9 replies to this topic

[ryn376]

Can anybody verify that the K-value for a pipe exit is zero if a fluid exits into an unconfined space? 

 

Ron Darby (of 3-K Method fame) indicates that it should not be used if a fluid exits into an unconfined space. In his book Chemical Engineering Fluid Mechanics on page 213 he writes:

 

A note is in order relative to the exit loss coefficient, which is listed in Table 7-5 as equal to 1.0. Actually, if the fluid exits the pipe into unconfined space, the loss coefficient is zero, because the velocity of a fluid exiting the

pipe (in a free jet) is the same as that of the fluid inside the pipe (and the kinetic energy change is also zero). However, when the fluid exits into a confined space the kinetic energy is dissipated as friction in the mixing process as the velocity goes to zero, so the loss coefficient is 1.0. In this case the change in the kinetic energy and the friction loss at the exit cancel out.

Edited by ryn376, 14 January 2014 - 01:38 PM.

[latexman]

All the examples I looked at in Crane TP410 which discharge to atmosphere have K = 1.0 for an exit loss.  My old college textbook, McCabe and Smith, was the same.  Cameron Hydraulic Data reports K = 1.0 for an exit loss with no qualifications whether the end destination is confined or unconfined.

 

I see no parameter in the Bernoulli Equation that translates whether "point 2" (the exit) is confined or unconfined.  Does anyone else?

 

I have used K = 1.0 for an exit loss into confined and unconfined areas my entire career.  No issues came of it. 

 

I think Mr. Darby may be outnumbered on this view.  Let's see what others think.

[PingPong]

I agree more or less with Mr. Darby, although I would formulate it differently.

 

If a fluid leaves a pipe without the fluid occupying the whole width of the space it enters then there is no frictional exit loss.

 

For example:

- if water flows from a pipe and falls into a tank (or whatever), the diameter of the water stream after the exit does not change, its velocity does not change, there is no turbulence outside the pipe-exit, and there is no exit loss.

 

- however if water from a pipe flows into a much bigger pipe (or tank) that is water filled, then there is an exit loss.

[katmar]

The widespread confusion over the exit loss is mostly caused by the sloppy way we conventionally use the terminology. In the overwhelming majority of cases where we talk of an exit loss, it is not an exit loss at all. If we go back to the Bernoulli Equation it is very clear that one of the terms takes into account the change in velocity (i.e. acceleration) of the fluid. However, in the Darcy-Weisbach Equation there is no term for acceleration and we compensate by adding a K=1 exit loss.

Why is it called an exit loss?  The acceleration usually occurs at the start of the pipe, so wouldn't it be more logical to call it an entry loss? At the pipe entry there are two phenomena that consume energy. The first is the forcing of the fluid into the reduced cross sectional area of the pipe entry and this results in flow separations, eddies and losses. This loss can be significantly reduced by using a bell mouth entry (page A-29 in the old Crane 410 manual). This is a side issue and has nothing to do with the exit loss.

The second phenomenon occuring at the pipe entry is the acceleration from the usually stagnant tank to the velocity in the first section of pipe. The reason we do not take this into account at the start of the pipe is that there will be more acceleration (or deceleration) along the length of the pipe each time the pipe ID changes. The velocity at the end of the pipe is the net result and is all that we need to account for and so we have come to talk of an exit loss. But it is important to remember that the loss is not actually occurring at the exit - that is simply the convenient point to calculate the net acceleration.

With a liquid discharging to atmosphere or into an open tank, or for a compressible fluid discharging to atmosphere, all the velocity head at the exit is lost and we can include a value of K=1 in the Darcy-Weisbach Equation when calculating the overall pressure drop.

The exception to this situation is for a compressible fluid discharging into a closed vessel. In this situation the velocity head can be partially recovered as pressure.  It will usually not be totally recovered because the situation is similar to the sudden enlargement situation. In my opinion this is the only situation where we can truly speak of an exit loss, but every writer on fluid mechanics uses the convention of the exit loss representing the acceleration loss and there is no way to change that now.

Luckily it is very rare for the exit loss to be significant in normal process piping calculations and any errors that we have made over the years are masked by all the other uncertainties and inaccuracies in the overall piping calculation.

On a related note, there is often also confusion over what losses the K value takes into account when a fluid flows through a reducer or diffuser in a pipeline. In line with the convention used above (where all the acceleration is accounted for at the exit) the K value for a reducer ignores the velocity head component and only considers the losses due to changes in flow path. In a situation where a fluid flows through a reducer from a small diameter pipe to a much larger pipe pressure gauges installed on both ends of the reducer could show that the static pressure has actually increased as the fluid flows into the larger pipe because the velocity head recovery can be larger than the friction loss.

