A Process Design Engineer’s Perspective on Using Equivalent Lengths of Valves and Fittings in Pipeline Pressure Drop Calculations

Reader / Author Question and Answer Added at End of Article

     One of the most basic calculations performed by any process engineer, whether in design or in the plant, is line sizing and pipeline pressure loss.  Typically known are the flow rate, temperature and corresponding viscosity and specific gravity of the fluid that will flow through the pipe.  These properties are entered into a computer program or spreadsheet along with some pipe physical data (pipe schedule and roughness factor) and out pops a series of line sizes with associated Reynolds Number, velocity, friction factor and pressure drop per linear dimension.  The pipe size is then selected based on a compromise between the velocity and the pressure drop.  With the line now sized and the pressure drop per linear dimension determined, the pressure loss from the inlet to the outlet of the pipe can be calculated.

Calculating Pressure Drop

     The most commonly used equation for determining pressure drop in a straight pipe is the Darcy Weisbach equation.  One common form of the equation which gives pressure drop in terms of feet of head is given below:

     Another common form of the Darcy Weisbach equation that is most often used by engineers because it gives pressure drop in units of pounds per square inch (psi) is:

     To obtain pressure drop in units of psi/100 ft, the value of 100 replaces L in Equation 2.

     The total pressure drop in the pipe is typically calculated using these five steps.  (1) Determine the total length of all horizontal and vertical straight pipe runs.  (2) Determine the number of valves and fittings in the pipe.  For example, there may be two gate valves, a 90^o elbow and a flow thru tee.  (3) Determine the means of incorporating the valves and fittings into the Darcy equation.  To accomplish this, most engineers use a table of equivalent lengths.  This table lists the valve and fitting and an associated length of straight pipe of the same diameter, which will incur the same pressure loss as that valve or fitting.  For example, if a 2” 90^o elbow were to produce a pressure drop of 1 psi, the equivalent length would be a length of 2” straight pipe that would also give a pressure drop of 1 psi.  The engineer then multiplies the quantity of each type of valve and fitting by its respective equivalent length and adds them together.  (4) The total equivalent length is usually added to the total straight pipe length obtained in step one to give a total pipe equivalent length.  (5) This total pipe equivalent length is then substituted for L in Equation 2 to obtain the pressure drop in the pipe.

     See any problems with this method?

Relationship Between K, Equivalent Length and Friction Factor

     The following discussion is based on concepts found in reference 1, the CRANE Technical Paper No. 410. It is the author’s opinion that this manual is the closest thing the industry has to a standard on performing various piping calculations.  If the reader currently does not own this manual, it is highly recommended that it be obtained.

     As in straight pipe, velocity increases through valves and fittings at the expense of head loss.  This can be expressed by another form of the Darcy equation similar to Equation 1:

     When comparing Equations 1 and 3, it becomes apparent that:

     K is called the resistance coefficient and is defined as the number of velocity heads lost due to the valve or fitting.  It is a measure of the following pressure losses in a valve or fitting:

• Pipe friction in the inlet and outlet straight portions of the valve or fitting
• Changes in direction of flow path
• Obstructions in the flow path
• Sudden or gradual changes in the cross-section and shape of the flow path

     Pipe friction in the inlet and outlet straight portions of the valve or fitting is very small when compared to the other three.  Since friction factor and Reynolds Number are mainly related to pipe friction, K can be considered to be independent of both friction factor and Reynolds Number. Therefore, K is treated as a constant for any given valve or fitting under all flow conditions, including laminar flow.  Indeed, experiments showed¹ that for a given valve or fitting type, the tendency is for K to vary only with valve or fitting size.  Note that this is also true for the friction factor in straight clean commercial steel pipe as long as flow conditions are in the fully developed turbulent zone.  It was also found that the ratio L/D tends towards a constant for all sizes of a given valve or fitting type at the same flow conditions.  The ratio L/D is defined as the equivalent length of the valve or fitting in pipe diameters and L is the equivalent length itself.

