First, and for the late-comers, I think they can still join us when they are ready. Okay, cool. Thanks, everyone, for coming. This is the second session of our developer gym. Michael, by the way, is recording. Okay, now, cool. So for this session, we will continue from our last session where I give you an introduction of what TDD is.  I saw some of the new faces in this session. So I will start with some of the kind of recap of what the TDD is. But it would be very brief. And the majority of this afternoon, you will spend by doing more coding and practice the skill by yourself. And actually, sorry, not by yourself, with your friends. So what is the code cutter? Code cutter is basically the sessions where you practice your basic programming skills like writing a simple program, writing test cases, make it run faster, make it run better.  So just similar to those Kung Fu masters, right? They become mastered by practicing those basic moves again and again. Those may look very boring if you just look at it by yourself. But in the end, it's those basic exercises and practices that make them Kung Fu master when they fight. Today, let's do some TDD practice or the exercise on the TDD. Let's do a quick recap. What is TDD? By the way, just to have a show of hands, how many of you are totally new to TDD? You are OK. And assume the rest. OK. Cool. But I will still give you some quick introduction what it is. So TDD or test-driven developments is our approach. It's approach or it's a style for you to write your code. So typically, when you develop your program, you will just write the logic to solve your problem. And you may or may not write test cases or sometimes you write test cases after you write your program. But then TDD is a kind of approach which encourages you or forces you to write the test cases in advance even before you write the first line of the code to solve your problem. The benefit is this test cases will help you to guide your design of your solution. As we will see later today, you will do some exercise and you will find out that those test cases help you to get focused on the step that you are moving and to help you to design your solutions. And also, in the end, if you practice this, you will accumulate a set of test cases which can be a safety net. Later, when you make further modifications to the system, et cetera, those test cases would be a layer of production for you. So there are typically three steps. If you follow the TDD style in writing your code, the first step is to write a test case, which obviously would fail in the beginning because you haven't got any logic or real implementation to make it a pass. The second step, you need to write the code, the production code, which makes this test case pass. And the secret here is you should just make enough implementation to make it a pass, no more than that. So for, you know, you should not think about error scenarios that is not being tested yet. You should not think about what will happen tomorrow or the day after tomorrow. You should not think about too much about the future. Just focus on the problem or the test case that is failing right now and make just enough changes in your code to make the test case pass. And then the third step, which is also very important, to do the refactoring on your code base. Later, I have another slide to talk about what kind of a refactoring you need to do. There are only four simple rules you need to follow today. And then after you make the code cleaner, then you finish one cycle, and then you can continue with the next cycle. Writing the next test case, make it a pass, and then make your code better. So just keep doing this kind of cycle. Okay, in the last session, we gave you a few rules or tips you should follow. So as beginners, I would suggest you to treat this as rules and basically to force yourself to follow those basic rules. And once you become more proficient, you might want to move in bigger steps, or maybe you can skip some of the rules. But in the beginning, try to follow them. So the first rule, okay, this rule you can't really violate the rule. This is TDD, it requires you to write the test case first, right? Before you write any production code, the first thing you should do is write a test case, which will fail in the beginning. This is the rule number one. Rule number two, you should only have one failing test case at a time. So some of the people, they may have a habit of writing a few test cases, which are all failing, and then they started to implement the code or the logic. That is, I would say it's a bad habit. So for TDD, you should really just follow the habit of having one failure test case, and then move on to the implementation to make this test case pass, then you have your second failed test case. So do not have more than one failed test case at a time. Again, the reason is because each test case is supposed to be a guide, right? You follow this failed test case, which makes you focused to just think about how to make it pass. If you have more than one failed test cases, your mind will be more kind of bloated, right? You're thinking about, what do I need to do in order to make those two test case pass or 10 test case pass, then you couldn't really get focused. So in order for you to get focused, just stay with one failed test case each time. Okay, the number three, you should write small test. Small meaning I have slide basically to describe what could be a small test. So one way to describe the small test is each test should only focus on one particular behavior or feature or maybe one rule or one pattern in the problem that you're trying to solve. And in terms of the time taken to make the test case pass, it should be very fast. So I put the tip number here like 10 minutes. But up to you, if the problem is very simple, maybe you should even achieve even shorter time to make the test case pass. But in reality, if you're solving a very complicated problem, maybe you can allow yourself to take, let's say, 20 minutes or so to make the test case pass. But still, the guideline is if your test case is small enough, you should not spend a lot of time to make it a pass. And typically, each test case should only have one reason to fail. And in some cases, this basically means that each test case, you should only have one assertion. If you're familiar with the unit test or those libraries, right, you know that you can use expect or assert, whatever. So you should try to have one assertion per test case. Then basically that means your test case only have one reason to fail. Number four, you should make the baby steps, meaning that you should write the production code to make the field test case to pass, no more than that, which is the one I already mentioned this one to you just now. Rule number five, so when you have a choice, right, the test case failed and when you try to make this test case pass, you may have a few choices. And among those choices, I would recommend you to pick the simplest one. To start with the simplest one first, and when you add more and more test cases, naturally your solution or your algorithm will grow more and more complicated. So do not run into the most generic solution or most complicated algorithm right away. Try to follow the simplest solution you can have. And then generalize your solution when you add more and more test cases. Rule number six, do not forget about refactoring. So remember there are three steps in the TDD cycle. And once you make the test case pass, the next step is always do the refactoring. Do not skip right away to your next test case. Look at your code. Are there anything you can improve? Are there any duplications in your code that you can simplify? Or are you choosing a more complicated algorithm? Can you simplify it? So think about along those directions to make your code better. Also do not forget to refact your test case. So once you have more and more test cases, you should also look into the duplications in your test cases. Are there anything you can make your test case more concise or more easier to understand, et cetera? So the test case in the end is kind of a live documentation for the code. If you make this test case easy to understand, whenever the test case fail, we have a better or have an easier life to troubleshoot why it fails. If the test case itself is very hard to understand, then in the end it becomes pretty useless. Rule number eight. So try not to do the refactoring when your test cases are failing. So this is kind of a pitfall. When you have failed the test case, you're trying to make some changes in your code to make the test case pass. But then you notice that some part of your code is ugly. And you feel that, oh, I should make this path better. And then at that time tell yourself, this is not my focus right now. My focus is to do something to make the test case pass. Even if there's some problem in this piece of code here, I should not worry about it right now. I should probably note it down somewhere. And then later, once you make the test case pass, there is a dedicated slot called refactoring for you to do the improvement at that time. So try not to mix the refactoring together with the second step, the implementation part. Okay, now let's talk about how do you do refactoring. So the term refactoring is referring to the changes you make in your code to make it better without changing the behavior. So it sounds very abstract. But then in the end there are four simple rules you need to follow or you can follow to make your code better. The first rule is your code needs to pass all the test case. This is rule number one. All right? Second one is your code needs to be clear enough to express your intention. So basically in the layman's term, that basically means that if you ask someone to read your code, he should be able to understand your code maybe in a few seconds without spending minutes or even hours to read and understand or even ask you, why? Why do you write in this part? What does it mean? So if someone needs to ask you what it means, then it's not clear enough. So you should try to write your code to express your intention so that it's so clear that the code explains by itself and nobody needs to ask around why what's the reason for this piece of code. Then the third rule is there shouldn't be duplications in your code. Whenever you see the code, two pieces of code, they look very similar, then you should try to do something to protect a common function or common methods to make it more reusable and avoid duplications in your code. The fourth rule, which is called minimal. So this rule takes a bit of time to understand and practice. So to make it easy to start with, you should just ask yourself, given a line of code, if I delete this line, would my code still work? If yes, just delete it. Or from the method or function level, I have a function here. Do I still need this function?  And from the, even on the higher level, let's say in your code you have a few classes, or let's say you have a few interfaces you define for some of the classes. Ask yourself, if I delete that interface, would my code still work? And if it doesn't say yes, then you should delete that interface. So basically try to go minimal. Write as little code as you need in order to solve the problem. So if you follow this direction, your code overall will become more simpler. So just remember this for lose to rewrite the code. Later during the exercise today, we will be following the TDD cycle. And during the refactoring, you should follow this for lose. Whenever you tell yourself, I'm going to do the refactoring now, then you should check your code. Does it pass all the test case? Or does it express your intention? And you will probably pair with some of your friends. And then just ask your friend, is this code clear to you? If it's not clear, then you need to refactor it to make the intention clearer. And then check, are there any duplications? Even your test cases, are there any duplications in your test case? What can you do to remove those duplications? Then are there anything you can delete? So ask yourself those questions. So to just check if your code can be further simplified. Okay, that's a quick crash course for what is TDD. And your last session, we spent a lot of time on theory. And I heard some people complain that the pace was a bit slow. So today, let's do more hands-on exercise. So you will hopefully enjoy this TDD style and learn something useful out of this. Today, the exercise we are going to do is write a program to convert Arabic numbers into Roman numerals. How many of you have heard about this exercise before? Okay. Number one, this is actually a pretty simple problem. So let's basically talk about the Roman numerals first. So you've probably seen some Roman numerals numbers before, so do you know the rules that form those numbers? If not, let's start with those rules first. So your task today, firstly, is to understand the rules. How do you form Roman numerals numbers? And later, your task is to write your code to convert the integers, the Arabic numbers into this Roman numerals form. So let's start with the Roman numerals rules. There are letters, right? Those funny letters, they have some meanings, like i stands for one and v stands for five and x stands for 10, et cetera. So there are a bunch of letters and each of the letters corresponding to one value. You can start with this first, okay? And then once you have those letters, you can combine them to