Copyright © Harvey Wilson - Katmar Software

2024

2024

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Virtually all fluid mechanics handbooks discuss the changes in pressure described by Bernoulli's Equation and also discuss the frictional losses in pipe entrances, exits and at changes in pipe diameter. However, very few of them give much attention to how the Bernoulli pressure changes and the frictional losses interact, and some of the handbooks that do include examples of this are somewhat misleading.

In its original formulation the Bernoulli Equation assumed frictionless pipe, so in a discussion based on the original formulation there can be no consideration given to frictional losses. But the Bernoulli Equation does give a rigorous treatment to the interaction between pressure, elevation and flow velocity.

On the other hand, most handbooks consider the Darcy-Weisbach Equation only in terms of frictional losses caused by the surface of the pipe and by changes of flow direction in various fittings.

In the real world we need to consider all of these factors and need to understand how they all fit together to enable accurate, working pipe designs. Here pressure drop calculations will be looked at from both of these viewpoints and some simple examples that illustrate the necessary calculation procedures will be considered.

An additional resource for understanding this subject is the set of videos produced by the YouTube channel "Fluids Explained". (Note: Katmar Software has no affiliation or commercial link with "Fluids Explained".) These videos show physical experiments that clearly illustrate the principles of fluid flow. The videos are a bit slow moving and they work towards a "modified" version of the Bernoulli Equation which includes the frictional losses rather that emphasizing that although Bernoulli's original work did not include these losses they do need to be dealt with using the Darcy-Weisbach Equation. Nevertheless, spending time going through these videos will give a very good understanding of the phenomena occurring. The videos are available at

https://www.youtube.com/playlist?list=PLgswUthJnnpDA5NLfqEZi6X46rkwwRec1

There are some handbooks that also include a friction term as part of the Bernoulli Equation in the way these videos do, but in the original work by Bernoulli it was assumed that there were no frictional or viscous effects and that approach will be used here.

In the sections that follow the Bernoulli and Darcy-Weisbach Equations will be discussed briefly and then their impact on the pressure changes at pipe entrances, exits and diameter changes will be examined.

Possibly the most frequently used form of the Bernoulli Equation is where each of the terms has the units of length, making each term a head associated with a different factor.

Here the first term is simply the static height, the second term is the static pressure expressed as a column of the flowing liquid and the third term is the velocity head. The locations of points 1 and 2 are totally arbitrary so Equation 1 can be simplified to

For all points along the pipe.

When working with the pressure losses in pipe and fittings it is more usual to express these losses in pressure terms and AioFlo follows this convention. The Bernoulli Equation can be recast in pressure terms by noting that for a column of fluid the pressure is given by

The form of the Darcy-Weisbach equation usually used to calculate the pipe friction and also friction losses in fittings (where the friction loss is in pressure terms) is

As mentioned earlier, the AioFlo software expresses the losses due to friction in terms of pressure rather than head and the rest of this section will use pressure terms to express losses.

The diagram above shows the typical resistance coefficients included in many hydraulics handbooks. These are intended to apply to the full range of turbulent flows, but in the AioFlo software they are modified according to the Reynolds number using the method proposed by Hooper (1981). The variation from the numbers shown above is small except for laminar flow and for this discussion the numbers in the table above will be used to keep the math simple.

The losses at pipe entrances are caused by eddies and turbulence resulting from the fluid abruptly changing direction as it passes around the edge of the entrance. The more this edge is rounded the less the resulting turbulence and the smaller the losses.

In the example shown below the pressures will be calculated before and after a pipe entrance to illustrate the various changes in pressure as described by the Bernoulli Equation and by the Darcy-Weisbach Equation.

The tank is filled to a level of 10 m above the center line of the outlet nozzle with a liquid of density 1000 kg/m3 and of viscosity 1.0 centipoise. The pipe entrance (or tank outlet nozzle) has a sharp edge (r/d = 0) and the outlet pipe has an inside diameter of 78 mm and a roughness of 0.05 mm. The flowrate out of the tank is 15 liter/second. The objective of the calculation is to determine the various pressure losses involved and to calculate the pressure that would be measured by a gauge at point C, 2 meter downstream from the entrance to the pipe.

