radiator size

I want to add heating to the cellar/basement, it measures 2.5m * 4.5m *

2.1m (L*W*H), there's a DG half glass door exiting to the garden and a single window 400mm * 650mm...

anyone know what size rad I'll need ?

system is microbore currently running 7 rads off a combi if that makes a difference ?

thanks LJ

Reply to
in2minds
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With a bit more info., a reasonable estimate could be made on the basis of treating the cellar walls as exterior. THe situation should be better than that so it will give you a worst case. More information is needed, though.

- Which wall is the exterior one, and what is its construction? is it the longer wall or the short? How thick?

- What is the construction of the interior cellar walls? How thick?

- What is the construction of ceiling, then floor of the rooms above?

- What is the temperature of the room above when the heating is running in degrees Celsius?

- What temperature do you want in the cellar?

- Is the cellar reasonably dry?

One issue that you could have is condensation on the walls since they will tend to be cold. It would be a good idea to insulate them with Celotex or equivalent as that would also reduce heat requirement a great deal.....

Difficult to say. Without knowing the size of the "rads", it is not possible to say whether the boiler has sufficient spare capacity. Usually with a combi it does unless it's a very small one. What make and model?

If you can answer the first set of questions then I'll show you how to do the calculations to get a reasonable idea.

Reply to
Andy Hall

short wall, has door, window and about 1.2m wall space, solid stone

450mm thick

there'll be insulated stud wall errected along half of 1 long wall, the rest of that wall is already rendered the other long wall is rendered stone, both are about 2.5m thick (yes, 8 feet) with neighbours cellar either side (it's a terrace btw) rear is a cavity wall of 100mm thermal block, 33mm kingspan (I had for free) and concrete block

ceiling will be insulated with rockwool (probably) and there's a tiled chipboard floor above, joists are 6"

it varies constantly, it's the kitchen

comfortable enough for children to play all year round, I think the minimum we're allowed is 12'c

yes, there's 1 damp corner but that will be behind the insulated stud wall with a 150mm air gab between, the air gap is that big because it's an alcove

the cellar is relatively warn, even on the coldest days it's about 8'c, unless you leave the door/window open that is

Worcester 28i there are 2 rads 1500mm the rest are around 800mm to 1000mm and the bathroom has a small towel rad

cheers LJ

Reply to
in2minds

Right. This one is going to be difficult to calculate with much accuracy, so I would tend to work on worst case.

Well the principle of calculating heat loss through surfaces is to use the formula:

Heat loss (Watts) = U x T x A

where

U is the U value of the material

T is the temperature difference through the material in degrees C

A is the area in square metres

The principle is that you calculate the heat loss through each component and add them up.

Strictly, where several types of material are involved in a component, the U values of each should be taken and an overall U value for the whole thing calculated. However, you may not have all the values, and sometimes one of them dominates anyway - e.g. a piece of insulation - and you can take its value and ignore the rest because they make little difference to the total.

As you don't know some of the numbers, certain further assumptions will be needed.

- You don't know the temperature of the rooms above. Let's be conservative and say that they are chilly and at 10 degrees because the heating is off upstairs.

- 12 degrees would be pretty nippy for the cellar for the kids unless they are running around. Minimum workplace temperature is 16 degrees AIUI, so for reasonable comfort, I'll use 18 degrees for the cellar.

You can change these numbers as you feel appropriate.

The norm is to work on the basis of an outside temperature of -3 degrees.

Window =======

Area is 0.4 x 0.65 Temp diff is 21 degrees U value from my table is 2.8

Heat loss = 15W.

Door ====

You didn't give the size, but assume it's 2m x 0.7m divided 50/50

Glass part

0.7 x 21 x 2.8 = 41W

Other part (wood)

0.7 x 21 x 3 = 44W

Wall (stone) ============

U value of 450mm stone is 2.2

Area is 2.5 x 2.1 = 5.25 sqm less window and door areas gives net of 3.9 sqm.

Heat loss = 3.9 x 21 x 2.2 = 180W

Total for outside wall = 280W

It would be better to insulate the lot, but let's assume you insulate half with 50mm Celotex. (U value = 0.3)

The thickest stone wall for which I have a U value is 610mm where U is

1.7. It's not linear with thickness, although thicker walls would have less heat loss and lower values, so I will use 1.7 as being the worst case number.

