Venturi puzzle
01-30-2011, 10:32 AM
#1
Gets Weekends Off
Thread Starter
Joined APC: Aug 2007
Position: 777er driver
Posts: 101
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
01-30-2011, 12:57 PM
#3
Moderator
Joined APC: May 2006
Position: ATP, CFI etc.
Posts: 6,056
Originally Posted by 698jet
...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.
Last edited by Cubdriver; 01-30-2011 at 01:16 PM.
01-30-2011, 03:32 PM
#8
Chief Jeppesen Updater
Joined APC: Oct 2005
Position: Executive Transport Driver
Posts: 3,080
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.
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.
01-30-2011, 05:10 PM
#9
Moderator
Joined APC: May 2006
Position: ATP, CFI etc.
Posts: 6,056
Originally Posted by FlyerJosh
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.
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.
, 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.
01-30-2011, 05:59 PM
#10
Chief Jeppesen Updater
Joined APC: Oct 2005
Position: Executive Transport Driver
Posts: 3,080
Originally Posted by Cubdriver
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.
, 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!)
Thread
Thread Starter
Forum
Replies
Last Post