Solution

The velocity in the pipe is (15 / 1000) / ((pi / 4) x 0.0782) = 3.14 m/s

The pressure at point A is taken to be 0 kPag. The pressure at point B will be

1000 kg/m3 x 9.81 m/s2 x 10 m = 98 100 Pascal gauge = 98.10 kPag

The frictional loss at the entrance can be calculated with the Darcy-Weisbach Equation using a resistance coefficient of 0.50

Entrance pressure loss = 0.50 x (1000 x 3.142 / 2) = 2470 Pascal = 2.47 kPa

The Reynolds Number in the pipe is

1000 x 3.14 x 0.078 / (1.0 / 1000) = 244 900

The relative roughness of the pipe is

0.05 / 78 = 0.00064

From the Moody chart this gives a friction factor of 0.019

The frictional loss due to the 2 m of pipe can be calculated with the Darcy-Weisbach Equation using this friction factor

Pipe friction = (0.019 x 2 / 0.078) x (1000 x 3.142 / 2) = 2400 Pascal = 2.40 kPa

The velocity at point B is essentially zero so the acceleration loss calculated with the Bernoulli Equation is

Acceleration loss = (3.142 - 0.02) x 1000 / 2 = 4930 Pascal = 4.93 kPa

The total loss from point B to point C is therefore

2.47 + 2.40 + 4.93 = 9.8 kPa

And we can expect a pressure gauge at point C to read 98.1 - 9.8 = 88.3 kPag

It is interesting to note that in this example, which involves fluid and pipe data that is typical of real-world processes, the losses at the entrance are dominated by the acceleration losses of the liquid. In some piping handbooks this loss is taken to be an "exit loss" but it is very clearly not related to frictional losses at the exit. This point will be discussed further in section 5 on pipe exit losses.

This example will be completed by looking at how it would be calculated with the AioFlo software. In the Fittings section the Start of Line would be set to a square, flush entrance and the End of Line would be set to Continues to Other Pipe with no diameter change. All other fittings are set to zero. Since the aim is to calculate the pressure change from point B to point C, which are at the same elevation, the Elevation Change has been set to zero. The rest of the input and results are summarized in this screenshot.

AioFlo results for Pressure Losses at Pipe Entrance

The small differences between these results and the manual calculation are caused by rounding errors in the calculation of the Reynolds Number and the pipe friction factor, and by correction for the Reynolds Number in the resistance coefficient for the entrance.

In most fluid mechanics texts the extent of the discussion on pipe exit losses is just one line:

Pipe exit resistance coefficient (K) = 1.0

This is "sort of" true, but not conducive to good design calculations. Here the exit loss will be examined in more detail and the reasons why this single line treatment is inadequate will be considered. Some examples will be examined that illustrate when the assumption of K=1 gives "correct" answers and when it does not.

As discussed in section 3 on the Darcy-Weisbach Equation, the K values or resistance coefficients are intended to provide for frictional losses in pipe fittings. These friction losses in fittings are caused by changes in flow direction which cause eddies and turbulence. There are some situations where eddies and turbulence do occur at pipe exits but in the case of free jets of liquid issuing into air or gas this does not occur, so there are separate situations that need to be considered.

The texts that do explain the situation beyond the simplistic K = 1.0 talk of discharges into "confined" or "unconfined" spaces. However, a better understanding can be obtained by considering 3 separate cases that can occur at the exit.

Exit Case 1 - Free discharge of liquid stream into atmosphere

When a liquid jet emerges from a pipe into air (usually described as an "unconfined space") there is essentially no turbulence between the jet of liquid and the surrounding air and consequently there are no frictional losses i.e. K = 0.

The kinetic energy in the liquid is lost (i.e. is not converted back to pressure energy) and it is equivalent to 1 velocity head. Losing this 1 velocity head is numerically equivalent to K = 1.0 in the Darcy-Weisbach equation and although these losses have nothing to do with Darcy-Weisbach frictional losses it is a fairly standard practice to include a frictional exit loss of K = 1 to cover the kinetic energy lost at the exit - even though the conversion from pressure energy to kinetic energy actually occurs at the pipe entrance and has no relevance to the exit as shown in section 4 above.

