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Friday, September 03, 2010

Calling All Math Wizards - Chimney Math

Ok.  So I have a math degree from MIT.  That doesn't mean dookie - it was my wife who was more helpful in calculating these numbers that I'll get to in a minute.  Just because I could implement fast fourier transforms 15 years ago doesn't mean I can do anything useful today.  But still, my point is that I'm not a freakin' moron.  So let's get to the topic of discussion.

My wife and I have been trying to get our chimney lined for about 10 months now.  The short version of the story is that the cross sectional area of the chimney liner needs to be sized relative to the opening of the firebox (aka, the fireplace in your living room.)  You need a certain number of square inches in the cross section of the chimney liner (which is a flexible steel tube, essentially) to get the proper draft and airflow so that smoke goes up the chimney instead of coming out into the room.

Well, the first liner that was installed was too small (we need roughly 60-65 inches of cross sectional area).  Smoke billowed out into our room.  We did the math to show the chimney guy this fact.  The liner had started life as a 10 inch round tube, and had been "ovalized" to fit our chimney flue.  The problem is that it had been ovalized too much - squashed down too much so that the cross sectional area was too small.  We had a discussion about this, and although I don't think I actually convinced him that an oval does NOT have the same cross sectional area as a circle (seriously - he didn't/doesn't believe me), he knows it didn't work as it was.  So now we're still talking to the guy about putting in a new one.   

He has proposed that they will try an eleven inch round liner, ovalized down to 5 inches.  I think the company he is talking to is using this chart.  You can see from the chart that if you take an 11 inch round and ovalize it down to 5 inches, that, by definition, implies a major axis length of 14.8 inches, which my wife and I agree with.  (Interestingly, we don't quite agree with the other numbers in that chart - but the 11 inch --> 5 x 14.8 actually matches up with what we calculated for that specific scenario).

NOW - the chart on the top of that page I just linked above also has cross sectional area for each scenario, and we simply can't figure out how they get to 65 square inches here.  This is where you math super-wizards come in.

For area, if "a" is the minor axis and "b" is the major axis of the liner, we're using [a/2 * b/2 * pi]for the area.  In this case, it comes out to 58, which is a far cry from the 65 in their chart.  

My chimney guy actually told me that the ovalized liner would be 5 inches by 16 inches (yielding a cross section of 63 inches).  I explained that I didn't think that was possible - my wife designed a spreadsheet showing the major axis length implied by each minor axis length (after all, circumference doesn't change), and 5 inches implies 14.8 on the other axis - which is what the chart shows.   Of course, when I told the guy that I didn't think it was correct, it didn't go over too well.  We (my wife) used ellipse math to calculate the implied dimensions of the cross sectional axis, starting with a circle of fixed diameter, and assuming the dimension of the minor axis.


So there are two issues.  1) I don't understand how they get from an 11 inch circle to a 5x16 oval. and 2) even if they get to a 5 x 14.8 inch oval, how do they calculate 65 inches of area?

Thoughts?  Anyone? Is there somehow a problem in the ellipse assumptions?  (And yes, we know that all ovals are not ellipses)


-KD

35 comments:

UrbanAnalyst said...

Kid,
Firstly let me say I do not myself have a chimney, so liners or any related woes are not something I think of on even an infrequent basis. However that said, on the math side you're correct to my back of the envelope calculus: 11 x 11 circle = 5 x 14.8 ellipse = 58 square inches of area. I did have two thoughts though.

1) While I assume such a liner would be stainless steel or aluminum, is there any chance the material has a degree of elasticity? Such a circumference expansion could alter the calculations.
2) Is there any chance the liner is about 4/10 of an inch think? If so they may be cunningly measuring the outer surface area rather than the actual flow-through.

That's all I've got. Best of luck!

Kid Dynamite said...

UA - i don't think elasticity has anything to do with it - i just talked to the guy who actually makes the liners, and he said it's "squoosh" not "Stretch"... if they are talking about outer cross section, not flow through, they are morons - I can't imagine they'd do that - wouldn't help anyone.

