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Open top display cooler - energy loss

  • Thread starter ttj
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ttj
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Homework Statement


I am working on a energy analysis in a supermarket. There is some open top display coolers, which is using alot of energy - I dont have the opportunity to check excatly how much energy they use, so I have to calculate it. I dont have problems in calculating it when it is closed, but I have no clue on how to do it, when there is a "open 'wall'".
I know there are some uncertain variables, temperature, humidity and ventilation. But I'm gonna make them constant:
Temp outside 22 degr C, humidity outside 50%, temp in cooler 2 degr C, humidity in cooler 85%.

I have been using the program called Coolpack (EESCooltools), and it is telling me, that my problem is the air infiltration! And I do agree :)

Homework Equations


The heat loss caused by infiltration can be calculated as

Hi = cp ρ n V (ti - to) (8)

where

Hi = heat loss infiltration (W)

cp = specific heat capacity of air (J/kg/K)

ρ = density of air (kg/m3)

n = number of air shifts, how many times the air is replaced in the room per second (1/s) (0.5 1/hr = 1.4 10-4 1/s as a rule of thumb)

V = volume of room (m3)

ti = inside air temperature (oC)

to = outside air temperature (oC)

Sources ^^ = http://www.engineeringtoolbox.com/heat-loss-buildings-d_113.html

Would it be possible to measure the windspeed in the cooler with a anemometer. Then check the speed every 3 cm from the products to the top, and so multiply it by the cross-section and breadth and take the average?


My other idea is to measure the temperature 10 cm above the products and then: 1) Create a frame of flamingo, cover it with a lid of acrylic 2) Measure the temperature here and then find the difference between an open and a closed.


All ideas will be appreciated, thanks in advance.
 
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Answers and Replies

  • #2
Spinnor
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