Water Flow Through a Pipe w/180° Horizontal Elbow

In summary, the conversation discusses a problem where water flows through a circular pipe with a horizontal elbow and exits through a nozzle. The diameter of the pipe and nozzle exit are given, as well as the density and mass flow rate of the water. The problem asks to determine the velocity at two different sections, the gauge pressure at one section, and the magnitude and direction of the force exerted by the water on the elbow. The conversation also includes equations and an attempt at a solution, with some confusion regarding the angle and density.
  • #1
teegfit
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Homework Statement



Water flows through a circular pipe with a 180° horizontal elbow and exits to the atmosphere through a nozzle as shown in Fig. Q3. The diameter of the pipe is 300 mm and the diameter of the nozzle exit is 160 mm. The density of water is 999 kg/m3 . The mass flow rate of water is 140 kg/s. Incompressible, inviscid flow may be assumed.

Determine:
(i) the velocity of the water at sections 1 and 2 respectively
(ii) the gauge pressure at section 1,
(iii) the magnitude and direction of the force Rx exerted by the water on the elbow

Homework Equations


Firstly I changed the mass flow rate to the volume flow rate by dividing by 1000 and then calculated the area by (pi((D^2)/4) giving A1 = .0707 m^2 and A" = .0201 M^2

The Attempt at a Solution


i) I found the velocity,V, by using Q(volume flow rate)/A(cross sectional area of the pipe at section 1 & 2) giving me V1 = 1.98 m/s and V2 = 6.965 m/s

ii) I assumed pressure at section 2 to be zero as it is out of the pipe and then used bernoullis equations and after cancelling unneeded coefficients found

P1 = (denisity)( (V2^2)/2 - (V1^2)/2 ) giving me a pressure of 22,273. 117 Pa

iii) Finally i used a formula given to us by our lecturer to calculate the force Rx

-(denisty)(V1^2)(A1) + (denisty)(V2^2)(A2)cos(theta) = P1A1 + P2A2cos(theta) + Rx

rearranging we get as P2 = 0 ,
Rx = -(denisty)(V1^2)(A1) + (denisty)(V2^2)(A2)cos(theta) - P1A1

i wasnt sure wheter theta was 180 or 0 but i put it in as 0 and got the answer

Rx = - 880 .011 N and there it is 880.011 N acting in the negative x direction

i feel due to the diagram that the answer to part iii) is wrong, could anyone show me where i might have gone wrong.

Thank you very much for your help
 

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  • #2
teegfit said:
Firstly I changed the mass flow rate to the volume flow rate by dividing by 1000 and then calculated the area by (pi((D^2)/4) giving A1 = .0707 m^2 and A" = .0201 M^2
The density of water is given, and it is not 1000 in any unit system (don't forget the units).
teegfit said:
i wasnt sure wheter theta was 180 or 0 but i put it in as 0 and got the answer
The angle is given in the problem statement. You can also check what happens if you don't have the nozzle (so velocity, cross-section and pressure don't change): a straight line will give zero, a bent line will give a non-zero value.
 
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  • #3
mfb said:
The density of water is given, and it is not 1000 in any unit system (don't forget the units).The angle is given in the problem statement. You can also check what happens if you don't have the nozzle (so velocity, cross-section and pressure don't change): a straight line will give zero, a bent line will give a non-zero value.

Im confused by what you mean in the last part? in some examples given we had (180 - angle given). That is where the confusion came from. Also are my calculation correct apart from the angle?
 
  • #4
teegfit said:
Also are my calculation correct apart from the angle?
And the density.
teegfit said:
Im confused by what you mean in the last part?
I showed how you can check the definition of the angle. Apply the formula to a simple problem where you know the correct answer by other means, and see if you get the right result.
 
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  • #5
mfb said:
And the density.I showed how you can check the definition of the angle. Apply the formula to a simple problem where you know the correct answer by other means, and see if you get the right result.

Thanks you very much for your help sir
 
  • #6
There are two different angles, one coming in and one going out. One angle is zero, and the other angle is 180 degrees. When the two momenta are moved to the same side of the equation, they reinforce, they don't partially cancel.

Chet
 
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Related to Water Flow Through a Pipe w/180° Horizontal Elbow

1. How does the 180° horizontal elbow affect the water flow through a pipe?

The 180° horizontal elbow creates a change in direction for the water flow, causing a decrease in flow rate and an increase in pressure drop.

2. What factors can affect the flow rate through a pipe with a 180° horizontal elbow?

The flow rate through a pipe with a 180° horizontal elbow can be affected by the diameter of the pipe, the angle of the elbow, the viscosity of the fluid, and the roughness of the pipe walls.

3. How can the pressure drop in a pipe with a 180° horizontal elbow be calculated?

The pressure drop in a pipe with a 180° horizontal elbow can be calculated using the Bernoulli's principle and the Darcy-Weisbach equation, taking into account the fluid properties, pipe dimensions, and flow conditions.

4. Is there a way to minimize the pressure drop in a pipe with a 180° horizontal elbow?

Yes, the pressure drop in a pipe with a 180° horizontal elbow can be minimized by using a larger diameter pipe, reducing the angle of the elbow, and ensuring the pipe walls are smooth and free from obstructions or bends.

5. What are some practical applications of studying water flow through a pipe with a 180° horizontal elbow?

Studying water flow through a pipe with a 180° horizontal elbow is important in various industries, such as plumbing, irrigation, and fluid transportation. It can also help in designing efficient piping systems and predicting potential issues that may arise from using elbows in pipe networks.

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