Also, this guy agreed with my 58 number, but said that it's a special smooth-walled liner that adds up to 20% to the flow effect, so it should be big enough...

Dan said...

This is obviously the end result of the "new math" movement. If your current Chim-a-ney is too small, then you deserve a stimulus payment for a new one.

When I tried to get mine lined and the liner would not fit, all they wanted me to do was either put in a gas insert or go with a wood burning stove. These Chim-a-ney guys are not much into customer service.

Best of luck!

Richard Smith said...

Your spreadsheet sounds wrong - possibly using an imprecise approximation for the circumference of an ellipse?

An 11 inch diameter pipe has a circumference of 11 Pi ins, or about 34.57 ins. But an ellipse with a long axis length of 14.8 ins and a short axis of 5 ins has a diameter of 31.64 ins - so your sheet is losing some circumference somewhere! You need to use the elliptic integral to get the exact value of the circumference; if you are using an approximation with something like PI()* SQRT((a^2+b^2)/2) in it, that is the source of the error - it's quite a way off, 10% or so in this case.

Those wrong figures give you a CSA of 58.2 ins.

But if you have a long axis of say 16.33 ins, a short axis of 5 ins, the circumference comes out about right and the CSA comes out at 64.1 ins. So they are being a wee bit generous with their chart perhaps but not wildly so.

Anonymous said...

My first thought is that the liner can expand. This would mean that a circle w/ a cirmference of X" won't change into a perfect oval of the same circumference. Instead the liner will expand to fit as much of the chimney as possible. (So it's more like a rectangle with rounded corners. This would explain the higher expected area of the cross section.) Your formulas are right, but that's assuming a regular oval. I've seen liners that have give i.e. they are accordion-like and can expand. That's why the circumferences of the 'oval' is higher than the circumference of the original circle.

Here's a question: what's wrong with overshooting and just getting the biggest liner possible? If an 11" would fit, why did he bother installing a 10" the first time around? Is there an issue of too much air? Is it cost? To not waste time of another episode of air billowing back in, why not overshoot and fit the largest one that can possibly fit?

Here's my last suggestion: if your main goal is to not only have the chimney work properly, but also not to waste any more of your time, then have the installation guy put his money where his mouth is. Tell him that you don't think that his plan will work (assuming you need 60-65"). Put up $50. That'll give him incentive to find the solution, regardless of what the cross sectional area really is. (Note: you might need signficantly more than 65"--I've had this problem before.) Focus on the binary result (either the smoke goes up the chimney or it billows out) and not on the math. If you do this, then beware of the installation guy cheating when it comes time to test the chimney. So word the bet to say that you'll pay if there are no smoke issues for the first three months of use. And mentally account for a loss of $50 on your end so that you aren't disappointed. Don't think of this as a bribe for someone to do the job that you're already paying them for. Think of it as saving yourself some wasted time.

Kid Dynamite said...

thanks for the comments guys. but i have two people telling me my math is correct, and another telling me that my math is wrong!

yes Richard, we're using an approximation like that (not sure if it's that exact one)- i'm surprised it has a 10% margin of error.

anon @ 10:47 - i'm guessing that the chimney guy tried to save money with the smaller liner. it was a huge mistake on his end because he's ended up spending 10x as much labor time as he needed to.

as for making a bet with him - the problem is that i've already paid him for the job, so i'm not going to introduce any other financial issues here...

again, i don't think it's a matter of "Stretch" that's accounting for differences - i'm guessing Richard is right about the approximation.

Anonymous said...

Check out the bottom left corner of the product spec sheet in the link that you sent. It says: “…other lining materials behave differently when ovalized and may not assume the same major diameters or cross sectional areas.” I think that this highlights the fact that these materials might not act in the manner you expect. (Saying that there’s a difference between actual and realized results.)

