Maths and physics challange

[Paul C]

I'm not sure you've used the right term when you say "heat capacity" - but 3.19kW heat loss seems about right. Basically, that's the output a heater would need to give out, to maintain a temperature (the average of the two figures you've used, 24 and 17). In other words, to maintain 20.5 deg C in the boat with the outside temperature of 2 deg C, you'll need a 3.19kW heater running.

[Doodlebug]

I'm not sure you've used the right term when you say "heat capacity" - but 3.19kW heat loss seems about right. Basically, that's the output a heater would need to give out, to maintain a temperature (the average of the two figures you've used, 24 and 17). In other words, to maintain 20.5 deg C in the boat with the outside temperature of 2 deg C, you'll need a 3.19kW heater running.

 

Yes sounds about right - think I should have said specific heat capacity? Or the thermal mass. Not sure.

 

3.19kw for a boat that small sounds excessive. I would like to think it will be less but need to do the calculations with proper figures.

 

The good thing about that equation is it factors in all heat loss without the need to work out the insulation value of the boat. It works out how much heat has been lost, it doesnt matter where it was lost.

 

It doesn't factor in the heat contained within the internal walls but this should be minimal. The areas where more heat might be stored are in cupboards against the wall which in a way don't count because they are colder than the actual air in the boat. Which is the thing that makes you feel warm.

 

(Those house calculations suggested that a 1.1kw heater would be fine)

[Paul C]

Google "Allen's rule", basically its to do with surface area vs volume. A house will naturally be more efficient, being larger, and a narrowboat is less efficient than the equivalent volume as a cube.

[Graham Bowers]

You have posed a very complex heat transfer problem that in my opinion could theoretically be answered but practically is in the too hard to do pile. This is because there are so many factors involved you would require many terms in the equations and have to make many assumptions. Also, to develop a reliable mathematical model would require experimental work to validate the model and the application of fudge factors. I used to do heat transfer sums for a living btw and have had plenty of exposure to mathematical modelling of various types of problems.

 

If all you really want to know is the potential effect of double glazing then I feel more practical and workable approach that can provide useful info would be to simulate the effect of double glazing by doing a practical experiment. Your scenario is overnight, so solar loading is not a factor. You'd need to find a series of nights that have similar weather in order to obtain meaningful results.

 

Others have mentioned U values. I'd take the alternate approach of estimating the R values (thermal resistance) of your current windows, estimate the R value of double glazed (ask the manufacturer) , then add insulation to the single glazed windows to simulate double glazed. This would probably be a foam (e.g depron) and sticky tape experiment.

The R value of a glass and foam sandwich is easily calculated by adding the two r values.

 

Mathematically, R=1/U and U values are easier to find.

 

Library U values:

Ref http://www.designingbuildings.co.uk/wiki/U-values

 

U single glazing = 4.8 W/m2K. R=.21 m2K/W

U double glazing = 2.8 W/m2K. R=.35 m2K/W

 

Check out what your proposed supplier R value is and adjust as needed.

 

R improvement needed .35-.21= .24 m2K/W

 

6mm depron R=.17 m2K/W

http://www.depronfoam.com/depron-foam/resource/Depron-White-Technical-Data-Sheet.pdf

 

I'd buy some 6mm depron and stick it on your windows overnight and experience the difference, double it up and try again. You could try other insulating material, you'd have to do the research.

Depron is commonly used for under floor heating insulation and I use it to make model aeroplanes !

 

Hope this helps

 

Graham

 

 

 

 

[Beaker]

I've got a headache.

[BlueStringPudding]

For anyone who wants to prove themselves master of maths.

 

It must be possible to calculate the amount of heat needed to keep a boat warm by taking the inside temperature drop over 12 hours vs the outside temperature.

 

So if I heat our boat to 24C and then switch the heating off, and go to bed with the outside temperature being 5C and then wake up 12 hours later to a boat which is 18C and an outside temperature of 0C you must be able to calculate the heat loss of the entire boat

 

And having done that you could work out the exact kw needed to keep the boat warm in any condition, and then work out how much gas coal or diesel you would be using in different scenarios.

 

Who knows the equation you would need to use.

 

Reason i ask is having insulated the boat and being about to install double glazing it would be useful to know how it will affect the heating capacity.

 

I know it would depend on the temperatures dropping constantly but i'm sure there are nights where that happens. It did last night anyway.

 

Cheers!

 

 

 

Or save the money you're spending on double glazing. Instead use your inventiveness to invent some sort of double glazing equivalent and spend the money you would have spent on double glazing on a toilet? Fewer trips outside to unfreeze / unblock /change the jerry can means better heat retention inside the boat.

 

[Graham Bowers]

I've got a headache.

Me too ;-)

[NBDensie]

Here we go.....

