Previous in Forum: Product Engineering for Less   Next in Forum: Angle Calculation
Close
Close
Close
Rate Comments: Nested
Commentator
India - Member - New Member

Join Date: Aug 2008
Posts: 61
Good Answers: 1

Air Cooler

10/06/2010 1:26 AM

Dear Members,

I have following Information available

- Air Quantity = 63.3 m3/sec.

- Temp. in/out = 48 / 50.4 oC

To calculate the heat load I have done the following :

U - m cp Delta T

m = 63.3 x 28.996/22.4 = 81.93 kg/sec.

cp = 1.009 @ 60 Deg C (kJ/kg K) - Got it from Internet.

Delta t = 2.4

So, based on above formula I have calculated Heat gain by Air = 198 kw.

Is my calculation is correct. OR do I need to consider other important factors.

Regards,

Register to Reply
Pathfinder Tags: Oil cooler Cooled by Air
Interested in this topic? By joining CR4 you can "subscribe" to
this discussion and receive notification when new comments are added.
Guru
Hobbies - Musician - Engineering Fields - Chemical Engineering - New Member Engineering Fields - Control Engineering - New Member Engineering Fields - Instrumentation Engineering - New Member

Join Date: Jan 2007
Location: Moses Lake, WA, USA, Thulcandra - The Silent Planet (C.S. Lewis)
Posts: 4216
Good Answers: 194
#1

Re: Air cooler

10/06/2010 11:08 AM

Hi Nilesh,

To convert the volumetric flow rate, I used the density of air (values from engineering toolbox on internet) at 50ºC, which was ~1.1 kg/m3. Consequently, the number I got was somewhat lower: 169 kW.

Everything else looks good!

__________________
"Reason is not automatic. Those who deny it cannot be conquered by it. Do not count on them. Leave them alone." - Ayn Rand
Register to Reply
Register to Reply

Previous in Forum: Product Engineering for Less   Next in Forum: Angle Calculation

Advertisement