# Consider an ice-producing plant that operates on the ideal vapor-compression refrigeration cycle and uses refrigerant-134a as the working fluid. The refrigeration cycle operating conditions require an evaporator pressure of 180 kPa and the condenser pressure of 1400 kPa. Cooling water flows through the water jacket surrounding the condenser and is supplied at the rate of 250 kg/s. The cooling water has a 10°C temperature rise as it flows through the water jacket. To produce ice, potable water is supplied to the chiller section of the refrigeration cycle. For each kg of ice produced, 333 kJ of energy must be removed from the potable water supply. (Take the required values from saturated refrigerant-134a tables.) Determine the mass flow rate of the potable water supply, in kg/s. The mass flow rate of the potable water supply is_______ kg/s.

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Consider an ice-producing plant that operates on the ideal vapor-compression refrigeration cycle and uses refrigerant-134a as the working fluid. The refrigeration cycle operating conditions require an evaporator pressure of 180 kPa and the condenser pressure of 1400 kPa. Cooling water flows through the water jacket surrounding the condenser and is supplied at the rate of 250 kg/s. The cooling water has a 10°C temperature rise as it flows through the water jacket. To produce ice, potable water is supplied to the chiller section of the refrigeration cycle. For each kg of ice produced, 333 kJ of energy must be removed from the potable water supply.

(Take the required values from saturated refrigerant-134a tables.)

Determine the mass flow rate of the potable water supply, in kg/s.

The mass flow rate of the potable water supply is_______ kg/s.

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Step 1

Assume an ideal vapor-compression cycle.

In an ideal vapor compression cycle, the refrigerant enters the compressor at state 1 as saturated vapor at 180kPa and is compressed isentropically to the condenser pressure of 1400kPa. The vapor then enters the compressor as superheated vapor at state 2 and leaves as saturated liquid at state 3. Heat is rejected to the water jackets around the condenser.

The saturated liquid is throttled to state 4 such that the enthalpy at state 3 is the same as the enthalpy at state 4, which is the state of entry of the refrigerant in the evaporator as a low-quality saturated mixture.

The T-S diagram of the process is drawn.

Here, QH is the heat lost by the condenser and QL is the heat gained by the evaporator. It is also the refrigeration effect in this case. Win is the work put into the compressor in pumping the refrigerant from the evaporator pressure to condenser pressure.

Step 2

Find the enthalpy in  hi,  where enthalpy is kJ/kg   and i denotes the corresponding points 1,2,3,4 using the standard pressure table for saturated refrigerant 134-a.

Enthalpy of saturated vapor h1 at pressure 180kPa is 242.90 kJ/kg.

Enthalpy of saturated liquid h3 at pressure 1400kPa is 127.25 kJ/kg.

Process 3-4 is a throttling process where enthalpy remains constant so the enthalpy of the saturated mixture h4 is  also 127.25 kJ/kg.

Step 3

To find the enthalpy at point 2, use the fact that the process 1-2 is isentropic.

So, S1 equals S2.

S1 is the entropy for the saturated vapor at 180 kPa, which is equal to 0.93979 kJ/kg∙k, evaluated using the standard pressure table for saturated refrigerant 134-a.

Hence, S2 is also 0.93979 kJ/kg∙k.&nb...

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