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# Need help on a exam question!! Fluid Mechanics

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1. The bent cone nozzle carries a water flow of 25kg/s. At point 2 the flow discharges into the atmosphere, and at point 1 the nozzle is bolted to a rigid pipe. Assume the flow to be frictionless. The pipe diameter at point 1 is 100mm and 70mm at point 2. Calculate the magnitude and the direction of total force exerted by the water on the nozzle, if the gauge pressure at point 1 is p1 = 65kPa.

The angle is 60 degrees.
2. (Original post by Lawalk97)
The bent cone nozzle carries a water flow of 25kg/s. At point 2 the flow discharges into the atmosphere, and at point 1 the nozzle is bolted to a rigid pipe. Assume the flow to be frictionless. The pipe diameter at point 1 is 100mm and 70mm at point 2. Calculate the magnitude and the direction of total force exerted by the water on the nozzle, if the gauge pressure at point 1 is p1 = 65kPa.

The angle is 60 degrees.
I didn't quite get the question. My assumptions:
1) The pipe and nozzle junction is horizontal and nozzle outlet is inclined at 60 degrees below horizontal.
2) Density of water is constant i.e. 1000 kg/m^3, throughout the operation.
3) As change in height is not indicated, change is potential energy is assumed to be negligible.
4) Isothermal condition is assumed.

If that's the case then, velocity of water at inlet is given by (25kg/s / (density of water X area of cross section at inlet) = 3.183098862 m/s.
By using continuity equation (A1 v1 = A2 v2) we get velocity of water at outlet = 6.496120126 m/s.
By using Bernoulli's equation we have pressure at outlet = 48966.27 Pa

Pressure Force on the inlet = 65,000 X area of inlet = 510.51 N horizontally towards right
Pressure Force on the outlet = 48966.27 X area of outlet = 188.44 N at angle 60 degrees above horizontal towards left.

For the force, Resolve these forces horizontally and vertically. which should give you total force of 447.13 N at an angle of 40.98 degrees above horizontal.

Hope this helps.
If it does not then change the assumptions as required and perform calculations in similar manner.
3. (Original post by tangotangopapa2)
I didn't quite get the question. My assumptions:
1) The pipe and nozzle junction is horizontal and nozzle outlet is inclined at 60 degrees below horizontal.
2) Density of water is constant i.e. 1000 kg/m^3, throughout the operation.
3) As change in height is not indicated, change is potential energy is assumed to be negligible.
4) Isothermal condition is assumed.

If that's the case then, velocity of water at inlet is given by (25kg/s / (density of water X area of cross section at inlet) = 3.183098862 m/s.
By using continuity equation (A1 v1 = A2 v2) we get velocity of water at outlet = 6.496120126 m/s.
By using Bernoulli's equation we have pressure at outlet = 48966.27 Pa

Pressure Force on the inlet = 65,000 X area of inlet = 510.51 N horizontally towards right
Pressure Force on the outlet = 48966.27 X area of outlet = 188.44 N at angle 60 degrees above horizontal towards left.

For the force, Resolve these forces horizontally and vertically. which should give you total force of 447.13 N at an angle of 40.98 degrees above horizontal.

Hope this helps.
If it does not then change the assumptions as required and perform calculations in similar manner.
Thanks for the reply!
The question is written exactly how it is in the exam paper. I found it hard to understand as well. From what you have written it has helped greatly. Thanks
4. (Original post by Lawalk97)
Thanks for the reply!
The question is written exactly how it is in the exam paper. I found it hard to understand as well. From what you have written it has helped greatly. Thanks
No problem

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