Ivor and Olive Lewis on thu 11 mar 04
Dear Michael,
You have changed the conditions of the experiment.
I cry "Foul"
We were discussing natural draught. Now you are using a blower (Fan)
to induce motion.=20
Which way do you want to go, that is, do you want a mechanically
induced draught or do you wish to stay with an open atmospheric
system.
If we follow your current argument many people will be scared
away so it becomes a private matter, and a topic I am not particularly
interested in even though I do have a big centrifugal blower to hook
up.
By the way, with your new system, if you double the cross sectional
area of your stack you halve the velocity of the gas flow. If you
double the length of your stack but keep to a 1 sq ft cross section
the velocity remains constant.
Back to the Wind Tunnel l!!!
Best regards,
Ivor Lewis. Redhill, South Australia
----- Original Message -----
From: "Michael Wendt"
> Dear Ivor,
> In a chimney of uniform cross section the velocity of the gas stream
can be deduced by inserting a thermocouple probe at different heights,
measuring the temperature and applying the Ideal Gas Law:
>
> PV =3D nrT
where:
> P =3D the pressure
> V=3D the Volume
> n =3D the number of molecules present
> r =3D the gas constant
> T =3D the temperature in degrees Kelvin (absolute scale)
>
> I recently did some flash calciner design work for a company and had
to run these very calculations so bear with me please: P in our case is =
so close to constant that it drops out V will be the variable sought n =
does not change regardless of temperature so it drops out r is a =
constant so it drops out T is a variable and one we can measure so it =
stays and determines V. Further, since V is proportional to T, a pair of =
equations allow the approximation of the volume at T2 if The volume at =
T1 is known.
Example:
My burner fan is rated at 50 cubic feet per minute. The gas flow =
was
about 3 cubic feet per minute. The starting temperature was 20 degrees C =
or 293 degrees Kelvin. The goal temperature was 1100 degrees C or 1373 =
degrees K.
> So...
> V2/V1 =3D T2/T1
> solve for V2
> get V2 =3D T2/T1 * V1 here V2 =3D 1373/293 x 53 =3D 248 cubic feet.
> if this much gas is produced each minute and flows through a 1
square foot chimney, then at that point, it flows at a rate of 248 feet =
per
minute =3D 4.13 feet per second.
> The exit point temperature was measured and found to be 1230 degrees K =
so the velocity at the exit point had to be V2' =3D T2'/ T1* V1 =3D =
1231/293 x 53 =3D 222 cubic feet per minute =3D 3.7 feet per second. =
This is slower than the hooter part of the calciner.
> At any point in the calciner (or chimney), the volume of the gas is
directly proportional to the absolute temperature so as the calciner =
cooled the gas stream, the volume of the gas shrank and the velocity =
dropped. To keep the velocity constant in a chimney, the cross sectional =
area of the chimney must take into account the temperature of the gas at =
various points. It does not matter if the cross section is constant but =
a tapered chimney requires less material to build it and so saves money.
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