Need to know BTU of heat loss 2

[glegee]

I need to know the BTU of heat loss I have in a room based on the following.
I have an 800 cu.ft. room with a 5200 btu heater operating.
The room temperature started at 65F and over a period of time the room temperature stabilized at 80F.
How many BTU of heat loss do I have? What would be the formula for this?

Thanks,

Gary

 

[MintJulep]

At your stabilized final 80F condition the heat loss is 5200 BTU/hr.

 

[zekeman]

Your heat loss depends on the difference between the room temperature and the outside temperature.
So at 80 degrees room temperature, the loss is 5200 BTU/hr but at , say 70 degrees inside temperature the loss would be
5200(70-T0)/(80-T0)
For example, if the outside temperature is 30, then
5200(70-30)/(80-30)=4160 BTU/hr.

 

[glegee]

Thanks for the replies!

I see you use the 70 and 80 in the equation for the room stabilized temperature at 70, If my starting temperature is the same and the stabilizing temperatures are different then the btu/hr heat loss would be greater at 70F than at 80F correct?

Here are two senarios, what are the btu/hr loss formulas for each?
1. room starts at 65f and stabilizes at 80F with 5200 btu heater.
2. room starts at 65f and stabilizes at 75F with 5200 btu heater.

 

[MintJulep]

Scenario 1: 5200 BTU/hr
Scenario 2: 5200 BTU/hr

If my starting temperature is the same and the stabilizing temperatures are different then the btu/hr heat loss would be greater at 70F than at 80F correct?

No.

Starting temperature is irrelevant.

By definition of "stabilized", heat in = heat out. So whatever temperature you stabilize at with a 5200 BTU/hr heater inside is the stabilization temperature of your structure with a heat in = 5200 BTU/hr.

So if you keep asking the question "there is a 5200 BTU/hr heater in a structure and it stabilizes at some temperature, what is the heat loss?" the answer is always going to be 5200 BTU/hr.

Maybe you are not asking the right question?

 

[glegee]

I am thinking it is a differential equation. When the room stabilizes the BTU/hr is equalized, BTU heat = BTU cooling which I understand. But if the 5200 BTU heater can only raise the air temperature by 5F rather than by 15F the heat losses must be greater and that is what I am trying to calculate.

Thanks for the replies.

 

[MintJulep]

Q=UAΔT

Q is heat loss (BTU/hr)
U is Thermal Transmittance (Btu/ (hr °F ft²))
A is surface area (ft ²)
ΔT is the temperature difference between inside and outside at steady state.

You are most likely interested in the quantity "UA"

 

[Twoballcane]

Wait, are you asking the heat loass from the heater to the room or from the room to outside via the wall and ceiling?

Tobalcane
"If you avoid failure, you also avoid success."
“Luck is where preparation meets opportunity”

 

[vpl]

Is this homework?

Patricia Lougheed

******

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[zekeman]

Maybe to understand this concept, look at the electrical analogy ( since you are an electrical guy). The network is a parallel R-C with one end grounded and the other end node being fed from a 5200 ampere current source.
At the beginning,all of the current goes into the capacitor and builds up voltage at the node ( similar to heating up the room) and eventually all of the current goes through the resistor; the node voltage becomes IR, 5200R corresponding to your stabilized room temperature. The thermal equivalent to the R is the thermal resistance to the outside temperature, 1/UA.

 

[glegee]

This is a special project I am working on.
The problem is I have a known 5200 BTU/hr heater in a room, the ambient temperature started at 65F, after a period of time the room ambient stabilized at 70F. The heater was only able to raise the room ambient by 5F, how many BTU/hr heat loss do I have?

 

[DRWeig]

glegee, your question has been answered many times in this thread.

If the temperature is stable, the heat loss from the room equals the heat gain from your heater (5200 BTU/hr).

Good on ya,

Goober Dave

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[satchmo]

Turn up the thermostat.

 

[glegee]

Lets look at these senarios.

1. I place a 5200 BTU/HR heater in a room and it is only able to raise the room temperature by 0F.
2. I place a 5200 BTU/HR heater in a room and it is only able to raise the room temperature by 5F.
3. I place a 5200 BTU/HR heater in a room and it is only able to raise the room temperature by 10F.

The heat loss BTU/HR in each of these senarios can not be the same.

