Next, I apply Bernoulli equation or net headloss around the closed loop is

equal to 0, so, starting from A, I go from A to B.

And I'm going in the same direction of the flow so that is a positive headloss.

It's plus hl1.

From B to C, the headloss there is hl2 which is also positive,

because of my assumed direction.

However, the headloss from C to D is negative because I'm

assuming that the flow is opposite to the direction I'm going round, so it's minus.

Similarly, the headloss from D to A is negative.

Those are my equations for loop I, and

then I have to repeat the process for loop II and any other loops.

So the procedure is that first of all

you just make some guess of the flow in each pipe, Q1, Q2, Q3, etc.

Then adjust them to satisfy the continuity equations.

Then calculate the headlosses and, of course,

as you go around the loop the headloss won't be 0, first of all.

So then you have to iterate until all of the equations are simultaneously

satisfied, and this method is known as the Hardy Cross method.

And obviously this is only something generally speaking for

a complex system we can only do with computers.

And it's very unlikely that they would ask

a generalized computational problem like this in the FE exam.

But rather, they would probably ask some conceptual

question based on those equations rather than a numerical one.

More likely, the calculations will be in simple parallel pipes.

For example, this diagram here from the manual shows two pipes in parallel,

and the essential thing here, the only thing that matters,

is that the head at each node point here is the same.

In other words, the head loss in each parallel pipe is the same.

You don't add them together.

They're equal to each other.

The headloss in pipe A, here is equal to fA L

over Dv squared over 2g is equal to the headloss in the second pipe,

pipe B, and that's the equation they give here.

So for pipes in parallel, you don't add the head losses,

they're the same in each pipe.

The continuity equation, if I have Q entering here, then this splits into two.

So the flow rate in pipe A is QA, in pipe B is QB.

So the flow Q is equal to QA plus QB.

There's a very simple analogy to this with electrical networks.

Suppose that I have some current flowing through here and

a voltage drop across these two resistances, R1 and R2.

Then the current splits into I1 and I2, and you can just think of most

of the current as flowing down the pathway of least resistance or lower resistance.

The analogy to the headloss here is the voltage drop and

the analogy to the flow rate is the current in each segment.

So, similarly, for

our parallel pipe network here, most of the flow is going to go down

the path of least resistance which will probably be the largest diameter pipe.

So lets do an example on that.

We have water flowing in the branched parallel pipe network shown andwe're given

that the pressure just upstream of A here at this point is 60 psi and

the pressure downstream of D here is 54 pounds per square inch.