Cooling load calculation.

  • Thread starter gaber2611
  • Start date
  • #1
gaber2611
14
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Hello everybody,

I have been here before, with different username and cause i didn't log into long time ago, the username doesn't work for m now, so here i am with the new user.

Just as a respect for the forum rules, to not post a topic once i register.

i am doing cooling load calculations, for cooling a 5 cubic meters volume water tank, drains 1000l/day, exposed to sun, installed on building roof, to cool water from 55 to 25 celcuis degrees, working hours of the coil is 12hrs during the day.

i am thinking to insulate the tank, so solar heat gain won't be taken into account for cooling load, and it will be only the heat transfer between the water and fereon cooling coil, which can be calculated using the formula:

q= m*c*delta t

i calculated it, and the result didn't make sense to me, so i need your help with the calculation result.
will appreciate your reply, as i am in urgent need to do that work.
Thanks in advance.
 

Answers and Replies

  • #2
Vadar2012
208
0
What you're calculating is the heat transfer rate of the water to change the temperature from 55 to 25 degrees C. What doesn't it make sense? It'd be helpful if you gave us the calculation that you did. If you wanted to incorporate other factors it will require other relations.
 
  • #3
gaber2611
14
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Thanks vadar for your reply,

here is my calulations,which doesn't make sense to me:


q= m * c * delta T
= (1*5)grams * 4.18 j/g.C * 30 C.
= 627 J (watt/sec.) , = 627*60 w/hr ,= 37.62 kw/hr.
1 TR = 3.5 kw, then q= 10.75 TR.

does this make sense to you?, when you buy a heat pump with that capacity?,isn't it too big?, or i did something wrong?,if wrong, what are the relations or procedures to select the right cooling coil for that water tank?
 
  • #4
gaber2611
14
0
anyone to help?
 

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