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mechanicaldup (Mechanical) (OP)
1 Jul 05 4:25
Hi every one!
Could some one perhaps explain to me how to indicate the residual head at the pipe exit (for example fire hydrants) on the system curve. As far as I understand it the system curve stop at the end of the pipe. The pump operate at the pressure which it "see". the pump therefore have to see the residual head otherwise it will automaticly adjust on the system curve and operate at a higher flow rate.

Do you add the residual head to the static head or do incorporate it with the exit losses and velocity pressure ( k.Q^2 + V^2/2g) or how?



Your advice would be appreciated.

have a nice day
Helpful Member!  katmar (Chemical)
1 Jul 05 9:34
When you develop your system curve you simply add the three components, i.e. static head, friction head and velocity head.  The pump develops a certain head and does not "care" how you use it.

A confusing aspect is that the velocity head is usually called an "exit loss", or as you have termed it here a "residual pressure". It becomes confusing because everyone looks to the end of the pipe to solve the problem, but the "exit loss" actually applies to the START of the pipe.

Part of the head developed by the pump goes into accelerating the liquid, and of course this occurs at the very start of the pipe.  If the exit discharges into a closed, pressurised system you can recover this kinetic energy that was put into accelerating the liquid and that is why if it is not recovered people call it an exit loss.  But with an open system like a fire hydrant there is no opportunity for pressure recovery.

An example with real numbers may help.  In a fire system you can have high flowrates, so let us imagine that we want to put 150 m3/h of water through a 50 m long pipe with an 80 mm ID and where the discharge point is 5 m above the pump. The velocity would be 8.3 m/s and the velocity head is 3.5 m (= 8.3^2 / (2 x 9.8) ). The friction head is 39.9 m and the static head is 5 m. You add all these heads together to get 48.4 m and plot this against 150 m3/h as a point on your system curve because the pump would have to develop this total head.

If you were to put a pressure gauge near the start of the line you would read only 44.9 m head because the 3.5 m has already been used to accelerate the water. But you must add the 3.5 m in to get the head developed by the pump.
mechanicaldup (Mechanical) (OP)
1 Jul 05 10:00
Thks katmar
in other words, the residual head = velocity pressure= exit losses=k.V^2/2.g where k=1

and to increase your residual head you have to decrease the nozzle diameter.
katmar (Chemical)
1 Jul 05 10:44
Yes and no. If the velocity in the pipe was higher then yes, the velocity head would be higher.

If you just put a smaller nozzle on the end it will consume head as any other fitting would. The extra friction head would decrease the flowrate, and the velocity and therefore decrease the velocity head.
mechanicaldup (Mechanical) (OP)
20 Jul 05 9:50
Am I correct when I say you have to add either the velocity head or the pipe exit losses to the total friction and not both?
katmar (Chemical)
20 Jul 05 13:12
Yes, you just add one of them if there is no pressure recovery at the exit.  If it is a free exit then the exit loss will be one velocity head.  If the exit is into a pressurized system (typically with gases) then the velocity head is recovered and there is no exit loss.
PZas (Mechanical)
21 Jul 05 15:59
Katmar, I may have misunderstood what you were saying about the nozzle decreasing the exit losses.

Just for clarification lets set the flow rate static.  Now if we decrease the exit area we will likely increase the friction head and the exit losses.  The friction head results from the fitting (nozzle) and the exit losses result from the larger velocity component of the dynamic pressure.  Is this correct?

Therefore if we are designing a system with a set flow rate decreasing the exit diameter will always result in a higher head load on the pump.  However an increase in the exit area will be dependent on the relationship between the exit loss decrease (lower velocity) and the increase in friction loss (fitting).

Correct?
mechanicaldup (Mechanical) (OP)
22 Jul 05 3:24
PZas

I understand it also in the way you have describe it above, in other terms the higher the outlet velocity, the higher the kinetic energy available and the more the pump head required.

Do you agree Katmar?
quark (Mechanical)
22 Jul 05 8:36
Katmar painfully and brilliantly explained a rather nontechnical term residual head. As far as I get it from his posts, like me, he doesn't believe in the term residual head.

When you are discharging the liquid into the atmosphere, against no back pressure, then you should calculate the dynamic losses at the required flowrate considering the nozzle characteristics. Then simply plot the system resistance curve by the parabola method originating from the intersection of flow and head axes.

