Monday, January 7, 2013

Calculating Heat Loss in a Truck Camper

To calculate heat loss, you need to know a few things:

1) The square footage of all exterior surfaces, the top, walls, floors--every surface needs to be added up.  I calculated my wall surfaces to be 552.5 square feet and windows to be at 25 square feet.  I included two roof vents as windows.

2) The insulation value of these surfaces.  Windows would lose more heat than an insulated wall.  I plan to insulate my walls at R9 and expect them to be effectively less than this, due to thermal conductivity.  So I'll reduce that to R8

3)  The temperature difference you want to be able to keep warm.  For example, I want to be able to keep the inside of my camper at 75 deg F in 20 degree weather.  That is a difference of 55 degrees.

4) You need to know how to convert R-Values into U-Values.  This is simple, U = 1/R.

Considerations for glass windows.  While the R value is 1 for glass, adding window tint changes this 1.125 or a U Value of .89.  Adding storm windows, another layer of glass increase the R Value to 2.125 (U Value for tinted windows and storm window is .47).  Consider also that the roof vents would loose heat faster than glass.  I will assume that I insulate these to at least as good a the windows for simplicities sake.

How to reduce heat loss further:
1) I could put solid R-6 insulation over the front windows.  This is 10 square feet of area which would reduce my BTU loss substantially.
2) Another heat loss area is the door.  I plan to improve the door seals.  These are jalousie windows.  I'll need to seal these with shrink film to reduce air leaks.  Just like the front windows, installing insulation of these windows will help also.

These could reduce heat loss by 167 BTU per hour.  So you can see, going to extra efforts does not radically reduce your BTU needs for heating.  It is important to stop drafts that let it too much cold air, and shrink wrap film does that.

Most of the surface area will be insulated to R-9.  I was worried heat would be conducted out of the camper on the wall studs.  To achieve R-9,  I need a thermal break like butyl tape or aerogel tape. This R value does not include wall coverings.  Even an 1/8" of wood paneling would add a small but significant added insulation value.  If I attached these panels without screws, the thermal break would be complete.

So the formula is T x U x sq ft for each windows and walls.

Here are my Calculations for heat loss:

Total Area:  552.5 sq feet insulated + 25 sq feet in glass

552.5 @ R 9  U-Value = (assume .111)
25 at R 1  U-value = 1

Delta T:  I want 75 degrees inside with 20 degrees outside. Delta T = 55 degrees.

552.5 x .111 x 55 = 3798.44

25 x .47 x 55 =  646.25
Insulating front glazing and door glazing
15 x .16 x 55= 137.5
10 x .47 x 55 =  258.5

4194 BTU per hour

About 100,000 BTU’s per day.

That is a bit more heat loss than I'd like, but it is acceptable.  

How to reduce heat loss further:
1) I could put solid R-6 insulation over the front windows.  This is 10 square feet of area which would reduce my BTU loss substantially.
2) Another heat loss area is the door.  I plan to improve the door seals.  These are jalousie windows.  I'll need to seal these with shrink film to reduce air leaks.  Just like the front windows, installing insulation of these windows will help also.

These could reduce heat loss by 167 BTU per hour.  So you can see, going to extra efforts does not radically reduce your BTU needs for heating.  It is important to stop drafts that let it too much cold air, and shrink wrap film does that.

Most of the surface area will be insulated to R-9.  I was worried heat would be conducted out of the camper on the wall studs.  To achieve R-9,  I need a thermal break like butyl tape or aerogel tape. This R value does not include wall coverings.  Even an 1/8" of wood paneling would add a small but significant added insulation value.  If I attached these panels without screws, the thermal break would be complete.

 If I am tending a fire, I will likely be sleeping on the dinette.  So I would be using the area over the cab for storage.  If I block off the cab overhang, and place rigid insulation across that area, my heat needs would decrease by 94.5 sq ft and 578 BTU's per hour.