 

1) The heat energy stored by the boat is some constant "a" multiplied by the temperature inside the boat T: H=aT

 

2) The rate of loss of heat is proportional to the difference between the boat temperature and the outside temperature - Fourier's law.

dH/dt = -b(T-T_out) where b is a constant, t the time in hours from start and T_out the outside temperature. dH/dt is the differential (calculus!) of the heat with respect to time.

 

3) T_out varies over time, assume linearly: T_out = 5 - 5t/12 so when t=0 T_out =5 and when t=12 T_out=0.

So dH/dt = a dT/dt = -b(T- (5-5t/12))

Rearranging gives: dT/dt = -bT/a + 5b/a -5bt/12a

 

4) This is a differential equation whose solution is:

T = c.exp(-bt/a) -5t/12 + 5+5a/12b where c is a constant.

 

5) But when t=0 T=24,

so 24=c + 5 + 5a/12b

c=19-5a/12b

So

T = (19-5a/12b)exp(-bt/a) - 5t/12 + 5 + 5a/12b

 

6)Now we have to determine a and b

We can find a by experiment:

Measure the boat temperature, turn on the heating and determine the fuel H₁ to raise the temperature by 1 degree. Then a = H₁.

 

6) We can get b/a as at t=12 T=18

 

So 18 = (19-5a/12b)exp(-12b/a) +5a/b

Its a bit messy to work out b/a from this : I will leave it to you to sort out.

But now we know a, b/a, and therefore b

 

7) The fuel needed to maintain the boat at T when the outside temperature is T_out is b(T-T_out) per hour.

 

I hope the maths is correct, my brain hurts!

[Graham Bowers]

 

Or save the money you're spending on double glazing. Instead use your inventiveness to invent some sort of double glazing equivalent............<snip>

 

That's not a bad suggestion. It depends on the OP's reasons for wanting to improve the thermal resistance of the windows and the bu99eration he is prepared to accept. Is it about price of heating, comfort in that favourite armchair by the window, or both.

 

Putting appropriately designed removable insulated "shutters" on the windows when it is dark in the winter will achieve the same heat savings or better and eliminate cold air down the back of the neck. You have to stow the shutters when not in use though and deploy them when needed.

Graham

[Richard10002]

 

 

(Those house calculations suggested that a 1.1kw heater would be fine)

I wasn't suggesting that the house calcs would give a specific, and correct, answer for a boat. But it would give an idea of the relative effect of DG versus SG.

 

If I go to bed at night with the boat nice and toasty, and pretty cold outside, after turning the air to the fire off - when I wake up about 8-10 hours later, it's bloody cold. If I had double glazing, I'd expect it to be less cold, but still cold.

 

Refering to your OP, If you're fitting the double glazing anyway, why not just do it, and see how it goes? Given the "no brainer" regarding double glazing in houses, it must have a similar positive effect on a boat

Edited January 6, 2014 by Richard10002

[Doodlebug]

You have posed a very complex heat transfer problem that in my opinion could theoretically be answered but practically is in the too hard to do pile. This is because there are so many factors involved you would require many terms in the equations and have to make many assumptions. Also, to develop a reliable mathematical model would require experimental work to validate the model and the application of fudge factors. I used to do heat transfer sums for a living btw and have had plenty of exposure to mathematical modelling of various types of problems.

 

If all you really want to know is the potential effect of double glazing then I feel more practical and workable approach that can provide useful info would be to simulate the effect of double glazing by doing a practical experiment. Your scenario is overnight, so solar loading is not a factor. You'd need to find a series of nights that have similar weather in order to obtain meaningful results.

 

Others have mentioned U values. I'd take the alternate approach of estimating the R values (thermal resistance) of your current windows, estimate the R value of double glazed (ask the manufacturer) , then add insulation to the single glazed windows to simulate double glazed. This would probably be a foam (e.g depron) and sticky tape experiment.

The R value of a glass and foam sandwich is easily calculated by adding the two r values.

 

Mathematically, R=1/U and U values are easier to find.

 

Library U values:

Ref http://www.designingbuildings.co.uk/wiki/U-values

 

U single glazing = 4.8 W/m2K. R=.21 m2K/W

U double glazing = 2.8 W/m2K. R=.35 m2K/W

 

Check out what your proposed supplier R value is and adjust as needed.

 

R improvement needed .35-.21= .24 m2K/W

 

6mm depron R=.17 m2K/W

http://www.depronfoam.com/depron-foam/resource/Depron-White-Technical-Data-Sheet.pdf

 

I'd buy some 6mm depron and stick it on your windows overnight and experience the difference, double it up and try again. You could try other insulating material, you'd have to do the research.

Depron is commonly used for under floor heating insulation and I use it to make model aeroplanes !

