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pressure drop relations between different fluids

pressure drop relations between different fluids

pressure drop relations between different fluids

(OP)
Dear colleagues!

I have again an other hydraulic challenge:

Whe perform pressure drop measurements with water at ambient
temperature.

Now I have to predict how the pressure drop would change if, for example, the temperature rises or a different liquid will be used with other fluidic properties like density, viscosity, surface tension and vapour pressure.

It should be ensured that there is always liquid - no two-phase flow / cavitation.

Are there formulas/relationships that I could use?

Thanks in advance
Stephan

RE: pressure drop relations between different fluids

I don't see the effect of surface tension and vapour pressure on pressure drop...

"We don't believe things because they are true, things are true because we believe them."

RE: pressure drop relations between different fluids

(OP)
Dear Sheiko!

I think you are right!

For the vapour pressure, I agree with you under the constraint that cavitation is excluded.

The effect of surface tension could also be neglected for a one phase flow.

Could we then state that viscosity and density are the only remaining fluid properties having an influence on the pressure drop?

----

Dear FOURE,

thanks for the interesting link, but I can't find anything on that page concerning my actual problem.

Greetings,
Stephan

RE: pressure drop relations between different fluids

You are correct, the basic fluid relationships are defined by Bermouli's Equation with friction.  The frictional coefficients are usually defined as a function of the dimensionless Reynolds Number, which is a function of Fluid Velocity, Flow Path Dimension, Viscosity and Density.  Hence the only relevant fluid properties here are Viscosity and Density.

RE: pressure drop relations between different fluids

(OP)
Thanks FredRoss!

Ok, from Bernoulli and mass conservation formula I get:

dm/dt is proportional to sqrt(rho *dp)

from that follows:

(dm/dt)**2 / rho is proportional to dp

I want to have

dp_water =! dp_other_fluid

So I get from the formula above:

const * (dm/dt)**2_water/rho_water =
const * (dm/dt)**2_other_fluid / rho_other_fluid

that results in:

dm/dt_water = sqrt(rho_water/rho_original_fluid) * dm/dt_original fluid


So I get the relationship between the mass flow for the water test and the test with the other fluid which both yield the same dp.

But, how do I fomulate the influence of viscosity?

Thanks in advance,
Stephan

 

RE: pressure drop relations between different fluids

The basic equations you present above are approximations that do not consider the Reynolds Number influence.

The friction coefficients (friction factor for pipe, Kloss for fittings, and discharge coefficient for orifices) are all functions of the Reynolds Number.

For many applications, ususlly at high Reynolds Numbers, the friction coefficients level out to be constants, and your math may work OK.  For conditions where the  Reynolds Number has influence on the loss coefficients, the fluid properties come into play.

For example, with sharp edged orifices and large Beta ratios, 4 in pipe, ASME lists discharge coefficients ranging from 0.66 at Rey = 18,000 to 0.60 at Rey = 1,000,000

Another example, common friction factor in smooth piping, f = 0.04 at Rey = 4000, and f = 0.01 at Rey = 2,600,000

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