Question on compressible flow through a piping system.

[dbecker]

Hello everyone,


I have a Crane's handbook in front of me. I am trying to calculate the flow rate of a system given upstream/downstream pressures.

From what I understand Crane's equation 3-7 relates the pressures, friction, specific volume and pipe diameter to flow rate.

However, for the f(L/D) term thats in the denominator, can I use the SUMMATION of all K values? Since f(L/D) is K, can I use equation 3-7 for bends? With their corresponding K values?

Please help,

Thank you

 

[Latexman]

Please describe the fluid and application (P'₁, P'₂, L, D, [Σ]K, T₁, T₂). It would help a lot.

Good luck,
Latexman

 

[katmar]

Adding the K values of the fittings to the f(L/D) term is certainly the way I would do it.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[dbecker]

Hello, thank you for replying.

The pressures and pipe geometry is unknown. I am building a calculator that is system specific but one can input different values for pressures and temperatures, pipe diameters etc... Right now, I want to understand if the equation 3-7 is still valid if I use K values from both elbows and straight pipe (sum them), since in the crane book it is referred to as "isothermal flow of gas in pipe lines" i.e i understand this as being valid for straight pipes only? Or am I wrong to use K values from different sources of losses?

Thank you again.

 

[katmar]

To see that what you are doing is correct, compare Crane eq 3-5 with Crane eq 3-7. These are both integrations of Bernoulli's equation (Crane eq 3-1) with the difference that 3-5 assumes the velocity is constant. In eq 3-7 the term that starts with 2log_e is there to take into account the acceleration of the gas due to the expansion as the pressure decreases. In equation 3-7a this term disappears under the assumption of negligible acceleration (see also the discussion on eq 1-6 and 1-7).

This means that in eq 3-5 and 3-7 the f(L/D) term is there to calculate the friction losses and it is reasonable to add the K's for the fittings to this term in both 3-5 and 3-7.

Note also that although the gas expands and accelerates down the pipeline the Reynolds number is constant (see mass based version of eq 3-3 and therefore the friction factor will be constant for the whole line, and K values will be unaffected by where they are in the pipeline.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[dbecker]

Thank you for the clarification.

I will sum the losses and use them with equation 3-7.

One more thing if I may, if I wish to calculate the flow rate through the system, I would use the smallest d in the system for equation 3-7? The result should be the system flow rate (assuming no compressibility at this point).

Thanks again!

 

[katmar]

If you base your calculation on the smallest d your answer will be conservative. You can get a good estimate by adjusting your pipe lengths and K values to a common diameter.

For the sake of example let us assume you have some 4" pipe and some 2" pipe. We want to base the calculation on 4" pipe so we have to convert the 2" pipe and fittings to the equivalent in 4" pipe and fittings.

In turbulent flow the pressure drop is roughly proportional to the fifth power of the diameter. For a given length the pressure drop in the 2" section will be ((4/2)^5) = 32 times that in the 4" section. So if you have 100 foot of 2" piping you could consider it as 100 x 32 = 3200 foot of 4" piping in you calculation.

The K values are applied to the V² term and V is proportional to D² so we need to adjust the K values by D⁴. In the Crane method a 2" standard elbow has a K value of 0.266. To convert this to an equivalent 4" fitting we multiply by ((4/2)^4) = 16, i.e. the K value is now 0.266 x 16 = 4.26. In order to base your calculation on a standard diameter of 4" pipe you would add a K of 4.26 for every 2" standard elbow.

Now you can apply eq 3-7, but you will still have to do a few iterations because you will have to guess an initial f value and correct it for the actual flow with each iteration.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[dbecker]

Thank you for the support!

 

[dbecker]

Hello again,

Something doesn't seem right with what I am doing.

I will lay it out step by step.

First, I have a bunch of K's in a system that I am analyzing. I am adding the K's to get some system K value.

Now, for me to calculate pressure loss for each component I need the velocity of fluid through the component (this is required by equation 3-5, deltaP equation).

I can get velocity if I have flow rate and pipe area (neglecting actual geometric flow area for a valve as that is too difficult to calculate right now).

mass flow = rho x V x A as we know

At this point I can establish V but i need A (have that) and Rho (have that too) and mass flow (eq 3-7).

I will use equation 3-7 to calculate mass flow for EACH COMPONENT to get the velocity value FOR EACH COMPONENT. Can someonae please direct me on how to assume the P1 and P2 values for the individual components as these are unknowns and will not equal the SYSTEM P1 and P2. This is where I am failing.

Thank you and sorry for the long questions!

 

[rmw]

I have a Crane and I use it often. I have had it out recently but insted I prefer to use my Camerons Hydraulic Data which gives values for equivalent length for fittings and common valve types. I recently did what I think you are trying to do in Excel by converting all the fittings and valves to equivalent length in their respective rows and then totalizing the result column arriving at an overall equivalent length for the whole string.

