Thermodynamics: Designing a hydrocooling unit

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SUMMARY

The forum discussion focuses on the design of a hydrocooling unit capable of cooling fruits and vegetables from 30°C to 5°C at a rate of 20,000 kg/h. The calculations involve the mass flow rate of water, specific heat capacities, and the heat transfer equations. The user calculated a water mass flow rate of 0.9774 kg/s and a heat transfer rate of 20.55 kJ/s, but received feedback indicating that the calculations may underestimate the required water flow due to the significant temperature drop needed for such a large mass of produce. The importance of accounting for the volume occupied by the fruits in the channel was also emphasized.

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  • Understanding of thermodynamic principles, specifically heat transfer.
  • Familiarity with specific heat capacities, particularly for water and fruits.
  • Knowledge of mass flow rate calculations in fluid dynamics.
  • Experience with refrigeration system components, especially evaporators.
NEXT STEPS
  • Research the principles of heat transfer in refrigeration systems.
  • Learn about the specific heat capacities of various fruits and vegetables.
  • Investigate the effects of channel geometry on fluid dynamics in cooling applications.
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Engineers and designers involved in food processing, refrigeration system designers, and anyone interested in optimizing hydrocooling processes for agricultural products.

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Homework Statement


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
the figure shows the unit

Homework Equations


m1*h1(at inlet)+m2*h2(at inlet)=m1*h1(at output)+ m2*h2(at output)
let's consider that
m1 is mass rate of water
m2 is mass rate of the fruits and vegetables
Cp of water= 4.2
from the steam tables:
for water:
h4=25.32 kJ/kg
h3=4.28 kJ/kg

3. this is the attempt at a solution
I want to know if I approached the problem correctly or not?? and if the answer is reasonable??

m1*h1(at inlet)+m2*h2(at inlet)=m1*h1(at output)+ m2*h2(at output)
I assumed that Cp of fruits is 3.7 ( as an average of Cp of fruits and vegetables)
then,
m(water)=m(fruits) *(h1-h2) / (h4-h3)
= m(fruits) *Cp * (T1-T2) / (h4-h3)

I got that m(water)= 0.9774 kg/s
Q( that transferred from fruits to water)= m(water) *(h4-h3) = 20.55 kJ/s
velocity of water = m(water)/ [(density of water) * width of unit* its height ] = 3.62*10^-4 m/s
 

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Hi,

Two comments:

Method looks OK, but the math can't be right. Cooling 20000 kg/h fruit by 25 degrees requires a lot more water if that can only increase 5 degrees in temperature.

Water speed is not through an otherwise empty channel: the area that the fruit occupies is not available for the water
 
Ok,
we can ignore the effect of the area that the fruits occupy,, Also, as you said we need a lot more water that's why I'm not believing that this is the answer..
Thanks,
 

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