 

Hope this helps

 

Graham

 

 

 

 

 

Thank you, i'll see if I can work through and understand it all!

 

Surely though using the equation above you can find the amount of heat loss from the air over whichever period? Obviously this will change with each change in temperature outside but its going to give a rough idea surely? I think it is relatively accurate because through all those long numbers I came to 3kw. If I was making a mistake you would expect it to be way off? And if 3kw is correct then it should enough to work out heating.

 

My toilets working perfectly! And theres no way you can say double glazing wont be a good idea!

 

Thanks all!

Here we go.....

 

1) The heat energy stored by the boat is some constant "a" multiplied by the temperature inside the boat T: H=aT

 

2) The rate of loss of heat is proportional to the difference between the boat temperature and the outside temperature - Fourier's law.

dH/dt = -b(T-T_out) where b is a constant, t the time in hours from start and T_out the outside temperature. dH/dt is the differential (calculus!) of the heat with respect to time.

 

3) T_out varies over time, assume linearly: T_out = 5 - 5t/12 so when t=0 T_out =5 and when t=12 T_out=0.

So dH/dt = a dT/dt = -b(T- (5-5t/12))

Rearranging gives: dT/dt = -bT/a + 5b/a -5bt/12a

 

4) This is a differential equation whose solution is:

T = c.exp(-bt/a) -5t/12 + 5+5a/12b where c is a constant.

 

5) But when t=0 T=24,

so 24=c + 5 + 5a/12b

c=19-5a/12b

So

T = (19-5a/12b)exp(-bt/a) - 5t/12 + 5 + 5a/12b

 

6)Now we have to determine a and b

We can find a by experiment:

Measure the boat temperature, turn on the heating and determine the fuel H₁ to raise the temperature by 1 degree. Then a = H₁.

 

6) We can get b/a as at t=12 T=18

 

So 18 = (19-5a/12b)exp(-12b/a) +5a/b

Its a bit messy to work out b/a from this : I will leave it to you to sort out.

But now we know a, b/a, and therefore b

 

7) The fuel needed to maintain the boat at T when the outside temperature is T_out is b(T-T_out) per hour.

 

I hope the maths is correct, my brain hurts!

 

Woah, ok, deffo have a headache and need to work through that one! Thank you!

[Graham Bowers]

 

 

 

Surely though using the equation above you can find the amount of heat loss from the air over whichever period? Obviously this will change with each change in temperature outside but its going to give a rough idea surely? I think it is relatively accurate because through all those long numbers I came to 3kw. If I was making a mistake you would expect it to be way off? And if 3kw is correct then it should enough to work out heating.

 

 

 

Not sure which equation you mean, but I'm going to stick by my guns and maintain that experiment is your only way to obtain meaningful information

If you wish to discuss further I suggest we switch to voice and IM me your phone number and contact times. In the spirit of the forum you could post a precis of the conversation.

Good luck

Graham

[Doodlebug]

Basically I have a small amount of money which I am happy to spend on improving the warmth of the boat. We have a pet which requires constant temperatures and my partner likes it to be warm in the morning when we get up.

 

I am almost deffo going to change the windows to double glazing - our windows leak heat massively and i'm sure it will help, but I fancy doing a before and after comparison.

 

Also we are on an incredibly small boat. We started off with only central heating. (Gas) and found we were using tons and tons of gas. However I then realised why this was the case - we were heating the canal just as much as the boat. I have now fitted a huge radiator which keeps the boat nice and warm. There is now 50mm of insulation under the floor and 25 on everything else except the ceiling.

 

We changed to solid fuel which does work well, but is a pain, and dirty, and takes a huge amount of space. So it occurred to me yesterday that I should run a test to see how expensive it now is to heat the boat. So my head has quite a few different ideas floating about. Hence one post about diesel vs gas, and this one about calculating heat loss.

 

I have no idea what i'll end up doing but am trying to learn as much about everything before I make decisions. Hope that clears things up! (And you don't all think i'm mental!)

Not sure which equation you mean, but I'm going to stick by my guns and maintain that experiment is your only way to obtain meaningful information

If you wish to discuss further I suggest we switch to voice and IM me your phone number and contact times. In the spirit of the forum you could post a precis of the conversation.

Good luck

Graham

 

That sounds like a good idea (and I will experiment), although as I say part of this is more about learning stuff than coming to a precise number. I know insulating, double glazing will help but i'm curious to put a number on it!

 

I'll pm you

[RLWP]

I've got a headache.

Good, you are starting to understand.

 

Richard

[luctor et emergo]

You could use the heat from a composting toilet? Or use the methane in your gas heating? You could simply conect a pipe from the top of your tank(s) to your gas appliance.

