Calculating Impact Pressure on a Panel from Vertical Water Flow

In summary, the conversation discusses calculating the impact pressure of a water jet on a surface at a certain distance, taking into account the water trajectory and volume flow rate. The formula for calculating the force of a jet striking a surface at right angles is mentioned, as well as the factors that affect the pressure on the impact surface. The conversation also mentions a test being done for IPx6 and the problem of a low volume flow rate. The formula for calculating the force of a jet is derived using the volumetric flow rate and density of the fluid.
  • #1
shash912
3
0
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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  • #2
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.


[tex]Force = \frac{{density{{(VolumetricFlowRate)}^2}}}{{(CrossectionalArea)}}[/tex]

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.
 
  • #3
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.


[tex]Force = \frac{{density{{(VolumetricFlowRate)}^2}}}{{(CrossectionalArea)}}[/tex]

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.
 
  • #4
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
 
  • #5
Thank you. I am not sure how this will work but thanks
 

1. How do you calculate surface pressure?

To calculate surface pressure, you need to divide the force exerted on a surface by the area of the surface. This will give you the pressure in units of force per unit area, such as pounds per square inch (psi) or newtons per square meter (N/m2).

2. What is the formula for calculating surface pressure?

The formula for calculating surface pressure is P = F/A, where P is the surface pressure, F is the force exerted on the surface, and A is the area of the surface. This formula assumes that the force is perpendicular to the surface.

3. What are the units of measurement for surface pressure?

The units of measurement for surface pressure can vary, but some common ones include pounds per square inch (psi), newtons per square meter (N/m2), and pascals (Pa). It is important to use consistent units when calculating and comparing surface pressure values.

4. How does surface pressure affect objects?

Surface pressure can affect objects in different ways depending on the material and structure of the object. For example, high surface pressure can cause deformation or damage to a soft material, while a hard material may be able to withstand higher surface pressures without any significant effects. Additionally, changes in surface pressure can affect the stability and movement of objects in fluids.

5. What factors can influence surface pressure?

There are several factors that can influence surface pressure, including the force applied to the surface, the area of the surface, and the properties of the material the surface is made of. Other factors may include the temperature, altitude, and atmospheric conditions surrounding the surface. It is important to consider all of these factors when calculating or measuring surface pressure.

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