VIP Group for the Project Course 👉 https://www.engehall.com.br/vpe-grupo... ⚡️ FREE CLASS: How to Do a Complete Electrical Installation from Scratch, Easily, Even if You Are Not a Professional Electrician: ➽ https://eletr.co/aula?utm_content=yt1... ------ 📣 FAAALA ENGEHALL COMMUNITY 📣 Join our community on WhatsApp to receive valuable tips and news in the electrical area, important information and much more. 😉 👉 https://eletr.co/fala-eng?utm_content= ------ Did you like this video? 😍 ✔️ Leave your comment and share with your friends ✔️ Follow our networks ➽ https://eletr.co/links - - - - - - - - - - - - - - - - - - - - - - - - - - - 👉 NR10 Course with 20% discount ➽ https://eletr.co/nr10?utm_content=yt1... ------ Hey my dears, how are you? So far, we've learned how to predict the load for lighting circuits and sockets, whether for common use or specific use, as was the case with the shower. We've also learned how to size cables and conduits, taking into account grouping, temperature, and occupancy rate factors. Anyway, this week is a ton of information, isn't it? And to end this week on a high note, we're going to learn how to size the general protection circuit breaker and we'll take the opportunity to talk about demand and voltage drops. Are you ready? So, let's get to class... Well, everyone, the first step to sizing the general protection of a QDC and, consequently, its feeder cables, is to survey the total installed power of all the circuits in the installation. To make it easier to understand, let's give an example: in a residence, after you do all the sizing as we taught before, your QDC will have 7 circuits. Adding up the power of all the circuits, we have a total installed power of 13,890 VA. And to know the current of a circuit we use the following equation: CURRENT = POWER VOLTAGE Therefore, the total current of this QDC would be 13,890VA / 220V = 63.13 A Therefore, using table 36 of NBR 5410 as a reference, we would have a general protection of 70A and feeder cables of 16mm² (this taking into account cables with PVC insulation and installation method B1). But we don't stop there, since we are talking about sizing a general circuit breaker we need to apply a new factor which is the demand factor. But what is this demand factor for, André? Well, we already said in a previous video that the demand factor is nothing more than a number, always smaller than 1, that the energy concessionaire adopts for some types of loads with the objective of assisting in the sizing of electrical installations, making the sizing more in line with the reality of the installation. How so, in line with reality? Well, we know that in a residential installation it is unlikely that we will use all the outlets at the same time, as well as some equipment with a “heavier” load, so to speak. So the demand factor is there only to adjust the load to this reality, that is, it is there to reduce this sum of loads a little, after all, as I said before, it is unlikely that everything will work together, right? And where do we find these factors, André? Each energy provider has a demand value. Let's use our local provider here in Belo Horizonte as a reference, in this case CEMIG. In table 11 of CEMIG's ND 5.1, it recommends a demand factor to be used only for lighting and outlets for each installed load range. In our example, we have 1 lighting circuit and 4 common-use outlet circuits. Adding up the powers, we have a total of 6,890 VA. Now with this value, we will consult table 11 of CEMIG's ND 5.1. Note that for this value we must multiply the demand value by 0.60. Therefore, our power demand will be 6,890VA x 0.60 = 4,134VA. The demand factor for lighting and sockets has been applied. Now it's time to apply it to the other circuits. The microwave and shower are left. Now let's consult table 14 of CEMIG's ND 5.1 to determine the demand of our heating appliances, in this case, the microwave oven and the shower. Adding the powers of these devices, we have a total of 7,000VA. Since there are 2 devices, we apply the demand factor of 0.92, giving us a power demand of 7,000 x 0.92 = 6,440VA.