make bigger numbers. For example, if you want to say two, two basically means one plus one, right? And that means you have an i, then followed by another i. Put them together, that basically means two. And then three is also very simpler. Three basically means one plus one plus one. So you just need to have three i's, put them together, that is three. Okay, so far the rules are pretty simple, but then the next one would be a bit more complicated. So the number four, it shows as r, v, the y. The rule here is v stands for five, i stands for one. So whenever you see i, which is one, before v, which is five, the i is smaller than five. In this case, it's not a summation, it's actually a subtraction. Meaning that in order to interpret this value, it is actually the value on the right minus the values on the left, which is five minus one, which is four. So this is the rule. Somehow they define it in this way, don't ask me why. So they probably feel this is funny, or maybe this is more efficient, but they design it in this way. So four is equal to five minus one, so you just need to put one in front of five, and you have four. And six, how do you express six? Six, okay, back to the normal rule. Six equal to five plus one. And then basically you have a v followed by i. So you ask me, okay, i should also, six is also equal to three plus three. Can you just put six i's there? The answer is no. So there are certain rules saying that those values, like i, maybe x, each of them, they can only be repeated maximally three times. You cannot repeat them more than three times. That's why for four, you can't really just put i, i, i, i. That doesn't work. That violates the rule. So that's why you need to express four as five minus one. That's why you have that i, v representation. And then for seven, it becomes easier again. Seven basically means five plus two, right? You have v plus i plus i. And eight, again, is easier. So eight basically means five plus three. You already know how to represent five, and you also know how to represent three. So you just need to put them together. So basically in this Roman numeral rules, right, most of the time is just a kind of submission. You have a few letters. In order to find the total value, you just need to summarize the individual values from left to right one by one. The only exception you see here so far is for the situation like a four, where you have to put one before five. In this case, it's a subtraction instead of submission. So far, that's the only exception. Otherwise, the rule is pretty simple. But then, again, when you look at nine, this subtraction rule comes in again. So in order to represent nine, you can't see it's five plus four. You have to say nine is equal to ten minus one. Then in order to represent ten minus one, what do you do?   Then that becomes ten minus one, which is nine. And again, you have... Well, this is just rules that you find. So try to remember, try to follow. Once you get used to this kind of rule, it will be easier. So these are the basic rules. Actually, let me give you another more example. So how do you represent this 982 into Roman numeral? The logic is very simple. Firstly, you need to break this number into the digits. You need to break this into 982. Rather, if you think about digit, there's a one-digit non on the handle's place and another digit eight at the tens place. And the last digit is two at the ones place. So it depends on how you think about it. So you either think about this one as 900 plus 80 plus two, or you think about this one as nine at the hundreds place, eight at the tens place, and two at the singles place or ones place. So otherwise, once you break this down into three pieces, the task is just to convert each individual digit or number into the corresponding Roman numeral. And then for each of them, you just need to follow the rules that we shared just now. A 900 happens to translate to CM because 900 is equal to 1,000 minus 100. 1,000 is M. 100 is C. So you need to put C in front of M to follow this subtraction rule. That's why 900 is equal to CM. And 80, 80 is basically equal to 50 plus 30. Just now, I told you that 80 is equal to 5 plus 3. For 80, it's the same. 80 basically means that it's 50 plus 30. 50 is L, and 30 is basically 10 plus 10 plus 10, which is three triple X. That's why you come out with this rule, which is this one. Then two is very simple, just I. So now, once you have the representation for every single digit in this integer, you just need to concatenate them into one. And this is the final representation of this Roman numeral. So if someone asks you to give you an Arabic number in order to convert this integer into the Roman numeral, you just need to follow these two steps. First step, break it down into digits. Second step, for each digit, convert it into some kind of Roman numeral representation. Using either the summation rule or subtraction rule, I told you just now. And then you get the individual representation, and then you concatenate them. So that's the final representation. There are further rules, which are basically, I showed you some rules just now, like for X or I, basically for those values, represent 10, 100, 1,000. Those values, they cannot be repeated more than three times. That's one rule. Another rule saying that the I, they can only appear in front of V or X. So the letter I, they cannot appear in front of, let's see, 100. So that's another rule. If you like, you can take a screenshot of this slide. But then this rule is actually pretty complicated to understand and remember. I simplified it a bit. Okay, where are my simplified rules? Oops, I forgot to save it. Okay, let's go through these rules first. Maybe my rules are in the other places. Yeah, so the third rule, one says basically here says that whenever you put a smaller number in front of a bigger number, the rule is basically subtraction instead of summation. Also, there is no number zero in Roman numerals and there's no negative numbers. Actually, if you follow the rules we defined just now, actually there's also an upper bound, the maximum number that can be represented using Roman numerals, using those few letters, which is 3,999. So that is the maximum number if you follow the rules we described just now and using those few available characters. Yeah, if you like, take a screenshot. But otherwise, later you will do your Cata, but in the first round of Cata, actually these few rules are not that important. Otherwise, if you really need them, you can just search online, search on Google, just search Roman numerals. There are quite a few websites which basically tells you those rules. Later, if you need the rules, I can either share the slide on the screen or give you the link. Okay, so that's about the rules. And then, let's see, we're about 30 minutes into this talk. So let's start writing some code. But you would not write the code by yourself today. You're going to pair up with someone. Like two of you will be shared the same computer and the same keyboard. Basically, find your friend and then use one of your computers. And you will write your code in a ping-pong style, meaning that if I'm pairing with you, sorry, what's your name? Ping. Oh, sorry. Ping. Okay. Okay, I'm Pong. Okay. Okay, I'm Pong. So ping will write the test case, which fails. And then I'm going to write the code to make the test case pass. And then I'm going to do some refactoring to make the code cleaner. And then I'm going to write a new test case. So after I write this new test case, then ping is Princeton. She will make the test case pass, make the code cleaner, and then she will write the third test case. And then it's my turn to make the test case pass. So you select the table tennis or ping-pong where each of you just get a chance to do something. Typically you write a test case, make it a fail, and then it's your pair to her responsibility to make it pass. So you follow this style. And then you follow the TTT style as well. Basically that means you always write a failed test case to start with. And then we will do a 45 minutes per session, meaning that this afternoon we have about two hours, but we will not write the code all the way along for those two hours. We will break it into 45 minutes sessions, so we can possibly have maybe less two or three sessions. So after 45 minutes session, I will call a break. And let everyone stop. Then we can start to probably share what you were doing in the last 45 minutes, what's your learning, do you discover something new? You can share with your classmates. And then we will start again. But in the next session, you're going to delete all the code in the previous session. Start from scratch again. That's the basics of Cata. You can't reuse whatever you have in the last session. You start from scratch again. But then you will notice that you're getting faster and faster because whatever you did in the last session you still have the memory and you just need to redo those things. But then you notice that you will do it much faster. So this is why we call it Cata. You just do it again and again. Okay. So we will start the round number one now and you find your pair and then decide which computer to use, decide which language to use. Oh, you probably want to find out, so start with you probably want to find out someone who knows the same language as you. But if you would like to learn something new, you can also pair with someone that you use some language that you would like to learn. But otherwise, find your pair. Sorry, before you start, another just another reminder. Do not discuss a lot about how do you, your ideas, your solutions. Just quickly set up the project, write the first test case, and then start doing it. Do not put too much time on the design or thinking how to solve it. Let the test cases drive you. Shall we pause first? Okay, cool. So just now when you guys shared your solutions, I emphasize that you need to tell other people your test sequence and why do you choose that sequence. It's because in TDD, when you try to solve a problem, there could be different solutions to that problem. And when you write test cases, you are basically guiding yourself to generate a solution using the test case. So for different sequence of test cases, sometimes it guides you to generate different solutions or different answers. So for myself, when I did this exercise, the first time, the sequence I chose is I start from one, two, three, four, five, six, similar to one of the group that you were using. And then I generated one solution. Then the next time, I was trying, the second time I was trying, but then I adopted a different sequence or different solution. So the second time I was trying to, from one, two, three, four, until 10, that's okay, same as the first time I tried. But then instead of testing the scenario for 11 or 12, I jumped to testing the scenario for 20, 30. The reason is because the second time, probably because during the first time I got some insights into the problem, and then I did a bit of analysis. And I found that actually, there are some patterns in the Roman numeral rules. And for example, the rule for one, ten, one hundred, basically for the same rule, you can do the exact Roman numeral character for that rule, for that digit. And then the rule for two, twenty, two hundred, two hundred, in this case, two thousand, they're all very similar. You just repeat the one once. If you repeat the one twice, you get two. If you repeat ten twice, you get twenty. If you repeat one hundred twice, you get two hundred. So those few digits or numbers, they follow one rule. And similarly for four, forty, and four hundred, to me, they follow the same rule. Basically it's just a five minus one or fifteen minus ten or five hundred minus one hundred. Those few digits, to me, they follow the same rule. So when I look at those patterns, I feel there's this kind of a rule. And then how do I kind of write test cases to make those rules or patterns more obvious to me? That's why I wrote the sequence in this way. Purposefully I wrote test case for ten, then followed by twenty, followed by thirty. And then when I look at my code, I feel that, oh, there's a pattern. The logic is very similar to the way that I generate one, two, three. Between one, two, three, ten, twenty, thirty, one hundred, two hundred, three hundred. So I see a pattern there. So basically I did some analysis in the problem. I saw there's a pattern. I purposely changed the sequence of the test case to make those patterns more obvious to me. And naturally it leads my solution to a different structure. So basically for those two tries, I have different solutions for the same problem. And the second time is different just because I noticed the pattern and I changed the sequence of the test case to help me to make those patterns more obvious. So that's why I was talking with some of you. So when you write test cases for TDD, you probably need some kind of a bit of analysis up front and ask yourself, what is my first test case? What is my second test case? So how do I structure my test case to review the patterns that is in this problem? So that's a technique you can try when you solve your problem again. Okay, so let's take 10 minutes break and let's come back at 4, 10. And after that I would like you to switch your pairs. Do not work with your same pair again, switch your pairs, and start from the problem from scratch again. And then try a different approach if you like. Okay, so let's take a break.


