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If you have a classmate that you think these videos would help Please send them a link. Getting discovered is hard. I appreciate personal recommendations so much.

It's probably not harmful to think of it that way, even if technically not fully correct. Similar to in electrical circuits, we think of current flowing from positive to negative, even though we now know its electrons that are moving, and they move the other way. There's a bunch of these sorts of things throughout engineering curriculums, that while not entirely correct, it's still kind of ok to think that way, because it may still be useful in understanding a bigger picture, even if the details are a bit off.

​@@BrianBernardEngineering Thanks for the reply. I agree. We should keep in mind as many possibilities as possible for some conclusion in order to be able to see what works to what extent. It should not be exclusive. Doubt about correctness leads to the right answer. I don't speak English very well, sorry if there are mistakes.

You're welcome. Good luck this semester!

You're very welcome. This is maybe the best example in any engineering class to demonstrate that if you understand the formula, you don't actually need to know the formula at all. You can just do it on your own.

Thanks so much. I'll try. You might also be interested - I just launched an additional channel, "Explained By Cats" that is still educational, but a lot more entertaining, not just solving homework problems.

Happy to help. Most of the course will be 2D problems. But there will be a little 3d sprinkled in here and there.

You're very welcome. I hope you do great!

That's very kind to say, thank you so much for watching.

I prefer to not really think of the joints as 'things that move', and only consider the links as moving objects. I usually view a joint only restricting the motion of 1 of the attached links, not 2. Consider a 4 bar linkage. The crank has its motion restricted in 2 directions by the joint connecting crank to ground. Then the coupler is restricted in 2 directions by the joint connecting coupler to the crank. So that joint, connecting coupler to crank, I don't think of it as restricting the cranks movement, because the crank is already restricted from its other end. This isn't the only way to think about these problems, but it's my way.

A little counterintuitive, but yes, it would drain faster with a pipe extension to make the exit lower. Same premise works for siphoning as well. If you want to empty out something like a fish tank using a siphon, don't have the exit of the siphon right below the tank, get a longer pipe so the exit can be all the way down on the ground, lower exit is faster exit velocity since you convert more potential energy to kinetic with greater change in height.

It's usually the difference in pressure that actually matters. Pressure inside something pushes outward, atmospheric pressure pushes inward. The difference is important since you are summing forces in the two different directions to get a net force due to gauge pressure. This is why you don't actually measure atmospheric pressure usually. Atmospheric pressures changes a bit everyday due to weather. Since we only care about the gauge pressure, it doesn't matter exactly what the reference points absolute pressure is.

Awesome. Unfortunate that it's rare, but glad I can be the exception. Have a great semester!

Thanks so much for the kind words. Also especially good for students to hear that this stuff can actually have a real world application, not just homework!

​​​But I think the calculation conversion you made from ft3/s to gal/min is wrong.😂 Correct me I been mistaken. But anyway, you did great explanation and easy to understand. Tq😂​@@BrianBernardEngineering

You're right - that is definitely the right way to do it. In practice I usually try to ensure all my units are consistent at the start so that I know the answer will work out to be the correct units, which allows me to get away with being lazy - but I fully concede that it is risky and lazy. I'm making a note to do better next time. Good luck this semester.

@@BrianBernardEngineering thanks! My semesters are long gone though! I am here for the joys of fluid dynamics!

Main difference with submerged exit is increased pressure. However far under the surface, that's extra pressure. Bernoulli Eqn from inlet to outlet, you would then have a lower change in pressure between inlet and outlet, same change in height between inlet and outlet, therefore velocity difference would have to be smaller. Flow rate would be slower.

thanks so much. that really means a lot!

youre welcome, that's awesome. Keep up the good work!

Yea, this is definitely one of those things that seems like a two sentence topic ... until you try it yourself and realize there just keeps being one more step, then just one more, then ... and it keeps going lol.

@@BrianBernardEngineering yes! Then add multiple pumps in parallel, but different sized pumps. Then it becomes fun 😆 add then controls to it!