In Case 1 the exit should therefore strictly be dealt with by applying a K-value of 0.0 for the exit frictional losses, and separately considering any pressure energy converted to kinetic energy at the pipe entrance and any kinetic energy lost at the exit (because it is not recovered) as described by Bernoulli. In common practice and in many published examples, Case 1 is covered by including K = 1.0 in the Darcy-Weisbach equation and ignoring the Bernoulli effects at the entrance and exit. Although this will often give the correct answer it is logically inconsistent and this will be illustrated in the examples further below.

Exit Case 2 - Discharge of liquid into a body of liquid

This situation occurs when a pipe discharges liquid into a tank or pond below the surface (i.e. a confined space). In this case there is significant turbulence between the liquid jet and the body of liquid it enters. The Darcy-Weisbach frictional losses for the exit are equal to K = 1.0 because of this turbulence. But at the same time the kinetic energy in the emerging jet is converted to pressure energy as the liquid decelerates and the kinetic energy is therefore not lost. The kinetic energy in the exiting liquid is one velocity head. This means that the frictional loss and the pressure recovery are each one velocity head and cancel each other.

The way in which the frictional exit loss cancels with the pressure recovery as the velocity drops to zero should be visualized as happening in an infinite number of infinitely small steps. If the actual static (not stagnation) pressure could be measured at many points as the liquid exits the pipe the pressure would not be seen to suddenly decrease by one velocity head and then later increase by this same amount. Both processes occur together and there is no pressure change.

The strictly correct way to calculate the pressure effects at the end of the pipe in Case 2 would be to calculate the frictional loss at the exit and also calculate the pressure recovered as the liquid decelerates, However, because these two phenomena always occur together in a Case 2 situation and because no pressure decrease or increase is seen in practice, it is convenient to accept that they offset each other and to ignore both effects. This will be illustrated in the later examples.

Exit Case 3 - Discharge of gas to atmosphere or to confined space filled with gas.

In all cases where gas emerges from a pipe exit the jet will rapidly expand and mix into the surrounding gas. This makes all situations with gas at a pipe exit effectively identical to Case 2.

The only difference from the situation with liquids is that the velocities at the start and end of the pipe are more likely to be different and care must be taken to apply the correct velocity and density when calculating the velocity head at each point. Nevertheless, the friction losses at the exit are still exactly cancelled by the pressure recovery as explained for Case 2.

From these 3 cases it is evident that for a liquid the behavior could be described as discharging into an unconfined space (Case 1) or into a confined space (Case 2). However, a gas would always be emerging into a confined space. This would also be true of a gas exiting into a body of liquid, although this is a relatively unusual situation.

Overall, it is seen that the frictional exit loss is either zero and there is no pressure recovery (as in Case 1) or the exit resistance coefficient has a value of K=1 but is exactly cancelled by the pressure recovery as the fluid decelerates into the body of fluid (as in Cases 2 and 3). In a practical sense the piping exit loss can be relegated to the same category as the roughness of wood stave pipes and the friction loss in Borda entrances - these are all of only historical or academic interest and although they are always included in fluids texts and software they can be ignored. In AioFlo the options for wood stave pipes and for Borda entrances have been retained simply for consistency with other sources, but no option to include a frictional exit loss has been included as it is never relevant and only causes confusion.

A few examples follow to illustrate these points.

This example is similar to that for the pipe entrance losses in Section 4, but the exit pipe is now longer and it discharges to atmosphere as a free jet. The aim is to calculate the flow rate that would result.

From the earlier example it is known that the pressure at point A is 0 kPag so this makes the pressure at point C also 0 kPag. The pressure at point B was calculated to be 98.10 kPag. The pressure difference between point B and point C is therefore also 98.10 kPa and the aim is to calculate the flowrate that this would drive from B to C.

As before, the entrance at B is taken to be a sharp edged flush entrance. The pipe is again taken to have an inside diameter of 78 mm and a roughness of 0.05 mm. The pipe exit is a free jet issuing into an unconfined space as described above for Case 1. This example has to be solved by trial and error because the friction factor in the pipe cannot be calculated until the flowrate is known, so AioFlo will be used with the Calculation Type set to calculate the flowrate from the pressure drop and pipe ID.

All the information and options have been entered into AioFlo in the screenshot shown below.

AioFlo Results for Exit Case 1 - Example 1

As in the Section 4 example, the velocity head change from point B to point C can be calculated from the velocities at each of these points. This confirms the value shown in the screenshot above as

Velocity Head Change = (5.5952 - 0.02) x 1000 / 2 = 15652 Pascal = 15.65 kPa

This velocity head has to be provided by the conversion of pressure energy as described previously in the discussion on the entrance losses. Note that it remains a pressure loss as there is no pressure recovery from a free jet. Note also that the Fittings Total K Value is just the 0.5 for the sharp edged entrance and there is no exit friction loss for a free jet. The friction loss at the entrance is included with the pipe friction losses to make the total of 82.45 kPa. This combined friction loss of 82.45 kPa is added to the pressure loss of 15.65 kPa consumed in the conversion of pressure to kinetic energy to make up the total pressure drop available of 98.10 kPa.

If the velocity head effects had been ignored and a frictional resistance coefficient (K value) of 1.0 for the so-called exit loss had been applied the exit loss would have been calculated as

Exit loss = K x (density x velocity2/2) = 1.0 x 1000 x 5.5952/2 = 15652 Pa = 15.65 kPa

This is numerically exactly the same as the result obtained above. So why not just use the standard value for an exit loss? Apart from being logically incorrect, there are situations where this procedure does not give the correct answer as will be shown in the next example.

Example 2 for Exit Loss Case 1 involves a similar setup to Example 1 but the pipe at point B is a continuation of an existing pipe rather than being an entrance as before. This means that the velocity at point B is non-zero and will be exactly the same as at point C because the pipe ID remains the same and the liquid is incompressible.

This is the arrangement that would exist after a pump delivery flange or simply any arbitrary section of pipe where it is possible to measure the pressure at two points.

The pipe is again taken to have an inside diameter of 78 mm and a roughness of 0.05 mm. The points B and C are 20 meters apart and are at the same elevation.

If the pressure at point B is taken to be the same 98.10 kPag as in Example 1 the flowrate can be calculated for comparison with the previous result. In AioFlo the Start of Line option on the Fittings page is changed to "Continues from other pipe - no diameter change" and the rest of the information remains as before. Here is the calculation result.

AioFlo Results for Exit Case 1 - Example 2

There are two differences from the previous result. The first is that the Velocity Head Change is now zero because the liquid flowing at point B already had the same velocity as at C and there is no velocity change and therefore no velocity head change. The second difference is that the Fittings Total K Value is also zero because there is no pipe entrance in the section that is involved and there are no other fittings. These differences mean that the full 98.10 kPa available can be used to overcome the pipe friction and the result is that the flowrate increases from the 26.73 liter/s found in Example 1 to 30.74 liter/s in this example.

The important conclusion that can be made from this example is that taking the commonly listed exit loss of K=1 to describe the overall situation for Case 1 would give the wrong answer if the start of the pipe is not a pipe entrance.

The example shown below for Case 2 (i.e. discharge into a body of liquid) appears to be a very different situation from the second example for the Case 1 situation but as the discussion for Case 2 above showed, the losses at the exit are the same as for Case 1 and in AioFlo they are grouped as the same option on the Fittings - End of Line page.

The earlier examples calculated the pressure at point C to be 98.10 kPag. For this example the pressure at point B is taken to be 150 kPag and the aim is to calculate the flow from B to C. This makes the total pressure drop available to drive the flow 51.9 kPa.

The pipe is again taken to have an inside diameter of 78 mm and a roughness of 0.05 mm. The points B and C are 20 meters apart and are at the same elevation. AioFlo was set up with the Fittings - Start of Line to be Continues from other pipe with no diameter change and the Fittings - End of Line to be Discharge to atmosphere or to tank. All the other settings are shown in the screenshot below.