[PingPong]

The frictional pressure drop at the entrance or the exit of a pipe is not related to Bernoulli.

 

When a fluid from a tank enters a pipe with a sharp edged entrance (K=0.5) there is a frictional pressure loss of 0.5 velocity heads. On top of that there is a pressure drop of 1.0 velocity heads due to acceleration (Bernouili), so the total pressure drop over the entrance will be 0.5 + 1.0 = 1.5 velocity heads. If the entrance is not sharp but well rounded (K=0.04) then the total pressure drop (friction + acceleration) will be 0.04 + 1.0 = 1.04 velocity heads.

 

When a fluid from a tank exits a pipe (K=1.0) into a wider fluid filled pipe or tank there is a frictional pressure loss of 1.0 velocity heads. On top of that there is a pressure gain of 1.0 velocity heads due to deceleration (Bernouili), so the total pressure drop at the exit will be 1.0 - 1.0 = 0.0 velocity heads.

 

So in most situations there will not be a pressure drop at the exit of the pipe because most of the time the frictional pressure loss is compensated by the pressure gain due to decelleration. Make a sketch when in doubt and it should be clear whether that is the case in a given situation.

 

Obviously it should be known to everybody that Darcy-Weisbach should always be used together with Bernoulli when determining the total pressure drop over a system from origin to destination.

Edited by PingPong, 15 January 2014 - 02:49 PM.

[katmar]

PingPong and I obviously disagree on the definition of the exit loss. I cannot see how there is a velocity head pressure recovery from deceleration when a pipe discharges freely to the atmosphere. Have a look at Crane's example 4-6 (in my metric 1988 version - I think it should be the same in most others). There is no mention of acceleration from the stagnant fluid from the tank into the pipe - because it is covered by the "exit loss".

This aspect is generally not well covered in the standard fluids texts - or at least not in the ones I have.  Quoting from 2 of those on my shelf:

IE Idelchik, "Handbook of Hydraulic Resistance", 3rd Ed, Pg 628 : In the case of free discharge of the flow from a straight section of a tube of constant cross section into a large volume, the total losses are reduced only to the losses of the velocity pressure at the exit.

RK Bansal, "Fluid Mechanics and Hydraulic Machines", 9th Ed, Pg 477 Loss of Head at the Exit of Pipe: This is the loss of head (or energy) due to the velocity of liquid at outlet of the pipe which is dissipated either in the form of a free jet (if the outlet of the pipe is free) or it is lost in the tank or reservoir (if the outlet of the pipe is connected to the tank or reservoir).
 

I still believe my earlier post gives the correct definition.

[PingPong]

I cannot see how there is a velocity head pressure recovery from deceleration when a pipe discharges freely to the atmosphere.

I never said there is in that situation. When a fluid discharges freely into the atmosphere there is no frictional exit loss (so K is not 1 but 0) but there is no gain due to deceleration either.

 

In post #5 I was referring to the situation that: when a fluid from a tank exits a pipe (K=1.0) into a wider fluid filled pipe or tank there is a frictional pressure loss of 1.0 velocity heads.

In that case there is a considerable frictional pressure loss and there is also a pressure gain due to deceleration (Bernoulli).

 

 

PingPong and I obviously disagree on the definition of the exit loss.

I am not sure whether we disagree on definition.

 

Post #1 in this topic is about the K-value of a pipe exit, not about Bernoulli.

The K-value of an exit, or any pipe fitting (bend, Tee, whatever) refers to frictional loss only, irrespective of any possible Bernoulli related loss or gain.

 

For a widening (enlargement) of a pipe the K-value depends on the value of the enclosed angle θ.

As θ increases so increases K due to flow separation and eddy formation.

A very gradual widening with (say) θ = 20^o will have a K = 0.45 while a sudden enlargement (θ = 180^o) is everywhere reported to have K = 1.

There is a formula for that in Crane, and a formula plus graph in Ludwig (see attached figure).

This means that as long as θ is small enough the value of K is less than 1 and the pressure gain due to deceleration (Bernoulli) will exceed the pressure loss due to friction.

 

According to that Ludwig figure 2-16 the maximum value of K occurs around θ = 60^o and could be as high as 1.2 and gradually drops to around K = 1 at θ = 180^o.

