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Bernoulli's Equation and pipe reducer

  • Thread starter Nirupt
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  • #1
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Homework Statement




For the pipe reducer shown in Figure 6.27 in your textbook, the pressure at point A is 45.8 psig and the pressure at B is 39.5 psig. Calculate the velocity of flow, in ft/s, of the water at point B.

State your answer to one decimal place. Example: 25.6



Homework Equations



http://img28.imageshack.us/img28/4754/7m2c.png [Broken]
Which is the equation we are using, P is pressure, Z is height (which is 0 for both A, and B), g is 32.2 ft/s^2 and gamma (specific weight, is 62.4 lb/ft^3 I believe)

Image of the problem: http://img802.imageshack.us/img802/9700/byqt.jpg [Broken] Ignore the book problem as the numbers have been changed.

Remember to divide pressure by gamma
Remember to calculate velocities at points A and B Velocity = flow rate/Area
Elevation at point A is zero and Elevation at point B is zero
Do forget to take the square root to calculate the velocity

The Attempt at a Solution



I know since B is a small tube the velocity at point B must be higher than point A. I have the answer outcome to be 31.6 ft/s but I'm not totally sure that is correct.
I have two unknowns though which is both velocities.. and it's confusing me because the only way to cope with that is the equation VaAa = VbAb which Velocity is V and Area is A, finding the area using the equation ∏*D2/4 I have converted in to ft and obtained both areas which are..

Aa = .0218 ft^3
Ab = .00545 ft^3

I have tried putting one unknown and plugging either velocity in for the other.. but it doesn't make sense to me. Plus the flow rate is not given, and I cannot find it because I have neither velocity numbers.. overall there appears to be 2 unknowns, flow RATE, and VelocityA, and we are solving for Velocity B.
 
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Answers and Replies

  • #2
rude man
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Your post is entitled "Bernoulli's Equatiomn". Why not invoke it?
 
  • #3
rude man
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Homework Statement




For the pipe reducer shown in Figure 6.27 in your textbook, the pressure at point A is 45.8 psig and the pressure at B is 39.5 psig. Calculate the velocity of flow, in ft/s, of the water at point B.

State your answer to one decimal place. Example: 25.6



Homework Equations



http://img28.imageshack.us/img28/4754/7m2c.png [Broken]
Which is the equation we are using, P is pressure, Z is height (which is 0 for both A, and B), g is 32.2 ft/s^2 and gamma (specific weight, is 62.4 lb/ft^3 I believe)

Image of the problem: http://img802.imageshack.us/img802/9700/byqt.jpg [Broken] Ignore the book problem as the numbers have been changed.

Remember to divide pressure by gamma
Remember to calculate velocities at points A and B Velocity = flow rate/Area
Elevation at point A is zero and Elevation at point B is zero
Do forget to take the square root to calculate the velocity

The Attempt at a Solution



I know since B is a small tube the velocity at point B must be higher than point A. I have the answer outcome to be 31.6 ft/s but I'm not totally sure that is correct.
I have two unknowns though which is both velocities.. and it's confusing me because the only way to cope with that is the equation VaAa = VbAb which Velocity is V and Area is A, finding the area using the equation ∏*D2/4 I have converted in to ft and obtained both areas which are..

Aa = .0218 ft^3
Ab = .00545 ft^3

I have tried putting one unknown and plugging either velocity in for the other.. but it doesn't make sense to me. Plus the flow rate is not given, and I cannot find it because I have neither velocity numbers.. overall there appears to be 2 unknowns, flow RATE, and VelocityA, and we are solving for Velocity B.
Given v and A, what is volume flow rate?
 
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  • #4
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Given v and A, what is volume flow rate?
Vb is what is being looked for, Va is unknown, however.
 
  • #5
rude man
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Vb is what is being looked for, Va is unknown, however.
My question was general: given velocity of flow v and area of cross-section A, what is the volume flow in m^3/sec. or whatever units you like?

So then can you write an equation relating volume flow at A and at B?

And again, what about Bernoulli's equation?

You should wind up with those two equations with unknowns vA and vB.
 
  • #6
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My question was general: given velocity of flow v and area of cross-section A, what is the volume flow in m^3/sec. or whatever units you like?

So then can you write an equation relating volume flow at A and at B?

And again, what about Bernoulli's equation?

You should wind up with those two equations with unknowns vA and vB.
So far I have Vb = SQRT(Va^2+(Pa-Pb/y)) Which I know all of except Va and I have Va=Vb*(Ab/Aa)

Which leaves me clueless.. even though I have 2 separate equations.. I guess I have the flow rate equation which can either be Va = Qa/Aa or Vb = Qb/Ab

but Q cannot be solved because I cannot find either velocity.. I guess I'm going through a loop.
 
  • #7
rude man
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So far I have Vb = SQRT(Va^2+(Pa-Pb/y)) Which I know all of except Va and I have Va=Vb*(Ab/Aa)

Which leaves me clueless.. even though I have 2 separate equations.. I guess I have the flow rate equation which can either be Va = Qa/Aa or Vb = Qb/Ab

but Q cannot be solved because I cannot find either velocity.. I guess I'm going through a loop.
What about Va Aa = Vb Ab? You don't need to solve for Q if you don't want to.