Hello and a warm welcome on what is a glorious night for football, perfect playing conditions. Derek Ray here in the commentary position and I'm joined as always by Lee Dixon. And we've got Premier League action coming right up. It is Liverpool taking on Manchester United. Hi Derek, thank you. Well both managers will be reminding their players how important it is to start off the game really quickly. Really exactly opposition. Put them on the back foot, see what comes of that. Hopefully we get a good game. Now Lee knows he can do a lot better on the shooting front. Well how many times has he had that chance in training and scored on that occasion all over the place? So the starting line up for Liverpool. Sadio Mane starts with Mohamed Salah out wide and leading the attack today as the ultimate pressing machine, Roberto Firmino. This is the team Manchester United will go with. The formidable Davide De Gea starts in goal. Harry Maguire plays alongside Victor Lindelof in central defence. Fred starts alongside Nemanja Matich in the centre of midfield and a must tactical set up. They have just the one player in attack. Liverpool obsessed with finding a forward gear. What can they do? They might still be able to do a bit of damage. But he did what he had to do defensively. This might pay dividends. What a shocking pass really. How can you not be entertained by Mohamed Salah? A magician with the boldest feet. Lee, in what way will he contribute here? I kind of think he's pretty unique in world football right now because he's scoring all these goals from a wide position and that's his key. He comes in from that wide area in between defenders and his goal ratio second to none. I don't think the goalkeeper will be the slightest bit bothered by that effort. Well, he won't want to watch that on the replay. I tell you, he was way, way wide there. Juan Bissaca. Maguire giving the ball away. Promising looking ball. And he's found a bit of daylight. On what is an ideal night for football you've got to say. I'm Derek Ray here on the commentary box and I'm joined for commentary by Lee Dixon. And we've got Premier League action coming right up. It is Liverpool versus Manchester United. Thanks Derek. Good atmosphere here. Everybody looking forward to what should be a really entertaining game. And this is how Liverpool start the game. Andrew Robertson starts with Trent Alexander-Arnold in the full-back positions. Mohamed Salah plays with Sadio Mane out wide. You've got to expect something bad to happen. It's back off the keeper. Now it's in the back of the net. So the game has restarted. Liverpool in front. Lost possession by Naldum. Juan Bissaca. Really good reading of the game to win possession back. Here Salah doesn't keep the ball and it's gone out of play. Goal kick coming up. Waited pass. Power. He should have thought Composier.


Well then, let's go ahead and start. I'd like to convene this special meeting of the board of directors of San Lorenzo Valley Water District of March 10th, 2022. Holly, would you call the roll please? President Mayhood. Here. Director Ackman. Director Falls. Here. Director Smalley. Your muted mark. Here. Okay. Great. Thank you. All right. We now have a section of the meeting called oral communications. If there are any communications from the public on topics within the purview of the district, but not on the agenda tonight, now is your chance. If you have a comment you'd like to make, please raise your hand. And seeing none, hearing none, we'll go on to the next item. Which is the president's report, which I'll just say as way of introduction at the end of February, Director Henry sent an email resigning effective March 1st, stating her desire to make room for someone younger with time and energy to do the job and vote. I've appreciated the extensive coverage that our local newspapers have given to Lois and her activities and within the long involvement with water in San Lorenzo Valley. And so that brings us to our first item of new business, which is border director vacancy and what to do about it. Gina, would you like to give a thumbnail sketch on what the process is? I think most of us are pretty familiar with it given that we've had what now five of these in the last three years. And Rick, did you want to say anything before I dive into the report? I appreciate if you give the staff report. Thank you. Okay, thanks. All right, so as the chair may have pointed out, the district has been through a number of these processes of filling vacancies that have occurred for one reason or another on the board. And so I think most of you and as well as the attendees are pretty familiar with the process. Typically there's three options for filling a vacant board seat. One is for the board to make an appointment of a new board member to fill the seat for the balance of the unexpired term after advertising in the community for individuals who may want to apply for the seat. Another is to send the seat to election. This has to occur on one of the established election dates for the code, which there's only certain dates that can be used for this purpose. It's not every date that you may see for a primary and so forth. Unfortunately here, because of the timing, if the district were to call for an election, the election would be November 8th, 2022, which is the same date that the seat will be up for election regardless of an appointment. And then the third option is that the board can use neither appointment nor an election and then allow the County Board of Supervisors to make the decision instead how to fill the seat. So those are essentially the three options. But as I mentioned, because of the timing of this particular vacancy, if the district doesn't appoint somebody to fill the seat, it'll just remain vacant until the election when it's gonna be filled one way or the other regardless of what the board does. And staff's recommendation is that the board fill the seat by appointment until whoever, whatever candidate is elected in the November 2022 general election is qualified and takes office for the next four-year term. And the reason for that recommendation is that, at least from my perspective, it can be challenging to have a four-member board with the possibility of split votes and also the fewer board members you have, the more difficult it can be to get a quorum, which sometimes isn't an issue, but sometimes if there's urgent business, it could be a problem for the district's functioning. Okay. Rick, did you have anything you wanted to add to that? No, I heard with the report and I would also recommend that we move ahead and appoint for reasons that council outline. Okay. So in the staff recommendation, there is a motion and that motion also includes in addition to filling the seat by appointment, it provides a schedule by which we would do this so that the applicant will be, applicants will be interviewed at the first regular board meeting in April, which is scheduled for April 7th. And in attachment, what was it? I guess B, there's the schedule of when things will happen. In other words, it will be posted tomorrow and closing date for receipt of applications would be the 31st of March. And then we'd interview the applicants on April 7th with the idea that hopefully we could make a decision that evening. If not, we still have time given the state laws about timeframes to make it at a subsequent meeting. So let's go ahead and just start with you, Jamie. Did you have any comment on this? Well, thank you. I just first wanted to start by, I know that director Henry couldn't be here with us tonight, but I wanted to say for the sake of the record, how valuable her service was and what a warmth and real humanity she brought to all of our meetings and her focus always on trying to make the customer experience the best it could possibly be was really appreciated in values. So I will miss her contributions to the board, although I didn't get a chance to know her well personally. So I just wanted to say that for the record, but I would agree that there's just too many important decisions that we have to make to allow the seat to stay vacant, especially God forbid something happens to one of us. There is still COVID out there. And if somebody's not able to make a meeting for some reason, that makes it even more challenging. So I would support the motion to move forward with the appointment process. And just let me comment that we will have a formal resolution of recognition of Lois at the next regular meeting, okay? And she'll, you know, she can attend that if she'd like. Bob. I'm glad you brought that up, Gail, because I was gonna ask about exactly that. I think that it's important that we as an organization make sure that everybody's service to our community is honored and they receive a lot of thanks for doing so. It's a time-consuming position. Lois has certainly been doing it for a really long time, except for like a two-year break from 2016 to 2018. So, you know, that's definitely something that needs to be recognized in the community. You know, there was, I think, over the course of the last few years, a similar situation at the school board, where there were a very large number of resignations that happened outside of the election cycle. And I think this is one of the reasons why I really would like us to escalate our lobbying. I don't know if it'll do any good or not, particularly in today's day and age, and provide us with more flexibility around conducting elections for this sort of thing. I think when you start having a large number of people appointed to a board, it starts looking for better or worse, and whether it's reality or not, but perception being what it is, it starts looking a little clubby. And I think even the school district said when they reached the fourth resignation and were about to appoint a fourth board member, said enough, we can't do this anymore. And they actually waited several months in order to go to an election, as it turns out, only one person applied, so they didn't have to have an election. But still, the concept of actually filling these seats with an election is really, really important to make sure that the issues that are facing the board and the community, particularly critical issues around finance and infrastructure are fully vetted and discussed in public forums so that the people that are voting on their representatives can understand the policies and positions that people want to have. So I'm really disappointed that we're