×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Heat Load Calculation

Heat Load Calculation

Heat Load Calculation

(OP)
A box 10ft.x7ft.x6.5ft.high on the back of a truck. Walls and ceiling insulated with R5 and floor insulated with R3. Outside temp 0ºF. Assume 10% infiltration. One mandoor and not ventilation. What is heat load to maintain 70ºF in box with 200cfm fan?

Secondly, if we assume the box is completely sealed and 0ºF inside. What is heat load to bring up to 70ºF with 200 cfm.

RE: Heat Load Calculation

If I understand your question properly then:

Q conduction = Area x 1/R x delta T
Q conduction = (291 x 1/5 x 70) + (70 x 1/3 x 70)
Qc = 5700 Btu/hr

Q infiltration = 1.08 x cfm x delta T
Qi = 1.08 x (450 ft3 x 10%) x 70
Qi - 3440 Btu/hr

The amount of air your fan provides is not really important, unless it is supplying outside air. At which point you need an additional 15,120 Btu/hr of heat. If the box is completely sealed, remove the value for infiltration. For both of your scenarios the delta T is 70, so the answer is the same.

What exactly do you mean with the fan at 200 cfm?

Important things to note: heat transfer could be altered if the truck is moving vs. stationary. I've ignored radiation heat transfer.
Hope that answers your question.

RE: Heat Load Calculation

(OP)
Thank you Chris.

I believe the second part of my question was incorrectly stated.

Based on the conduction and infiltration requirements, I would need a 10000 BTU heater to maintain the temperature. What I was trying to ask is this.....

If the heating load is 5700+3440 BTUH = 9140BTUH, based on maintaining an interior temp of 70ºF. What is the heating requirement to bring that original volume of air (455FT^3) from 0ºF (which is what it would be after a period of time without any heat input)to 70ºF? Does this require knowledge of the heater capacity being used and the air flow being recirculated?

I apologize if I am confusing the issue with my question formulation.

Rgds,

Webber

RE: Heat Load Calculation

Your second posting makes more sense.
As to the amount of heat to raise the internal temperature of the box: As long as you add more energy to the space than is being lost through convection/infiltration then the temperature of the box will increase. If you add 10,000 Btu/hr the temp will increase until the heat loss = heat gain and temp will maintain steady.

If elapsed-time is more a serious issue then you should probably start posting on the Heat Transfer forum because it isn't an easy thing to determine. The heat transfer from the box when it is at 0 deg F is 0 Btu/hr, and when it is at 70 deg F is 10,000 Btu/hr. However, the heat transfer rate from 0-70 isn't linear and determining time is some reasonably serious calculus.

If time isn't a factor then a 10,000 Btu/hr (or even 15,000 Btu/hr to cover your tail) furnace will heat the space from 0-70 and also maintain it at 70 (when it is 0 outside). The value of cfm of the furnace is irrelevant except you'll want to ensure that the air temperature of the furnace isn't too high. If the furnace is 15,000 Btu/hr the supply air temp with 200 cfm will be:

15,000 = 1.08 x 200 x delta T
delta T = 70 deg

T supply = T return + delta T
Ts = 70+70 = 140 deg F

140 is reasonable hot, so depending on the application you may want to increase the airflow (say to 300 cfm) so that your discharge temp is lower.

Clear?

RE: Heat Load Calculation

(OP)
I was hoping there was a simple factor to add for the initial heating of the air volume. I do understand, however that the calculation could be quite complex. As the air warms up, the inlet air temperature to the heater changes and thus the outlet, etc., etc..

Thank you very much for your input.

Best regards

 

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources