Heat Used - Physics Question

ChumpusRex

You have only calculated the heat energy in the air - not the bricks - and I don't think you've done it right.

Specific heat x volume x density x temp rise
1.0 kJ kg-1 K-1 * 500 m3 * 1.27 kg m-3 * 20 K= 12700 kJ (3.5 kWh) for a 20 C rise.

So, let's assume a typical house with 16k bricks of 2.5 kg each - and that on average only the inner half the bricks will be heated

Specific heat x mass x temp rise
0.8 kJ kg-1 K-1 * 20000 kg * 20 K = 320,000 kJ (89 kWh)

Different types of insulation will alter the effects of the bricks - cavity wall insulation will heat about half the bricks - the outer skin won't be heated because of the insulation.
Internal insulation will remove most of the bricks from being heater, and external insulation will mean most of the bricks contribute.

These have different effects on house heating - external wall insulated houses will take a long time to heat up, but stay warm for longer. Internal insulated houses will heat up very quickly, but cool down very quickly when the heating stops.

Stone has a much higher density and specific heat than brick, so the effects are even more dramatic in stonebuilt houses/cottages, especially where the internal walls are also solid stone.

zeupater

Hi

Quick rule of thumb ..... 1kW for 1Hour raises 1Tonne 1degreeC ...

Now, all you need to know is the weight of the structure inside the insulated envelope (assuming cavity & loft), add the weight of the air (~1.2kg/cu m) and you have a rough idea for your property .... so, using water as a weight baseline at 1Tonne/cubic metre for a quick guess allow for a specific gravity of something like 2 for solid walls (brick/mortar/concrete), 1 for plasterboard and 0.5 for wood ... (at the top end of the scale, as we don't know the construction material - if you live in an insulated shipping container or battleship, steel comes in at just under 8 .. )

... anything more accurate and you'll need a pretty complex spread-sheet and access to a materials database ...

HTH
Z

nxdmsandkaskdjaqd

I wondered if someone would be so kind as to do a sanity check on my methodology for heat transfer and losses in the outlined below building.

HOUSE DESCRIPTION
10 meters x 10 meters x 5 meters = 500m3
2 Floors
Central Heating = 100 liters
Boiler and Structure Warm up Time from Cold = 1hr

HOUSE STRUCTURE
Brick Walls: 200m2 – Specific Heat = 0.8 (kJ/kg K)
Concrete Floor: 100m2 – Specific Heat = 0.8 (kJ/kg K)
Plaster Ceiling: 100m2 – Specific Heat = 1.0 (kJ/kg K)
Air: 500m3 – Specific Heat = 1.0 – Density = 1.27 (kJ/kg K)

HOUSE INSULATION
Brick Walls: 200m2 = 0.6 (W/m2K)
Concrete Floor: 100m2 = 0.5 (W/m2K)
Ceiling: 100m2 = 0.3 (W/m2K)
Windows: 50m2 = 2 (W/m2K)

OUTSIDE AIR TEMPERATURE FOR DEC/JAN/FEB
Average Min Temperature = 2C
Average Max Temperature = 9C

FORMULAS
Specific Heat (kJ/kg K) x Mass (kg) x T Rise (C) = KJ
Specific Heat x Volume (m3) x Density (Kg/m3) x T Rise (C) = KJ

AMOUNT OF HEAT REQUIRED TO RAISE THE TEMPERATURE OF THE STRUCTURE 1 DEGREE CENTIGRADE
Brick Walls = 0.8 x 30000 x 1 = 24000 KJ
Concrete Floor = 0.8 x 12000 x 1 = 9600KJ
Ceiling = 1 x 830 x 1 = 830KJ
Air = 1 x 500 x 1 = 500KJ
Total = 34930KJ
Total = 9.7kWh

AMOUNT OF HEAT LOSS THROUGH THE STRUCTURE 1 DEGREE CENTIGRADE
Brick Walls = 0.6 x 200 x 1 = 120KJ
Concrete Floor = 0.5 x 100 x 1 = 50KJ
Ceiling = 0.3 x 100 x 1 = 30KJ
windows = 2 x 50 x 1 = 100KJ
Total = 300KJ
Total = 0.083kWh

AMOUNT OF HEAT REQUIRED TO RAISE THE TEMPERATURE OF AIR 1 DEGREE CENTIGRADE
1.0 x 500 x 1.27 x 1 = 635KJ or 0.176kWh

AMOUNT OF HEAT REQUIRED TO RAISE THE TEMPERATURE OF CENTRAL HEATING SYSTEM 50 DEGREE CENTIGRADE
4.2 x 100 x 50 = 21000kJ or 5.83 kWh

ENERGY REQUIRED TO HEAT THE BUILDING
HEATING ON ONCE A DAY
16 hours duration
16 Air Changes
Morning/Evening Outside Temperature = 2C
Day Time Outside Temperature = 9C
Inside Temperature At Switch On = 16C
Final Temperature = 20C

Heating Up Cental Heating = 5.83kW
Heating Structure = 9.7kWh x 4C = 38.8kW
Air Changes = 0.176kWh x 16 = 2.8kW
Initial Heating Air = 0.176kWh x 4C x 1h = 0.7kW
Heat Loss Through Structure during Morning Switch On = 0.083kWh x 18C x 2h = 2.99kW
Heat Loss Through Structure During Day = 0.083kWh x 11C x 7h = 6.39kW
Heat Loss Through Structure During Evening = 0.083kWh x 18C x 7h = 10.5kW
Total = 68kW

So my calculations say that it costs 68kW to raise the temperature and maintain the heat within the building for a 16 hour period.

Based on my own home the energy expended sounds about right, but is my methodology correct.

I appreciate that with all of this type calculation there are assumptions being made (i.e the actual specific gravity of materials), and that these are generalised approximations based on researh.

Many thanks for your help if you are able to validate my method.

Regards

Robert

ChumpusRex

You have confused power and energy when computing heat loss.

nxdmsandkaskdjaqd

ChumpusRex wrote: »

You have confused power and energy when computing heat loss.

Could you let me have some more information as to what I have done incorrectly?

ChumpusRex

Heat loss is computed as a rate (W or kW).
You've written an energy (kJ or kWh). It looks as though there is a "per hour" missing from somewhere.

zeupater

Hi

Early observations .... how many hours are there in a day ?, what's the mass of the GCH components (radiators/pipes etc) and all of the house contents ? ....

... anyway, your sanity check is to compare to your actual usage .... if what you're saying is that for a given (measurable) ambient temperature your property's daily start-point internal temperature is 16C (and this is pretty consistent) and you use a consistent daily amount of energy to raise and hold the temperature at 20C, then your daily structure heat loss balances to what you use ... if your start-point temperature reduces over time but the average external ambient hasn't really changed then the heat provision cycle is too short ... you can compensate for external temperature changes by simply expressing your energy input/losses in terms of kWh per degree difference (Int-Ext).

HTH
Z

nxdmsandkaskdjaqd

ChumpusRex wrote: »

Heat loss is computed as a rate (W or kW).
You've written an energy (kJ or kWh). It looks as though there is a "per hour" missing from somewhere.

AMOUNT OF HEAT REQUIRED TO RAISE THE TEMPERATURE OF THE STRUCTURE 1 DEGREE CENTIGRADE
Brick Walls = 0.8 x 30000 x 1 = 24000 KJ
Concrete Floor = 0.8 x 12000 x 1 = 9600KJ
Ceiling = 1 x 830 x 1 = 830KJ
Air = 1 x 500 x 1 = 500KJ
Total = 34930KJ
Total = 9.7kWh

In this area of the calculation I have 34930kJ which if I divide by 3600 gives me 9.7kWh. Is that incorrect?

Pincher

I think you should just go on an EPC inspector's course, if you really want to know.


For your own house, all you have to do is maintain the house at say 18 degrees for a couple of days, and see how much gas you used.


If you have cold bridges(s), like damp walls, it will lose a lot more heat like wearing a wet T-shirt when it's really cold. How are you going to model that?

[Deleted User]

I'll go along with ChumpusRex when he says you're confusing power with energy, but another misconception is that you need to do a dynamic calculation not a static one. The temperature of the building takes many hours to change, so the heat loss is continuously varying throughout the period of calculation.

You need to draw out the thermal circuit for the building, and then the problem will yield to conventional circuit analysis techniques. Piecewise linear analysis is a less tedious option than calculus in this age of computers. I constructed a spreadsheet to calculate mine, but be warned it's quite a lengthy job, and laborious if you don't enjoy that sort of thing, even if you model distributed elements as lumped approximations for simplicity.

It is interesting though, mine plots the temperature of the air and brickwork over the course of a day, and calculates the difference in energy consumption between running on a timer and leaving the heat on continuously. I was rather pleased to see the overall thermal conductivity of the building came out the same as I get calculating it from my energy bills: 300W/K for my three bed semi (~240m^3).