Yes, there is a hidden assumption I made here - that this 4 bar linkage is in the "open" configuration. There is a 2nd possible set of angles that would also work, sometimes called the "crossed" configuration, where the rocker points downward instead of upwards. So, though there are technically 2 possible solutions, if you choose which one you want, then there's only one! I predict that the vast majority of your homework and test problems, you will use the open configuration, so it should be your default choice. If for no other reason, it's often easier to draw.

Thanks so much. Have a good semester.

Thanks so much. I'm happy to help.

Think of it as "diminishing returns". The first tiny bit of fin added is hugely beneficial. Then the next bit a little less, and next bit even less. a longer fin is always better (more heat transfer) than a shorter fin, but each bit of extra length you add does a little bit less than the piece before it. Eventually you've essentially maxed out, so extra length added does practically nothing extra for heat transfer, so that lowers the efficiency. A long fin, that gets longer, you add very little heat transfer, but a lot of extra length, so this is inefficient.

@@BrianBernardEngineering thanks!! That's a great way to put it!!

Nope. Only need miter line to connect top/side views. From top to front is just straight down. Right to front just straight horizontal. I just double checked ... I have 3 different videos on missing views (easy to find in my 19 video engineering graphics playlist if you haven't seen it yet), and not one of them has an example with a missing front view. Always missing top or missing side view. Seems like an oversight I might want to create a new video for. Thank you for the idea.

@@BrianBernardEngineering Yeah, thanks for the explanation. What's funny is that the question on the exam said to draw a mitre line (in bold). My exercise bundle for the course only had missing top/side views only as well.

This really is one of those "easy ... once you know how" topics that looks really complex at first.

oh no, I think you're right. When looking up numbers on the Moody Diagram, it should be shifted left, one section, which changes the f values from about .015 to about .017. Final plot for the system curve would then be above the orange line, but below the blue line. I didn't catch any mistakes with the process. If you follow this process, but don't make division mistakes like I did, you should still be golden.

Sorry, it's been awhile since I worked on more than basic animations, so don't think I can help. Good luck though!

@@BrianBernardEngineering Not a big deal :D

I wasn't at any airports today, sorry, wasn't me. Woulda been an amazing coincidence.

Glad I could help. Keep up the good work!

Awesome! Glad I could help.

You are very welcome

good luck on your exams!

Good news, is that this is one of the harder momentum problems you are likely to see, so if you can follow this, good chance your homework and tests might be similar or even a step down in difficulty.

Yes, Pressure at the exit DOES make a difference. This video focused on examples where both ends are at atmospheric pressure, since that's the normal scenario we see most often, like when draining water out of a large aquarium. But if the exit is above atmospheric pressure, then the siphon height CAN be higher than the max I stated in this video. This is a really good sign, when you are able to learn something, and you then can find edge cases like this where it's a little bit different - shows you are understanding the bigger picture and not just memorizing details. Good work!

@@BrianBernardEngineering Great thanks a lot for the reply and the educational content :) Does the water pressure at the entrance make any difference? Or in other words, the water depth of the end of the siphon (the higher end, the one where water will enter from).

@@solank7620 Yes, you're right, both inlet and exit pressures matter, it's really the difference in pressure between inlet and outlet that is important. Regular siphon, same atm pressure both ends, the difference in height inlet to outlet is what determines the velocity in the siphon. When inlet and exit pressures are different, that difference also contributes to velocity in the siphon, just like difference in height. If inlet lower pressure than outlet, velocity will be lower than it would be based on height alone. Once you know velocity, then you can get max height. And you should get same height whether you measure above inlet or above outlet.

@@BrianBernardEngineering​​⁠​⁠​⁠that is so interesting because it is a bit counterintuitive - that the siphon height increases with higher pressure at the outlet of the siphon. But looking at the equation, that is true. But still I can’t get my head around it!

Good luck!

g is not being considered as rate of change of velocity. If you let the video play a few seconds more, you'll see that the right hand side of the equation has another term in nit. And think of the overall problem, if g = acceleration, then the rocket would be in free fall. Since the rocket is accelerating upwards, not freely downwards, dv/dt will definitely not equal g. However, you are correct that technically g is not constant. This is why I needed to specifically state in my assumptions at the top that g = constant. To realistically model the rocket's motion all the way up to a zero g environment, you would definitely need to have g rewritten as a function of height.

