Coleman cycle, air enters the compressor at a pressure of 2....
    Solution:

    Given parameters in the problem are,

    - Inlet pressure of the compressor, P1 = 2.5 bar
    - Inlet temperature of the compressor, T1 = 100°C
    - Outlet pressure of the compressor, P2 = 6.25 bar
    - Temperature after cooling, T2 = 25°C
    - Final pressure after expansion, P3 = 2.5 bar

    The COP of the Bell Coleman cycle can be calculated using the following formula,

    COP = T2 / (T1 - T2)

    where T1 and T2 are the temperatures at the inlet and outlet of the cooler, respectively.

    Calculating the temperatures using the ideal gas law,

    - T1 = 100 + 273 = 373 K
    - T2 = 25 + 273 = 298 K

    Now, let's calculate the temperature at the outlet of the compressor using the isentropic compression process.

    - Isentropic compression process: P1V1^γ = P2V2^γ, where γ = Cp/Cv = 1.4 (for air)
    - V1 = (RT1)/P1 = (0.287 x 373)/2.5 = 43.176 m^3/kg
    - V2 = V1 (P1/P2)^(1/γ) = 43.176(2.5/6.25)^(1/1.4) = 25.776 m^3/kg
    - T2' = P2V2'/RT2' = (6.25 x 25.776)/(0.287 x T2') = T2'(K)
    - T2' = 507.28 K

    The temperature after cooling is given as T2 = 298K

    Now, let's calculate the temperature after the expansion process.

    - Isentropic expansion process: P2V2^γ = P3V3^γ
    - V3 = V2 (P2/P3)^(1/γ) = 25.776(6.25/2.5)^(1/1.4) = 61.472 m^3/kg
    - T3' = P3V3'/RT3' = (2.5 x 61.472)/(0.287 x T3') = T3'(K)
    - T3' = 469.77 K

    The COP can now be calculated using the formula,

    COP = T2 / (T1 - T2)
    COP = 298 / (373 - 298)
    COP = 3.12

    Therefore, the COP of the Bell Coleman cycle is between 3.1 to 3.5.
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    In a Bell Coleman cycle, air enters the compressor at a pressure of 2.5 bar at a temperature of 100C. It is then compressed to a pressure of 6.25 bar and then cooled in a cooler to a temperature of 250Cin the cooler before being expanded in the expansion cylinder upto a pressure of 2.5 bar. The COP will be ______ .Correct answer is between '3.1,3.5'. Can you explain this answer?
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    In a Bell Coleman cycle, air enters the compressor at a pressure of 2.5 bar at a temperature of 100C. It is then compressed to a pressure of 6.25 bar and then cooled in a cooler to a temperature of 250Cin the cooler before being expanded in the expansion cylinder upto a pressure of 2.5 bar. The COP will be ______ .Correct answer is between '3.1,3.5'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about In a Bell Coleman cycle, air enters the compressor at a pressure of 2.5 bar at a temperature of 100C. It is then compressed to a pressure of 6.25 bar and then cooled in a cooler to a temperature of 250Cin the cooler before being expanded in the expansion cylinder upto a pressure of 2.5 bar. The COP will be ______ .Correct answer is between '3.1,3.5'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a Bell Coleman cycle, air enters the compressor at a pressure of 2.5 bar at a temperature of 100C. It is then compressed to a pressure of 6.25 bar and then cooled in a cooler to a temperature of 250Cin the cooler before being expanded in the expansion cylinder upto a pressure of 2.5 bar. The COP will be ______ .Correct answer is between '3.1,3.5'. Can you explain this answer?.
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