[Richard10002]

I only use gas warm air heating to take the chill off occasionally. Way too expensive to use as the only heating source. The stove is my heat source of choice. Once you have one, it takes the space it takes, unless you remove it. Bags of coal can sit on the roof, in the cratch, on the stern if cruiser.

 

I guess if the gas heated water and radiators, it might be more cost effective, but it's hard to get away from the stove on the boat.

 

Diesel warm air heating was good on my previous boat.

[Willber G]

I've got a headache.

You should check your CO alarm...

[Paul C]

You should check your CO alarm...

 

Or open a window..........no hang on.....it would change all the calcs and you'd need to do them again.......

[Willber G]

Or open a window..........no hang on.....it would change all the calcs and you'd need to do them again.......

Good, I don't think we had enough variables to start with!

[PeterF]

Right having thought a bit more about it I think the way to calculate the heat lost is to work out the overall specific heat capacity of the boat in kwh trying to include the objects in the middle of the boat but not including the walls) and then work it out again after 12 hours of loosing heat.

 

So here is the formula to work out the specific heat capacity of the air within the boat (which is 2m by 2m by 4m approx) with the density of air being 1.225kg per m3 with a specific heat of 1005 and the temperature inside being 24 and outside average being 2.

 

q = (2 x 2 x 4m2) x (1.225) x (1005) (24-2)

 

= 433356Kj

 

=120.37kWh

 

Now we do the same equation assuming that after 12 hours the temperature in the boat has now dropped to 17

 

q = (2 x 2 x 4m2) x (1.225) x (1005) (17-2)

 

= 295470Kj

 

=82.07kWh

 

So we have lost

 

120.37-82.07 = 38.3kwh

 

Which when converted to kw over the 12 hours is

 

3.19kw.

 

So to keep the boat at 24 degrees you would need a 3.19 heater (in those conditions with that outside temperature)

 

So now what I need to do is actually measure the temperature over that period to work it out. 3.19kw seems like a lot. I swear we don't use that much. I'll get the figures to you tomorrow!

 

I think you have come up with the right answer even though your calcs contain a large error.

 

You start with calcualtion of the heat capacity of the air in the boat, but the value for the specific heat capacity of air is 1005J/kg°C so your answer is wrong to be stated in kJ, is is just in Joules, out by a factor of 1000.

 

q = (2 x 2 x 4m2) x (1.225) x (1005) x (17-2) = 295470 J

 

Therefore the heat loss from the air is worth 0.003 kW

 

The majority of the heat capacity in the boat is not the air, the air is miniscule in comparison, it is in all the 100s of kg of wood in the walls and furniture, the metal in the stove and cooker and other contents of the boat. The wood of the walls is closer to the inside temeprature than the outside because the insulation keeps the walls hotter.

 

The boat is lined with 12mm ply, cabin is 4m long and 2m x 2m as per your calc, assume would density is 700kg/m³ and Cp is 1700J/kg°C

 

q = (4 x (2 + 2 + 2 + 2)) x 0.012 x 700 x 1700 x (17-2) = 6854400 J which is 23 times the amount of heat stroed in the air. Then you need to start adding in all the other contents of the boat and this then gives you the thermal mass.

 

It is the same as the old myth when looking in the fridge, "do not keep the door open long or the warm air will get in and mean a lot more electricty it used and the contents will get hot. The thermal mass of air is tiny compared to the thermal mass of solids because the air is 700 to 1000 times less dense than wood / water / butter and 8000 times less dense than steel.

[Dharl]

..and just to add to the headaches, you need to input the value of the temperature of the water and also the temperature the steel hull is as well!

 

My favourite answer, and most mathematical correct one so far has to be : 42.

 

Oh and one more thing...as an added variable if on a Canal (no or little movement of water) or on a River with a flow.

[Stilllearning]

I can't be bothered to read all of the replies, but in case no-one else had mentioned it, shame about the spelling. Now I'm far too reasonable to say that all scientists and the like can't spell/write good English, but others might.

[geoffwales]

Easy formula

 

Volume (m3) x delta t (temp rise in c) x ins val (1 is excellent to 5 poor) x 4 = btu then divide by 3412 = kws

[bassplayer]

I can't be bothered to read all of the replies, but in case no-one else had mentioned it, shame about the spelling. Now I'm far too reasonable to say that all scientists and the like can't spell/write good English, but others might.

So long as the point is understood I doubt the spelling matters.

 

What's more worrying is the lack of common sense or ability to listen to other points of view.

 

Scientists and mathematicians need managing for those reasons. I know this being a musician. We get carried away with ourselves too....

[carlt]

None of these equations work for me as they all fail to factor in the discrepancy between what I feel is "warm" and the temperature that the OH is happy with.

 

One is 2 bags of coal a week and the other is a Town Class butty load a day.