It may be easier than that. Consider the implications of minimum fuel rate. I can think of 2. First, your fuel tanks would be empty and you only account for the weight of the rocket itself. If you have extra fuel, that requires a higher fuel rate just to lift the fuel itself. Second, your acceleration would be 0, or something incredibly small like .00001, like the rocket is just barely able to overcome gravity. Any faster fuel burn rate than this would cause a faster acceleration. I think you may be able to just do the same FBD = kinetic diagram type analysis, where the force due to burning fuel only has to exactly counter the force of weight of the rocket, and all the other terms are assumed to be zero, in this minimum fuel rate scenario.

Sorry, but I think the method in this video won't help you with that problem. In your givens, you are already provided with both flowrate and head. This video is used when you only know one of those, in order to find the other. You'll need to find a different video with more information about pump power - and I don't have one of those sorry. Good luck to you.

heat transfer is a very inexact subject with lots of assumptions and estimates, so you may see this sort of approach in other parts of the course too.

Glad it was helpful!

You're welcome!

So "One" could say a well designed vessel is like a well balanced meal..? LoL... Your explanation for 80% of mass near the lower 20% of the bottom of vessel would create unfavorable conditions? Small Ocean going/ full displacement/ live aboard...trawler? Norhavn/ Katie krogen style ... Thank you You realize why some boats have a perfect design, when you would otherwise think ill change it... LoL And some GOOD boat designs are best left alone (for the most part) Smart ppl designing smart things (like yourself) Thank you

Thanks so much. Textbooks make this seem really convoluted. Break it down and it's not scary at all. Just gotta stay organized. Have a good summer!

we're up to 6 now, Let's goooooooooo

This chart came from the textbook authored by "Moran". It's definitely the paid textbook I recommend most for Thermodynamics (though I actually use free textbooks for my classes this year). You should be able to buy a used version for WAY cheaper than a new one. I see used copies for previous editions on Amazon for around $10 including shipping. Way better than buying new, if you don't need the homework problems from the new book. Very good colorful tables, even if you go back a few editions.

Thanks so much and I'm really glad I could help. About exposure, my channel is growing slowly, but it is growing. Every new viewer that sends a link of one of my videos to one of their classmates to help their friends out, all those links help recruit even more new students to watch and that adds up over time. Thanks again!

welcome back :)

You have a really good page it is odd that you don’t get the viewership you deserve… has anyone ever looked into your analytics? You are hitting all of the check marks, good thumbnail that’s simple explains the point, good content, visual backgrounds are good. I can’t put my finger on why you are not getting more views. One element is engagement, but that’s really it. There should be more people commenting on these. FYI I found you bc I’m studying for my FE. Maybe if you include FE in your caption you’d get more engagement? cheers

That comment was me I was on my personal page

@@eded4889 this is me lol

There's a lot of competition, and since I'm at a small school, I face a big disadvantage. At another school, a professor may teach 3 sections of the same class that each have 200 students in them. That's 600 students watching his videos, so that gives youtube a lot of information that this is good (even if its not). But for me, I only teach 1 section of Statics, with just 15 students. So, there's not as much up front indication to RUclips of how good it is, so it takes a lot longer before my videos start to rank higher in search and get suggested. Even though 50% of my channel's videos have been released in the last year, 9 of my top 10 videos this month are more than 2 years old. They are all hockey stick graphs, slow slow slow, until a critical audience is finally reached, then they sharply increase. Usually happens 2-3 years after a video is released before it starts getting traction.

Absolutely - do a youtube search for "How to Read Steam Tables - 5 Interpolation Example Problems" and you will probably find it. Or you can go to my channel page, it's one of my oldest videos so the production quality isn't as good as my newer videos, but I think it will still be useful for you. Good luck on finals!

we're happy to help!

Yes, LOVES them. They are definitely his favorite treat. Not really cat food though so just on special occasions.

Thanks so much for watching, but sorry to say that civil FE is not coming up soon. I'm working on Heat Transfer videos now, and next on my plan is a series of Mechanical videos for Design of Machinery. Good luck on your final!