facing this again. I was sad to see the comment that Lois made in the press banner where I believe it was applied for the position and potentially run for the board in November. I really wish that Lois could have stayed with us all the way through to November, even if she wasn't going to run because I think that would have been the best way to honor the community, the commitment she made to the community and for us to honor her service as well. So given where we are, where we don't have that election flexibility, I guess my only ask is, 20 days seems a bit short. With Jamie, I think we did 26 and I forget what it was with Mark, but I think it was longer than less than three weeks. I would ideally like to see this run 30 days. I think a month is the bare minimum to allow people to decide whether or not they want to do this. 20 days makes it kind of seem like, well, maybe there's already been things set up. I don't, it just is too short. We still have time to be able to appoint somebody with both a regular meeting and if necessary, a special meeting before the deadline if we were to allow for 30 days, at least I'm assuming my counting was correct. Even if it's 29 days, it's close enough for 30. Had we had this meeting on the third, which we could have done, that would have given us a lot of opportunity to give people a 30 day window. I think given that we delayed a week to have this meeting, we shouldn't shrink the application time down to 20 days. That's just too short. I'm pleased to hear that we will get a chance to say to Lois, or form of advice during a meeting thank her for her service. I agree with the staff's recommendation to proceed with the appointment process. I also had a concern with the short timeframe that we're providing for applicants in order to be able to get the word out. I count 20 days from the day the ad runs to when applications are due. So I want to echo what Bob said about the timeframe on this. I think waiting two weeks in April from the seventh to the, what is that, the 21st is appropriate to give applicants and to give the district a bit more time to advertise a second time and get the word out, not through only the formal channels, but the informal laws to folks out there. So that's my thoughts. Two members of the public or comments. Let's see, is anybody like to speak? Ah, there we go. I'm just worried whether, okay, Jim Mosher, I just want to make sure sometimes I'm not sure when I don't see a hand or whether nobody wants to speak or whether that function isn't working, but go ahead, Jim. Jim Mosher from Felton. And I just want to support the staff recommendation and also to say that I appreciate Lois' service and I'm glad that the board is going to recognize her at the next meeting. I don't really have a position on whether 20 days is enough or 30 days. I think that it's the counter-failing issue is just to get somebody on board as soon as possible. And so whatever the board decides on that is I think would be fine. Thank you. Okay, thank you, Jim. Any other comments? Larry, board. Thank you very much for the opportunity to speak. I also support the staff recommendations and would echo what Jim Mosher just said. Thank you very much. Okay, any other comments or questions? I'll just go ahead and just make a comment. I think that the timeline that we came up with this was in part driven by wanting to get somebody on board sooner rather than later just because as processes go on the later somebody gets added it just isn't as helpful. And also if we're thinking about assigning somebody to a committee it's just well they're potentially depending on who we choose might involve some changing around of committee memberships. But I don't have a particularly strong feeling about that just to comment to Bob's question of why we didn't do this on the third. We could have but I actually was concerned that there would be people saying that we by doing it on the third the public wouldn't have much chance to know about the meeting and that we could be accused of rushing the decision about whether to a point or leave it open. And I was actually part of the reason we have this special meeting pop was I was afraid you were gonna object to the timeframe of the third. So, but like I said, I don't have strong feelings about this if we wanna put off the choosing the person till it would be the 21st of April and adjusting the other deadlines kind of accordingly that would be fine with me. I don't really care. Jamie, you have any thoughts about that? No, I agree with you. And I had not occurred to me how tight the timeframe was until Bob and Mark so rightly brought it up. But yeah, I can be persuaded that that makes sense. Okay. So then how about if I just change the motion that is Gina, did you? Sure, may I suggest April 13th that the second sentence say applicants will be interviewed at the second regular board meeting in April. Oh, I'm sorry. Let me take a step back here. I think what we would do, and this isn't in the motion is that I would suggest we change the timeline so that applications are due by 3 p.m. on April 13th which is the Wednesday before the board packet goes out for the meeting on the 22nd and it provides 30 days for applicants. First, right? 14 plus seven is 21, April 21. Yeah, and so then the second sentence of the motion would read applicants will be interviewed at the second regular board meeting in April which is scheduled for April 21st. Right. Okay. All right, so I'll just read the motion as revised just now. This is we are moving to proceed with filling the vacancy on the board by appointment. Applicants will be interviewed at the second regular board meeting in April which is scheduled for April 21st. The appointee would serve until the seat is filled in connection with the November 2022 general election. So I'm moving that. Is there a second? Second. Okay. Any further comment among members of the board? And let me just go out. We've got a couple of people that have joined us. So let me just go out and make sure that there isn't any more comments from members of the public. Seeing none, hearing none. Holly, would you take a roll call vote please? President Mayford. Aye. Director Ackerman. Yes. Director Fals. Yes. Director Smalley. Yes. Okay. And passes unanimously. All right. So this will be posted tomorrow with a change in the deadline to be the Wednesday before the meeting on the 21st. Okay. Now that brings us to the next item of business which is Lois was our vice president. So on her resignation, that position is open. So we need to vote in a new vice president. And I hope that you will indulge me in as president and allow me to nominate Jamie Ackman as vice president. Thank you very much, Gail. I appreciate the confidence. Okay. I accept the nomination. All right. I'll second that. Okay. Any discussion? Not among the board. Then how about among the members of the public? No discussion there. I don't see any. Okay. So Holly, could you go ahead and take the roll call vote please? President Mayford. Aye. Director Ackman. Yes. Director Falls. Yes. Director Smalley. Yes. All right. So Jamie, you just have to hope that I don't get hit by a beer truck in the near future of them. Gonna need you to stay very, very careful and wrap yourself in bubble wrap. All right, okay. All right. So finally we get to our one piece of unfinished business. California governor's office of emergency services, designation of applicants agent. Rick. Thank you. This item is directly related to our FEMA and Cal OES grant applications from the CZU fire requiring the district adopting and approving a designation of applicants agent resolution for non-state agencies. Basically housekeeping for us to move ahead with receiving grants from the two agencies, a requirement of designation of applicants agent resolution. The district needs to adopt the attached resolution. Basically housekeeping, moving forward. This means that the state and feds are getting closer to start reimbursing the district. And so this process is needed for that reimbursement. You'll see their standard resolution and the district's resolution to be adopted. Okay, so basically we just have a motion to adopt the attached resolution thereby adopting and approving and designating our agents for these non-state agencies. Any, so moved. Is there a second? Second. Okay, second. Any discussion? I feel sorry for Mark. There's only three lines for Boardman. There's only three lines for Boardman. Well, we could add him. I mean, let's add him. Okay. No, no, no, no, Mark, no, Mark. We're adding, we're adding you. If there was five people, we'd have five. Come on. Do not amend. Wait, we wondered about that this afternoon when we were getting ready for the meeting. How we ended up with those, but we decided yes, we needed one. Alpha order, I'm assuming, but yes. Yes. Board chair first and we went by alphabetical order. Yeah. Sorry Mark. I do have a question though. Rick mentioned getting ready to issue reimbursements. Reimbursements for. We have several of our category work obligated in the category B projects. And once obligation comes, that's mean they've been approved and move ahead and start the reimbursement process. So they're not the, this is work that's already been completed that the district expense funds that's been reviewed and FEMA now will now as obligated to make reimbursements. But we're getting through the process, but we still have a long ways to go because they haven't even started the permanent repairs, which is called category F, which, you know, a five mile pipe, one of the bigger projects. And they're still getting through one of the bigger claims that category A are emergency response, which is a little over an eight, three million plus dollars. So reimbursements for seizing your related damages. Exactly. Yes. Okay. Thank you. That's good news though. It is. Money money. Yeah. Closer anyway. So any other comments or questions? Okay. So I guess we need to vote on this, Holly. I think the public, the public. Well, I would, okay. That I was looking at, let me formally ask, are there any members of the public that would like to comment on this? Seeing none, hearing none. Holly, would you take the roll call though, please? If may have. Aye. Vice President Ackman. Yes. Director Fultz. Yes. Director Smalley. Yes. Okay. Passes unanimously. With that, we come to the end of the meeting.  This was fun. Thank you all for attending this special meeting. Good night. Okay. Good night everyone. Thanks.


Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, find the local maxima and local minima if any of the following functions. Find also the local maximum and the local minimum values as the case may be we are given fx is equal to sin x minus cos x where x is greater than 0 and less than 2 phi. First of all let us understand that if we are given a function f defined on open interval i then let function f be continuous at critical point c in interval i then if f dash x changes sin from positive to negative as x increases to c then c is a point of local maxima that is if f dash x is greater than 0 for x slightly less than c f dash x is less than 0 for x slightly greater than c then c is a point of local maxima and fc is the local maximum value. Now if f dash x is less than 0 for x slightly less than c f dash x is greater than 0 for x slightly greater than c then c is a point of local minima and fc is the local minimum value or we can say if f dash x changes sin from negative to positive as x increases through c then c is a point of local minima. We will use this discussion as our key idea to solve the given question let us now start with the solution we are given fx is equal to sin x minus cos x where x is greater than 0 and less than 2 phi first of all let us find out f dash x differentiating both the sides of this expression with respect to x we get f dash x is equal to cos x minus sin x you know derivative of sin x is cos x and derivative of cos x is minus sin x now simply find we get f dash x is equal to cos x plus sin x now we will find all the points at which f dash x is equal to 0 so we will put f dash x is equal to 0 this implies cos x plus sin x is equal to 0 now dividing both the sides of this expression by cos x we get 1 plus sin x upon cos x is equal to 0 we know sin x upon cos x is equal to tan x so we can write 1 plus tan x is equal to 0 now subtracting 1 from both the sides we get tan x is equal to minus 1 now we know tan 3 pi upon 4 is equal to tan 7 pi upon 4 is equal to minus 1 and clearly we can see values of these two angles lie between 0 and 2 pi so we get x is equal to 3 pi upon 4 or x is equal to 7 pi upon 4 now let us consider the case when x is equal to 3 pi upon 4 when x is slightly less than 3 pi upon 4 f dash x is greater than 0 if x is slightly greater than 3 pi upon 4 f dash x is less than 0 so using key idea we get x is equal to 3 pi upon 4 is a point of local maxima clearly we can see as x increases through 3 pi upon 4 f dash x changes sign from positive to negative so x is equal to 3 pi upon 4 is a point of local maxima and local maximum value is given by f 3 pi upon 4 now f 3 pi upon 4 is equal to sign 3 pi upon 4 minus cos 3 pi upon 4 now we know sign 3 pi upon 4 is equal to 1 upon root 2 and cos 3 pi upon 4 is equal to minus 1 upon root 2 now simply mind we get f 3 pi upon 4 is equal to 2 upon root 2 rationalizing we get f 3 pi upon 4 is equal to root 2 now let us consider the case when x is equal to 7 pi upon 4 now when x is slightly less than 7 pi upon 4 then f dash x is less than 0 when x is slightly greater than 7 pi upon 4 f dash x is greater than 0 now clearly we can see as value of x increases through 7 pi upon 4 f dash x changes its sign from negative to positive so using key idea we get x is equal to 7 pi upon 4 is a point of local minima and local minimum value is given by f 7 pi upon 4 f 7 pi upon 4 is equal to sign 7 pi upon 4 minus cos 7 pi upon 4 sign 7 pi upon 4 is equal to minus 1 upon root 2 and cos 7 pi upon 4 is equal to 1 upon root 2 so substituting their corresponding values in this expression we get f 7 pi upon 4 is equal to minus 2 upon root 2 now rationalizing we get f 7 pi upon 4 is equal to minus root 2 so we get local minimum value is equal to minus root 2 so we get f 7 pi upon 4 is equal to minus root 2 so we get local minimum value is equal to minus root 2 local maximum value is equal to root 2 so our required answer is local maximum occurs at x is equal to 3 pi upon 4 and local maximum value is equal to root 2 local minimum occurs at x is equal to 7 pi upon 4 and local minimum value is equal to minus root 2 so this is our required answer this completes the session hope you understood the solution take care and have a nice day.


Before we move on to other correlation I would like to take some questions I think before we take questions professor Arun wants to add something. One question which was there in Moodle was in the data handbook there are multiple correlations which is the one that I use for a given problem. There is no unique answer for that see there might be families of correlation for flow over a flat plate couple of them three or four of them somebody might have done for some fluid etcetera etcetera etcetera. What you have to do is look at the range of conditions of applicability of that correlation see if it is done if it is valid for the range of properties that you are having and then use it. And what should a good correlation is the one which should be valid for a wide range of fluids and overall what you say property values etcetera. Now if I do an experiment for with say R 1 2 3 or something and develop a correlation explicitly for that I cannot use it for flow with water or anything like that. So and please keep in mind that empirical correlations have a certain amount of uncertainty involved in them caused by the experimental uncertainty also ok. So these correlations are valid with plus minus 20 percent 15 percent 10 percent whatever it may be. So you get an answer with one particular correlation another correlation is also equally valid you get a slightly different answer does not matter ok. So this is something we need to emphasize and students will come and say this answer is wrong that answer is correct both answers could be equally correct or equally wrong ok. Erode any questions there is a question mark. In case of transport of crude oil from one country to another country the pipelines are taken underneath of the sea. In that case there is a chance for the pipeline to go through the high specs. So in that case the thermal boundary layer what what happens to thermal boundary layer because the pumping power will be high in case of if the thermal boundary layer is more and sometimes there is a chance for blockage of the crude oil flow. So what is your suggestion. See the question asked by one of the participants is that the crude oil transportation from one country to another country is generally taken through sea and sometimes it goes through icebergs that word I did not hear clearly I hope it is icebergs basically what pressure wants to impress upon us is that it is going through lot of temperature gradients. So whether the whether the crude oil is going to get affect the movement of the crude oil is going to get affected. See the answer I would think as like this see the boundary layer see there is going to be temperature gradient no doubt about that and instead of heating it is cooling the oil and in fact like in semi infinite medium assumption we had said that how below we should be taking yes if the temperature freezes my oil if it is below the freezing point definitely nothing can be done but I do not think it is going to be frozen. So yeah see professor says that seawater is generally warm at the base so it is not going to freeze or solidify the oil which is there inside that is that might be one of the reasons maybe we can just go back and check what would be the temperatures at the sea bed. So the point is I do not think they would lay the pipes intentionally where the oil is going to get frozen. So that is the answer I would think temperature is generally greater than 4 degree Celsius. So I do not think it is going to freeze my oil at all the freezing point of the oil would be much lower. I hope I have answered your question professor. So what happens to the thermal boundary layer and hydrodynamic boundary layer in such cases? The question asked is what happens to thermal boundary layer and the hydrodynamic boundary layer. Thermal boundary layer or the hydrodynamic boundary layer is like this let me it is going to be like whatever conventional of course this is little premature question but still I would go ahead and answer this because now that we have been asked so this is a pipe. So now parental number of the oil I would think it is much much greater than 1 so that means delta is much much greater than delta t. So thermal boundary layer thickness is very much lower than delta so that means delta is thicker that is this is delta and delta t is thinner this is delta t. So of course we are not introduced this concept because you asked me so we are telling this this is the hydrodynamic developing length and this is hydrodynamic developing length and this is thermal developing length. So the thermal developing length in case of oil will be higher than that of developing length the hydrodynamic developing length but I do not think all this is of any importance because my pipeline is much much much much larger developing length if the is typically of the order of 100 times the pipe diameter but my pipeline size is running to few kilometers. So developing length concept has no relevance much relevant but Nusselt number and friction factor are as good as which are derived for any other Newtonian fluid as long as we can assume crude oil as Newtonian. In such cases the pumping power will be extremely high then there will not be any flow to take place now. So question asked is in such cases for high Prandtl number pumping power will be very high yes pumping power will be very high because viscosity is very high for pumping power it does not know C p and all k C p it does not know for friction factor what is important is the viscosity. Viscosity is very high pumping power will be very high in fact this is one of the cases where we say classic example laminar flow will exist usually the flow is not turbulent in pipeline supplies it is laminar flow why because viscosity is very high I need to give that kind of pumping power if I have to supply the oil I cannot say that I cannot pump it has to be pumped it is laminar and the pumping power will be significantly larger than that of water or air. Ok professor. Will there be any provision to keep the oil well below the firing point of the oil in order to make the flow easier. See I really do not know the question is will there be any provision to keep below I really do not know how exactly the crude oil pipes are kept whatever I am answering is based on fundamental I would request you professor now that you have asked so many questions can you take this as a homework and answer this question over Moodle how exactly crude oil pipes are supplied if you do not answer definitely we will put it across but I would request you to having taken your three questions I think you can put the answer I think it is fair enough to ask you to put the answer for this question on the Moodle yourself I think ok. Ok Panvel. Hello Nasell number equal to say R e raise to m P r raise to n these are the constants of correlations. So, what are the different methods to get it what are the different methods to find out these constants how to obtain these value of constants. Yeah see one of the participants question is that we said that N u equal to A N u equal to C R e to the power of m and P r to the power of n how do I get this C m and n yeah it is a very good question anyway we are going to answer just little while from now we have solved it for laminar flat plate case, but it is not always possible to get the closed form solution if I cannot get the closed form solutions with mathematics for example, for turbulent flow one would take the recourse of experiments we would find the experiment we would conduct experiments and say that through experiments at least we know the functional form how to represent the results. So, I will find C m and n through experiments the turbulent shear stress is deducting from the daughter shear how it is. See one of the participants question is do we deduct the turbulent shear stress from the total stress no we do not deduct see as I have been saying since morning we have laminar sub layer and we have buffer layer and we have turbulent boundary layer I am not deducting in the turbulent boundary layer what is there is only minus rho u prime v prime bar as the turbulent shear stress and in the in the buffer