Parallel branch Head loss calculation 1

[Bigmac7]

Hi

I am busy with a pumping sizing exercise for a Sugar mill. A pump needs to pump liquid of density 1030kg/m3 at flow rate of 584m3/h through a parallel network of heaters and then into a split branch discharging into two flash tanks at 14m static level.

The system pumps from a 100m3 tank with a 2m static head above pump centreline. It delivers through 14" delivery pipe over 115m length before it splits into two 10" parallel branches. (See attached diagram)

Each branch has 3 heaters in line with linking 10" pipes of various lengths. The pressure drop over the heaters is 27.8kPa, 123kpa; &58kPa, respectively in each branch. The highest branch is 18m above pump centreline.

Assuming the the flow in each branch is equal the total Head loss in branch A = Ha and in branch B = Hb. Please can someone confirm that the equivalent headloss (He) can be determined from the following equation:

He = (Ha x Hb)/(Ha + Hb). i.e.similar toequivalent resistance in parallel resistor network calculation

The Headloss H = Hf +Hv. I have determined Hf from the Darcy Equation Hf = (4fL x v^2)/(2gd); Hv = K x v^2/2g
Where K = fitting losses. Both the suction and discharge sides are at atmospheric pressure.

 

[BigInch]

1st problem is that you mention the highest branch. If one branch is higher than the other, the flow will not be the same.

Assuming there is no real appreciable difference in elevations between branches, according to your diagram which appears symetric, the flow in each branch will be the same. Now that you know that, you don't really need to know anything about "equivalent head loss", of which I know of no such thing anyway, because you already know the flow in each branch and with that information, you can calculate the head loss from beginning to end of each branch right now, without any additional information.

If you are really asking about theoretically replacing both branches with ONE "equivalent pipe" that has the same flow as the sum of the flows in each branch and the head loss equal to the head loss of the 2-branch assembly, then ...

The head at the beginning of each branch (H1) is equal and the head at the end of each branch (H2) is equal, the head loss of EACH branch is H1-H2, so I would say you already have the "equivalent head", if there is such a thing, because they are both "equivalent" now. If you know the head loss in each branch is equal, and you have the same pipes and equipment in each branch, you get the same flow in each branch. If you have pipes and equipment with different hydraulic characteristics, you will get different flows in each branch, although the head loss from the beginning to the end of each branch are ALWAYS equal, no matter what equipment you have in each branch, different hydraulic characteristics ... or not.

Now since Ha and Hb (from your equation) are always equal, He = (Ha x Hb)/(Ha + Hb)
is the same as writing, He = (N*N)/(N+N)
which equals 1/2. So yes, if there is such a thing as equivalent head, its always equal to 1/2, whatever that means?


From the most general type of flow equation HL = Q x K^n
head loss, flow and K gives the proportionality of HL to Q
for each branch, and "n" is some exponent, you can see that by solving for the flow, Q, in each branch, and knowing the head lost in each branch is equal, each flow would proportion to each branch by the ratio of ,

HL = Q1 * K1^n1 and HL = Q2 * K2^n2
Q1 * K1^n1 = Q2 * K2^n2

Q1/Q2 = K2^n2 / K1^n1

In your branch system, K1 and K2 and n1 and n2 must be equal, so flow in each branch simply must be equal.

Hydraulics and Electricity can be thought of as equivalent, but you must be careful. In hydraulic to electrical relationships it is usually thought that flow = current as Amps, head = potential as Voltage and that flow can be proportioned by a relationship similar to electricity of,
Amps = Voltage/Ohms, but you note that this is linear. Hydraulic relationship between flow, potential and resistance is not linear, so thats similar to electrical impedence.

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[Bigmac7]

Thanks Big inch

Just to clarify is the following correct in determing the Total dynamic head (TDH) to size the pump for the system.

If the branches hydraulic characteristics differ, I will use that branch with the greater head loss(Ha or Hb), to determine overall discharge head(Hd).

