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Venturi puzzle
If anyone can help I have a puzzle I'm stuck on . it's a venturi problem My son asked me for his class and even I am having a tuff time figgering it out the puzzle is . the airflow in a venturi tubbe has a velocity of 200ft/sec in the wider part where the cross sectional aera is 3 sq ft and the tube narrows down to an aera of 2sq ft. the air density is 0.002378 slugs/sq ft what is the velocity at 2 sq ft. section and what is the dynamic pressure at this point. I have been stuck for two days so now looking to see if anyone knows a formula . thanks
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the key is "constant mass flow" the mass of the air through the pipe per second is constant. so figure out the mass per second at the big end, then solve for velocity in the little part.
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Originally Posted by 698jet
(Post 938835)
...The airflow in a venturi tube has a velocity of 200ft/sec in the wider part where the cross sectional area [S] is 3 sq ft and the tube narrows down to an area of 2sq ft. the air density [rho] is 0.002378 slugs/cu. ft. What is the velocity [V] at 2 sq ft. section(?) and what is the dynamic pressure [q] at this point....
S1V1=S2V2, V2=300 ft/sec To find dynamic pressure (q) Use Bernoulli's equation: po + .5*rho*V^2= PT (constant) where PT= total pressure (constant) po= static pressure dynamic pressure q= .5*rho*V^2 So: po1 + .5*rho* V1^2 = po2 + .5*rho*V2^2 po1-po2= .5*rho*(v2^2 -V1^2) If po1 is normal atmospheric pressure (2116 lb/ sq ft) then po2= 2116 lb/ sq ft - 59.45 lb/sq ft= 2056.55 lb/sq ft Since po1 + .5*rho* V1^2 = po2 + .5*rho*V2^2 as stated earlier and dynamic pressure is defined as q= .5*V2^2, it is simple to solve for q2, and I get 166.46 lb/ sq. ft. or 1.156 psi at the outlet side. |
That's what I got too. :)
Well done, sir. |
Hate to bust anyones chops including mine, but that problem would be worth maybe ten points on a typical 100 point test. It's first year aero or mechanical engineering.
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I just became violently ill.
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That reminded me of a T-Shirt I saw in one of the billion catalogs we get:
"I was an English major, you do the math". :p |
I'm just a pilot but doesn't this make sense?:
3 sq ft area * flow rate of 200 ft/min = 600 cu ft/min total flow in the wide section 600 cu ft/min total flow through the 2 sq ft area = flow rate of 300 ft/min in the narrow section. |
Originally Posted by FlyerJosh
(Post 938971)
I'm just a pilot but doesn't this make sense?:
3 sq ft area * flow rate of 200 ft/min = 600 cu ft/min total flow in the wide section 600 cu ft/min total flow through the 2 sq ft area = flow rate of 300 ft/min in the narrow section. |
Originally Posted by Cubdriver
(Post 939031)
Kind of, except for the units. Sanity check: does it make sense for a flow of this type which is perhaps a carburetor venturi or a wind tunnel flow, to be going only 300 feet in an entire minute? A regional pilot can run more than 500 ft per minute which is of course seriously slow by human standards :), so this air would be going a lot faster; how about V2= 300 ft/sec. You can treat this problem as a two dimensional flow.
I'd like to see a regional pilot run 500' a minute! (Unless he's trying to make a commute home!) |
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