Water Flow Through a Pipe w/180° Horizontal Elbow

[teegfit]

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

 

[mfb]

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).

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.

 

[teegfit]

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?

 

[mfb]

Also are my calculation correct apart from the angle?

And the density.

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.

 

[teegfit]

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

 

[Chestermiller]

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