When you are discharging the fluid into a pressurized chamber (for example, a chamber filled with water and you are discharging the nozzle from the bottom) then at any flowrate, you should provide this head into the pump curve. This is done by starting your system resistance curve at head equal to the pressure head of the chamber.

Regards,

katmar (Chemical)
26 Jul 05 16:17
MechanicalDup,

I'm afraid that I gave you bad advice in my earlier post. I misunderstood the term "residual head", which I have since discovered has a well-defined meaning in the sprinkler and fire fighting industries. These industries define the residual head as the pressure that a gauge placed immediately before the sprinkler would read if the rated flowrate was flowing. I suppose by "residual" they mean "how much head is left to drive flow through the nozzle after all other losses have been considered".

I previously gave the example of a fire fighting system with a flowrate of 150 m3/h through a pipe of 80 mm ID and length of 50 m and where the discharge point 5 m above the pump. Let us now modify this example by saying that there is a fire fighting nozzle on the end of the line and that this nozzle requires a "residual head" of 10 m to achieve this flowrate.

As before, the friction loss through the pipe (excluding the nozzle) is 39.9 m, the static head is 5 m and the velocity head is 3.5 m. Now, on top of this we need 10 m of residual head at the end of the pipe to drive the nozzle. So the head to be delivered by the pump is 39.9+5+3.5+10 = 58.4 m.

To get back to your original question of how does the residual head affect the system curve - the residual head is simply the way to define the head loss through the nozzle. So it is just part of the overall head loss and has to be added to all the other losses to get to the system curve.

Jammer Dup
Katmar

katmar (Chemical)
26 Jul 05 16:37
PZas,

I'm afraid we are not getting on the same wavelength here. In my realier post I said "If you just put a smaller nozzle on the end it will consume head as any other fitting would. The extra friction head would decrease the flowrate, and the velocity and therefore decrease the velocity head."

I should have been more specific. When I said the velocity would decrease I meant that because the flowrate had decreased the velocity at the *start* of the pipe would decrease. Chances are the velocity through the nozzle would increase if the orifice was smaller, but the pump can't "see" this. All the pump sees is the friction head of the line attached to it (of which the nozzle is a part), plus the work it does in accelerating the fluid into the line.

Try not to think of the exit loss as something stuck at the end of the pipe. Read again what I wrote earlier about the exit loss actually occuring at the start of the line where the fluid is initially accelerated (assuming the line is of constant diameter - lets not make it more complicated than it needs to be!).

If you look at the elements that make up the Bernoulli equation it may be clearer.

Katmar
mechanicaldup (Mechanical) (OP)
27 Jul 05 3:14
thks Katmar,

how do you propose to incorporate the residual head on the system curve for the pump to see it.
do you add it as constant or as a flow variable?
katmar (Chemical)
27 Jul 05 4:08
For sprinklers, the manufacturers usually give the flow characteristic as a "K" value to be used in the relationship
Q = K * sqrt(Residual Head)
If you are familiar with valve sizing, the K is the same concept as the Cv value of a valve.

In some catalogs it seems that rather than give a "K" value, a table or chart of varying RH vs flowrate is given.  But either way, as the flowrate varies the pressure drop (residual head) through the sprinkler will vary.

If you are drawing a system curve onto a pump curve the Residual Head will vary as the flowrate varies.  The value of K, as well as the applicable units for Q and RH will have to be obtained from the spec sheet for your sprinkler or nozzle.
mechanicaldup (Mechanical) (OP)
28 Jul 05 4:08
how is the residual head of a fire hydrant been desipated?
JohnGP (Mechanical)
1 Aug 05 0:51
mechanicalup,

For firewater systems the "residual head" is usually measured as part of a flow test to determine the available pressure at a particular point (possibly another hydrant) when the system is discharging at a measured rate.

If there is no hose connected to your hydrant, then the residual head will be dissipated in the form of the "exit loss" discussed by katmar above. If there is a hose and nozzle connected to the hydrant, then there will be a certain pressure drop across the valve - the magnitude of which will be dependent on the flow rate and flow characteristic of the valve, and so the residual head before the hydrant will be dissipated in the form of frictional head losses in the hydrant/hose/nozzle system along with the nozzle exit loss.

Residual head is usually stated as a minimum pressure requirement at a hydrant (or other device) for satisfactory performance, under system flowing conditions, and so should be used in the calculation to determine the minimum head to be generated by the pump to ensure that the required residual head was available at that point.

Katmar used an example above to basically explain this more simply than I have, so I hope that I haven't added to your confusion.

Regards,
John

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