Practice Problem - Head Loss Due to Flow

All right, hello everybody. So today, let's look at a head loss question. Here's what it says. This head loss for six meter, so the length is given, a six meter pipe with the property and dimensions given in the figure below. That's my question. I'm trying to find the head loss. All right, the pressure are given in each section, and you can see here that it just leaves to the atmosphere with gauge pressure is zero, and at this particular cross section, my pressure is 20 kilopascals. Now, in this kind of question, we have two options. Okay, I want to talk about these options so you can judge which one to use. And this question is using one of the approaches. The next question I'm going to solve will use the other approach. Okay, the first approach is using the extended Bernoulli's equation, and this equation is given in the reference manual to you. Okay, and the second approach is, basically, there's a definition for h f, which is the head loss. It's going to be f L over D V square over 2g. Okay, so in order to differentiate which one I should be using in this equation or this equation, will be determined by looking at these parameters, whether I can I can obtain them. One, this f is a function of the Reynolds number and epsilon over D. Epsilon over D is only for turbulent, so f is a function of Reynolds number and epsilon over D. So, Reynolds number. Reynolds number is rho V D over mu. Okay, density and viscosity, I should be able to get that if the particular fluid is given to me. But this V is not given, right, in this question. D is not given, so I will not be able to use the second approach here. This approach is going to fail. So then, I better use the first approach. Okay, so that is my approach for these kind of questions, to understand what am I going to use. Okay, so after obtaining this, actually the rest is not that difficult. So we're just going to use the extended Bernoulli's equation. Let's write this. P1 over specific weight plus V 1 square over 2g plus z1, and there's an h p over here. The manual doesn't say this is the input power, but zero in this particular case. It's simply a pipe, right. V2 square over 2g plus z2 plus h f, and the question is asking you on this. Right, so let's go back over here. What is V1? Let's start by that. I typically start by that. What is V1? What is V2? See that I was not given enough information. I don't know V1. I don't know V2. But here's why that what I do know. From the continuity equation. V1 A1 will be equal to V2 A2, correct. And did you realize that this is a constant diameter pipe? So then, these A1 and A2 will be the same. So from here, I will get V1 is equal to V2, right. So then the terms will drop out. Okay, if I go ahead and find a datum like this, right, so that’s that is aligned with the center of the cross section number two, and if this this becomes a datum, then what will happen is the z2 will be zero, right. So that will be gone. And what will be z1 while we are there? The length of the pipe is given as 6 meters, so this is going to be 6 times sine of 30. Right, so this will be 6 times sine of 30, which is 3. Right, sine of 30 is one half, so this is 3. That's the number. So then I should be good to go. So let's look at the P1. P1 is 20 kilopascal. That's given to me. So 20,000. Don't forget to convert to pascals. Specific weight is, let's use units as well. Pascal, and this is 9810 newton per meter cube. That is given to me in the reference manual. Plus 6 times sine 30, which is actually 3, right, will be equal to, what is P2? We said that it has been exposed to atmosphere at the exit, so that is gone. The only thing left on right-hand side is h f. So then, you can see here, we don't need, actually, we don't really need a calculator for this question. You can see this is, you know, think of this like close to ten thousand. So it's like two plus three. So that’s the answer is D, right, for this particular question. Thank you for watching this video.