I don’t think that using formulas here and expecting the computed result is different b/c the materials are funky (and might not even be labeled correctly). I’ve found that pipes that I’ve purchased at Home Depot often are not the size that they are listed as being. (Yea, ugh.) Understand that measurements are often approximations. I’m thinking that the original circle size won’t be 11” or so—it’ll be bigger. So maybe the 5”x16” is actually what you’ll get, which would imply a bigger than 11” original circle.

Anonymous said...

Hi KD,

I'm the 10:47 anon. I still say consider making a bet. He'd have to pay you $50 if the 2nd effort fails. No better way to get him to think hard about whether his calculations are right than having that person have the possibility of taking money out of his wallet if he screws up again. I find contractors value their time less than they do their money or their customers' time.

Anon

Anonymous said...

Whoops, meant to say that contractors often don't value their own time or the time of their customers much. The prospect of them having to refund you should be enough to ensure that they get it right this time.

Richard Smith said...

Oops, meant "a circumference of 31.64 ins" not "a diameter of 31.64 ins".

D said...
This comment has been removed by the author.
wcw said...

I read with interest about elliptical area approximation, but my advice is a little more real-world:
- one, use a wood stove. stoves are, to a first order approximation, one squidillion more times efficient than fireplaces
- two, insulate. there is the right way (pay the contractor for that), and there is the cheap way my father used: buy bulk vermiculite (by far the cheapest he found was an ag supply place) and fill the void space once you have a working flue
- three, eliminate other deleterious effects on drawing capacity. can't hurt, might help.

My mathe degree is from Cal, not that it matters.

Yangabanga said...

10:47am anon - way to go! KD how can you deny a pragmatic, free enterprise, incentive aligned solution like that? Not to mention the gambling angle...

I know the imprecision bothers you but really anon has the right answer to solve the problem in front of you. Then you can have the math debate in front of your working chimney.

D said...

I think the approximation for the elliptic integral is a problem, and that 58 is too low, but I wouldn't trust this guy that it will actually be 65 either. Seems to be closer to 61.

a) Squash to a rectangle. then its 12.3x5 and an area of 61.4
b) Use Ramanujan's approximation (source: Wikipedia) pi*[3a+3b-sqrt[(3a+b)(3b+a)]] I get 15.6x5 and an area of 61.3

Steamwhistle said...

Looking the Spec sheet for the companies product they do say that the product may behave differently when ovalized. I am visualizing more of a rugby ball shape versus a football shape. Possibly the rigidity of the product does not allow it bend into an ellipse and the corners are pushed out a bit more. This would explain why the liners major dimension is 14.8 vs. the 16 that the formula gives.

Anonymous said...

Let me weigh-in and say that I don't think cross-sectional area is all that is important to achieving the result that you want. Smoke is essentially a fluid that is in contact with the liner over the length of the chimney and subject to friction at the boundry. Perhaps the manufacturer is taking that into account in his table, explaining the trouble you are having with your calculations. It certainly dovetails with the material comment on the chart, although I would agree they are implying geometry is the only consideration.

Nemo said...

Richard Smith has it exactly right.

The area of an ellipse is easy because an ellipse is just a "stretched circle". Start with a circle of radius a/2 (area = pi*a^2/4), stretch it in the x-direction by a factor of b/a to get your ellipse, and this obviously scales the area by the same amount (so pi*a^2/4 times b/a equals pi*a*b/4). Easy.

But the circumference is much, much trickier. Read the Wikipedia article for a sampling or just take Richard Smith's word for it.

P.S. It is true that "all ovals are not ellipses", which leads to a somewhat interesting question: For any given cirumference and minor axis, is the ellipse the shape with largest area?

Kid Dynamite said...

Nemo - I had the same question "For any given circumference and minor axis, is the ellipse the shape with largest area?" i THINK the answer is yes, but i'm not sure.

D said...

"For any given circumference and minor axis, is the ellipse the shape with largest area?"

