The last topic I want to cover on flow in pipes is flow in pipe networks. This is a common problem that arises especially in civil engineering where we have multiple pipes which are connected and possibly multiple fittings such as accoutrements, valves, etc. Reservoirs, pumps, inflows and outflows, etc. How do we go about analyzing complex systems like this? Well, a first one is to reduce it to its essential elements. And here I've got a simple example of a number of pipes which are connected with nodes in between them such as here, and at various points, inflows and outflows to some of these nodes. So to analyze a system like this, the first step is to label each node point. And here I've arbitrarily labelled them A, B, C, D, E, F. Next we label the inflows and outflows so for example I have an inflow here which I'll call QA at A, an outflow at B, QB, etc. Next I'm going to label each pipe and I'll just arbitrarily number them as shown, 1, 2, 3, 4, 5, etc. The next step is to choose and label all of the internal loops. So here, I've just got two internal loops, I and II, and also, you see, I've put an arrow on here which I'm going to define as the positive area where I move around the loop in a minute. The next step, we have to guess at the direction of the flow in each pipe. And initially, we don't know what these directions are, so we just guess them. And these are my guesses for this particular system. So that defines our system and the basic equations we have are continuity and Bernoulli. And the first one, continuity, says that the net flow, at any junction, is zero. In other words, the sum of the flows which are approaching this node or junction must be equal to the rate at which flow is moving away from that point. There's no flow or water accumulating or depleting at that point. The next one is that al I go around any loop, for example, loop I here starting at A and I go all the way around this loop and back here, then the head at the end of that loop must be equal of course to the head at the beginning of it. In other words the net headloss around the closed loop is 0. So I stop then, I analyze each loop separately. So first of all, for loop I, I start at node A and my continuity equation is that the sum of the flows at that point is 0. Now, also I have to assume a sign convention. I'll assume that a flow away from a node is positive and towards a node is negative. So in this case, we have QA coming in here, so this is negative, but the flows in pipes one and four are away from that node, so they are both positive, +Q1 +Q4. Next, I move to node B and I get the same equation and then C and D, etc. 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. The flow rates and diameters are given in that little table there, so the flow rate in Pipe ABD is twice the flow rate in ACD. The diameters are the same and the lengths are the same. Which of the following statements is true? Pressure drop in ACD is 4 psi, etc. So, here are the calculations. The pressure drop in ABD across here is equal to 60- 54 = 6 pounds per square inch. The pressure drop in ACD, and we only have one value of the pressure at the node point so the pressure drop in ACD is exactly the same. It's 60- 54 = 6 psi, so the answer is C. A similar problem, we have water flowing in a branched network, parallel pipe. Pressure just upstream of A is 60 etc. The flow rates, diameters, and lengths of the two branches are as follows and now we're given the friction factor f1 and f2 in the two pipes where the pipe ABD is labelled pipe 1 and ACD I'll label as pipe 2. So which of the following statements is true? Well, our basic equation is that the headloss in both of those pipes is equal. So, the head loss in pipe 1 is f1 L1 over D1, etc. Head loss in pipe 2 is f2 L2 over D2, etc. The velocity in pipe one is the volumetric flow rate divided by the cross-sectional area. The volumetric flow rate is 2Q, and the area, I'll call A. Similarly, the velocity in pipe 2 is volume flow rate over A. Volume flow rate is Q over A. And because the diameters of both of these pipes are the same, the area is the same. So now substituting into this equation, I get the headloss f1 L over D, and L and D are the same, times V squared is 2Q over A squared is equal to f2 L over D Q over A squared. So almost everything cancels here. L over D cancels, L over D cancels, Q over A cancels, Q over A cancels, 2g cancels. So the answer is, 4 f1 = f2 and the correct answer is B. This finishes our discussion of pipes and pipe networks.
