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Determining the amount of material sprayed over time
2

Determining the amount of material sprayed over time

Determining the amount of material sprayed over time

(OP)
Given that
I know the shape the material forms when sprayed
I know the rate of flow out of the gun
I know the viscosity of the material being sprayed

RE: Determining the amount of material sprayed over time

Please explain your question better.

Why not, Rate of flow x time = Total flow (Amount)?

RE: Determining the amount of material sprayed over time

(OP)
This is what I came up with as a solution please tell me if my premise was wrong.

I just need to know if I spray for 2 seconds, 3 seconds, etc on any spot how much material have I applied.

So, if my flow rate is 2ft^2 per second then for each second I have applied 2 ft^2 of material.  
Therefore at t=1 2ft^2 of material; at t=2 4ft^2 of material and so forth?

This is the problem the flow rate coming out of the spray nozzle will most likely be represented in cm^3/t, but the surface that I am spraying the mixture (dry glue and water) on is flat with no depth and would be represented as cm^2/t.

So, how can I get this in terms of
dA/dr; area changing with respect to the radius and then dA/dt; area changing with respect to time.
When it is obvious that I the mixture is coming out of a hose/nozzle with some volumetric flow rate; i.e. - cm^3/t?

I hope this isn't too confusing, but I'm begining to confuse myself.

Thanks in advance for you all's help.

RE: Determining the amount of material sprayed over time

tegefe!

Actually the coverage area mainly depends upon the geometry of the nozzle and the height of nozzle tip from the sprayed surface. If you fix these two variables, the sprayed area remains constant and thickness of coating increase over a period. If your nozzle is stationary, the simple equation is that as suggested by ken.

However, here is a good technical write up on spray technology.

http://www.spray.com/pdf/b498_An%20Engineers%20Guide%20...

Good Luck,

Believe it or not : Though he couldn't prove, Einstein never agreed with Heisenberg on his Uncertainty Principle.

RE: Determining the amount of material sprayed over time

(OP)
Ok, I'm going to give this a shot.
I can obviusly fix the geomerty of the nozzle.
I can also fix the height of the nozzle tip from the sprayed surface.

The only issue is going to be that the time spent on a given area of the surface will have to be pre-determined for coverage amount given the previous variables.

Time and mixture viscosity will be the only changing variables (I can control the viscosity).

Time will be represented in the form of speed of travel as the spray gun traverses down the part.

I will follow up soon.

RE: Determining the amount of material sprayed over time

This may be similar to an agricultural problem I've worked on.

The nozzles were rated based on a linear coverage at a certain distance (example, 0.5 meters at a nozzle distance of 1 meter).  The area covered is then:

Linear Coverage *  Nozzle Velocity * Time.

Dividing this by the volume of material sprayed would also give you the thickness of the coat.  This does not, however, take into account dynamic stopping/starting effects, so it may be oversimlified for your application.

We used this approach to work out field coverage (in mL of chemical per hectare) as a function of flow rate and tractor velocity.

RE: Determining the amount of material sprayed over time

(OP)
This is close to what I am dealing w/ start and stop dynamics are not at issue for I will start before and end spraying after the target surface.

When you say time are you speaking of speed of travel of the spray gun or time spent spraying at one particular spot?

RE: Determining the amount of material sprayed over time

Sorry for the delay, I've been sick in bed for a week . . .

Time refers to the total time spent spraying.  As an example:

A particular nozzle sprays a footprint 0.5meters in width.  The tractor (and thus the nozzle)  moves at 3 meters/second.  Thus, over a 5 second period, the area sprayed is 0.5m X 3m/s X 5s = 7.5m^2.

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