Calculate pressure of pumped liquid

In summary, the pump is rated to 250L/h, not L/s. A 10.8 mm pipe wouldn't be able to handle that flow rate.
  • #1
madmike159
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I have a enclosed system for cooling bearing housings with oil. I need to calculate the rough pressure that will be produced when the oil pump is switched on. The oil will be pumped from a small tank, through the system and back into the tank.

Pipe diameter = 10.8mm
Pipe length = aprox 2m
Flow rate = 250L/s
Viscosity (this is where I get confused about Kinematic vs Dynamic)

I am stuck on how to even get started on this. Is there a simple equation/rule of thumb that would allow me to get a rough value?
 
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  • #2
You'll need to find out more information about your pump. At the minimum, you'll need to know the output head the pump is capable of producing.
 
  • #3
What do you mean by output head?
 
  • #4
You are trying to calculate the pressure drop for flow of a fluid through a pipe. You can look up the equations for this in Mechanical engineers Handbook, Chemical Engineers Handbook, or on the internet.
 
  • #5
madmike159 said:
What do you mean by output head?

A pump works by changing the pressure of the fluid it is pumping. The output head refers to the amount of pressure the pump is capable of producing.

See: http://en.wikipedia.org/wiki/Pump

under the section 'Laraib Pumping Power' at the bottom of the article.
 
  • #7
Baluncore said:
The key to using this reference is to get the friction factor f, which is a function of the Reynolds number Re. You need to first calculate the Reynolds number, and then check to determine whether it is above or below the critical value for transition to turbulent flow (2100). In your case, it is almost certainly above the critical value. You then use the relationship between f and Re that is applicable to turbulent flow. This relationship can be found in the references I cited in my earlier post.

Chet
 
  • #8
Thanks guys. I had already found the Hagen-Poiseuille Equation, but I used the flow rate incorrectly (m^3/h rather than m^3/s). I now have a reasonable answer and know a limit of the size of pipe I should use.
 
  • #9
madmike159 said:
Thanks guys. I had already found the Hagen-Poiseuille Equation, but I used the flow rate incorrectly (m^3/h rather than m^3/s). I now have a reasonable answer and know a limit of the size of pipe I should use.
Don't forget that Hagen-Poiseuille applies only to laminar flow (Re<2100). If the reynolds number is higher than this, you need to use the turbulent flow correlation for the friction factor. Otherwise, you will be underestimating the pressure drop. So, check the Re.

Chet
 
  • #10
Fundamentally, a cooling system should be operating at the lowest pressure possible that achives the required flow. If the flow is limited by pipe size then power will be wasted moving the oil through the tube. That will heat the oil which makes cooling the gearbox more difficult.

I think the OP may have been based on assumed data. If 250 L/s is required to achieve the required cooling then the 10.8mm pipe diameter is far too small to be in laminar flow.

There are nomographs for deciding the pipe diameter needed in hydraulic applications.
See for example; http://www.ryco.com.au/index.php?id=196
This shows you need something closer to a 50mm diameter pipe for that flow rate.

Your 10.8 mm tube would support only 25 L/s.
 
Last edited:
  • #11
Sorry madmike159,
I just found your intermediate post with the corrected flow reduction by a factor of 3600 = 70 mL/s.
 
  • #12
Chestermiller said:
Don't forget that Hagen-Poiseuille applies only to laminar flow (Re<2100). If the reynolds number is higher than this, you need to use the turbulent flow correlation for the friction factor. Otherwise, you will be underestimating the pressure drop. So, check the Re.

Chet

Thanks, I didn't think to check the Reynolds number. As Baluncore said, my flow rate is far to high which explains some higher than expected oil temperatures. I won't be blindly choosing the first pump I find next time.
 
  • #13
Baluncore said:
Your 10.8 mm tube would support only 25 L/s.

Just realized my OP was incorrect. The pump is rated to 250L/h not L/s...
 

Related to Calculate pressure of pumped liquid

What is the formula for calculating the pressure of pumped liquid?

The formula for calculating the pressure of pumped liquid is P = ρgh, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height or depth of the liquid.

What units are used for pressure and how are they related?

The most commonly used units for pressure are Pascals (Pa) and pounds per square inch (psi). These two units are related by the conversion factor of 1 psi = 6894.76 Pa.

How does the density of the liquid affect the pressure?

The density of the liquid directly affects the pressure, as seen in the formula P = ρgh. A higher density liquid will result in a higher pressure, while a lower density liquid will result in a lower pressure.

What is the significance of acceleration due to gravity in calculating pressure?

The acceleration due to gravity (g) is a crucial factor in calculating the pressure of pumped liquid. This is because the pressure is dependent on the weight of the liquid, which is determined by the force of gravity pulling down on it.

Can the pressure of pumped liquid be negative?

Yes, the pressure of pumped liquid can be negative. This usually occurs when the liquid is at a lower elevation or depth than the reference point. Negative pressure can also occur when there is a decrease in the density of the liquid, such as when it changes from a liquid to a gas.

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