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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