This would drop my total BTU needs to: 3616

Why am I making all these calculations?  I want to chose a wood stove that will be sized such that it will keep my camper comfortable--if there is such a thing that is small enough.  I may have to build a stove to suit my purposes.  I want to keep the cost for heating it to a minimum, whatever the heat source is.

So it seems a 4000 BTU stove will keep it close to 75 in 20 degree weather, and a 5000 BTU stove will keep it at 75 degrees inside in weather colder than 20 degrees.  If I take extra steps I can get by with a smaller stove, or else handle much colder weather.  All this is good news.  Even if I made a few mistakes on these estimates, it is clear that I can survive, even in sub zero weather, although the inside temperature goal may not be achievable.  I should be able to handle weather down to 0 degrees F, as 55 deg F temperatures are cold but bearable.

Some practical considerations:  Ever time I open the door, I will let in cold, cold air and all my heat out in a burst.  My idea is to have a second door inside the camper.  Either a pocket door that would close off the bathroom and shower area in the back, or else design a bathroom door to close two ways, one would close for privacy in the bathroom.  The other would open the bathroom door and close off the hallway between the bathroom and shower dividing the kitchen dinette area from the bathroom shower area.  This is called an airlock door.  Only a smaller amount of heat is lost when the door is opened or closed.  Stepping out quickly and closing the door quickly would help.  Heat loss through the door would be an important consideration.  Some sort of tarp hanging on the outside of the camper would help trap warm air inside the camper.  Speaking of curtains.  Insulated curtains could help reduce heat loss at windows.  I know this all sounds extreme, but staying warm in a very cold situation demands these kinds of thoughtful considerations. 

Bottom Line:  What size BTU output would I need to keep my camper warm in cold weather?  A small 3500 BTU heater would work ok.  A bigger heater would work better but at the risk of making it too warm inside.  I am setting a maximum size for my wood stove at 5000 BTU's.  I expect there will be ways to lose excess heat.  Opening doors, vents, and cracking windows.  Looking at the market place there does not seem to be a stove sized this small.  The Sardine stove seems to be the smallest I've seen.  Some tent stoves I've seen are very light, but take full sized logs and while the BTU output is not listed, they are clearly stoves that can put out more heat than I need.

Lets take an extreme case.  I am traveling in Northern Canada in the winter and get caught in a cold snap.  Temperatures are -10 degrees F.  But I'm prepared. I have my shower compartment filled to the ceiling with wood, charcoal briquettes and coal.  Running my stove at maximum output, Lets say 5000 BTU's, what temperature could I keep the camper using all of the above tricks, but leaving the kitchen and dinette windows alone, and using the one ceiling vent, and one lower vent available for ventilation. 


Equation 1:  458 x .111 x T = X    50.89 x T = X

Equation 2:  11 x .47 x T =  Y        5.17 x T = Y

Equation 3:  X + Y=5000      Substituting Equations 1 and 2 into equation 3 we get:


50.89T + 5.17T = 5000

56.06T = 5000

T = 5000/56.06 = 89.19

This means I could keep my camper at 80 degrees inside in -10 degrees outside temperatures.  And it gives me a cushion for heat loss due to opening doors.  Or windy conditions that increase heat loss. 

I could scale back on the amount of fuel I put in for less cold conditions.

So under these conditions, how much fuel would I need and what would it weigh?  Also, a small stove like this would need constant tending.  What would be the burn time.    While I don't have these questions answered yet, my search for a small stove lead me to multifuel stoves.  These stove will burn wood and coal.  I starting things that coal might be a good fuel for me.  I would not need to much, but it would give me longer burn times, and perhaps last over night.  An overnight burn would be nice.  While I don't mind tending a wood stove during the day, I don't like the idea of getting up frequently to stay warm.

