Pressure in a parallel pipe system

Yes, i think you are right .

Wrong. The pressure downstream is dependent on the flow rate and the pressure gradient.

With no flow the pressure will be equal in all places in a set of horizontal pipes, split or not splir.

Let flow commence and the pressure will drop as you go with the direction of flow. With large tubes and low viscosity fluids the drop will be small. With small tubes it will be greater. Take a garden hose and and make a Y connection. put 50 feet of hose on each arm of the Y and open the tap.

Flows will be equal. Change one hose for a 100 foot hose and the flows will not be equal

You say that I am wrong.

The system I am working with has minimal frictional losses due to material and the length of pipe (actually rubber tubing.) Tell me... does the pressure drop when neglecting losses from P1 to P2 or P3 when all three sections have the same cross section area? If there are no losses, and the system is on a horrizontal plane, I cannot see why the pressure would differ from 1 to 2 to 3.

Please elaborate focusing on pressure instead of flow rate.

i think that the pressure relative with the crosssection area that's why change in flow with make change in pressure

P=F/A

Q=AV

P=F.V/Q

any change in flow will be effect in pressure

Yes You are correct.

If the two parallel paths are similar, flow rates also will be 50% in each path.

all cross sections are the same

please modify response to account for the pressures

Of course you are correct.Pressure never drop aside from losses as you said as areas are the same.

If areas change, pressure will change too.

Fluid in motion in a pipe will have a higher pressure at the beginning than at the end. A longer, duplicate diameter pipe will have less presssure and flow at the end than the shorter one. Equal pressure at the end of each pipe and each pipe being the same diameter will be equal flow. This is true with Electrical Transmission lines too.

Tell your boss that the electrical equivalents for volume flow and pressure are electric current (amps) and electric potential (volts). The electrical equivalent for resistance (ohms) are the losses due to friction, sudden changes in direction, and changes in pipe cross section. In a simple DC electrical parallel resistance network, the total current in the network is equal to the sum of the branch currents. The steady state voltage at the input of the network is constant at some value as is the steady state voltage at the output of the network. The voltages at either end of the network are not equal, differing by the voltage drop across the resistance network.

I took the time to describe the equivalent DC electrical network because the steady state flow in a simple parallel piping network behaves in EXACTLY the same way although the calculation of the flow resistance is a bit more involved. This is because while electricity is electricity, not all fluids are created equal and the differences must be incorporated in the analysis.

Good Luck.

Need to bear in mind that for most practical purposes current flow is proportional to applied voltage difference, but liquid flow varies as √(pressure difference)

You're right.

Thanks, "Codey"

If the areas of each of the two output pipes are the same as the sourcing pipe, the velocity will clearly halve. According to Bernoulli, this will result in an increase in the pressure following the split (ignoring losses, of course). So P2 = P3 but both P2, P3 > P1! This shows that the electrical analogy is at best dubious.

Incidentally, the dimensions of Volumetric flow rate (L^4/T) and Pressure (M/L/T^2) are completely different. If your boss has any engineering input at all, and still thinks they are the same after considering this, you might do well to consider looking for another position.

I think some may have misunderstood part of the initial conditions/claims here...the area of EACH of the output pipes is not the same as the source (i.e. he's not splitting a 1/2"id tube into two 1/2"id tubes), so the flow area does NOT double, and therefore the velocity does not drop in half.

He's saying "what if" you split the flow of the source tube into a pair of tubes selected such that the total flow area remains the same (not possible with standard size tubing or pipes, but this is more of a theoretical question, after all). In this case, the velocity stays exactly the same, half the flow goes into each side, and the rest of your analysis, applied to this case, results in exactly the same pressure...IF you accept the theoretical condition stated that requires a system of zero friction losses (which of course are always present in a real system, and which are more significant with smaller tubing sizes or more viscous fluids, or greater changes in direction of flow, which create turbulence and areas of non-laminar flow).

One way to visualize this, if you don't see it already, is to think of sliding an infinitely thin divider up into a pipe so that exactly half of the flow goes down each side of the divider. Does the flow or pressure change? No. It's "the same" pipe, as far as the flow knows, it's just going down two channels. Now if you tried this in practice, with a divider of actual thickness and physical properties such as surface friction, you've just increased the surface area seen by the flow, so there will be a friction loss to overcome, meaning an increase in the initial pressure required to keep the flow constant at the end of the pipe.