That way when I change the flow (using the value of dp per 100 ft of pipe) or viscosity it recalculates the whole pipe string for me. If I changed the fitting count or valve count or valve type that would be accounted for in the overall equivalent length as well. And, yes, I did have to have a component for the static head as well (that is a constant in my problem).

rmw

 

[katmar]

The mass flow through each component is the same, so you do not want to calculate it individually for each component. The strength of the Crane technique is that by combining the pipe friction and fitting resistances into one term you can treat the entire pipeline as one lumped system. In this case you are only concerned with the overall pressure drop and not that across each component.

This is the same procedure described by rmw - using equivalent lengths for fittings is identical to the Crane K procedure. For example, on page A-29 of Crane the K value for a 1.5 r/d pipe bend is given as 14[ƒ]_T - the constant (14) is the equivalent length used by rmw.

You need to replace the [ƒ]L/D term in either eq 3-5 or 3-7 with ([ƒ]L/D + [Σ]K). L is the total length of straight pipe (including the adjusted lengths for the sections with a diameter different from the base size). D is the base pipe diameter size (see my earlier post) and [Σ]K is the sum of all the pipe fitting resistances (also adjusted to the base pipe diameter where necessary).

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[dbecker]

Thank you very much for the support. Please understand I am trying very hard to build this model as autonomously as possible but I still have an inquiry.

Can I choose any pipe diameter to make everything 'equivalent too' ?? Or is there convention for choosing this, i.e some 4" and 2" in a pipeline, I will make everything equivalent to 3" given that there is the same amount of both 4" and 2" (averaged or does it matter?).

I want to use equation 3-7 but I am not clear on adjusting.

For example:

I have a system with 100 feet of 4" pipe, and 100 feet of 2" pipe and a reducer in between.

I want to calculate the flow rate and pressure drop for each component in this system, but the system has two different diameter pipes and a loss component (reducer).

My approach is as follows:

First, I will calculate the value K for all components and this is dependant on the pipeline component geometry (diameter and such).

K for 4" pipe = f x (L/D) = 0.02 x (1200in / 4in) = 6
K for 2" pipe = 0.02 x (1200 / 2) = 12
K for reducer = 0.5(1-B^2) = 0.5(1- 1/4) = 0.375

Sum of K = 18.375

Now, I will produce the equivalent lengths for the above system for 4" pipe (arbitrarily chosen at this point).

K for 4" pipe remains the same as does diameter and Length.
K for 2" pipe = 12 = 0.02 x (L/4) Solve for L = 2400 in = 200 feet for 4" pipe.
K reducer = 0.375 = 0.02 x (L/4) therefore L = 75 in = 6.25 feet.

OK, I have this information. Now do I just substitute the sum of K's and diameter into equation 3-7 to calculate Flow Rate?

Please tell me im doing this right.

Thank you very much!!!!!!!

 

[katmar]

It is probably best to choose as your base diameter the pipe size of the section of pipe that is responsible for the greatest part of the pressure drop. But you don't know which section that is (unless it is very obvious) until you have done the calculation. Be prepared for a bit of trial and error.

For this type of "single equivalent diameter" calculation it is easier to treat the non-base size pipe as an equivalent length of base sized pipe than as a K value. The pressure drop for turbulent flow in straight pipe varies with the 5th power of the diameter. So you just multiply the length of 2" pipe by (4/2)^5 and add its equivalent length to that of the 4" pipe. In this case you will assume that there is 3200 ft of 4" pipe to represent the 100 ft of 2" pipe.

Remember that a K value is associated with a particular velocity. The K value you have calculated for your reducer (which you have treated as a sudden contraction) is based on the velocity in the smaller diameter. This is what Crane calls K₁. To convert the K₁ to a K₂ i.e. based on the larger diameter, you must divide by (d/D)^4. Compare Crane equations 2-10.1 and 2-11 and my previous post. The K value for the reducer, based on the velocity in the 4" pipe, becomes 0.375 x 16 = 6.0. Or if you want to convert this to an equivalent length of 4" pipe by using the [ƒ] x L/D = K relationship then L = 6 x 4 / 0.02 = 1200" = 100' (This is the method described above by rmw).

This makes the [ƒ] x L/D + [Σ]K terms either
0.02 x 3300 / (4/12) + 6 (leaving the reducer as a K value) or
0.02 x 3400 / (4/12) (treating the reducer as an equivalent length)

In both cases the value is 204

Substituting this value into Eq 3-7 will give you the flowrate.

If you want to calculate the pressure drop across each component you would have to recast Eq 3-7 to calculate P₂ (too difficult for me) or use the Goal Seek function in your spreadsheet to calculate the value of P₂ that would give the known flowrate.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com