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.

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!

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 KZbin 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!

You're welcome!

My first thought is that there are 2 ways to approach this. First is what you suggested, to start off with 2 guesses instead of just 1. I guessed velocities, but yes you can also start off by guessing Re instead that would work too, and each iteration you update your 2 guesses. Other method with will often work but depends on exactly how the pipes are connected and what you know and where you know it, is to treat the 3 pipes first as just 2 pipes in parallel with each other, and then separately, treat that combination of 2 parallel pipes as 1 combined inlet/outlet ... and have this combined inlet/outlet in parallel with the 3rd pipe. Which of the two methods is better would depend on the exact construction.

@@BrianBernardEngineering Gotcha, thanks so much for the quick reply too

Thank so much. I'll give credit to TA Serenity and TA Indiana for any interesting parts. It was their idea to insert video of slimy slugs anytime I say the slug units.

Your TA Indiana is just a regular kitty - no formal training at all - entirely self taught. For me - I was a nuclear power officer on submarines in the us navy. That was my first job after college. Nowadays, I have a phd in mechanical engineering, and I teach at Schreiner University in Kerrville, TX. Thanks so much for watching.

@@BrianBernardEngineering i would love some examples of an isentropic nozzle and its relation with static air temps at different points in the nozzle. How do the temps differ? especially, being isentropic (trick question?) Also, tie in the mach number at a specific point IN the nozzle. Want to see an example??

Statics is risky business alright. Glad I could help!

Glad you liked it. I couldn't have done it without your TA Indiana's help. He's a hard working kitty.

you're welcome!

​@@BrianBernardEngineering rewatching again for finals to refresh memory, best video i have seen for dof.

@@Koncholos Good luck on your final exam.

Oh - that's a great problem. I made a note to make a video about that in the future. Won't be anytime soon, but I like it. First, I would make a simplifying assumption that each hole out the side is either full or empty. It becomes WAY too complicated to consider the case where the water level is midway through the hole. So, water would flow out of each hole until it immediately stopped. You would need to solve for mass/volumetric flow rate out of each hole based on height. And then change in height would be a function of the sum of each hole. The interesting thing is that every hole would be on a Bernoulli streamline with the top of the tank. So - for example, the initial velocity out of the bottom hole is not affected at all by the other holes. It's still just a regular bernoulli equation. Each hole is independent of the others, at any given moment in time. They only relate to each other in the sense that the change in height will be faster since they all contribute to it. This may need to be solved somewhat discontinuous, solving for the amount of time to drain down to the bottom each of each hole, then for the next calculation, omit that hole that is now above the waterline when calculating time to the next hole. However, I think in the end if you plot height against time, this would still be a smooth curve, without corners or jumps, because flow out of each hole would approach zero as height approached its level. Good problem.

I think Ve is already a relative velocity in this problem. For a rocket, the particles emitted for thrust would always have their velocity measured relative to the rocket itself, not with respect to ground. Sort of like the experiment where a car drives 1 direction, and you fire a bullet from a gun out of the car shooting backwards. If the speed of the bullet and the speed of the car are the same, then a video camera on the ground would show the bullet just dropping straight down, not moving backwards or forwards at all. Same thing here, the particles exiting the rocket are moving very fast backwards relative to the rocket, but relative to the ground, they might be stationary, or might even still be traveling forwards, just not as fast forward as the ship.

@@BrianBernardEngineering My reasoning is the following. If the rocket burn the fuel at a constant rate, then the velocity at which the gases scape the rocket through the nozzle (lets call it Vj) should be constant in regard to a reference frame moving with the rocket. However, the reference frame itself is accelerating with a velocity U = U(t) and the velocity of the gases at the nozzle exit is constant in time. Therefore, the relative velocity is (Vj - U(t)), which is not constant and should be dealt with in the integral of dU.

This could potentially also be a Fluid Mechanics problem, not Thermodynamics. With inlet velocity and area, and outlet area, you can solve for outlet velocity using "Continuity Equation", ie conservation of mass. Then with outlet velocity, you can find outlet pressure using Bernoulli Equation (or Energy Equation if you want to account for head loss in the nozzle), then that outlet pressure could be used to find what you are looking for?

@@BrianBernardEngineering thanks Brian for clarifying this concept. Indeed I have seen that in order to simulate a nozzle, the solutions available are all about some CFD code.

awwww. TA Serenity thought she was being sweet. I'll give her the bad news :(

Nvm rewatched and realized it is our zeta1 from the second biot calculation

Great work, glad I could help! This must be how TA Indy feels when I thank all the time when all he does is play and try to distract me.

I feel ya. Hard to go straight from a derivation to application without a little extra help connecting the two.

185 seconds was completely arbitrary. I think I actually went backwards. Note that the final answer is exactly .500 m. That's too neat of a number to happen by chance. I probably solved for how much time it would take to get that 0.5m height ... and it turned out to be 185 seconds, so then I reversed the problem and asked what height would be after that time, now that I knew it would work out to a nice round final number. Nothing special about 185 seconds at all.

Yep, thats TA Indiana. He's always helping. Couldnt do it without him.

Good luck on your path to becoming a lecturer. I used to operate powerplants before i got into teaching, but am so happy to have switched, i really like this a lot.

Of course, they are very welcome to watch. If you have the ability to embed youtube videos in your course's LMS, if you click the "share" button, then in the popup, you can select "embed", and that will give you text so that you can paste the video directly in your course website instead of just the link. That's how I put the videos in my course websites so they can watch it directly there.

thanks so much

excellent memory, i never even showed my face in the cartoons. thanks for being such a long time fan.