This shows, by the way, that the electrical analogy is valid, for the limited theoretical conditions stated. Pressure overcomes resistance, and flow is how much of the medium can get through at those conditions. The units of measure noted should immediately show your compadre the qualitative difference between volumetric flow rate and system pressure, as they are NOT measuring the "same" thing, but rather two characteristics with related parameters.

The changes from initial flow & pressure and resulting pressure/flow at the end of various configurations of piping, as described in various posts above, are all due to the impact of friction and other losses in a "real" system.
The material, size, and fluid of a specific viscosity must be taken into account if you wish to apply or calculate the results in an actual system.

perhaps I did not state clearly... D1=D2=D3 (diameters)

Thank you guest questioner for your clarification. My intent was to highlight one of several variables that could invalidate the electrical analogy. Codemaster's comment (based on turbulent flow) highlights another - and not many practical fluid distribution systems conform to the low-momentum constraint that is needed for the electrical analogy to work.

Reply to #8 - I always thought flow dimension was L³/T....

You are correct of course, thank you. It just shows that I can neither type nor proof-read. Unfortunately, the system no longer allows me to correct that entry.

Codemaster pointed out a mistake in my original message (thanks). Volumetric flow rate should have read L^3/T. Apologies

Inform your EE that the situation is just like Ohm's law. Pressure is the equivalent to voltage. Flow rate is equivalent to current. Branching into two seperate tubes is like two resistors in parallel, with the flow rate in each branch being dependant on the diameter of the branches (or the resistance). I frequently use hydraulics in teaching ohms law because it is something more tangible to the student.

Unfortunately, this is exactly what is not true for many practical situations, as the analogy breaks down unless both:
a) the material is incompressible and
b) the velocities are sufficiently slow that momentum is irrelevant (see contributions ##12 and 14 on turbulent flow and #8 on Bernoulli's principle).

Perhaps the problem that "guest" is having with his boss is that he was taught in this way by someone he respected, but who never got around to the necessary caveats.

Hi all,

You are point is absolutely correct if area of the parallel pipes connected are same, you will have same pressure in each line. Pascal's law could prove this.

Pressure will not necessarily be equal in the two branches. However, pressure drop across the two paths will be equal. So. If you want to determine how much flow will go through each pipe you can do this by rearranging the Bernoulli equation to solve for Q and setting Delta P to be equal. Because flow will take the path of least resistance, as Q increases in one pipe the pressure differential will altar and so more or less flow will travel down the other path. Assuming that everything is exactly identical the pressure drop in the two paths should be the same and flow will be split in half. This is highly unlikely because any small defect in the manufacture of the pipe, valves, welds, etc. will play an impact on the roughness of the pipe and thus change the pressure loss due to friction. The way that I size my piping systems I set up one equation for each section of piping to solve for Delta P. I do this in Excel and set it up so that all I have to do is change the value of Q and it will calculate a new Delta P. Then, say, I have 100 GPM of total flow. I split this among the two paths until the Delta P is equal. And ta-da, I now know exactly how much pressure drop and flow I have among each path. You could re-arrange and set it all up as one big equation. However, doing them each individually gives a little more control if you are designing a new system and want to modify the Le/D values for a particular path. (Le = Equiv. Length).

so in that parallel pipe system (no matter D2=D3 or D2 not equal with D3), can we say that:

Q1 = Q2 + Q3

P1.V1 = P2.V2 = P3.V3, is this rite ?

Please response in account for pressure.

thanks :)

The first equation I could understand and accept (with limitation that the fluid is incompressible)

But from where did you get the second equation

responding to the intial guest comment. i agree with you. if we disregard losses in a pipe as a result of fittings or frictional loss the pressure will remain the same.

Tell your EE boss that Pressure is equivalent to Voltage, Flow is equivalent to current and K (frictional resistance factor) is equivalent to R (electrical resistance).

Assuming that your boss is working with laminar flow, I think the problem is communications.

If the pipe splits into two sections that are both equal to the original, the pressure gradient will indeed approximately halve (if the flow is laminar). If the flow in all regions is fully turbulent, the pressure gradient will reduce a factor of four. Intermediate cases will give intermediate results