Thermodynamics: Designing a hydrocooling unit

In summary, the 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
  • #1
Ahmed XMEDOX
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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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  • #2
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
 
  • #3
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,
 

1. What is thermodynamics?

Thermodynamics is the branch of physics that deals with the study of heat, energy, and their relationship to work and the behavior of matter.

2. How does a hydrocooling unit work?

A hydrocooling unit works by using a refrigeration system to cool water to a low temperature and then circulating this cold water through a heat exchanger. The heat exchanger is in contact with the product that needs to be cooled. As the cold water flows through the heat exchanger, it absorbs heat from the product, and the warmed water is then circulated back to the refrigeration system to be cooled again.

3. What factors should be considered when designing a hydrocooling unit?

Some important factors to consider when designing a hydrocooling unit include the desired cooling rate, the type and size of product being cooled, the flow rate and temperature of the water, and the type and efficiency of the refrigeration system being used.

4. How does thermodynamics play a role in hydrocooling unit design?

Thermodynamics plays a crucial role in hydrocooling unit design as it helps determine the efficiency and effectiveness of the unit. The laws of thermodynamics govern the transfer of heat and energy, which are essential in the cooling process. Understanding these laws allows for the design of a more efficient and effective hydrocooling unit.

5. What are some common challenges in designing a hydrocooling unit?

Some common challenges in designing a hydrocooling unit include balancing the cooling rate with the product's sensitivity to temperature changes, ensuring proper circulation and flow of the cooling water, and selecting the most suitable refrigeration system for the specific application.

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