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A heat pump operates on the ideal vapor-compression refrigeration cycle and uses refrigerant-134a as the working fluid. The condenser operates at 1000 kPa and the evaporator at 200 kPa. Determine this system’s COP and the rate of heat sup- plied to the evaporator when the compressor consumes 6 kW.

Respuesta :

davidlcastiblanco

Answer:

QL= 24.51 kW

COP = 5.085

Explanation:

h₁ = 244.5 kj / kg, h₂ = 278.07 kj / kg,  h₃ = h₄ = 107.34 kj / kg

P₁ = 1000 kPa , P₂ = 200 kPa

W = 6 kW

To determine the rate heat can use the equation

QL = m * ( h₁ - h₄)

QL = W * ( h₁ - h₄) / (h₂ - h₁)

QL = 6 * ( 244. 5 - 107.34 ) / (278.07 - 244.5 )

QL = 24.51 kW

To determine the COP is the rate hate relation with the work done

COP  = QH / W

COP = 1 + QL /Q

COP = 1 + [ 24.51 kW / 6 kW ]

COP = 5.085

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