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Old Feb 13th 2010, 08:20 PM   #1
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equation

Hello
I have a question please i am not sure about. I work with this machine as a operator but would like to learn more.

1. I have attached a picture of a juice separator tank. The juice enters the separator where the liquid goes to the bottom of the tank and the vapour rises to the top of the tank where it leaves to a condensor.The pump at the bottom of the separator tank pumps out the liquid so as to keep the tank level at 52% between the 2 sensors ...The 52% tank volume is calculated between the 2 sensors and not actually 52% of the actual tank volume...Liquid exists in about only 10% of the tank .The distance between the 2 sensors is 2.5metres.The pump with the bottom sensor is about 1.1metres beneath the bottom of the tank

What formula would i use to calculate the pressure...P2 is the top pressure sensor and P1 is the bottom sensor. I also think that static pressure is part of the equation. Is there a name for this equation of P2/P1 and Static?
How does it calculate 52% between these 2 sensor points? The pump at the bottom of the tank speeds up and slows down so as to keep the volume between the 2 sensors at 52% in the tank
Any advice appreciated

Regards

Pete
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Old Feb 14th 2010, 05:31 AM   #2
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If it can be considered static then the bottom pressure would be
P1 = liquid density * height of liquid surface above the sensor + P2

I assume that the P2 sensor is there because the pressure of the vapour may may be different from atmospheric?

If it's flowing fast enough to not be static, then it depends on the geometry of the pipe, flow reigime, etc.
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Old Feb 15th 2010, 04:16 AM   #3
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Empiler1...thank you

The equation ended up being P1 (liquid + Vapour) - P2 (vapour) = P (Liquid)

The P1 ended up as Liquid Height plus Vapour Pressure on the Liquid Surface and P2 as Vapour Pressure.

If the density of the liquid is 1.13 and the height of the liquid is 1.3metres
what would the pressure be?
What equation would you use to calculate?

Is weight of the liquid then used to somehow control the 52% volume between P1 and P2?

Thanks again for your help....Pete
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Old Feb 15th 2010, 04:41 AM   #4
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Originally Posted by petdem View Post
Empiler1...thank you
If the density of the liquid is 1.13 and the height of the liquid is 1.3metres
what would the pressure be?

Is weight of the liquid then used to somehow control the 52% volume between P1 and P2?
P1-P2 would be 1.13kg/m^3 * 1.3m

But I think you have some other units for density, so make sure they're consistent.

Yea that value P1-P2 would be proportional to liquid height above the P1 sensor. You could obtain the height by going backwards:

height = (P1-P2) / density

Not sure why it's expressed at 52% instead of a height. Maybe that's just a more meaningful parameter for this particular process??
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Old Feb 15th 2010, 04:46 AM   #5
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Oops I forgot gravity

P = rho g h

not

P = rho h like I said.
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Old Feb 15th 2010, 05:17 AM   #6
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empiler1...again thank you

The density is 1113kg/m^3

As an example for the P1 : Is this correct -- 1113kg/m^3 * 1.3m * 9.81
What would be the final answer to the equation above please?
Do i need to do some conversions of units?

Thanks.....Pete
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Old Feb 15th 2010, 06:11 AM   #7
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It would be:

P1 - P2 = 1113kg/m^3 * 1.3m * 9.81N/kg

In general you can check units by simply including them in the calculation. Rearranging you get:

P1 - P2 = 1113 * 1.3 * 9.81 * kg * m * N / (m^3 * kg)
= 14000 N/m^2
= 14000 Pa

Are you working this out to get the upper hand over some know-it-all coworkers?



Originally Posted by petdem View Post
empiler1...again thank you

The density is 1113kg/m^3

As an example for the P1 : Is this correct -- 1113kg/m^3 * 1.3m * 9.81
What would be the final answer to the equation above please?
Do i need to do some conversions of units?

Thanks.....Pete
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Old Feb 15th 2010, 06:36 AM   #8
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Hi empiler1

yeah sort of trying to the upper hand...are you a engineer...you have all the answers...thanks for your help

Finally did you just cancel out the kg from each side and m^3 - m = m2 which left N/m2 .... is 1Pa = to 1N/m2

Are there some good tutorials or books on this stuff

All the best....Pete
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Old Feb 15th 2010, 06:51 AM   #9
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Yea engineer, but I don't work in production anymore.

Most of what I said about pressure is in
Pressure - Wikipedia, the free encyclopedia

I've seen plenty of turorials on the internet about this kind of stuff, don't remember anything in particular.

You can cancel units just like you can numbers. So yea
m / m^3 = 1/m^2
kg / kg = 1

This even works with imperial units, but you generally have to use conversion factors to get things into the same units before they can be cancelled.



Originally Posted by petdem View Post
Hi empiler1

yeah sort of trying to the upper hand...are you a engineer...you have all the answers...thanks for your help

Finally did you just cancel out the kg from each side and m^3 - m = m2 which left N/m2 .... is 1Pa = to 1N/m2

Are there some good tutorials or books on this stuff

All the best....Pete
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