 

[satchmo]

The room temp raises by 0, 5, or 10F, then stays there. At that point your heater is pumping 5200 BTU/hr into the room and the temp is not changing. Therefore, you are losing all of the 5200 BTU/hr at that point.

 

[MintJulep]

The heat loss BTU/HR in each of these senarios can not be the same.

Or you do not understand what heat loss means.

 

[DRWeig]

Look at it from the load side, glegee.

The heat loss from your room varies with the room temperature and the outdoor temperature.

If it's 0°F outdoors, and your indoor temperature stabilizes at 70°F, the capacity of your heater is exactly equal to the room heat loss when stable.

Same room, same heater: If it's 10°F outdoors, your room temperature will become stable at about 80°F. The heat loss at that point is also exactly equal to the heater capacity when stable.

Heat loss depends on the difference in temperature between outdoors and indoors. Imagine one step further: What if the temperature outdoors rises to 100°F? How hot can your heater make the room until it stabilizes? I bet it's around 170°F. Again at that point, the heat loss is exactly equal to the heater capacity when stable.

At all three conditions, your heater is maintaining a 70°F temperature difference between indoors and outdoors. As long as the temperature difference stays at 70°F, the heat loss equals 5200 BTU/h, same as the heater capacity.

Final stab at explaining this: Your heater puts heat into the room at a rate of 5200 BTU/h. The temperature in the room does not change. Where is the 5200 BTU/h going? It's going out through the walls, windows, roof, floor, and doors, and maybe through some warm air leaving the room and being replaced by cold air from outside.

Whenever your heater is running at maximum output and the temperature is not changing in the room, heat loss = heater output.

For purists, I left out internal loads and the variations in infiltration, etc... over broad temperature ranges. Trying to keep it simple.

Good on ya,

Goober Dave

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[glegee]

Thanks for the great replies! I understand what you are saying and perhaps I am using the wrong term for what I am looking for.

If my 5200 BTU/hr heater is stabilizing the room temperature at 70F then the heat loss is 5200 BTU/HR. But if I want the stabilizeed temperature to be 75F I will need more heating BTUs to achieve this with the current heat losses that exsist. Now that I have explained it that way I see that I was using at the wrong term.

 

[IRstuff]

yup...

TTFN
faq731-376

 

[DRWeig]

Measure the outdoor temperature at which your room reaches 70°F and becomes stable (i.e., the heater can't do any better and is putting out its full 5200 BTU/hr).

Subtract the outdoor temperature from 70°F. The result is the existing design delta-T between the space and outdoors.

Now add 5°F to the design delta-T you just computed. The result is the new design delta-T that you'd like to have (in order to achieve 75°F in the room at the same outdoor temperature).

Divide the new design delta-T by the existing delta-T. The result will be a number greater than one. This is the size factor by which your heater capacity needs to increase.

Multiply the size factor by the original heater capacity (5200 BTU/hr in your case). The result will be the new heater size needed.

For example, if your 70°F case occurs when the outdoor air is 20°F, the existing design delta-T is 50°F. You'd like a 75°F room, which means the new design delta-T is now 55°F. The sizing factor is 55°F/50°F or 1.1. The new heater size needed is 1.1 x 5200 BTU/hr or 5720 BTU/hr.

Cautions:

1) I've assumed a simple case, as if you were heating a typical small apartment. If your heating load varies significantly with factors other than the outdoor temperature, correction may be needed. An example would be anything that changes the amount of outdoor air reaching the space. If there are lots of windows and doors, the wind speed and direction could be as much of a factor as the outdoor temperature.

2) If you're going to design based on measurements, and your room has a roof load, be sure to measure on a cold night that is also clear (no clouds). Radiation to the night sky can be significant if no radiation barrier is in place. The attic can get cooler than the outdoor air under some conditions.

3) For energy conservation's sake, look at changing the load before up-sizing the heater. Insulate the walls and roof better, caulk around all penetrations, etc...

4) 5200 BTU/hr is suspiciously close to the heat output of a typical plug-in 1500W electric heater for residential use. If this is the case, please make sure the electrical circuit in the room can handle any additional heating capacity you want to add.

Advice:

Study some basic HVAC concepts and thermodynamics. There are good resources available to help you learn to size heating and air conditioning systems.

Good on ya,

Goober Dave

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