Then Hd = Hl + Ha (or Hb);
where Hl = head loss in single lines

If the Suction head is Hs.

Then the TDH = Hd - Hs.

Thanks again
Bigmac7

 

[BigInch]

No.. no.. no... no... no.

The branches have the same head loss, there is not one branch with a greater head loss than the other. The flows in each branch proportion themselves to MAKE the head losses in each branch ALWAYS EQUAL. Flows can vary, if the system is not symetric to make it so. Since your system IS symetric, flows are equal. Heads losses in any branchs that connect together at both their inlet and their outlet points are ALWAYS EQUAL.

A pump sees only one or the other head loss. Like two sets of stairs to the same floor, the pump doesn't care if one stairway has three ramps and 50 little steps and the other stairway has 2 ramps and 37 steps. All it cares about is having the capability to go up to the next floor. It will pump water up one stair, the other stair, or both. It doesn't care what the water does when it leaves, or what route it takes, as long as it leaves with enough energy to make it to the next floor, the pump's job is done.

You should spec a pump with,

HEAD = HL_pump to tee + HL_1 ONLY for one_of the 2 branches + HL_pipe from 1st equipment group to 2nd equipment group + HL_2 the head loss in ONLY ONE of the two branches in equipment group 2 + greatest head loss in either of the two last outlet pipes.

and the system flowrate,

FLOWRATE Q = Q_BRANCH_1+ Q_BRANCH_2

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"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[HEC]

Just to stick my paddle in! The head loss down each branch is identical, I like to use a hydraulic gradient to visualise this. The flow down each branch will change such that the pressure loss is the same. As head loss varies with the square of flow (velocity) I doubt the electrical analogy of two resistors in parallel is applicable in this case. The best solution is to plot the two head loss vs flow curves for each branch. From this plot for a given head loss the flow in each branch can be determined. As BigInch states the branches are symetrical so theoretically the fows will be symetrical. (rember the diference between theory and practice is generally more in practice than in theory!)

There is a potential problem with the proces here. You are pumping what I presume is a sugar solution through a heater. There is likely therfore to be some burn on or fouling occur in the heaters. This will cause resistance changes and therefore unbalanced flows through the heaters lower flow will lead to more burn on, more fouling less flow and so on! IMHO I would be adding flow measurement on each leg and flow balancing valves to maintain equal flow down each leg. This may appear at first lazy engineering, but will pay off in operation.


Mark Hutton

 

[Pholonz]

BigMac7

Parallel pipes are analogous to parallel circuits, but you have the concepts off a little.

The resistance to fluid flow acts like the resistors in parallel circuits and follows the equation: 1/R(total)= 1/R1+ 1/R2 + 1/R3 + ..... If you were to make the parallel circuits/branches into one equivalent curcuit/branch.

Head Loss, however, is not the resistance. Head Loss is like the Voltage drop in ciriuts. It is the same for each branch which is the same as the drop from before to after all the branches.

The total head loss/voltage drop for parallel branches/circuits is the same as each individual branch/circuit.

In fact, the head loss is the only thing that will remain equivalent for each of the parallel branches. The flow rate and resistance to flow can be different for each branch but the head loss will always be the same between them.

 

[df72]

If the two end branches were to be used at the same time, how would the flow be calculated for them? I understand the theory of parallel pipes when they split and then reconnect again but what if they do not reconnect? Thank you.

 

[BigInch]

You should start a new thread, but this should be a fast answer.

To calculate any flow in any pipeline you need two of the following 3 primary variables, flow, inlet pressure or outlet pressure. So, at each branch end, you must know or set the outlet pressure and calculate the flow. Lacking that, you must define the branch's flow and calculate the end pressure. The inlet pressure's at those branches would normally come from loop calculations. Google "Hardy Cross Method" to find some examples of how loop flows were calculated by hand in the old days, if you're interested.



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