Calculating Impact Pressure on a Panel from Vertical Water Flow

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To calculate the impact pressure on a panel from a vertical water flow, the discussion centers on a nozzle with a 12.5mm diameter and a flow rate of 100l/min, reaching a panel 3m away. The key formula for force is derived from the volumetric flow rate and the cross-sectional area, considering the density of water and the velocity of the jet. It is noted that a higher flow rate (1.5 liters per second) can maintain a stable jet, while a lower flow rate (0.1 liters per second) may not. The pressure on the panel is determined by the force divided by the impact area, and the distance does affect the pressure calculations. The conversation emphasizes the need for clarity in deriving formulas and understanding the dynamics of fluid flow.
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Hello can some one please help me I am trying to get my head round this.

I have a nozzle of 12.5mm diameter and there is a volume flow rate of 100l/min. water from the outlet reaches a surface that has an area of 640mm x 20mm panel at a distance of 3m. I want to calculate what would be the impact pressure on the surface of the panel... considering the water is from vertically downwards rather than horizontal since there would be a water trajectory.

once i have calculated the impact pressure i would need to simulate this. i have a nozzle of 12.5mm but the volume flow rate is 7l/min. what i want to find out is would the distance be a factor for pressure? or can we calculate pressure at the impact surface for this and adjust the distance such that they are same? we have to keep the volume flow rate constant in both cases.

can some one tell me how can this be done or a clue on how to go about ? cause i am very confused.
 
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Good afternoon shash and welcome to Physics Forums.

I think that a flow of nearly 1.5 litres per second vertically downwards for 3 metres would remain as a tight stream, (called a jet).
Can you confirm this or is this a thought experiment?

However I do not think that about 0.1 litres per second flow would be enough to sustain a stable stream as a jet, particularly from such a large orifice as you have described.

The force of a jet striking a plate at right angles to the flow has an easy formula.


Force = \frac{{density{{(VolumetricFlowRate)}^2}}}{{(CrossectionalArea)}}

I have not accounted for the increase in velocity imparted by accelerating 3m downwards under gravity.

The pressure on the plate will depend upon the diameter of the stream at impact and be the force divided by the impact area.
 
Studiot said:
Good afternoon shash and welcome to Physics Forums.

I think that a flow of nearly 1.5 litres per second vertically downwards for 3 metres would remain as a tight stream, (called a jet).
Can you confirm this or is this a thought experiment?

However I do not think that about 0.1 litres per second flow would be enough to sustain a stable stream as a jet, particularly from such a large orifice as you have described.

The force of a jet striking a plate at right angles to the flow has an easy formula.


Force = \frac{{density{{(VolumetricFlowRate)}^2}}}{{(CrossectionalArea)}}

I have not accounted for the increase in velocity imparted by accelerating 3m downwards under gravity.

The pressure on the plate will depend upon the diameter of the stream at impact and be the force divided by the impact area.
Hello Studiot.

To make things easier I am trying to do a test in accordance with BS EN60529 for IPx6 for the second numberal 6.but the problem is I have a pressure washer with 12.5mm diameter but the the volume flow rate is only 420 liter's/ hour. i.e 7 liter's / min. can you tell me how did u derive the formule for the impact force what you have mentioned above?

Thank you for taking time to reply.

Cheers.
 
volume of water striking plate per second = Xsection area x fluid velocity= av = V (volumetric flow rate)

mass of water striking plate per second = volume x density = ρav

and v = V/a

initial velocity normal to plate = v
final velocity normal to plate = 0
change of velocity = (v - 0) = v

Force = F = rate of change of momentum = mass per second x velocity = (ρ) x (av) x (v)

substitute for av and v

F = ρ (V) (V/a) = ρ V2 / a
 
Thank you. I am not sure how this will work but thanks
 
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