Hi Guys, my name is Grant.
I am trying to find out how to work out a pressure over area formula.
I work tire industry with raw rubber that has to be cured (vulcanized in other word), I know that raw rubber needs 10kg of pressure for every square cm and a heat of 140degrees Celsius to cure.
The problem I have is that if you have a bigger area the pressure has to be more this is what I have been thought, I currently use a curing system with air bags but no matter how big the hole that is filled up with raw rubber the pressure stays the same, sorely it should be more if the size is bigger
I am trying to build a press that uses hydraulic rams to apply the pressure needed, I would like to know how I know how much pressure I must put into the hydraulic ram to get 10kg/cm2 over area, I know ALMAX vulcanizing presses use a similar system.
I don’t know if you can help thanks
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Pressure
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4 posts
• Page 1 of 1
by Guest » Sat Dec 22, 2007 9:53 pm
I believe you are confusing force and pressure. The pressure (force per unit area) to vulcanize may be constant, but if you have a larger area, the hydraulic ram must exert more total force to create this pressure. P= F/A or F = P*A
By the way, the kilogram is a unit of mass. Using it as unit of force requires asuming "standard gravity." Force should really be expressed in newtons. At standard gravity, 10 kg has a weight of about 98 newtons, which could perhaps be rounded to 100 N. Your area of 1 sq cm is 10^(-4) square meters, so your pressure is about 100 N/10^(-4) m^2 or 1 MPa. (the pascal is the metric unit of pressure and is 1 newton per square meter). If my rounding is unwarranted, and the process is that "fussy," the figure is 0.98 MPa.
By the way, the kilogram is a unit of mass. Using it as unit of force requires asuming "standard gravity." Force should really be expressed in newtons. At standard gravity, 10 kg has a weight of about 98 newtons, which could perhaps be rounded to 100 N. Your area of 1 sq cm is 10^(-4) square meters, so your pressure is about 100 N/10^(-4) m^2 or 1 MPa. (the pascal is the metric unit of pressure and is 1 newton per square meter). If my rounding is unwarranted, and the process is that "fussy," the figure is 0.98 MPa.
- Guest
Pressure
by grant » Sun Dec 23, 2007 3:07 pm
Hi
Thanks for the information, so if I understand you correct the pressure that is needed over a bigger area stays the same.
If I have a conveyor belt 1500mm wide and you do a splice length of 1000mm, 20mm thick.
Let’s say you have 6 beams with 5 rams in each of them, how would you work out how much hydraulic pressure to put into each ram.
Obviously you would have t pieces in between each ram with a quick coupling system, so in this way you should have equal pressure though out all 30 rams.
P=F/A is your formula 10kg x 122cm/2 = 1220.
How do I know how much pressure to put in to the rams???
Thanks for the information, so if I understand you correct the pressure that is needed over a bigger area stays the same.
If I have a conveyor belt 1500mm wide and you do a splice length of 1000mm, 20mm thick.
Let’s say you have 6 beams with 5 rams in each of them, how would you work out how much hydraulic pressure to put into each ram.
Obviously you would have t pieces in between each ram with a quick coupling system, so in this way you should have equal pressure though out all 30 rams.
P=F/A is your formula 10kg x 122cm/2 = 1220.
How do I know how much pressure to put in to the rams???
- grant
by Guest » Sun Dec 23, 2007 6:12 pm
From area of joint and required pressure, determine total force, F = PA. If you use n rams, the force per ram is F/n
Now for the hydraulics, the hydraulic pressure drives a piston, which has an area. Ignoring losses in seals, that piston also obeys F = PA. I really need subscripts, but here we are talking the hydraulic poston area and hydraulic pressure. Using F/n from above, you need hydraulic pressure of P = F/(n*A). You need to know the effective area of the pistons in your rams.
Now for the hydraulics, the hydraulic pressure drives a piston, which has an area. Ignoring losses in seals, that piston also obeys F = PA. I really need subscripts, but here we are talking the hydraulic poston area and hydraulic pressure. Using F/n from above, you need hydraulic pressure of P = F/(n*A). You need to know the effective area of the pistons in your rams.
- Guest
More info
4 posts
• Page 1 of 1