layer both of this minus rho v prime u prime bar and mu del u by del y they are of the same order both are of the same order now in the laminar sub layer it is mu del u by del y we are not deducting in the laminar sub layer for example, u prime is 0 and v prime is 0. So, there is no question of having rho u prime v prime bar. So, this is not there this is 0. So, we are not deducting anything from anywhere it is just whatever is measured is just being reported I think what professor Arun says is that may be your question is why is it negative sign no minus rho u prime v prime is actually positive that is a good question always I used to forget this, but I used to give this as a homework for student, but actually it is not it is like this if u prime is generally if it is positive v prime is generally v prime negative. Why because where does the energy come from wherever if it is lost in one direction it has to be compensated in the other direction. So, if u prime is positive v prime is negative. So, minus this minus is essentially to make this positive if one is positive and the other is negative means product is negative minus of this product is again negative sorry the product of all of this is positive. So, total thing is positive. So, we are not deducting this minus sign is essentially to make it positive any other question sir what about moving plate boundary part about what boundary layer for moving plate the question is by one of the participants what will happen for a if I have a flat plate in addition to the velocity if this is moving if this is moving no issue the question here is always it is the relative velocity which generates the boundary layer. So, if it is moving the velocity at the edge of the boundary layer will be equal to the velocity of the plate, but there is as long as there is a relative velocity between u infinity and the plate velocity there is going to be boundary layer that is what is their covetive flow. In fact we are going to deal with this when we come to internal flow, but we are not going to solve for moving plate and this is essentially what we study in various covetive flow. Point is for moving plate also boundary layer is going to be there and that boundary layer depends on the relative velocity between the fluid velocity and the plate velocity. I have a question regarding the energy equation which we derived in the morning the left side of the energy equation is convection and the right side contains the diffusion the pressure work and the dissipation. So, is it appropriate to call the left side as convection or should we call it should we be calling it as advection instead of convection. One of the participants question is that if we see the energy equation in the energy equation we have we have this term what we have called this as convection this is conduction pressure work and viscous dissipation. One of the participants question is why do not we call this as advection rather than convection, but if you really see it is a combination of both advection and convection see why because why I say convection because convection is nothing but conduction within the boundary layer del t by del y and del t by del x I am taking within the thermal boundary layer, but why is it there it is because of the convection velocities u and v. So, it is very difficult to differentiate both the advection and the convection component or advection and conduction within the boundary layer. So, that is the reason why we usually call this as convection knowing very well or in the back of our minds that it is the conduction within the thermal boundary layer that is all the answer is for this. Baramati any questions please. Sir my question is about the Eckert number what is the limit for the velocity when you consider Eckert number above the certain velocity value. See the question is what in the Eckert number now that we are anyway here only in the Eckert number what is the velocity limit above which we should consider or neglect Eckert number actually it is there here itself Eckert number equal to Eckert equal to v square divided by C p into T s minus T infinity. Now of course it depends on the fluid also, but if I assume that my fluid is air C p for fluid is 1050. So, for the Eckert to become 1 let us say for Eckert to become 1 what should happen one of you should help me. So, that is if Eckert number is 1 v squared equal to 1050 into delta t or I should I should have my delta t such that my temperature gradient should be such that my v v v squared should be of the order of delta t or this number v squared and delta t should be such that my Eckert number should be around 1 when will that happen when the velocity is very very high. Because I cannot write delta t just like that because it depends on the velocity I cannot write I cannot fix delta t and go on increasing v or I cannot fix v and go on decreasing delta t. It is they are coupled based on my velocity the temperature gradient is high what they say is that if my velocity is supersonic or hypersonic delta t is of the order of T s itself is of the order of T s itself then perhaps then it becomes that is if the temperature increases so high so high virtue of kinetic energy that is all almost all the kinetic energy is transformed into thermal energy. So, that is what happened with Kalpana Chawla its crash was essentially because one of the tile had gone up come out the insulating material on the what is that rocket had come out and the temperature had picked up because velocity. So, that is why the there was the trash or whatever accident had occurred my next question is sir one more question is there sir is it a incompressible fluid is in ideal or it is in existence. The question is is incompressible flow ideal or existing incompressible flow what is incompressible flow what is incompressible flow that cannot be compressed are there fluids which cannot be compressed are very difficult to compress yes water is a incompressible flow very difficult to compress water. So, if you take the velocity of sound in water it is very high it is in thousands actually. So, but if the velocity of sound in air is 330 meters per second both are linked. So, point is water can be assumed or water can be considered why assumed it can be considered as incompressible flow. So, incompressible flow is not an ideal situation, but inviscid is an ideal situation that means there is no fluid which is thoroughly inviscid. So, this is an ideal situation, but this is not a ideal situation. So, one more question is there I have not understood the statement when a certain number is 1 heat transfer across the layer is only by conduction. I think we will take this question when let me answer this question now that this question we had expected this question. The question asked is Nusselt number equal to 1 means conduction is equal to convection. See we need to understand conduction upon convection. See we need to understand one thing here we have written h l by k actually it is not conduction over the length l or actually what is that we are equating it is conduction within the boundary layer conduction is equated with the. So, what should I take here actually I should take h delta t by k h delta t by k, but do I know delta t today only we derived for flow over a flat plate for Prandtl numbers less than 1 delta t is of the order of r e to the power of minus half and p r to the power of minus 1 by 3, but for all situations for a moment delta t is of delta t is of the order of r e to the power of minus half and p r to the power of minus 1 by 3 that is I cannot get this type of solution for all cases for that cases I do not know delta t delta t is unknown. So, we can definitely say this one this thing is right when only when I say h delta by k in that form only one has a meaning that is conduction within the thermal boundary layer is equal to convection if I write h l c by k that equal to 1 has no meaning actually, but I write h l by k because I do not know delta t I give all correlations in terms of characteristic length when I write in terms of characteristic length h l by k equal to 1 need not mean that conduction is equal to convection it is equal to 1 only when conduction within the thermal boundary layer is equal to conduction convection any other questions. Sir one more question in temperature gradient d t by d y at y is equal to 0 is this definition or this means is it exactly on the surface part or on the layer which is in no slip condition. See one of the questions is is this minus k delta t by del y is h equal to h equal to minus k delta t by del y at y equal to 0 is this only on the first layer or subsequent layer. Now let me answer this way delta t by del y at y is equal to 0 is it not getting affected because of the velocity distribution in the temperature distribution within the thermal boundary layer. Yes for argument for argument sake we can say that it is conduction in the first layer why because there is velocity in the upper in the subsequent layer, but I think we can conveniently say that actually it is conduction within the thermal boundary layer, but this del t by del y is affected because of the velocity distribution and the temperature distribution within the thermal boundary layer. How do we get the temperature distribution after the solution of the velocity distribution.  So this both u and v have come from the velocity distribution which was obtained from the momentum and the continuity equation. So k d t by d y is definitely affected by the velocity above the layer. So I cannot have k d t by d y at y equal to 0 independent of what is happening above it. So it is definitely affected. Everything is interlinked. Sir having two questions for you one is when you are talking about the Nusselt number is function of x star r e and Pantel number is it a better idea that when x star is increasing that your r e is also increasing over the flat plate and if you. The question asked by one of the participants is that we said that N u equal to is a function of x star and r e and P r e is with the increase of x is r e also increasing. Yes if you take N u equal to what is our relation point and let me take the relation because I do not remember 0.332.  So that means what is r e x rho u infinity x by mu. So as x increases that means x increases for a given fluid and for a given velocity definitely r e x is increasing definitely r e x is increasing. So there is no doubt about that r e x is increasing what is the second question please. So in case of the flat plate suppose we are calculating say if you are having a 1 meter length if you are calculating the r e over the plate say at different locations 0.1, 0.2, 0.3, 0.4. So we need to calculate x or r e based on the L. The question asked is as we move along the length let us say for a flat plate of 1 meter if I calculate x equal to 0.1, x equal to 0.2 and x equal to 0.3, x equal to 0.4 will I calculate Nusselt number at x equal to 0.1 on the basis of L equal to 1 meter or x equal to 0.1 meter. Right from the beginning we have been talking about this local heat transfer coefficient and average value. This Reynolds number being affected by the length gives you a local value of the Nusselt number local value of the heat transfer coefficient. So at x is equal to 0.1 I am going to get the value of heat transfer coefficient at x equal to 0.1, x is equal to 0.4 it will be a different heat transfer coefficient based on the Reynolds number at that location. So if I have this this is a common x book problems if I have electronic chips which are being cooled by force convection the chip at the first chip from the leading edge versus the 10th chip versus the 100th chip. So each of this the heat transfer coefficient is going to be different it is going to be decreasing with respect to increase in x and that quantity is a local quantity. Now if you are asked to calculate what is the total heat dissipation from the plate or something then it is over the entire length that case you will take a average h value in which case your Reynolds number will be based on the length of the plate not the local distance your Nusselt number will be based on the length of the plate not the local Nusselt number. So whatever is the subscript that you use for Nusselt number x that is the same as subscript that you use for Reynolds number if that is x this is x if Reynolds number is l this