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A room is kept at -SC by a vapor-compression refrigeration cycle with R-134a as the refrigerant. Heat is rejected to cooling water that enters the condenser at 20C at a rate of 0.13 kg/s and leaves at 28C, The refrigerant enters the condenser at 1.2 MPa and 50 C and leave as a saturated liquid. If the compressor consumes 1.9 kW of power, and c, of water is 4.18 kJ/kg K, determine a) the refrigeration load, in kw and the COP. [3 marks] b) the rate of flow mass of the refrigerant and the COP, of the reversible Carnot cycle [3 marks] c) Is the cycle reversible or irreversible? Explain why? [1.5 marks]

Sagot :

  • The refrigeration load would be 1.7 kJ/s.
  • The COP would be 0.89.

The Carnot cycle is a theoretical model of a reversible heat engine, which consists of four processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. It is the most efficient type of heat engine, meaning that it can convert the maximum amount of heat into work. However, in practice, actual heat engines are never 100% efficient and are often irreversible, meaning that some heat is always lost to the environment.

The refrigeration load, in kW, can be determined using the formula:

Q = m x c x ΔT

Where Q is the rate of heat transfer, m is the mass flow rate of the water, c is the specific heat capacity of water, and delta T is the change in temperature of the water.

In this case, the refrigeration load would be:

Q = 0.13 kg/s * 4.18 kJ/kg K * (28 C - 20 C) = 1.7 kJ/s.

The COP, or coefficient of performance, is a measure of the efficiency of the refrigeration cycle. It is defined as the ratio of the rate of heat removal to the rate of work input to the compressor. In this case, the COP would be:

COP = Q / W = 1.7 kJ/s / 1.9 kW = 0.89.

Learn more about Carnot cycle, here https://brainly.com/question/16448102

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