Overal heat transfer coefficient: Refrigeration Heat Exchanger

In summary, the estimated area and coefficient would be based on a condenser coil size that you measure.
  • #1
recreated
50
1
Hi There

Does anyone out there know a good estimate or any real values for the total heat transfer area for the hot and cold sides of a domestic refrigeration system? i.e. the sum of the areas of the evaporator and the condenser.

For the Evaporator:
QL=UL.AL.(TL-T1)

AL=heat transfer area of cold side (Low side)=?
UL=overal heat trans. coefficient of cold side.=?
QL=heat removed from cold side = 7.93 KW
TL=refrigerator temp. = 3degs C + 273.15=276.12 Kelvin
T1= Refrigerant evaporating temp. =0degs C + 273.15= 273.15 Kelvin

I am trying to estimate this area so I can calculate the overal heat transfer coefficient which I need for other calculations. So far the outcome of estimating the area and finding the heat transfer coefficient based on it has not been realistic.

If it helps:
The aim of my project is to find the optimal heat transfer ratio between hot and cold side, there is theory on this which I am following, I just need some realistic values for area and heat transfer coefficient. I think it is more logical to estimate the area and find the h.t. coefficient based on that rather than the other way around.
 
Physics news on Phys.org
  • #2
recreated said:
Hi There

Does anyone out there know a good estimate or any real values for the total heat transfer area for the hot and cold sides of a domestic refrigeration system? i.e. the sum of the areas of the evaporator and the condenser.

Don't know about the evaporator, it's unfortunately hidden, but you can estimate the area of the condenser coil in a refrigerator by measuring the length of one left-to-right run, multiplying by the no. of runs and then by pi*D where D is the diameter of the coil. Add 10% or so for the bends.
 
  • #3
Thank you. I will try that now. Do you know if the condenser is usually larger or smaller than the evaporator? Or any estimated proportion to use?

I assume condenser area > evaporater area as it has to get rid of heat from the refrigerator & the compressor, but maybe in practise it's different.
 
  • #4
recreated said:
Thank you. I will try that now. Do you know if the condenser is usually larger or smaller than the evaporator? Or any estimated proportion to use?

I assume condenser area > evaporater area as it has to get rid of heat from the refrigerator & the compressor, but maybe in practise it's different.

That seems to be the case. I would estimate the evap coil at 1/7 the area of the condenser. The evap coil is usually small enough to fit in the freezer compartment.

cf. http://www.ebay.com/bhp/refrigerator-evaporator-coil

for examples.

For both coils, extensive heat sink fins are used to dissipate heat (condenser) and draw in heat (evaporator). You would need to take those into consideration. Also, evap coils almost always use a fan to blow the cooled air into the refrigerator. Apparently the newer models put the condenser coil at the bottom & then a fan is probably required there too.

All in all, you're faced with a difficult analysis I think.
 
  • #5


I understand the importance of accurate and realistic values in any calculation or experiment. In order to estimate the total heat transfer area for the hot and cold sides of a domestic refrigeration system, we need to consider several factors such as the type of refrigerant, the design of the system, and the operating conditions. These factors can greatly impact the overall heat transfer coefficient and the heat transfer area.

To accurately estimate the heat transfer area and the overall heat transfer coefficient, it is important to conduct experiments or simulations using real-world values and conditions. This will provide more accurate results and help in understanding the optimal heat transfer ratio between the hot and cold sides.

I would suggest consulting with experts in the field or referring to published literature on similar refrigeration systems to gather realistic values for the heat transfer area and overall heat transfer coefficient. Additionally, using advanced simulation software can also help in estimating these values with more precision.

In conclusion, it is crucial to use realistic values and accurate methods in determining the total heat transfer area and overall heat transfer coefficient for a domestic refrigeration system to ensure the success of your project.
 

1. What is the overall heat transfer coefficient?

The overall heat transfer coefficient is a measure of how much heat is transferred through a material or system, taking into account the combined effects of conduction, convection, and radiation.

2. How is the overall heat transfer coefficient calculated?

The overall heat transfer coefficient is calculated by dividing the total heat transfer rate by the temperature difference between the hot and cold sides of the system. This can be expressed as U = Q/(ΔT), where U is the overall heat transfer coefficient, Q is the total heat transfer rate, and ΔT is the temperature difference.

3. What role does the refrigeration heat exchanger play in the overall heat transfer coefficient?

The refrigeration heat exchanger is a key component in the overall heat transfer coefficient, as it is responsible for transferring heat between the refrigerant and the surrounding environment. The efficiency and design of the heat exchanger can greatly affect the overall heat transfer coefficient.

4. How can the overall heat transfer coefficient be improved?

There are several ways to improve the overall heat transfer coefficient, including increasing the surface area of the heat exchanger, improving the flow rate of the refrigerant, and using materials with higher thermal conductivity. Regular maintenance and cleaning of the heat exchanger can also help improve its efficiency.

5. What factors can affect the overall heat transfer coefficient?

The overall heat transfer coefficient can be affected by various factors such as the type and properties of the materials used, the design and size of the heat exchanger, the temperature and flow rate of the refrigerant, and external factors like wind and humidity. Changes in any of these factors can impact the overall heat transfer coefficient and may require adjustments to maintain optimal performance.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
3
Views
835
  • Engineering and Comp Sci Homework Help
Replies
5
Views
2K
Replies
14
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
12
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
5K
Replies
1
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Mechanical Engineering
Replies
34
Views
4K
  • Differential Equations
Replies
3
Views
1K
Back
Top