A hydrocooling unit can cool fruits and vegetables from 30 to 5 C at a rate of 2

Question

A hydrocooling unit can cool fruits and vegetables from 30 to 5 C at a rate of 20,000 kg/h under the following conditions: the channel is 3 m wide by 90 cm high. The water is circulated and cooled by the evaporator section of a refrigeration system. The refrigerant temperature inside the evaporator coils is to be -2 C, and the water temperature is not to drop below 1 C and not to exceed 6 C. Assuming reasonable values for the average product density, specific heats, and space taken up by the fruits, recommend reasonable values for (a) the water velocity through the channel and (b) the refrigeration capacity (how much heat is removed) of the refrigeration system.

Explanation / Answer

the key for heat exchangers is that inflows=outflows

example m1h(in1)+m2h(hin3)=m1h(out2)=m2h(out4)

you can solve for your enthalpys and find m2.

If this problem is a control volume that has an evaporator with a pipe, then

First step is to node your closed system. the initial temperature of the water is 30 c and the phase is saturated liquid. The enthalpy and saturated pressure can be found on the temperature tables for water. The pressure going out the evaporator can be the same, but the temp is going to be 5 c, this will help you determine your phase. After finding the phase, the enthalpy can be determined.

For the velocity, h(initial)+mgh=h(final)+.5mv^2

for the heat, q=(h(out)-h(in))(mass flow rate)

I hope this helps, but if it doesn't, please ignore it.

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