AioFlo Results for Exit Case 2 Example

The important results to note here are that the Fittings Total K Value and the Velocity Head Change are both zero. The total K value is zero because there is no pipe entrance involved and the exit loss has been offset by the pressure recovery as the liquid decelerates into the body of liquid.

Fluids texts and handbooks routinely give equations to calculate K values which allow the frictional losses at changes in pipe diameter to be evaluated. It is important to note that these K values only describe the frictional losses caused by turbulence resulting from changes in flow direction as the flow converges or diverges at the diameter change. These losses do not include any changes in pressure resulting from changes in the average flow velocity. The Bernoulli Equation must be used to evaluate the conversion between kinetic energy and pressure.

The most common geometries for changes in diameter are sudden (square) changes, conical convergences and typical pipe reducers with radiused ends. The principles for performing complete evaluations of the pressure changes are the same for all 3 geometries, although the equations for the K values are different for each of them. The principles will be demonstrated here using a sudden change in diameter.

For this example the flowrate is taken to be 15 m3/h. This makes the velocity in the 50 mm section 2.122 m/s and in the 100 mm section it is 0.531 m/s. The aim of the calculation is to calculate the pressure change from A to B, but for the sake of clarity the friction in the pipe sections will be ignored.

An equation often used for the K value for a sudden enlargement for turbulent flow is

Solution

The velocity in the 50 mm pipe = (15 / 3600) / ((pi / 4) x 0.0502) = 2.122 m/s

The velocity in the 100 mm pipe = (15 / 3600) / ((pi / 4) x 0.1002) = 0.531 m/s

Resistance coefficient based on velocity in D1 = (1 - (50/100)2)2 = 0.563

Frictional pressure drop through enlargement = 0.563 x 1000 x 2.1222 / 2

= 1267 Pascal = 1.267 kPa

Change in velocity head = 1000 / 2 x (0.5322 - 2.1222) = -2110 Pa = -2.11 kPa

Overall pressure drop from A to B = 1.267 + (-2.110) = -0.843 kPa

The important point to note is that in flowing from A to B the velocity decreases and since there is no change in elevation, and because the sum of the terms in the Bernoulli Equation is constant this decrease in velocity head must be matched by an increase in the static pressure.

When expressed as a pressure drop this gain in static pressure would be negative and when it is added to the frictional pressure drop calculated from the K value it generally happens (and is indeed the case here) that the overall pressure drop is negative and there is a higher static pressure at B than at A.

As D1/D2 tends to zero, the frictional pressure loss and the kinetic energy pressure recovery become more closely matched. As was discussed in the section for Case 2 Pipe Exits (discharge into a body of liquid) the pressure lost to friction and the pressure recovered from deceleration become so similar that they can be offset against each other.

The other possibility for a diameter change is for the diameter of the pipe to decrease. This scenario is examined in the following example, again using a sudden change.

For this second example the flowrate is taken to be 30 m3/h. This makes the velocity in the 100 mm section 1.061 m/s and in the 50 mm section it is 4.244 m/s. The aim of the calculation is to calculate the pressure change from A to B, but for the sake of clarity the friction in the pipe sections will be ignored.

The equation recommended by Hooper (1988) and slightly simplified here for clarity for the K value for a sudden contraction for turbulent flow is

Solution

The velocity in the 100 mm pipe = (30 / 3600) / ((pi / 4) x 0.1002) = 1.061 m/s

The velocity in the 50 mm pipe = (30 / 3600) / ((pi / 4) x 0.0502) = 4.244 m/s

Resistance coefficient based on velocity in D1 = 0.6 x (1 - (50/100)2) = 0.45

Frictional pressure drop through contraction = 0.45 x 1000 x 4.2442 / 2

= 4053 Pascal = 4.053 kPa

Change in velocity head = 1000 / 2 x (4.2442 - 1.0612) = 8443 Pa = 8.443 kPa

Overall pressure drop from A to B = 4.053 + 8.443 = 12.496 kPa

In this example the velocity increases in the direction of flow and the energy to cause this acceleration has to come from the conversion of pressure to kinetic energy. The pressure drop from this conversion will always be positive and since the frictional pressure drop is also always positive the overall pressure drop will be positive.