I would assume that that graph is based on actual measurements.

 

It seems to me that it is to simple to state that the generally used K=1 for a sudden enlargement is due to loss of velocity, as you seem to do in post #4, as that would not explain why K > 1 for θ = 60^o.

Note also that for a 90^o mitre bend, or a Tee Branch, K is also greater than 1 for sizes smaller than 4" (see Crane) although there is no velocity loss there.

 

Frictional pressure loss at a pipe exit (sudden enlargement), or a (mitre) bend, or a Tee Branch, or a pipe entrance, is caused by turbulence, irrespective whether the fluid velocity stays the same, or increases, or decreases. I therefor suspect that a sudden enlargement (θ = 180^o) just happens to create so much turbulence that K is ~1, just as that happens to be the case for a 4" 90^o mitre bend.

Attached Files

•  Figure 2-16 from Ludwig Vol 1.jpg   179.74KB   20 downloads

Edited by PingPong, 16 January 2014 - 10:01 AM.

[latexman]

I believe post #1 came about for the simple reason that Mr. Darby did mix frictional loss and kinetic energy changes (Bernoulli) together without being perfectly clear verbally/written.  I do appreciate PingPong's discipline to keep the two separate, and, therefore more simple.  I'm still struggling to understand this confined/unconfinded concept that Mr. Darby did not fully define to my satisfaction, so I am following this post with ineterest.

[katmar]

PingPong, I agree with latexman that the way you have described the relationships between the friction losses and the changes in kinetic energy is accurate in terms of what is actually physically happening in the pipe - and you have certainly explained it more clearly than Prof Darby did. Thank you for bringing about this clarification.  But the way you have described it is not the way engineers conventionally employ the exit loss in their calculations.

You (and Prof Darby) have stated that for a fluid issuing into an unconfined space the exit loss is zero. Apart from the two of you I know of no other reference that states this.  Everybody from Crane to Hooper gives a blanket statement that the exit loss K value is 1.  Your procedure requires a rigorous application of Bernoulli to every change in pipe diameter along the length of the pipeline and then a careful consideration of the exit situation. What everybody does is to combine all of this into the K = 1 exit loss and be done with it.  The net result of the two procedures is the same answer. That was why I said right at the beginning that the conventional procedure is sloppy and disregards the actual phenomena occurring.

Your statement that K values refer to frictional losses only describes how things should be, but in conventional usage the exit loss of K = 1 does describe the Bernoulli equivalence between static and velocity head - like I said, sloppy terminology.

If we add the PingPong clarifications into the quoted section from Darby's book included in Post #1 we would get something like (my additions in bold)

A note is in order relative to the exit loss coefficient, which is listed in Table 7-5 as equal to 1.0. Actually, if the fluid exits the pipe into unconfined space, the loss coefficient is zero, because the velocity of a fluid exiting the pipe (in a free jet) is the same as that of the fluid inside the pipe (and the kinetic energy change is also zero giving no recovery or conversion of kinetic head to pressure). However, when the fluid exits into a confined space the kinetic energy is dissipated as friction in the mixing process as the velocity goes to zero (and the kinetic energy is converted to pressure), so the loss coefficient is 1.0. In this case the change in the kinetic energy and the friction loss at the exit cancel out.

Edited by katmar, 16 January 2014 - 03:13 PM.

[PingPong]

Being a process design engineer for the past 30 years I have had to do or supervise many hydraulic calculations to determine required pipe diameters, pump heads and control valve pressure drops. Believe it or not, but most of those calculations were done by me or my coworkers without us knowing the length of the pipe or the number of bends in it. However our piping department will not start work on the piping layout until they have the line sizes. Catch 22.

By the time they are done it is very difficult to change any pipe diameter. So we have to make sure that we, process engineers, design conservative by (hopefully) overestimating pipe length and number of bends. You will understand that I therefore do not really care what the exact K value of any fitting, bend, tee or exit is. I have to make sure it will work later on, and so I will always use a K=1 for an exit loss, whether I feel it is applicable in a particular case or not. Better safe than sorry, and moreover I do not want to have this kind of discussions with my young coworkers if I would instruct them otherwise and they would come up with a design manual or textbook that says ..........

 

Bernoulli should not be applied for each change in pipe size, but only for the difference in kinetic energy, pressure and height between origin and destination.

 

The discussion in this topic is rather academic, but so was the question of the topic starter. In actual practice it will make no real difference as the exit loss will usually be small in the whole picture anyway.