You have two equations in 2 unknowns: Va and Vb.

One equation is Bernoulli, the other is volume flow conservation.

You have both equations in front of you. This is just high school algebra.

What's that "y" doing there?
 
  • #8
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Y is the specific weight of water, which is either 9.81kN/m3 or 62.4 lb/ft3 in this case I am using the 62.4.
 
  • #9
rude man
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Y is the specific weight of water, which is either 9.81kN/m3 or 62.4 lb/ft3 in this case I am using the 62.4.
Unusual units.

The Bernoulli equation is usually given as ρv2/2 + ρgh + p = constant

ρ = density (for water, = 1000 kg/m3
v = velocity, ms-1
g = 9.81 m s-2
h = height above an arbitrary reference
p = pressure, Pa
p can be absolute or gauge as long as yo're consistent.

I imagine your equation and your units are OK too, so as I said twice before, you have 2 equations and 2 unknowns: v1 and v2.
 
  • #10
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Unusual units.

The Bernoulli equation is usually given as ρv2/2 + ρgh + p = constant

ρ = density (for water, = 1000 kg/m3
v = velocity, ms-1
g = 9.81 m s-2
h = height above an arbitrary reference
p = pressure, Pa
p can be absolute or gauge as long as yo're consistent.

I imagine your equation and your units are OK too, so as I said twice before, you have 2 equations and 2 unknowns: v1 and v2.
I'm reworking it but I'm only using the units I gave simply because the answer is in ft/s, so I've converted the Area from inches to ft as well, or rather the diameter. I will respond my attempt shortly. However, I do understand that under normal circumstances 9.81 m s-2 is used, so I assume that my equations are setup properly, just not my units somewhere.

Such as, I found this equation..

p1-p2/ɣ = v22-v12/2g


v = velocity
p = Pressure in psig
g is the gravity acceleration constant (9.81 m/s2; 32.2 ft/s2)
ɣ = Water has a weight density of 62.4 pounds per cubic foot in Earth gravity (32.2 feet per second squared)

p1 = 45.8psig
p2 = 39.5 psig
v1 = unknown
v2 = solve; also unknown

1 pound per square inch =
144 pounds force per (square foot)

Then my 2nd equation

But first my area

∏*D2/4

D1 = 2 inches
D2 = 1 inch
A1= .0218 ft2
A2=.00545 ft2

v1 = v2*(A2/A1)
v1 = v2*(.00545 ft2/.0218 ft2)
v1 = .25v2

However we need v12
v12 = .0625v22
Plug it in..
45.5-39.5 psig*144 lb/ft2/62.4 lb/ft2 = v22-.0625v22/2*32.2 ft/s2

14.54 = .9375v22/64.4ft/s2
936.376ft/s2 = .9375v22
√998.801ft/s2 = v22
v2 = 31.6 ft/s

I finally figured it out while typing this.. I completely missed where the psig was in lb/in2 and I should have converted it to lb/ft2 multiplying it by 144... but at least I figured it out?
 
  • #11
rude man
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I'm reworking it but I'm only using the units I gave simply because the answer is in ft/s, so I've converted the Area from inches to ft as well, or rather the diameter. I will respond my attempt shortly. However, I do understand that under normal circumstances 9.81 m s-2 is used, so I assume that my equations are setup properly, just not my units somewhere.

Such as, I found this equation..

p1-p2/ɣ = v22-v12/2g


v = velocity
p = Pressure in psig
g is the gravity acceleration constant (9.81 m/s2; 32.2 ft/s2)
ɣ = Water has a weight density of 62.4 pounds per cubic foot in Earth gravity (32.2 feet per second squared)

p1 = 45.8psig
p2 = 39.5 psig
v1 = unknown
v2 = solve; also unknown

1 pound per square inch =
144 pounds force per (square foot)

Then my 2nd equation

But first my area

∏*D2/4

D1 = 2 inches
D2 = 1 inch
A1= .0218 ft2
A2=.00545 ft2

v1 = v2*(A2/A1)
v1 = v2*(.00545 ft2/.0218 ft2)
v1 = .25v2

However we need v12
v12 = .0625v22
Plug it in..
45.5-39.5 psig*144 lb/ft2/62.4 lb/ft2 = v22-.0625v22/2*32.2 ft/s2

14.54 = .9375v22/64.4ft/s2
936.376ft/s2 = .9375v22
√998.801ft/s2 = v22
v2 = 31.6 ft/s

I finally figured it out while typing this.. I completely missed where the psig was in lb/in2 and I should have converted it to lb/ft2 multiplying it by 144... but at least I figured it out?
I haven't checked your math, and I can't imagine why you were given a problem in nonstandard physics units (usually SI aka 'rationalized mks'), sometimes cgs) but it looks like you got the point of solving a system of two equations in two unknowns.
 

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