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Determining the amount of material sprayed over time 2
- Thread starter tegefe
- Start date
- Thread starter
- #3
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.
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.
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1
- #4
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.
Good Luck,
Believe it or not : Though he couldn't prove, Einstein never agreed with Heisenberg on his Uncertainty Principle.
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.
Good Luck,
Believe it or not : Though he couldn't prove, Einstein never agreed with Heisenberg on his Uncertainty Principle.
- Thread starter
- #5
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.
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.
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1
- #6
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.
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.
- Thread starter
- #7
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?
When you say time are you speaking of speed of travel of the spray gun or time spent spraying at one particular spot?
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.
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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