Where the wall has Celotex, the effective U value with this thickness wall, would be 1/( 1/0.3 + 1/1.7) which is 0.25 - not a lot less than 0.3. Therefore for this approximation, one can take the insulation only and ignore the wall.

So for one half we have:

2.25 x 2.1 x 1.7 x 21 = 169W

for the other

2.25 x 2.1 x 0.3 x 21 = 30W

For the whole wall 200W.

Notice the difference the insulation makes.....

On the same principle this would be

4.5 x 2.1 x 1.7 x 21 = 340W

This all assumes that the neighbour's cellar is at outside temperature, which it may not be, but again worst case.

U value for this is 0.4 for the block and 33mm Celotex are both about

0.4

Heat loss is 2.1 x 2.5 x 0.4 x 21 = 44W

OK, I'll use the number for 50mm Celotex again, but you can substitute if you like.

Earlier, I used 10 degrees for the room above and 18 for the cellar as an example. This gives a temperature drop of 8 degrees.

Heat loss would be

4.5 x 2.5 x 8 * 0.3 = 27W.

For the floor of the cellar, you already mentioned insulation.

For a ground floor solid floor, there are U value corrections to account for the effect of the edges where heat loss is higher.

THat won't apply here in the same way, but again, to get a worst case, use 21 degrees as the drop and 0.3 for the U value of the 50mm Celotex.

This gives 4.5 x 2.5 x 21 x 0.3 = 71W.

If you add this lot up, you get to total surface losses of 962W.

YOu now have to account for heat loss through air changes.

It's difficult to know what these will be, but 1.5 per hour is the conventional figure for a habitable room that's not a bathroom etc.

The formula for heat loss is similar, but there is a volumetric constant of 0.33 used for the calculation.

Formula is volume x temperature difference x ventilation rate x 0.33

This gives us 23.6 x 21 x 1.5 x 0.33 = 246W.

The grand total, with all of these assumptions is 1208W, so something like 1500W would be reasonable, in all likelihood.

There are some dynamic effects as well. If you leave stone walls, they will have the additional effect of delaying the rate at which the room heats and cools. If you insulate everything, then that will largely not happen.

Strictly, you should also calculate the heat output of the existing radiators. This depends on size as well as number of panels and fins. You can get the numbers from manufacturer data sheets. Don't forget the correction factor for radiator data sheets. When the boiler is run at 82 degrees, the numbers in the main tables of radiator data sheets should be multiplied by 0.9 to get the real output since the test temperature is higher. So if you measure exsting radiators you need to multiply by 0.9 for what the boiler needs to supply.

I would think that the boiler is plenty capable, but it would be as well to do the sums.

For the new radiator(s) for the cellar, you know the heat output required (1500W). For this, nominal radiator capacity of 1670W, say

1700W would be appropriate.

These should be comfortable minimum figures. If you want to heat the room more quickly, you can oversize the radiator and control it with a TRV or zone thermostat.

Having done all of this, because of the uncertainty of what the U values really will be (although insulating makes that more predictable), I would do a sanity check.

The first thing I'd do is to plug the above formulae and numbers into a spreadsheet so that you can easily play around with the U values and temperatures depending on circumstances.

Then use something like a 3KW fan heater with known settings of (e.g.)

1,2 and 3kW. Run it and measure the inside temperature when stable and the outside temperature. Plug those temperatures into the spreadsheet and see how reasonable the results are.

OK. It would be an idea to try to do something about that, but I am not sure what would be practicable. Perhaps some waterproof sealer? At any rate, I would use pressure treated timber throughout and put DPC strip where it contacts masonry.

Reply to
Andy Hall

thanks Andy

I'll try an electric heater first (as you suggested) see what results I get.

do you know if there're any useful sites with details on the running cost of an oil filled (or other) electric heater ?

LJ

Reply to
in2minds

Not really, but it should be reasonably predictable.

What you could do is get a plug in wattage meter and measure it directly. These are quite cheap from places like Maplin.

Reply to
Andy Hall

No, but if you have gas heating an electric heater is a no brainer. No only is it the most expensive fuel it is also the dirtiest, that is mos Co2 emissions of all domestic heating use in the UK. When our Tony wa signing us gas fitters up for condensing techknowlodgy he should hav taken a look at the electricity generating plants if he had any rea interest in Co2 emissions

-- Paul Barker

Reply to
Paul Barker

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