I think not... I think the optimal shape may be more rectangular. I have two reasons to believe this:

a) I've compared using the Ramanujan #2 approximation (see wikipedia linked above) to compute your ellipse's area and its lower than that of a rectangle which is 5x12.27. I get 61.393 for the rectangle and 61.300 for the ellipse. Can someone please check my arithmetic?

b) I'm imagining a balloon being blown up between two parallel opposing walls. It starts as a circle, and consider the first moment it touches the wall.

There are two ways it could expand from here...
1) expand as an ellipse with a single point of contact with the wall. In this case the expansion is constrained by the tension of the elastic balloon, which is a function of the circumference.
2) expand flush up against the wall in a more rectangular shape. Here the inward pressure is the tension of the elastic at those parts not in contact with the wall, and a normal force when it is.

My belief is that (2) happens, so that must have a lower inward pressure than (1). But since both shapes incorporate a tension that is a function of the circumference it would seem that (2) must have the smaller tension, and therefore smaller circumference.

In other words: The pressure against the wall is P, but since the balloon is not expanding, the pressure at points not touching the wall is P, and the pressure from the tension of the elastic materials is T=P. Now T is an increasing function of C, and P is a decreasing function of A.

Therefore IF an ellipse were of the same A but smaller C then T could be reduced by snapping into that shape. The failure to change into an elliptical shape seems to suggest its not optimal.

I'm obviously ignoring things like the friction between the walls and the balloon surface, but am I missing something else?

Nemo said...

On further reflection, I think the answer is "no".

First we need to define "minor axis". It is clearly not just two points through which the curve happens to pass, because if the perimeter is big enough (greater than pi*a I think), the maximal-area figure passing through two points is piece of a circle and its mirror image (looks kind of like a figure-eight).

Consider the two parallel lines perpendicular to the minor axis that pass through each of its endpoints. To be a "minor axis", I would say the curve cannot cross either of these lines.

Now, suppose you have an ellipse with large (> pi*a/2) perimeter and minor axis a. Assume (for the sake of contradiction) that this is the maximal-area shape with this perimeter and minor axis.

Draw a line L parallel to the minor axis cutting the ellipse such that the segment of the ellipse outside of L has arc length pi*a/2. (This is always possible if the total perimeter is greater than pi*a.) Fix the two points where L cuts the ellipse, and then let that arc deform itself into the area-maximizing shape for the segment. That shape is not an elliptical segment; it is a semicircle. But this is then a transformation of the ellipse that preserves the perimeter but increases the area: A contradiction.

I believe this argument shows that if the perimeter is larger than pi*a, the area-maximizing shape is a rectangle of side-length a capped by two semi-circles of diameter a.

Let me know if I need to upload a picture. :-)

- Nemo (who also happens to have a math degree from MIT)

Nemo said...

Now, suppose you have an ellipse with large (> pi*a/2) perimeter and minor axis a

Of course I meant "> pi*a", not "> pi*a/2". That is, the perimeter is larger than that of a circle with diameter on the minor axis.

P.S. I see D beat me to it by seconds.

Kid Dynamite said...

Nemo, D - sorry, my brain is fried from spending all day contemplating the dialogue Nemo and I had yesterday on a different subject. I can see your lips moving but I can't hear what you're saying.

I believe you guys though.

Nemo said...

OK my proof does not exactly work. Although it's close. Let me try once more.

Given ellipse of perimeter > pi*a and minor axis a, assume it is maximal area with that perimeter and minor axis.

Take a line L through the minor axis and start sliding it toward one end of the ellipse. Let L intersect the ellipse and A and B, and consider the length of AB (call it "l(AB)") and the length of the arc of the ellipse between A and B (call it s(AB)).

Now, when L starts at the minor axis, s(AB) > pi/2*l(AB). And when L gets very near the end of the ellipse, s(AB) < pi/2*l(AB). So somewhere along the way, s(AB) = pi/2*l(AB). At that point, we can replace the arc from A to B by a semicircle of the same length, and thereby increase the area.

