 Hi friends, my name is Meirubalai, I am assistant professor in WIT, Swalapur. Today we are going to discuss a topic that is quantity balance method in this video. So what are the learning outcomes? Student will be able to understand the quantity balance method for the analysis of pressure in the distribution system. See, what is basically quantity balance method? It is generally known as Hardy-Cross method. This method is applicable to the system in which pipes from the closed loops, the outflows from the system are generally assumed to occur at node junctions. So next point is for the given pipe system with the known outflows. It means we know the discharge. The Hardy-Cross method is an iterative procedure. It means it is a trial and error method which is based on initially iterated flows in the pipe. It means we are assuming the known outflows or the discharges. We are assuming the discharges for the start of this procedure. So there are two principles for the studying of this Hardy-Cross method. The first one is continuity equation. What does it mean? The flow entering in a junction or network must be equal to the flow leaving the same. Here it means that at this junction, we had seen the word the junctions or the node in the previous slide. Here let us see a junction A. At this junction A, the QA is the flow which is coming into the loop and QB and QC are the flow which is coming out from this node in the loop. So what I can write? At this junction, the inlet flow that is our QA is equal to QB plus QC. So what is the second equation? It is energy equation. What does it mean? The algebraic sum of the pressure drops or head loss around a closed loop must be zero. What does it mean? Here I am writing closed loop. Fine. Therefore, there can be no discontinuity in the pressure. What is discontinuity? There are two words which have to understand very much neatly. Let's see the first one, closed loop. If you see in this diagram, this is our closed loop. This is open loop. This one is the open loop. This middle one is our closed loop. This quantity balance method is only study for this closed loop. Fine. We don't study for this loop. We don't study for this loop because they are open loop. We study only for this closed loop. Fine. Now you are understanding the meaning of closed loop, right? Let's see about the don't know discontinuity meaning. Let's see the meaning of the summation of head loss around the loop is zero. What does it mean? It means that in the closed loop, there are individual pressure drops, okay? In these four pipes, there can be individual pressure drops. But when we took the summation of all these four pipes, it must be coming to the zero. What we can say in the other words, the head loss in this loop is always zero. So there should be no discontinuity in the flow of order or what we can say the discontinuity in the pressure. Fine. So what we assume in the quantity balance method. See, this is the general formula on which the quantity balance method is best. Here, the head loss that is HL is always equal to K into Q raised to N, okay? The K is constant here and Q is discharge for us. Then we calculate the value of N, okay? N is also a parameter or it is numerical parameter for the quantity balance method. So by two formulas for the calculating of discharge, we use Darcy's Wage-Wage Equation and Hayes and Williams formula, okay? So when we convert Darcy's Wage-Wage Equation into this general quantity balance method, we will get, okay, N is equal to 2 and K is equal to 8 FL divided by G means gravity force multiplied by passcode multiplied by small d raised to 5, okay? Here small d is diameter of pipe, okay? Similarly for Hayes and Williams formula also, when we convert the discharge formula of Hayes and Williams into this general formula, we will get the value of K and N as N is equal to 1.85 and K is equal to 1.68 into L, L is the length of pipe divided by C which is coefficient drag raised to 1.85 multiplied by small d raised to 4.87. Here small d is diameter of pipe, okay? So let's see how we can derive the equation for quantity balance method. So in the closed loop, we are assuming, okay, the flow rate, okay? We are starting with the assumption of flow rate and this is the actual flow which is going to be happen in the practice, okay? So there has to be some error, okay, which we have to calculate and we have to minimize that error so that we can get a proper designed flow rate which must be equal to the actual flow rate, okay? And for this calculation of error, we have to convert into HL is equal to K into Q raised to N, fine? Now we have a head loss in the pipe. This Q is the actual pipe, we are putting Q is equal to QA plus error, okay, in this general equation. What we are getting? HL is equal to K which is constant multiplied by QA plus delta or error, I can say delta raised to N, fine? Now this is in the form of A plus B raised to N. When you derive A plus B raised to N, you are getting the formula of A raised to N plus N into A raised to N minus 1B, okay? It is a continuity equation, okay, fine? If you see, this is the equation what we are getting. Here you can say QA raised to N. If you see, we had assumed the flow rate which is very higher in the rate and if you see the N value, N is mostly 1.85 for Heisen-Williams and Darcy's it is 2. So it is going to increase. If you see here this part, N is a integer, positive integer, but QA N minus 1 is also a greater number, relatively greater number. But if you see here, it is a very, error is a very lesser number, the delta is very in the lesser number. But it is still significant. But when you are coming to the third part, here you are squaring the delta. It means it is giving out a very, very lesser number, okay? When you multiply it by this number, you are getting a very lesser number. It is mostly coming to be 0 and after that the delta is going to increase, right? Here it is delta raised to N, here is delta raised to 2, another one is delta raised to 3. So automatically, the further parts are going to be valued 0. So we remove these parts and what we can write, HL is equal to K in the bracket QA raised to N plus N QA raised to N minus 1 delta, fine? Now we use the second assumption that is our energy equation. What is the second assumption? The summation of HL in the loop, in the closed loop is always 0. In the previous equation, what we can write, HL is 0. HL is 0 means the summation of K in bracket QA raised to N plus N QA raised to N minus 1 delta is equal to 0. When you shift this part here, you are putting it here negative and this equation will form. Here you have to understand here that delta is always same for all pipes. In the closed loop, this is a error which is going to happen in all the pipes, okay? So for that, you can't put for the summation because it is going to be same, fine? It is not coming under the summation. For the next successive pipe, it is not going to be double or it is not going to increase. So it is same, it is like a constant. So we are taking the delta as a constant outside this equation and we can rewrite it as delta is equal to summation of K Q raised to QA raised to N divided by summation of K dot N Q raised to N minus 1, fine? Now here you see, since the delta is the given same sign, so delta must be having the same sign value, positive or negative in the loops. So the denominator of this equation must be taken as an absolute sum. I am taking here the absolute value of the all sum or individual items in it, okay, fine? So how I can rewrite that here as we know the general equation of quantity balance method as K Q raised to N is our HL. So this is our HL, fine? And similarly, how we can rewrite this, we can rewrite is that N summation of HL divided by QA because this is also coming out to be our K value, fine? K into QA raised to N minus 1, we can take QA at the bottom and we can write QA raised to N. So K QA raised to N will form HL and the downward part that is, this is our QA, fine? Now let's see some review questions. Hardy-Cross method is not a controlled trial and error method, true or false? In a Hardy-Cross method, flow through pipe is assumed in such a way that the principle of continuity is satisfied, principle of energy is satisfied, both A and B are none of the above. According to Hardy-Cross method, summation of head loss in a loop is always 0, less than 0, more than 0, or none of the above. So let's see their answers. First is Hardy-Cross method is always trial and error method, so it is false. In a Hardy-Cross method, flow through pipe is assumed in such a way that the principle of continuity is satisfied. According to the Hardy-Cross method, summation of head losses in the loop is always 0, fine? This is one assignment, okay? This is the question where you have to do, you have to do only two trials, fine? It is a closed loop, if you see, if you can see, there are two loops in this pipe network. This is one loop one and this is loop two and these are the discharge values and for every pipe the values of K is given and here you are using a Darcy's Wage-Wage formula, that's why the N is written to be 2, fine? So you have to solve this assignment, okay, after this video and check whether you can, coming out to be right or not or what are the problems occurring, please share with me. So these are the references I have studied, thank you.