Determining exactly equations for piping system

[Luike]

Dear Coleagues:

It is my first time in this forum, I'm a begginer in this field but I need , due to job requirements, to clarify a doubt in order to establish equation for piping (with accessories as bends, elbows, etc).

In my proper special case, I've a piping just at pump discharge zone that goes up to 5 meters in order to avoid an obstacle (it must to pass through a plant roof) and after, it must goes down up to reach same height of outlet zone of pump, the rest of piping system has to be instaled at same heigh, it is to said with geometrical difference equals to cero.

How I must consider this hill and fall just at discharge point? If I apply bernoulli equation, the altitude difference (between hill and after, fall) will not appear, nevertheless, the pump must be able to generate the suitable Head. Please, could you help in this point advising me how I must to proceed?.

Thanks in advance and sorry for caused disturbs

Luike

 

[MortenA]

For priming the pipe you will need the "high" head - but during normal operation if suction and discharge is at the same level then there is no static head. Normally you could perhaps accept that the flow was lower during priming? Then maybe the pump would do anyway (depending on the max head of course.

Best regards

Morten

 

[pleckner]

You can only take credit for the gain in liquid head (high point to low point) if the pipe stays full. Otherwise you must indeed take the highest point as the full liquid head against the pump discharge. The pipe will stay full if there is a significant horizontal run, a control device or a natural pocket after the vertical drop.

 

[BRIS]

MortenA has given you the answer - the flow will siphon over the hump and your static head will be zero. You may need to release air from the top of the hump on starting the system.

 

[vpl]

On a tangent, I would recommend installing a high point vent valve in the piping. Depending on the service, it might be necessary to periodically vent air out of the line. Otherwise, the assumption that the static head of the pump will be zero will not hold true.

Patricia Lougheed

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[katmar]

The chances are good that Morten and BRIS are correct and the syphon effect will cancel the 5m head, but you should consider pleckner's note of caution. To be confident that the syphon will work you need to calculate the pressure in the pipe at the top of the down-leg and if it is less than the vapour pressure of the fluid you will not get the full syphon credit. As pleckner pointed out, if the pipe is long the pressure at this point is likely to be high enough but you should calculate it to be sure.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Luike]

Thanks for your promptly answer at all:

Probably, I must explain better the situation, we are talking about a piping sistem for to distribute R-134a refrigerant from a storage tank until the distribution points that are distant to about 50 ms.

The tank has a pipe that goes until the pump, this one impels refrigerant until a height of 5 meters, just avoiding a roof of an industrial ship and returning to descend 5 meters such, from here to the distribution points piping will be installed on same horizontal plane.

I understand that, if the pipe remains full, there would not be problem since the pump simply would impel the liquid in the horizontal and the head delta z would be equal to zero.

But it is not recommendable to place a vent valve since there are severe environmental restrictions with respect to coolants of this type.

So, if I assume that I have this situation, for to perform hydraulic calculations (determining pipe system equation) I must consider a delta z = 5 meters?

Thanks for your best efforts

Luike

 

[Artisi]

You static dischard head will be + 5 metres only if the pipe doesn't run full, if it remains full the 5 meters will be cancelled. However, you will might require some form of pressure control to keep the pump at the required flow and so it doesn't run way out on the curve resulting from the low head if there is a syphon effect.

 

[BigInch]

The lowest pressure you can have at the high point (or any point) is the vapor pressure of the fluid.

http://virtualpipeline.spaces.msn.com

 

[REFINERYPROJECTS]

You need develop your piping system curve against your pump curve to find the best efficiency point for you pump.

The pressure at any point in a liquid can be thought of as being caused by a vertical column of the liquid which, due to its weight, exerts a pressure equal to the pressure at the point in question. The height of this column is called the static head and is expressed in terms of feet of liquid.

The static head corresponding to any specific pressure is dependent upon the weight of the liquid according to the following formula.

HEAD IN FEET= PRESSURE IN PSI X2.31/SPECIFIC GRAFITY

Centrifugal pump imparts velocity to liquid. This velocity energy is transformed into pressure energy as the liquid leaves the pump, the head developed is equal to the velocity energy at the impeller This relationship is expressed by the following formula:

H=V SQUARE /2g

H = Total head developed in feet.
v = Velocity at periphery of impeller in feet per sec.
g = 32.2 Feet/Sec2

You need to calculate the following
Total Static Head, Friction Head (hf)and Velocity Head (hv),

You need to show all different levels of your piping system, preferred to be in isometric drawing to have a clear picture about you piping system behavior.

ZUHAIR ABOL-OLA

 

[zdas04]

See thread124-177112 . In that discussion someone contributed the equation 10.2*ID^2.5 as the minimum flow rate (in gpm) required to keep a pipe full. I believe this empirical equaiton was developed for water, so you probably need to divide the constant by the SG of your fluid, but I'm not certain.

If your flow rate is greater than that, and the pressure at the top is more than the vapor pressure then the static head is zero and you only have to account for friction losses. If either is not true then you have to account for the pump head to get out of the building.

David