Calculating Pressure Drop Across Orifice 3

[KimWonGun]

I am using an on-line calculator to estimate the pressure drop of a light motor oil across an orifice given two different operating scenarios:

The first assumes 200F oil temperature with the corresponding oil viscosity, Reynolds number, and flow coefficient.

The second assumes -40F oil temperature with the corresponding oil viscosity, Reynolds number, and flow coefficient.

Question 1: If I wish to observe the impact of temperature on the pressure drop across the orifice, is it correct to hold the flow rate constant?

Question 2: And if the flow rate is held constant, does it make sense that the pressure drop at -40F across the orifice would be much less than at 200F due to the higher fluid viscosity?

Question 3: If my objective is to determine what the transducer type (differential pressure or point) and range should be, is there a more effective way to estimate those values?

 

[ione]

For the first two questions


1) Q = Cd*A*[SQRT(2G*deltaP)]/SQRT(1-(D1/D2)^4)

Geometry must be consistent, that is keep the same A, D1/D2. The acceleration of gravity is a constant as well.
You do have to keep flow rate constant to evaluate the effects of temperature throughout the coefficient of discharge Cd solving the equation above for deltaP.


2) You can easily check how temperature modifies the coefficient of discharge Cd, and check the effects on pressure drop for a given flow rate.

Coefficient of discharge can be calculated using equation reported on ISO 5167 (check here

http://www.brighthub.com/engineering/civil/articles/63291.aspx)

 

[ione]

KimWonGun,

I have seen this is not your first thread on this topic, and so you could find the attached paper useful.

 

[KimWonGun]

So if the discharge coefficient has a factor of [+ 0.0029B^2.5(10^6/Re)^0.25] (the only one containing the Reynolds number) and I know that at -40F the Reynolds number is very low while at 200F it is quite high, then the discharge coefficient is inversely related to temperature. Therefore deltaP ((rho/2)(Q/CA)^2 should decrease with a decrease in operating temperature, correct?

Question: Even if that is true mathematically, I still struggle to comprehend it physically. Can anyone help?

 

[ione]

You are right: increasing Re implies reducing Cd.

You can see it this way.
The matter is flow regime. The higher the Reynolds number (above 4000) the more turbulent is the flow regime. Turbulence implies flow develops in a chaotic way with random trajectories. And when a fluid moving such a way encounters an obstacle (the orifice plate in your specific case) the dissipated energy (since pressure loss is dissipated energy) increases.

 

[sheiko]

ione,
For fittings, pressure losses are more important in laminar flow than in turbulent. You can check it in Perry Handbook - in the section dealing with resistance coefficient (K) in laminar regime, or on a Moody diagram.

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

 

[ione]

sheiko,
Thanks for your input. I’m aware of this. Pressure losses dependence on Re becomes stronger with low Re (Hooper’s 2-k method and Darby’s 3-k method).

 

[ione]

sheiko,
a work carried out by Miller (1983) has reported the following relationship between the coefficient of discharge Cd and Re (orifice geometry and specifically D1/D2 ratio plays a role).

Cd = C?*b/Re^^n

Where C? is the coefficient of discharge for Re -> ?. Coefficient b and exponent n are reported in “Chemical Engineering Fluidmechanics” Second Edition written by Ron Darby (chapter 10). Darby has also included a couple of interesting tables on this topic, with the range of applicability and the accuracy of the formula.

The correlation above, which is valid in the turbulent region (Re>4000) and applies to orifice meters, as well as to venturi and nozzles , shows how Cd decreases as Re increases.

 

[sheiko]

Thanks ione,

Do you know what happens if the Reynolds number is less than 4000?

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

 

[ione]

Hello sheiko,

The same source quoted in my previous post also reports a qualitative graph (Coefficient of discharge vs Re for given D2/D1) which includes the laminar regime. That graph shows that the coefficient of discharge increases with Re in that region.

The graph I’m talking about is kind of that reported in the link below

http://www.flickr.com/photos/8507083@N04/1752206443/

 

[KimWonGun]

Reacting to the discussion between ione and sheiko in this thread, does this at all alter either my earlier assumption about the inverse relationship between temperature and delta pressure across an orifice plate as well as the application of the algorithm for discharge coefficient ione suggested?

Expressed another way, if at 200F the Reynolds number is over 15,000 (turbulent) and if at -40F the Reynolds number is under 100 (laminar), does it no longer hold that the pressure drop at 200F is greater than the pressure drop at -40F because of the change in flow regime?

 

[ione]

KimWonGun,

First of all I (and supposedly sheiko too) didn’t mean to hijack your thread, anyway apologies for this.

Well if the change in viscosity is that great to affect the flow regime, turning it from turbulent to laminar, the response of the orifice plate should qualitatively be as per graph reported in my previous post. So for turbulent regime Cd decreases with Re, whilst for laminar flow Cd behaves inversely.

 

[sheiko]

Thanks ione for the clarification.

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

 

[KimWonGun]

ione:

No apology is warranted; on the contrary I am grateful that you and sheiko "hijacked" the thread for the important content provided.

So if I calculate a Reynolds number of less than 4, would it be inappropriate to extrapolate a discharge coefficient value from the graph you supplied, assuming the floor to be 0.2?