Michael Wendt on sat 13 mar 04
Dear Ivor and chimney enthusiasts,
I dug out my old physics text:
"Physics for Scientists and Engineers"
second edition
Raymond A. Serway
Saunders College Publishing
ISBN 0-03-004534-7
See page 320
Mechanics of solids and fluids.
There is a detailed explanation of why lighter materials float in fluids and
gasses.
As to why a taller chimney generates more draw than a shorter one,
The pressure at the top is less than the pressure at the bottom.
If the chimney is twice as tall, the pressure differential doubles.
Consider: F= Ma
here: F stands for force
M stands for mass
a stands for acceleration.
If the pressure differential doubles, that is the force causing draw.
Of course the mass of the gas doubles if the chimney is twice as tall,
so on the face of it you would think it a wash, BUT...
the length of the chimney is doubled so it turns out that the
DURATION of acceleration accounts for the increase in velocity.
V = at
Velocity = acceleration x time
If the chimney is taller, the accelerating force has a longer time to act
on the gas and so it does speed up more. Additionally, when we try to
slow it with a damper, inertia of the taller column is greater and it is
harder to slow the more massive column of gas.
I hope it is more clear now,
Regards,
Michael Wendt
Wendt Pottery
2729 Clearwater Ave
Lewiston, ID 83501
wendtpot@lewiston.com
www.wendtpottery.com
Ivor and Olive Lewis on sun 14 mar 04
Dear Michael,
Yes, I know.
Posted my results to Phil in El Vee a couple of weeks ago. <The chimney stack problem. You want the answers. Seems to be a linear
function.
Rough values are (could be equal to or less but not more than)
1m high V = 4.43 m/s; 5 m high V = 9.99m/s; 10m high V = 14.01 m/s;
20m high V= 19.81m/s. I want to back check before I put this to the
Group. Only holds true for atmospheric aspiration, not for blown kilns
04.03.04>>
Been tormenting you since then. Just knew if I prodded enough you
would come up with the goods
Started of with Acceleration due to Gravity at 9.81 M/Sec/Sec,
Initial velocity as zero, various heights of chimney from fire port to
stack top. Calculated time and final velocity.
These values may have to be corrected for air and effluent density,
fuel type, ambient atmospheric pressure, ambient temperature,
geographic elevation and so on but we can get ball park values to play
with now.
What is even better, as a team we have cracked the nut and reached the
kernel.
Well done friend, great work.
Best regards,
Ivor Lewis. Redhill, South Australia
Bruce Girrell on sun 14 mar 04
>...so it turns out that the
> DURATION of acceleration accounts for the increase in velocity.
Bless you Michael,
I knew we had to get time into it at some point, but I wasn't sure how.
Bruce "Ding, ding, ding! - We have a winnah!" Girrell
Ivor and Olive Lewis on mon 15 mar 04
Dear Bruce,
"Time" was there, all the time. Cannot be any "velocity" without
"time"
You have missed the point which I brought to Phil's attention in a
private post on the 2nd, with values for the maximum potential
velocities at a variety of stack heights. What was not accounted for
was "force". Where did the impetus came from? Gravity drives this heat
engine we call a Chimney. Withough Gravity there is no impetus, just
buoyancy with the less dense gas resting on the greater density gas.
But when the denser gas becomes rarefied it has to be replaced, lifted
out of the way by more dense gas. Gravity is the only extant force
which can provide the impetus, the force which induces motion in this
system. Potential energy resides in the height differential from the
stack exit to the fire ports .
Had I stopped pestering, would you have presented a solution?
Best regards,
Ivor Lewis. Redhill, South Australia
----- Original Message -----
From: "Bruce Girrell"
To:
Sent: Sunday, 14 March 2004 6:22
Subject: Re: chimney height, references to physics text
> >...so it turns out that the
> > DURATION of acceleration accounts for the increase in velocity.
>
> Bless you Michael,
>
> I knew we had to get time into it at some point, but I wasn't sure
how.
>
> Bruce "Ding, ding, ding! - We have a winnah!" Girrell
>
>
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Ivor and Olive Lewis on mon 15 mar 04
Dear Bruce,
You say <...so it turns out that the DURATION of acceleration accounts
for the increase in velocity.
Sorry to contradict you but it is the MAGNITUDE of the acceleration
which controls the velocity. See later.