     In Equation 4, ¦ therefore varies only with valve and fitting size and is independent of Reynolds Number.  This only occurs if the fluid flow is in the zone of complete turbulence (see the Moody Chart in reference 1 or in any textbook on fluid flow).  Consequently, ¦ in Equation 4 is not the same ¦ as in the Darcy equation for straight pipe, which is a function of Reynolds Number.  For valves and fittings, ¦ is the friction factor in the zone of complete turbulence and is designated ¦_t, and the equivalent length of the valve or fitting is designated L_eq.  Equation 4 should now read (with D being the diameter of the valve or fitting):

     The equivalent length, L_eq, is related to ¦_t, not ¦, the friction factor of the flowing fluid in the pipe.  Going back to step four in our five step procedure for calculating the total pressure drop in the pipe, adding the equivalent length to the straight pipe length for use in Equation 1 is fundamentally wrong.

Calculating Pressure Drop, The Correct Way

     So how should we use equivalent lengths to get the pressure drop contribution of the valve or fitting? A form of Equation 1 can be used if we substitute ¦_t for ¦ and L_eq for L (with d being the diameter of the valve or fitting):

     The pressure drop for the valves and fittings is then added to the pressure drop for the straight pipe to give the total pipe pressure drop. 

     Another approach would be to use the K values of the individual valves and fittings.  The quantity of each type of valve and fitting is multiplied by its respective K value and added together to obtain a total K.  This total K is then substituted into the following equation:

Notice that use of equivalent length and friction factor in the pressure drop equation is eliminated, although both are still required to calculate the values of K¹.  As a matter of fact, there is nothing stopping the engineer from converting the straight pipe length into a K value and adding this to the K values for the valves and fittings before using Equation 7. This is accomplished by using Equation 4, where D is the pipe diameter and ¦ is the pipeline friction factor.

     How significant is the error caused by mismatching friction factors? The answer is, it depends.  Below is a real world example showing the difference between the Equivalent Length method (as applied by most engineers) and the K value method to calculate pressure drop. 

An Example

     The fluid being pumped is 94% Sulfuric Acid through a 3”, Schedule 40, Carbon Steel pipe:

 Notes:

1. K values and L_eq/D are obtained from reference 1.
2. K values and L_eq are given in terms of the larger sized pipe.
3. L_eq is calculated using Equation 5 above.
4. The reducer is really an expansion; the pump discharge nozzle is 1” (Schedule 80) but the connecting pipe is 3”.  In piping terms, there are no expanders, just reducers.  It is standard to specify the reducer with the larger size shown first.  The K value for the expansion is calculated as a gradual enlargement with a 30^o angle. 
5. There is no L/D associated with an expansion or contraction.  The equivalent length must be back calculated from the K value using Equation 5 above. 

     The line pressure drop is greater by about 4.5 psi (about 62%) using the typical equivalent length method (adding straight pipe length to the equivalent length of the fittings and valves and using the pipe line fiction factor in Equation 1).

     One can argue that if the fluid is water or a hydrocarbon, the pipeline friction factor would be closer to the friction factor at full turbulence and the error would not be so great, if at all significant; and they would be correct. However hydraulic calculations, like all calculations, should be done in a correct and consistent manner.  If the engineer gets into the habit of performing hydraulic calculations using fundamentally incorrect equations, he takes the risk of falling into the trap when confronted by a pumping situation as shown above.

     Another point to consider is how the engineer treats a reducer when using the typical equivalent length method.  As we saw above, the equivalent length of the reducer had to be back-calculated using equation 5.  To do this, we had to use ¦_t and K.  Why not use these for the rest of the fittings and apply the calculation correctly in the first place?

