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Calculate Room Temperature

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HotChilli

Mechanical
May 21, 2017
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Can someone help me calculate the indoor temperature of a room in the summer. See conditions below and assume there is no ventilation in the space.

Room area = 600 sq.ft, roof area = 600 sq.ft, window = 36 sq.ft, perimeter wall = 150 sq.ft, wall height = 10 ft, outdoor temperature = 85 F, total heat gain (walls, roof, window, lighting, people) = 3700 btuh, wall R-value = 20, Roof R-value = 30 and window R-value = 3.7.
 
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Simple.
Heat gain = heat loss.
Heat gain = solar + transmission + internal heat gains
Heat loss = (room wall area)*wall U-factor* final room temp. (call it Tfinal)- ambient temp.) + (roof/ceiling area)*(roof/ceiling U-factor)*(Tfinal-ambient temp).

equate the 2 equations and solve for Tfinal (This assumes Tfinal is > ambient temp).

Good luck with that homework problem.
 
This is not homework, it's for an actual project I'm working on. I'm having a brain fart and can't seem to figure this out. The numbers I provided are not 100% correct but are close.
To clarify, I am providing some ventilation in the space but not enough to cool it down to 75F. What I am trying to figure out is the temperature rise or the room temperature if I only provide ventilation that will cover 1/4 of the cooling load (calculated cooling load = 3700 btuh).
 
It seems that you have all the info needed to calculation. To have a cooler room temperature, the insulation can be helpful in additional to the ventilation.
 
I'm actually looking for help to solve this question. I have all the info, just need to figure out how to calculate the room temperature.
 
Take a look at the formulas/examples in ASHRAE Handbook 2013 - Fundamentals, Chapter 1: Psychrometrics, section "Typical Air-conditioning Processes". You basically need to convert the heat transfer to a temperature change and use a psychrometric chart to plot the conditions.
 

I tried trashcanman equation but I disagree that heat gain equals heat loss.

FL Engineer, I will take a look at ASHRAE fundamentals, this may be the solution.

I guess my question was not so simple as no one had a simple answer :)

Thanks everyone
 
palitaali,

It is pretty simple, just maybe not explained very well. I'll give it another go.

You need to follow the energy and temperature difference.

If you start at say 85F, then your heat outflow is zero as there is no temperature gradient to get heat out through the walls, windows and roof.

Your heat input to the room though is fixed at 3700btu/h - whatever cooling you can supply (you mention 25%).

This excess heat energy needs to go some where, which is heating up the air inside.
As the air temperature increase, the heat outflow through the walls and roof starts to increase as a temperature difference starts to appear.

When that energy outflow matches the heat input - cooling then the air temperature has peaked. That's the equation mentioned above by trashcanman.

At that point heat entering your space minus cooling equals heat outflow through the walls and roof ( a bit simplistic but hey, that's what you're looking at). Now how long that's going to take is a different question and needs to include some element of how much energy it takes to heat up a roomful of air 1 degree F. That will give you an idea if it's going to be minutes or hours / days. My calc shows it's going to be hours. One hour of 75% of 3700 btu in a room of 6000 ft3 raises the temp 25F (no heat outflow to make it simple).

Sounds like it's going to be a hot stuffy room.....



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Wouldn't there be some sort of differential equation to figure this out?
this sounds like a physics problem i did in high school. forgive me since it's been a while
 
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