"Time" was there, all the time. Cannot be any "velocity" without
"time"
You have missed the point which I brought to Phil's attention in a
private post on the 2nd, with calculated values for the maximum
potential
velocities at a variety of stack heights. What was not accounted for
was "force". Where did the impetus came from? Gravity drives this heat
engine we call a Chimney. Without Gravity there is no impetus, just
buoyancy with the less dense gas resting in equilibrium on the greater
density gas, while ever temperature differences are constant.
But when the denser gas becomes rarefied it has to be replaced, buoyed
out of the way by more dense gas. Gravity is the only extant force
which can provide the impetus, the force which induces motion in this
system. Potential energy resides in the height differential from the
stack exit to the fire ports .
Had I stopped pestering, would you have presented a solution?
Best regards,
Ivor Lewis. Redhill, South Australia
terryh on tue 16 mar 04
Ivor Lewis wrote:
>You say <...so it turns out that the DURATION of acceleration accounts
>for the increase in velocity.
>Sorry to contradict you but it is the MAGNITUDE of the acceleration
>which controls the velocity. See later.
dv(t)=a*dt: this is the definition of acceleration.
the change of velocity is determined from the magnitude and
direction of acceleration and from the duration of the
acceleration that is dt.
i don't know anything about chimney.
if the acceleration is nearly constant, the different duration
will account for the variation of the velocity increase.
if the duration happens to be nearly constant, the magnitude of
acceleration is the key factor that controls change of velocity.
if neither of tehm is nearly constant, a tough luck.
the change of velocity itself is a product of acceleration and duration.
terry, just interupting without knowing anything about teh discussion :)
terry hagiwara
terryh@pdq.net
http://www.geocities.com/terry.hagiwara
Ivor and Olive Lewis on wed 17 mar 04
Dear Terry Hagiwara,
Thanks for your note. Good to get other people interested in the other
arts associated with claywork
We are speaking of air falling the height of the stack to the
fireports to provide the impetus to move buoyant exhaust gases.
Surely a = dv/dt. I would have thought dv(t)=a*dt defined the velocity
under an acceleration of a for t seconds where s is undefined. But I
am no physicist so I could have got it all wrong.
Tell you what. Let us all know what the velocity is in a stack which
is sixteen feet high when the damper and the fire ports are wide open.
Thanks, and best regards,
Ivor Lewis. Redhill, South Australia
terryh on mon 22 mar 04
Ivor wrote a few days ago,
>Tell you what. Let us all know what the velocity is in a stack which
>is sixteen feet high when the damper and the fire ports are wide open.
ivor,
sorry for interupting while not knowing what the discussion
was all about. i just thought both of you were talking about
the velocity in terms of acceleration and the duration, and
i got confused. (and i was celebrating st. pat two days earlier:)
i don't know anything about chimney.
i'm going to regret writing this,
but if this were an exam, i may write:
chimney height 'H' generates pressure differnece 'dp' at the
chimney due to buoyancy (sp?) of warm air, of
dp=(rho_air - rho(T))*g*H
where rho_air is the density of air outside and rho(T) the
density of air inside chimney (at warm temperature T in
averaged sense).
maybe you were talking about this? or have talked this already.
so, looks like taller the better.
what is the minimum height?
maybe enough height to avoid turbulent flow around the hole
so that laminar flow is realized in the chimney?
if so, i don't know how to estimate the minimum height.
what is the maximum height?
since cooling of updrafting air is ignored to considerable
height, the taller chimney should work better.
i know i re-phrased the question that i can say something.
terry
terry hagiwara
terryh@pdq.net
http://www.geocities.com/terry.hagiwara
pdp1@EARTHLINK.NET on mon 22 mar 04
Hi Terryh, Ivor and all, of these fun Chimney wonderings...
Nice mentions Terryh...I enjoyed your post here...
Too...
Should we not expect turbulance or it's inducements to
interfere with what the exhaust
gasses might otherwise do, hence effecting the final
velocity with
which they may do it..? At least at these kinds of
pressures...
That the shape of the opening where they
enter the Chimney at it's base, the angle from which the
primary direction of the incipiently egressing gasses are
comeing, their interior confluence with whatever aspirations
are entering...and the
interior shape of the Chimney as well as it's ultimate
height, I believe must be regarded as playing some role in
the final potentials of the exhaust Gasses' velocity...
Yes?
Phil
Las Vegas
----- Original Message -----
From: "terryh"
(snip)
> i don't know anything about chimney.
> i'm going to regret writing this,
> but if this were an exam, i may write:
> chimney height 'H' generates pressure differnece 'dp' at
the
> chimney due to buoyancy (sp?) of warm air, of
> dp=(rho_air - rho(T))*g*H
> where rho_air is the density of air outside and rho(T) the
> density of air inside chimney (at warm temperature T in
> averaged sense).