Final Thoughts - K Values

     The 1976 edition of the Crane Technical Paper No. 410 first discussed and used the two-friction factor method for calculating the total pressure drop in a piping system (¦ for straight pipe and ¦_t for valves and fittings).  Since then, Hooper² suggested a 2-K method for calculating the pressure loss contribution for valves and fittings.  His argument was that the equivalent length in pipe diameters (L/D) and K was indeed a function of Reynolds Number (at flow rates less than that obtained at fully developed turbulent flow) and the exact geometries of smaller valves and fittings.  K for a given valve or fitting is a combination of two Ks, one being the K found in CRANE Technical Paper No. 410, designated K_Y, and the other being defined as the K of the valve or fitting at a Reynolds Number equal to 1, designated K₁.  The two are related by the following equation:

K = K₁ / N_RE  + K_Y (1 + 1/D)

     The term (1+1/D) takes into account scaling between different sizes within a given valve or fitting group.  Values for K₁ can be found in the reference article² and pressure drop is then calculated using Equation 7.  For flow in the fully turbulent zone and larger size valves and fittings, K becomes consistent with that given in CRANE. 

     Darby³ expanded on the 2-K method.  He suggests adding a third K term to the mix.  Darby states that the 2-K method does not accurately represent the effect of scaling the sizes of valves and fittings.  The reader is encouraged to get a copy of this article.

     The use of the 2-K method has been around since 1981 and does not appear to have “caught” on as of yet.  Some newer commercial computer programs allow for the use of the 2-K method, but most engineers inclined to use the K method instead of the Equivalent Length method still use the procedures given in CRANE.  The latest 3-K method comes from data reported in the recent CCPS Guidlines⁴ and appears to be destined to become the new standard; we shall see.

Conclusion

     Consistency, accuracy and correctness should be what the Process Design Engineer strives for.  We all add our “fat” or safety factors to theoretical calculations to account for real-world situations.  It would be comforting to know that the “fat” was added to a basis using sound and fundamentally correct methods for calculations.

 REFERENCES

1.  Crane Co., “Flow of Fluids through Valves, Fittings and Pipe”, Crane Technical Paper No. 410, New York, 1991.
2. Hooper, W. B., The Two-K Method Predicts Head Losses in Pipe Fittings, Chem. Eng., p. 97-100, August 24, 1981.
3. Darby, R., Correlate Pressure Drops through Fittings, Chem. Eng., p. 101-104, July, 1999.
4. AIChE Center for Chemical Process Safety, “Guidelines for Pressure Relief and Effluent Handling systems”, pp. 265-268, New York, 1998.

Reader / Author Question and Answers

1.  "Could you please give me in layman terms a better definition for K values.  I know that K is defined as "the number of velocity heads lost"...But what exactly does that mean???"

Well, I'll try to give you the Chemical Engineer's version of the layman answer. Velocity of any fluid increases through pipes, valves and fittings at the expense of pressure. This pressure loss is referred to as head loss.  The greater the head loss, the higher the velocity of the fluid. So, saying a velocity head loss is just another way of saying we loose pressure due to and increase in velocity and this pressure loss is measured in terms of feet of head. Now, each component in the system contributes to the amount of pressure loss in different amounts depending upon what it is. Pipes contribute fL/D where L is the pipe length, D is the pipe diameter and f is the friction factor. A fitting or valve contributes K. Each fitting and valve has an associated K.

2.  "It appears that the K values in CRANE TP-410 were established using a liquid (water) flow loop.  Is this K value also valid for compressible media systems?   (Can a K value be used for both compressible and incompressible service?)"

Crane also tested their system on steam and air. Now, this is where things get sticky. As per CRANE TP-410, K values are a function of the size and type of valve or fitting only and is independent of fluid and Reynolds number. So yes, you can use it in ALL services, including two-phase flow.  However, as I point out towards the end of my article, there is now evidence that shows using a single K value for the valve and fitting is not correct and that K is indeed a function of both Reynolds number and fitting/valve
geometry. I reference an article by Dr. Ron Darby of Texas A&M University which can be found in Chemical Engineering Magazine, July 1999. Dr. Darby just published a second article on the subject which can be found in Chemical Engineering Magazine, April 2001.

I don't believe there is any question as to the proper way to use K values in pressure drop calculations. The only question is whether industry will accept the new data.

