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How to find heat loss by experiment

How to find heat loss by experiment

How to find heat loss by experiment

(OP)
I'm trying to verify the accuracy of some software that calculates heat transfer through buildings. This is to be done by some practical experiment, but I'm not quite sure how to set it up, so I want peoples ideas on this. I'm thinking of using a room, its u'values, heating devices, and the adjacent rooms for some measurements. On paper I can calculate the heat loss in Watts using basic equations, but how can I somehow support the calculations by doing an experiment?

RE: How to find heat loss by experiment

Maybe use an IR temperature sensor to measure wall temperatur(s) and confirm it with your calculations?

RE: How to find heat loss by experiment

Get a resistance heater which is surrounded by a medium which will not increase its direct radiation to walls.

With a known I^2R there is a new equilibrium room temp (and hopefully, negligible increases in the surrounging temperatures).
Theta = measured temp-calculated temp
Plot the theta-vs the added heat generation.

Regards

RE: How to find heat loss by experiment

the problem are the windows... that calculation may be almost "easy" to verify... when there are windows... it's a whole different game...
in any case... you do all the calculations, very carefully... considering all factors and alternatives... and then the HVAC OEM says: we have available: 5, 10 or 15 ton equipment, make your pick.

saludos.
a.

RE: How to find heat loss by experiment

(OP)
Problem is that I don't really have any equipment except temp.gauges and heating devices. If I heat the room to a certain temp and notice how long it takes for the surrounding rooms to reach this temp I will have a period of time. Can I maybe calculate something from this?

RE: How to find heat loss by experiment

It's not hard if the outside can maintained at constant air-conditioned temperature(s).  Set up as sailoday28 described plus an internal room circulating fan (for uniform T).  Program the heater thermostat in a step function pattern, with both higher & lower temps.  Record both T & heater output.

The heater output at steady state during each isothermal hold gives the heat flux for a given ΔT.

The increased or decreased heat input needed to change the room T can be used to approximate the room's heat capacity.

Re windows.  No problem in doing the heat flux measurements as described; maybe more correction for room heat capacity.  
Repeat the measurement cycle with one window blocked off or replaced with polystyrene insulation.  
Maybe then repeat with a second window insulated off.
Ditto if a third window.
Perhaps start with the polystyrene sheets loosely stacked in the room so always contribute to the heat capacity.

RE: How to find heat loss by experiment

The method suggested by sailboy will work, but it provides only an indication of the composite UA of the structure.  You learn nothing about individual components.

There are many building energy software packages out there, and some have been validated against models.  Some of these have been publically funded, so the details of the validation method should be available to the public.

Alternatively, validate your software against some already validated and accepted software.  Much easier.

RE: How to find heat loss by experiment

My friend uses a thermo camera, scans the entire building and get the temperature distribution with very distinct hot spots or thermal bridges recognition. Try to borrow one, you will get maximum information in shortest time and use your thermometers to get the room temperatures.
m777182

RE: How to find heat loss by experiment

I did miss the point about window radiation/fenestration.
If the inflow/outflow of air to the central room can be cut off---
Then with additional I^2*R bring the room up to its original temperature. Add  this I^2*R to other heat generation in the room to get the total heat flow.

Regards

RE: How to find heat loss by experiment

Based on my suggestion, Maybe there will have to be heat removed by some cooling device rather that heat added.

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