> maybe you were talking about this? or have talked this
already.
> so, looks like taller the better.
> what is the minimum height?
> maybe enough height to avoid turbulent flow around the
hole
> so that laminar flow is realized in the chimney?
> if so, i don't know how to estimate the minimum height.
> what is the maximum height?
> since cooling of updrafting air is ignored to considerable
> height, the taller chimney should work better.
> i know i re-phrased the question that i can say something.
> terry
>
> terry hagiwara
Ivor and Olive Lewis on tue 23 mar 04
Dear Terry,
I know what you are getting at there. I did that as soon as the
original proposition was put forward (Go into Clayart Archives for the
full thread, though not the solution). You quickly prove that the
taller the stack the faster the flow, accepting that the speed up the
flue is constant if the area is constant. So I ask "What is the
velocity in a sixteen foot stack (chimney)when the temperature is
approaching cone ^6. Take the measure as being from the fire port to
the top of the stack.
By the way, make as many assumptions as are needed provided they are
stated and acceptable
I see Phil has touched down again in El Vee and is stirring the pot a
little more!! I hope he has his stack "in" the head along with the
rest of his plans for the future.
Best regards,
Ivor Lewis. Redhill, South Australia.
With five weeks left to legal open air firings.
Time I started slaying bricks ! ! !
terryh on wed 24 mar 04
Ivor wrote,
>So I ask "What is the velocity in a sixteen foot stack (chimney)
>when the temperature is approaching cone ^6.
the taller the chimney is, the better it works.
so, why care for the velocity?
.....
i still don't understand why we are talking about
the velocity. what does this prove?
nevertheless, my answer sheet goes like this:
if buoyancy is the mechanism behind chminey, and heat
loss and all aerodynamical complication are ignored,
the velocity is determined by energy conservation:
loss of potential enegy '{rho_air-rho(T)}*g*H' is equal
to gain of kinetic energy 'rho(T)*v(H)^2/2'.
so, the velocity is given by,
'v(H)=sqrt{2*g*H*(rho_air-rho(T))/rho(T)}'.
H=15ft=5m, g=10m/s^2; sqrt(2*g*H)=10m/s.
rho_air at ambient temperature 300K (27C+273).
cone 6^ temperature 2250F=1230C=1500K.
so, rho(1500K)=300/1500*rho_air(300K)=1/5*rho_air;
(rho_air-rho(T))/rho(T)=4.
so, v(H)=20m/s.
in terms of acceleration and duration:
the buoyancy force on heated air is
'(rho_air-rho(1500K))*g=rho(1500K)*a'.
this leads to a constant acceleration,
'a=(rho_air-rho(1500K))/rho(1500K)*g=4*g'.
it takes t=0.5 sec to reach the top,
either from 'H=a*t^2/2=(4*g)*t^2/2'; or 'v(H)=a*t=(4*g)*t'
terry
terry hagiwara
terryh@pdq.net
http://www.geocities.com/terry.hagiwara
Ivor and Olive Lewis on thu 25 mar 04
Thanks Terry,
Just need that example on the table.
Well Done even if it is an approximation.
Now we need people to time the passage of a sample through their kilns
to validate our feelings.
Come on you lot of layabouts, get the oily rags out and have a go.
Best regards,
Ivor Lewis. Redhill, South Australia
Ivor and Olive Lewis on thu 25 mar 04
Dear Terry,
Walking up to the kiln I had another think about your result
You give a time of 0.5 seconds for a height of 5 metres. so the
velocity is 10 metres per second. You say the velocity will be 20
metres per second. Where does the difference come from?
Best regards,
Ivor.
> H=15ft=5m, g=10m/s^2; sqrt(2*g*H)=10m/s.
> rho_air at ambient temperature 300K (27C+273).
> cone 6^ temperature 2250F=1230C=1500K.
> so, rho(1500K)=300/1500*rho_air(300K)=1/5*rho_air;
> (rho_air-rho(T))/rho(T)=4.
> so, v(H)=20m/s.
>
> in terms of acceleration and duration:
> the buoyancy force on heated air is
> '(rho_air-rho(1500K))*g=rho(1500K)*a'.
> this leads to a constant acceleration,
> 'a=(rho_air-rho(1500K))/rho(1500K)*g=4*g'.
> it takes t=0.5 sec to reach the top,
> either from 'H=a*t^2/2=(4*g)*t^2/2'; or 'v(H)=a*t=(4*g)*t'
>
> terry
>
> terry hagiwara
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