So the ellipse is definitely not the area-maximizing shape.

I am still pretty sure the area-maximizing shape is a rectangle capped by two semi-circles. This argument does not quite prove it, though.

D said...

Nemo... Proof beats intuition any day.

The really nice thing is that this proves the workman is being honest, the 11in pipe should work (unless its too rigid).

You have your two semicircles with diameter 5in which is area 19.63 in^2, and you have 18.84 in of circumference remaining to use as the top and bottom of your rectangle. Your rectangle is then 9.42x5=47.12in^2 for a total of 66.75in^2.

Kid Dynamite said...

you guys are studs, but i have to intervene on that last comment, D - this has nothing to do with the workman being honest. he has NO clue. If the "chart" told him that an 8x8 square had an area of 100 inches, he'd say "it's on the chart.".... that's all... and that's the problem - I have to do the math myself because he already proved his complete incompetence by thinking that an oval and a circle had the same cross sectional area.

D said...

KD -- Fair, honesty is not the right word. The workman's use of the chart probably won't lead to your room being full of smoke.

Nemo -- I think you do have a proof. You have shown that the end is capped by a semicircular shape. All that remains is to show that the beginning of the semicircle coincides with the moment when the perimeter leaves the wall, and the problem becomes an unconstrained optimization. I think this is a classic result, but I've never seen a proof.

rjs said...

KD, im a late to this party, and havent read every word of the commentary, but the thought that comes to mind is that youre not really dealing with a ellipse or oval; youre dealing with an approximately parallel sided "squished sphere" with a side to side of 5 inches (to fit the chimmey. right?); so you should be figuring for the area of the internal rectangle plus the semi-circles on either side...
thus i figure a draft area of approximately 9.8x5 + 2.5 squared x pi....
thats almost 57, less than your ellipse, and less than the 65 you were quoted...

D said...

Nemo -- here is the rest of the proof. Suppose that the semi-circular cap does not meet the parallel boundary. Then it must be inside the parallel constraints. Since the semi-circle is parallel to the parallel lines at its ends, then this figure must not be convex, and is obviously sub-optimal.

Kid Dynamite said...

RJS - i think you did it right but you just had an addition error... your numbers don't add to 57.

rjs said...

KD, it doesnt surprise me that i made an arithmetic error...im pretty tired...

my point was that i thought the "VOV612 Ovalizing Machine" probably produced a parellel sided liner to fit chimneys of various widths..

Kid Dynamite said...

Rjs - yep - you got the hard part right!

interestingly, the guy i talked to on the phone today who actually does the "ovalizing" said that the area of the 11 inch circle mashed down to 5 inch small axis is 58 sq inches! so who knows. maybe he's using the same bad spreadsheet approximation that I am!

rjs said...

it has been a lot of years since i had my liner put in, but i seem to recall a chart wherein the draft potential increased with the chimney height; you might ask about that...

KP said...

Discussions like this one are why I love this blog.

Well, that and the occasional dog doing merengue.

Kid Dynamite said...

RJS - yes - higher chimney height is better for draft. as is an interior chimney (one that doesn't have an outside wall)... and a oval or circular liner provides much more draft than a rectangular one, since the corners get "Wasted" in a rectangle, due to fluid/air dynamics - but i'm just repeating what they told me there...

KP - thanks, although to be honest we don't have a lot of advanced math discussions here!

Anonymous said...

If all this math does not solve the problem once the new liner is in, try this: preheat the chimney before use. To preheat the chimney, twist some newspaper into a torch-like shape and light one end. Hold the lit newspaper up in the flue until you need to let go. Repeat as necessary. This is most useful on winter days when the air in the chimney is cold and therefore more dense. This cold dense air is difficult for the warm air from the fireplace to displace, resulting in it taking the path of least resistance and coming into the room. Preheating the chimney heats the air in the flue and it becomes less dense, which allows the fire to draft up the flue.