is l one other word of caution if you have a plate of 1 meter and there are 10 chips the 10th chip which is at x is equal to l Reynolds number local will be r e l but Nusselt number for that will still be a local because you want to calculate heat transfer at that location it is again play with English only at a given location if you are asked it is a local quantity over a given length if you are asked that is an average quantity ok. So heat transfer at x equal to 1 meter means how much heat is lost at that location heat transfer over 1 meter length of the plate would mean what is the heat transfer over the full length of the plate there you will use the h bar value ok. V N 19 Agpur. Sir what is the effect of pressure on thermo physical properties? This is a good question the question is what is the effect of pressure on thermo physical property there is going to be effect of pressure but the effect of temperature is more significant than effect of pressure but definitely pressure is also going to have an effect we cannot say that there is no effect of pressure but for example thermal conductivity if we take thermal conductivity or viscosity for that matter if we take both of them are more affected by temperature rather than pressure. So usually that is the reason in most of the textbooks also they do not list out the effect of pressure on thermo physical properties because it is influence on the thermo physical properties is significantly lower compared to that of temperature. Amal Jyoti College. Hello sir we have actually three questions one is that sir in the derivation of the energy equation you have not considered the heat generation term so why it is so and second question is that in suppose you are going to measure the velocity in case of a laminar sorry in case of buffer layer we are placing a hot wire anemometer so the will the anemometer affect anything to the flow conditions so will it cause a very serious problem to the flow condition. Now third question is that why it is that the Reynolds transport theorem is called as a transport theorem. Okay first let me list out question by question one first question is I before I forget first question is why volumetric heat generation is not used first question is why volumetric heat generation why volumetric heat generation not considered in energy equation number one not considered in energy equation and second question is in the buffer layer in the buffer layer in the buffer layer in the buffer layer if you if I introduce hot wire anemometer if it will it affect velocity that is the second question and in Reynolds transport theorem in Reynolds transport theorem why is Reynolds transport theorem called as Reynolds transport theorem.  So that is if you take a huge plate let us say of 1 meter and 2 meters wide then you would perhaps get a boundary layer let us say of let us say of 10 centimeter then in that 10 centimeter if you introduce a probe of 3 mm then the intrusion levels are low that is one way of looking at it otherwise there are today non-intrusive way of measuring things that is laser Doppler anemometer or particle image velocity particle now particle image velocity so these are non-intrusive non-intrusive how do how does it work how does it work in a non-intrusive for example particle image velocity I am just going to hand wave it is quite difficult to make you understand but what is done is it is a pulsed laser that is one laser to the second laser it is going to be there is a time difference of nanosecond that is 10 to the power of minus minus 9 seconds so 10 to the power of minus 9 seconds delta t I have a pulsed laser now if I introduced a particle and I have tracked this particle let us say I take this particle and I take picture between 2 nanoseconds that is sorry between 1 nanosecond and if I were to track this nanoparticle within 1 nanosecond a distance delta x same particle mind you this is a and a but this a is after time t so I have got delta x by delta t I get velocity this is non-intrusive so it is we are not introducing any probe or anything so people I have measured with particle image velocity symmetry and laser Doppler animometer where in which they are non-intrusive and people have found that in fact when velocity distribution was figured out they were all done with either Prandtl probe or hot wire animometer but later on these universal velocity distribution has been confirmed even through non-intrusive measurement one thing we should be able to appreciate is that any measurement is going to be having some uncertainty no measurement is a perfect measurement so every measurement is going to have problem near wall nevertheless people all over the world in different places have measured and find that universal velocity distribution that is in the buffer layer is what we have got so that is the answer for the second question third question why is it called Reynolds transport theorem see whatever loss we have studied for example conservation of mass mass of the system is conserved that is system system is not moving here but what is happening if I take my fan as my control volume air is coming in and getting out what am I tracking I am tracking a fluid particle which is moving in and out of a control volume so whatever I apply to a control volume into which my fluid particle is going in and going out so that so that is Reynolds transport theorem why transport because my fluid particle is getting transported why Reynolds may be perhaps I this I do not know why may be Reynolds only has derived this I do not know but perhaps either he has derived or in his honor it has been named as Reynolds transport theorem one of the easiest or best examples of this is that if you have a balloon in which you are trying to blow and there is a leak so your mass is coming into the control volume the control volume is changing shape and at the same time mass is leaving the control volume. So, this Reynolds transport theorem basically looks at this whole thing now in m out and the change of shape of the control volume because you have transformed from a system to a control volume whose dimensions can also change that is also there is that ok. I have left out another question q dot not considered ok. So, if q dot is there that can be taken as a source term in the energy equation see why for example, in combustion if there is combustion there is net chemical energy release we add that as a chemical energy release that is source term on the right hand side of our energy equation. So, that is how even volumetric heat generation also can be accounted there is we have not taken volumetric heat generation because we want to be as generic as possible that is all the same way we took the energy generation term in the heat diffusion equation if you see here one important thing what I have missed out is if you see the energy equation and if I put u equal to v equal to 0 what is that I get I get the heat diffusion equation only. So, that is essentially the answer for your question to take the volumetric heat generation is that ok. So, for Reynolds transport theorem can we apply to mass diffusion case or is it only for fluids or can we apply to any larger areas further areas. Will you repeat can you apply to this Reynolds transport equation can we apply only to the fluid can we apply to fluid cases heat generation cases heat cases or in any other cases so suppose there is a mass diffusion case that is mass transfer cases also can we apply this type of equation. See the question is can we apply the Reynolds transport theorem for mass diffusion also yes Reynolds transport theorem can be applied for mass diffusion also if you are handling combustion you have to keep track of each species. So, for each species you are going to write Reynolds transport theorem only that is again conservation of mass and conservation of momentum. So, Reynolds transport theorem can be applied for mass diffusion also what we need to realize is that Reynolds transport theorem can be applied as long as my fluid as long as my fluid particle is moving ok. So, let us start off with the tutorial problems. So, we will take up one problem I know there will be lot of questions today I would encourage you to ask us as many questions as possible. So, we will solve less of problems, but more of questions today till 5 30. So, let us take up problem number 27. So, what is problem number 27 saying? So, I need to experimental measurements of the convection heat transfer coefficient for a square bar in cross flow for a square bar in cross flow air is v velocity v and l is characteristic length l is 0.5 meters. So, now in the first case h 1 bar equal to remember bar means average heat transfer coefficient 50 watts per meter square Kelvin and when v 1 is 20 meters per second and h 2 bar h 2 bar equal to 40 watts per meter square Kelvin when v 2 is 15 meters per second ok. I think it is natural heat transfer coefficient has decreased with the decrease of velocity ok. Now, the question is a Nusselt number is of this functional form a nu bar equal to C r e to the power of m, P r to the power of n. What will be the where C m and n are constants without telling we have been thinking that since morning I mean we have been telling that since morning. What will be the convection heat transfer coefficient for a similar bar with l equal to 1 meter when v equal to 15 meters per second that is the question asked. So, how do I go about this problem? First of all what is that I need to find that is when v equal to 15 meters per second and l equal to 1 meter and bar is similar to this there is no doubt about that. So, now assumptions I think we do not have to write we will just state steady state by now we know steady state and we are taking average heat transfer coefficient that is a nu bar or h bar and properties are assumed to be constant.  So, now what is that we are saying h bar l by k equal to h bar l by k equal to C r e to the power of m P r to the power of l.     So, what is h 1 bar given h 1 bar is given to be 50 and l 2 equal to l 1 no.   So, it will get cancelled out and l 1 equal to l 2. So, that is also getting cancelled out. So, we are left out with only v 1 by v 2. So, what do I get now? So, I get v 1 by m I get that is 20 by 15 to the power of m that is m equal to 0.7825 0.78 yeah this is how one can if you know if you somehow experimentally measure the heat transfer coefficient we can get this power m that is I think this is the this is we are not reached yet. Our question now asked is what will be the value of now we can write h bar equal to I think we will write what is h bar when v equal to 15 meters per second and l equal to 1 meter let us write this. So, that is l equal to 1 meter per second. So, now let us write h bar upon h 1 h bar b for velocity equal to 15 meters per second for 15 meters per second velocity equal to and l equal to 1 meter. So, h bar nu for velocity equal to 15 meters per second l equal to 1 meter.  So, now h bar nu equal to what is h 1 bar 50 into l 1 is 0.5 where did I get this 0.5 from see the question in the question it is given as l equal to 0.5 this 50 is for 0.5 in the diagram in the problem.  