Room Temperature and Heating Capacity

[HotChilli]

Hi everyone,
I have a couple of questions about calculating room temperature. I think I already know the answer to the first question but would like your opinion. The second question is more of a curiosity.

Outdoor temperature = -28°F
Room setpoint = 72°F
Heat loss = 40 mbh
RTU = 2000 cfm @ 82°F supply air
Roof and wall area = 10000 sqft
U value = 0.04

My first question is, can the RTU keep the room temperature around 72°F by running longer or will the room temperature drop to ~63.5°F (see calculations below)? I am also thinking if the room temperature drops to 63.5°F then the RTU supply air temp will drop based on the return air temperature, therefore, if outside air stays -28°F then the room temp will drop even further over time. Is my thought process correct?

1.08 x 2000 x 18.5 = ~40 mbh
82 - 18.5 = 63.5°F

My second question is, if heat loss = 40 mbh and heat gain = 20 mbh from electric or hydronic baseboard heaters, how would you calculate the room temperature?

Thanks for all your input

 

[GT-EGR]

For your first question - BTU/h is a rate, so you are matching the heat lost through the walls with the heat added in the RTU. Running longer won’t matter because time isn’t part of the equation. The only way time would help is if over time it gets warmer outside and your heat loss through the walls isn’t as much.

As far as your 63.5 room temperature, your answer is mathematically correct. The issue is that unless you do some differential equations, you have to do more iterations. If your unit was able to make 82 degree air from 72, it can probably make 75 deg air from 63.5 (you’d need manufacturer information to know). So even though the air coming out of the unit is colder, the lower entering air temp allowed it to exchange more heat.

On the opposite side of this equation, if your room is 63.5, you no longer have 40 MBH heat loss, it’s not 36.6 MBH since you don’t have such as big a difference between indoor and outdoor temp.

The above answers your second question, you have to do iterations with the room temp to find the exact temperature where the output from your RTU is equal to your loss through the walls.

The extreme example is if the room was -28 you’d have no heat loss.

Also keep in mind doors open, walls leak, so you might want to consider more than just conductive loss through the walls.

 

[HotChilli]

Hi GT-EGR and thank you for your response.

I figured as much for time but thought I would ask anyway. The one thing I didn’t think of was lower heat loss when the room temperature drops. However, even with a lower heat loss of 36.6 mbh, and even if the unit was able to supply 75°F, the new room temp will be ~59°F. The point I am making is so long as the outside air is -28°F, the indoor temp will keep dropping until things balance out.

For question two, are you saying we calculate room temperature when heat gain = heat loss? If that’s true, and please correct me if I am wrong, then the room temperature will be 22°F (10000 x 0.04 x 22 - -28 = 20000 mbh)? I did come up with 22°F before posting my questions but I thought it was too low to be correct. Your feedback is appreciated.

Regarding the total heat loss, I am considering more than just the conductive loss through the walls but I wanted to simplify the questions.


Thanks again

 

[GT-EGR]

It wont be 22 because the other side of your equation, the heating capacity of the unit, increases as the room temperature drops. It’s counterintuitive, but the heating capacity of your RTU goes up as the entering temperature drops. So if you put 28 deg into the unit you’ll get more than 20 MBH out of it - you’ll need manufacturer data to know how much. I assume that’s where you got the 2,000 CFM entering 72 leaving 82 number?

Sounds like your unit isn’t big enough to do the job. So if this isnt already installed, you want a higher capacity option.