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What is the velocity of the air flowing though the pipe?

  • Thread starter Siebo
  • Start date
  • #1
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Homework Statement


The horizontal pipe is very large and long: 100 feet in diameter and 1500 feet long with an infinite supply of air on both ends. The pressure difference between the two ends is 0.6 psi. I assume friction is near zero as the pipe is so big so I ignore it. What is the velocity of the air that flows through it.
The density of the air is 1.2 grams/liter at 20 degrees C.(I converted all units to same system)
kinematic viscosity of the air is 1.64 x 10-4 ft2/sec
dynamic viscosity of the air is 3.82 x 10-7 lbf sec/ft2

Homework Equations


Which viscosity do I use for the formula. I thought the dynamic viscosity because the units work out but that gives an even more ridiculous answer.

The Attempt at a Solution


I have tried to use Poiseuille's equation of v= (P1 - P2)(3.124)r4/8L and I get an outrageous answer of 800,000,000 cubic feet per second. What am I doing wrong?
 

Answers and Replies

  • #2
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Did you calculate the value of the Reynolds number that this implies?

Chet
 
  • #3
6
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I looked at that and I need to know the velocity or flow rate of the air.
 
  • #4
19,918
4,094
I looked at that and I need to know the velocity or flow rate of the air.
You calculated the velocity and the flow rate. What Reynolds number does your result imply?

Chet
 
  • #5
6
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I don't understand your response. I don't know the velocity or the flow rate. That's what I'm trying to find out. And I can't calculate the Reynolds number without the velocity or flow rate.
 
  • #6
19,918
4,094
I don't understand your response. I don't know the velocity or the flow rate. That's what I'm trying to find out. And I can't calculate the Reynolds number without the velocity or flow rate.
According to the Poiseulle equation, you calculated the flow rate correctly. Now, for a flow rate of 800000000 cubic feet per second, what would the Reynolds number be equal to?

Chet
 
  • #7
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
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1,666

Homework Statement


The horizontal pipe is very large and long: 100 feet in diameter and 1500 feet long with an infinite supply of air on both ends. The pressure difference between the two ends is 0.6 psi. I assume friction is near zero as the pipe is so big so I ignore it. What is the velocity of the air that flows through it.
The density of the air is 1.2 grams/liter at 20 degrees C.(I converted all units to same system)
kinematic viscosity of the air is 1.64 x 10-4 ft2/sec
dynamic viscosity of the air is 3.82 x 10-7 lbf sec/ft2

Homework Equations


Which viscosity do I use for the formula. I thought the dynamic viscosity because the units work out but that gives an even more ridiculous answer.

The Attempt at a Solution


I have tried to use Poiseuille's equation of v= (P1 - P2)(3.124)r4/8L and I get an outrageous answer of 800,000,000 cubic feet per second. What am I doing wrong?
It's not clear what you mean when you say you converted all units to the same system. The units for the dynamic and kinematic viscosity of air in the OP are not SI. How did you account for this fact in your calculations?

Using your calculated value for the air flow, did you also check the velocity of the air inside the pipe?
 
  • #8
6
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Hi Steamking. I just tried another way to solve it and I think it worked using Bernoulli's equation. Please check this if I am right.
By using Bernoulli's equation we assume: no friction, no shear stress, only steady flow, and density is constant. Given that the length and diameter of pipe is irrelevant.
Known data:
A horizontal pipe with a 0.6 psi or 4114 pascal pressure difference from each end and the question is what is the velocity of the air flowing though the pipe.
Bernoulli's Equation: P1/p + V"1/2 + gz1 = P2/p + V"2/2 + gz2 where P1 = pressure at high pressure side of pipe = 4114 pascal
p = density = 1.2 grams/liter
V"1 = velocity of particle at supply side of air vessel
g = gravity
z1 = height of particle at entrance side of pipe
P2 = pressure at low pressure side of pipe and equal to atmospheric pressure or 0
V"2 = velocity of particle at exit side of pipe
z2 = height of particle at exit side of pipe
Without the ability to draw a picture I will try to describe with words:
Since pipe is horizontal, gz1 and gz2 are equal and can be subtracted from the equation.
Also, since V"1 starts at nearly zero (assuming the particle starts in an infinite sized vessel at a point barely influenced by the situation) it can be nearly zero:
therefore the equation reduces to this:
P1/p = P2/p + V"/2
solving for V"2 produces: V"2 = square root of ((P1 - P2)/p x 2
by entering the data it becomes: V"2 = square root of (( 4114 Pa - 0)/1.2kg/m3) x 2)
= 82 meters/sec
I apologize for not being able to draw diagrams or enter the proper nomenclature but it seems to be correct. What do you think?
thankyou
 
