Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum Feb 13th 2010, 07:20 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

Regards

Pete
Attached Files separator.pdf (64.0 KB, 2 views)   Feb 14th 2010, 04:31 AM #2 Member   Join Date: Feb 2010 Posts: 66 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.   Feb 15th 2010, 03:16 AM #3 Junior Member   Join Date: Feb 2010 Posts: 6 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   Feb 15th 2010, 03:41 AM   #4
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 Originally Posted by petdem 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??   Feb 15th 2010, 03:46 AM #5 Member   Join Date: Feb 2010 Posts: 66 Oops I forgot gravity P = rho g h not P = rho h like I said.   Feb 15th 2010, 04:17 AM #6 Junior Member   Join Date: Feb 2010 Posts: 6 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   Feb 15th 2010, 05:11 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 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   Feb 15th 2010, 05:36 AM #8 Junior Member   Join Date: Feb 2010 Posts: 6 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   Feb 15th 2010, 05:51 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 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  Tags equation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post guaiamum Thermodynamics and Fluid Mechanics 7 Aug 1st 2016 04:26 AM webconnector Advanced Electricity and Magnetism 6 Jul 9th 2015 01:18 PM ipinkbaby Kinematics and Dynamics 1 Jun 14th 2009 12:31 AM asi123 Quantum Physics 2 May 29th 2009 04:23 AM AAKhan07 Quantum Physics 1 May 28th 2008 02:27 PM 