3.  "When answering my first question, you stated:  'Velocity of any fluid increases through pipes, valves and fittings at the expense of pressure.'  When you say this, you are talking about compressible (gas) flow right?  For example, in a pipe of constant area, the velocity of a gas would increase as the fluid traveled down the pipe (due to the decreasing pressure).   However, the velocity of a liquid would remain constant as it traveled down the same pipe (even with the decreasing pressure).  Is this a correct statement?

Sorry for the confusion. Yes to both of your questions. If you look at the Bernoulli equation, you will see that velocity cancels out for a liquid as long as there is no change in pipe size along the way and pressure drop is only a function of frictional losses and a change in elevation.

However, the K value of a fitting is still a quantifier of the head loss (frictional loss) in that fitting and this head loss is still calculated as the velocity head of the liquid (V^2/2g). So in essence, you still achieve a
liquid velocity at the expense of pressure loss; the velocity head just happens to be constant. Read section 2-8 in CRANE TP-410. They define the velocity head as a decrease in static head due to velocity.

The big thing is not to get too hung up on the definitions and just remember you can't have flow unless you have a driving force and that force is differential pressure. Also, in a piping system there is frictional losses which comes from the pipe and all fittings and valves. The use of K is just a way of quantifying the frictional component of the fittings and valves. You can even put the piping friction in terms of K by using fL/D for the pipe and multiplying that by V^2/2g.

I hope this helps. If you are still confused, let me know and I'll just explain it again but I'll try to do it in a different way. Sometimes, a concept just needs to be re-worded and I'm willing to spend as much time on this as you need.

4.  I'm reading the Crane Technical Paper #410 and I have the following
questions/comments:

Page 2-8 of TP 410 states that:
"Velocity in a pipe is obtained at the expense of static head".  This makes sense and Equation 2-1 shows this relationship where the static head is converted to velocity head.  However, there is no diameter associated with this.  So is it correct to say based on equation 2-1 that if you had a barrel of water with a short length of pipe attached to the bottom that discharged to atmosphere, and in this barrel you had 5 feet of water (5' of static head), the resulting water velocity would be 17.94 ft/sec (regardless
of the pipe diameter).

Maybe the real question is how do you use equation 2-1.  Do you have to know the velocity and then you can calculate the headloss?  And why does equation 2-1 and equation 2-3 seem to show headloss equaling two different things?

Also, why does it say that a diameter is always associtated with the K value, when as I mentioned above there is no diameter associated with equation 2-1?

Maybe I'm trying to read into all of this too deeply, but I still do not feel that I fully grasp what page 2-8 is trying to reveal.

You need a diameter to get velocity. Velocity is lenght/time (for example, feet/sec). Flow is usually given in either mass units (weight/time or lb/hr for example) or in volumetric units (cubic feet per minute for example). To get velocity, you need to divide the volumetric flow by a cross sectional area (square feet). To get an area, you need a diameter. So the velocity is always based on some diameter.

As I show in my paper, equation 2-1 is just the basis of the velocity head. To get the frictional loss, you need to know the contribution of each component in the system; pipe, fitting and valve. To get that contribution, you use 'K' (equation 2-2). Each component has an associated 'K' value. You multiply the velocity head by the appropriate 'K' value. Equation 2-3 is just another way of expressing the same thing. As you can see, this means you can calculate a 'K' for a component such as a pipe using the formula fL/D as shown in Equation 2-3. Again, I explain this in my paper so I would suggest you re-read it.

I would also suggest you look at the examples in CRANE. There are many of them in Chapter 4.

'K' is associated with the velocity and therefore the diameter. Look at the values for 'K' in CRANE (starting on page A-26). You will see that for the most part, K is a function of a constant times the friction factor at fully turbulent flow. This friction factor changes with pipe diameter as shown on page A-26. Again, re-read my paper and look at the examples in Chapter 4.

By Phil Leckner, Chief Content Manager (read the author's Profile) pleckner@hotmail.com