  • #9
19,918
4,094
Hi Steamking. I just tried another way to solve it and I think it worked using Bernoulli's equation. Please check this if I am right.
By using Bernoulli's equation we assume: no friction, no shear stress, only steady flow, and density is constant. Given that the length and diameter of pipe is irrelevant.
Known data:
A horizontal pipe with a 0.6 psi or 4114 pascal pressure difference from each end and the question is what is the velocity of the air flowing though the pipe.
Bernoulli's Equation: P1/p + V"1/2 + gz1 = P2/p + V"2/2 + gz2 where P1 = pressure at high pressure side of pipe = 4114 pascal
p = density = 1.2 grams/liter
V"1 = velocity of particle at supply side of air vessel
g = gravity
z1 = height of particle at entrance side of pipe
P2 = pressure at low pressure side of pipe and equal to atmospheric pressure or 0
V"2 = velocity of particle at exit side of pipe
z2 = height of particle at exit side of pipe
Without the ability to draw a picture I will try to describe with words:
Since pipe is horizontal, gz1 and gz2 are equal and can be subtracted from the equation.
Also, since V"1 starts at nearly zero (assuming the particle starts in an infinite sized vessel at a point barely influenced by the situation) it can be nearly zero:
therefore the equation reduces to this:
P1/p = P2/p + V"/2
solving for V"2 produces: V"2 = square root of ((P1 - P2)/p x 2
by entering the data it becomes: V"2 = square root of (( 4114 Pa - 0)/1.2kg/m3) x 2)
= 82 meters/sec
I apologize for not being able to draw diagrams or enter the proper nomenclature but it seems to be correct. What do you think?
thankyou
This calculation inherently assumes that the entire 0.6 psi pressure drop occurs within the infinite sized vessel leading up to the pipe, and none of it occurs between the beginning and end of the pipe. Is this what you meant to assume?

Chet
 
  • #10
6
0
According to the Poiseulle equation, you calculated the flow rate correctly. Now, for a flow rate of 800000000 cubic feet per second, what would the Reynolds number be equal to?

Chet
Hi Chet - something isn't right. At a flow rate of 800,000,000 cfm, the velocity of the air blasting out of the pipe would be over 69,000 miles per hour! That's just not realistic at all. Since then I used Bernoulli's equation and got a better result of 256 ft/sec
 
  • #11
19,918
4,094
Hi Chet - something isn't right. At a flow rate of 800,000,000 cfm, the velocity of the air blasting out of the pipe would be over 69,000 miles per hour! That's just not realistic at all. Since then I used Bernoulli's equation and got a better result of 256 ft/sec
I know that, but did you see what I wrote in post #9?

Chet
 
  • #12
6
0
This calculation inherently assumes that the entire 0.6 psi pressure drop occurs within the infinite sized vessel leading up to the pipe, and none of it occurs between the beginning and end of the pipe. Is this what you meant to assume?

Chet
Hi Chet - yes that would be correct. Bernoilli's equation assumes no shear stress which would be involve friction. So it doesn't matter how long the pipe is or where the pressure drop occurs even though it would have to be in the transition between the vessel and the pipe.
 
  • #13
19,918
4,094
Hi Chet - yes that would be correct. Bernoilli's equation assumes no shear stress which would be involve friction. So it doesn't matter how long the pipe is or where the pressure drop occurs even though it would have to be in the transition between the vessel and the pipe.
So, suppose you did have a velocity of 256 ft/sec in the pipe. That flow would certainly be turbulent. What pressure drop would that velocity correspond to in turbulent flow through the pipe (where shear stress at the wall is a physical reality)?

Chet
 

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