Coal:  Coal produces 15,000 BTUs-per-pound.  I would need one third of a pound an hour, at most.  This seems like a modest amount of weight for the heat produced.  So if I wanted to camp in 10 degree weather for one week, I would need only 56 lbs of coal?  Deal.  I'll take that.  Realistically, I could burn wood during the day, and toss in paper bags with pre-measured loads of coal.  To burn overnight, I'd need, at full output, 2.67 lbs of coal.  I expect I could get a good fire going, toss in a nice supply of coal, dial my stove back some and get though the night, or perhaps get up no more than once if I ran it wider open.   I have little experience with small stoves and no experience with coal fired stoves.  I found the oak is about 6200 BTU's per pound.  If this was a wood stove I expect I'd need to feed it every hour or two.  I'd need about a .8 pound of oak an hour to stay warm, 135 lbs for a week heat.  This does not seem like enough for that output, but these are the numbers I'm getting. 

Water Heating:  Some stoves come with a water coil for heating hot water.  I like this idea tremendously, although it added complexity and the possibility of disaster if the water boils into steam. 


Benefits:  
1)  It reduces the heat output of a stove because a portion of the heat goes to heating the water tank.  That heat is not lost.  It is stored and will re-radiate back into the camper.
2)  A water coil and a water tank serve to moderate the heat in the camper.  My pipes won't freeze if the stove runs out of fuel--not right away in any event.  
3)  If it is not too cold out and I want a clean burn with little smoke, I can put some of that heat into the water tank and still have enough heat output while producing a fun to watch fire.

Downsides:
1) Complexity needed to ensure the water does not boil and turn into steam would  could cause and explosion. 

By definition, a BTU is the amount of energy required to raise one pound of water one degree F.
But you have one gallon of water, which weighs approximately* 8.34 pounds. So, you'd need 8.34 BTU to increase one gallon of water one degree F.  Assume a typical 6 gallon RV water tank, and ignore water in the pipes and stove.  To raise the entire tank 1 degree would take 50 BTU's.  That requires So if 2000 BTU's go to heating water.  2000 BTU's would raise the water temperature 40 degrees in one hour.  Assuming the water starts at ambient temperature--say 70 degrees, the wood stove could take it to hot shower levels in one hour, and three hours would put it at 190 degrees.    Time to shut off the stove because in a short while you will have to vent steam or purge your tank and refill with colder water to continue.  Scary?  If not you should be.

So I need to consider some fail safes.  One option is to heat the main water tank, which will be around 50 gallons as well as the hot water tank.  That would give me 56 gallons to be heated.  No to raise this one degree would require 467 BTU's.  So to raise this quanitiy of water from 70 degrees to 105 degrees (comfortable shower temperature) would require 16346 BTU's and 8 hours.  I'd say that is safe provided thermostats and alarms can be attached that will warn of any problems.

No pump would be needed if the water tank was above the stove.  My plan calls for putting the water tank below the stove.  So a pump failure would be a disaster.  All this is interesting, but it is unlikely I'll be heating water with a wood stove unless it is a tank sitting on top without any plumbing.

Availability: 

The smallest marine stoves I could find put out 6500 BTU's on low. A diesel heater this size burns 1.29 gallons a day.  At $4.50 a gallon for diesel that would cost  $5.80 per day, and 38 gallons or $174 a month if I ran it continuously.  That is pretty cheap!  And easy.

The smallest wood/solid fuel stove puts out between 3000 and 8000 BTU's and used very small pieces of wood, or a handful of charcoal briquettes.

A propane vented stove would produce 5500 BTU's and burn fuel at a cost of $4.80 day.  About one 20 lb tank every 140 hours.

So it seems I over analyzed things. The best benefit of insulation seems to be for cooling not heating as most stove put out more heat than I need.  It is clearly of benefit to be well insulated as my heating needs will be modest, even in cold weather.

At these insulation values, I'd  have to consider heat produced by people and dogs in the camper, heat generated by cooking and lighting.  The net result is very little heat will be needed to heat this camper.  A candle or two might be enough to keep it warm in mild conditions or a propane cook stove left on would keep it warm in mild weather.




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