So, now h nu equal to h nu equal to h nu equal to 15 into 0.5 by 1 into 15 into 1 meter 1 upon 20 into 0.5 to the power of 0.7825. So, h bar is equal to 34.3. So, second part we are not doing because the same procedure you can apply all that is as these length is different and velocity is different. So, answer is h bar equal to 59.06. So, we are not going to solve this problem second part because it is a similar exercise question is what the results would the results be same if the side of the bar rather than it is a diagonal were used as the characteristic length. If the side of the bar is used as the characteristic length no only thing is that we can use side of the bar as the characteristic length no issue only thing is that my constant would have changed otherwise my constant would not have changed that is all that is all c m n will be n is independent of the characteristic length why because n is parental to the power of n which is a fluid property, but c and m depend on the characteristic length. So, the way I choose the characteristic length it will depend on the characteristic length. So, we will take another we will start taking questions again back. So, we can go to may be K. K. Vag you have any questions. Yes sir why Reynolds number critical Reynolds number is taken as 5 into 10 s to 5 for this laminar flow a flat plate and why it was 2000 for a pipe. See the question asked is for flow over a flat plate for flow over a flat plate we are taking critical Reynolds number as 5 into 10 to the power of 5, but for 5 into 10 to the power of 5, but for pipe R e pipe we take as 2300 why because these two are different why they appear because we have taken different characteristic lengths in both the cases R e is equal to rho u infinity l c by mu l c for flat plate we have taken l c equal to characteristic length which is the length of the flat plate for the pipe we generally take diameter, but what we should take what we should take we had specified this in Nusselt number also H delta t by K that is what we told. Similarly, for Reynolds number also we should take rho u infinity delta hydrodynamic boundary layer upon mu. If I take rho u infinity delta by mu no matter what class of problem I choose that is external flow, internal flow, natural convection, jet flow or any other flow you can think of they are all going to be of the order of 20 to 30. So, because we are taking different characteristic lengths not taking appropriate hydrodynamic boundary layer thickness we get different critical Reynolds number, but why do we do this knowingly why because hydrodynamic boundary layer thickness is not known. What we know is the characteristic length that is the engineering parameter that is either the diameter or the flat plate length that is the reason why we take flat plate length or the pipe diameter because of which I get the critical Reynolds number different. So, there is another question this is related to why we are considering only the fluid properties in deciding the flow whether it is a laminar or turbulent why not the surface properties. The question asked is for deciding whether the flow is laminar or turbulent or transitional why we are taking only the fluid properties why not the solid properties. Now who is offering resistance for the flow to take place because of the fluid viscosity it is not whether it flows on a copper plate or it flows on a stainless steel plate or it flows on an aluminum plate who is creating resistance it is not the solid property material by virtue of which it is being resisted it is because of viscosity. If there is no viscosity then my fluid would have simply slipped over the plate whether it is stainless steel or whatever it is because of viscosity only it is getting no slip condition we are coming across because of which only we have hydrodynamic boundary layer or the thermal boundary layer that is no slip and the no temperature jump condition. So, there is no bearing or there is no connection whatsoever between the solid properties and the boundary layer boundary layer is only because of the viscidity or the viscosity of the fluid. One more last question sir it is not expected to have a hydrodynamic boundary layer but the thermal boundary layer is expected for the heat transfer hydrodynamic boundary layer is not expected thermal boundary layer whether it is desirable sir for heat transfer. See there are the participant is asking me two questions whether hydrodynamic boundary layer is expected or thermal boundary layer is expected I do not know what do we mean by expected but what I feel by the question is that when I say expected we are not supposed to have the boundary layer that is when I have a thermal boundary layer I am not supposed to have the hydrodynamic boundary layer that is not possible it is expected both are expected why because there is resistance for the flow and there is resistance for the heat transfer to take place. So, because of the conduction in the thermal boundary layer there is del t by del y and because of the resistance offered by the flow there is within the boundary layer that is hydrodynamic boundary layer there is velocity. Now is thermal boundary layer a desirable thing no not at all a desirable thing because that is the resistance what is f for fluid dynamics that is Nusselt number for heat transfer what is delta for hydrodynamics that is for fluid flow that is delta t for heat transfer. So, both delta and delta t are expected but are unwanted but they are there they are there nothing can be done. So, we have to overcome them then the only way to overcome them is either to have a fan or a pump. So, that is why our pumping power otherwise this pump should have never been invented if there were not to be any boundary layer it would have been just going by slipping they will just go by because of the slippage it just by you give the head then they will just flow once you generate the flow they will continue to flow there is no question of any resistance or getting the flow stopped somewhere this is because the resistance is there inherently for the heat transfer or for the fluid flow to take place boundary layers are unwanted but they are there and they are going to limit us the amount of the heat transfer or amount of the what is that fluid flow why today laptop or anything miniaturization is not possible let us come to application essentially because of thermal analysis because we are not able to cool the chip the heat fluxes are going up but my chip temperature is expected to stay at 30 to 40 degree Celsius otherwise it is not going to behave the way it is designed to behave the heat fluxes are going up but my temperature is expected to be at 50 degree Celsius people generally take this example if you take the sun's temperature sun's heat fluxes are of the order of 50's or 100's kilo watts per meter square what is the sun's temperature 5500 Kelvin but same amount of the heat fluxes we are expecting on my laptop but what is the temperature I am expecting to maintain it is 50 degree Celsius not an easy task that is why we are not able to make my laptop as small as possible that is why their life of the laptop for example is much lower than that of desktop because I am able to cool the desktop even why desktop also if you do not put properly in an AC room if you just put in an outside its life gets affected because of cooling okay ma 90 Bhopal have a question how can we correlate the bondy layer thickness developed over the fin and the effect of this thickness of the layer be on the distance between the two adjacent fin okay I think can we postpone this question why because we are not told the concept of developing flow so if we complete internal flows we will be able to the question asked is if on the fins how does the boundary layer start and stop and in between fins how does it grow actually we have to understand the concept of developing flow so I think we will take up this question after we cover the internal flows all that I can say is if you know if you are very good at internal flows this is the answer it is continuously developing flow it is never a fully developed flow but after sometime it may be reached periodically fully developed flow you will not be able to understand what I have told you will understand I will after I complete internal flow any other question please quickly see generally I just want to make one clarification morning we had one question we have not taken unsteady terms in the non dimensionalization of both the momentum and the energy equation even if you take the non dimensionalization for example del u by del t if you take so what will you get u instead of u you will get u infinity u star you get u infinity outside del u star u infinity outside and del by del t for t you take it as t star equal to t upon l upon u infinity so you get u infinity squared by l so that is yeah so if you if you see non dimensionalization that is del t by del t if you will get rho c p rho c p u infinity rho c p u infinity by l so you are going to get del t star by del t star so you are not going to get any other new term that is all I want to add here see you are going to get you want to write that is del t star rho c p del t by del t equal to rho c p del t star t star sorry t infinity minus t s into del t star upon del t star where t star equal to t star is equal to t upon t upon u infinity by l no l upon u infinity characteristic length l upon u infinity yes l upon u infinity l upon u infinity let me so yeah so l upon v so now if I substitute this in the top for t star t star so into t star into l by v that is in the denominator l upon v so I get this is equal to rho v c p upon l rho v c p rho v c p upon l t infinity minus t s del t star by del t star we are dividing throughout by this term so what am I going to get I am going to get del t star by del t star nothing new so no new non dimensional number is going to come either in the momentum equation or in the energy equation for unsteady flows as well that is the that is the point I wanted to professor Sevetkar had asked this question in the morning I had not given such a convincing answer but this is the convincing way to look at this so whether I take unsteady flows or steady flows I am not going to get a new non dimensional number both in momentum and energy say lump will thermal boundary layer separate like boundary layer separation in case of hydrodynamic boundary layer separation like boundary layer hydrodynamic boundary layer separation will thermal boundary layer also separate the question asked is will thermal boundary layer separate like hydrodynamic boundary layer yes if there is sudden obstruction if I am asking it to go through ribs, if I just draw ribs let us say, let us say I take flow over a flat plate with two ribs let us say two ribs. So, here velocity boundary layer is separating and then I have again regrowth of the boundary layer. This is same thing is happening both for not only for velocity boundary layer, but also for thermal boundary layer. So, thermal boundary layer also separates wherever velocity boundary layer separates that is how I think we can look at it ok. Sir, my question is that suppose there is a flow in a flat plate and there is a flow of a that is gases flow over our flat plate have high velocity gases and temperature of the gases also high. Suppose we injecting a nozzle through nozzle of water what is effect of that flow in that plate. The question is the question is there is a flat plate over a flat plate high temperature gases are flowing. Now, all of a sudden or gradually I have injected water mist or just water over this. So, we are what we are asking we are asking for two phase flow what are we studying we are studying single phase flow. So, when in a two phase flow it is actually interaction between the gases and the liquid. So, that we are not covering in any of the fundamentals. So, but anyway to answer the question what is happening is then there is heat transfer between the gas and the plate and water and the plate and gas and water also. So, it is the interaction between the gas plate water plate and gas water. So, it is not going to be easy task to answer this question and there is no closed form answer for this. So, but then it is a two phase flow problem all that I can say is it is a two phase flow problem wherein which I need to take the interfacial shear stresses and interfacial heat transfer coefficients also. So, you are saying gas is very high may be my water will get evaporated also. So, I really do not know and if my water and the plate temperature is very less may be my water will get condensed also sorry. So, either ways whichever has got evaporated back again it will get condensed. So, it is a very difficult task it all depends on the numbers what I choose. So, it is essentially the sum and substance of this question is that it is two phase flow problem. Sir, can we apply the N S equation over here? N S equation and then energy equations. Question is can I apply the energy equation over here? Yes, but I have to apply the energy equation separately for gas separately for water and then couple them through that is what I have been saying interfacial shear stress, interfacial shear stress these two get connected through interfacial shear stress and interfacial heat transfer coefficients. That is what happens in casting that is what happens in melting. So, wherever two phases are there interfacial shear stresses and interfacial heat transfer coefficients come into picture, but energy equation, momentum equation, continuity equation separately for both the phases can be taken.