Holy shit. Thank you Dr. Morrison! This completely cleared up how to solve these style problems for me. Never even put together that you're really solving two parts.
The density of the working fluid in a manometer must be higher than the fluid above it or the fluids would mix and switch positions. Carbon tetrachloride has a density of 1.5867 g/cc, which is higher than water, density_water=1.0 g/cc.
I don't get the static assumption at 5:28, I thought that the venturi effect caused the pressure to drop whenever a flow speeded up, so the pressure between point 1(free fluid) and the surface of the liquid inside the manometer can't be calculated just as a static column of fluid. The only way I can understand that is that the pressure inside the tube, even if the fluid is moving, is constant( the speed doesn't change because the area remains constant) so we can make that static assumption, is that correct? Thank you and sorry for my english.
On 5:20 we have manometer. In left side is it open or close to the pipe? When the diving cylinder is from firm metal and is under water pressure also acts to a gas in bottle. Please answer me. Thanks
Dear Dr. Morrison, I'm currently studying for a final exam in environmental facility design. Would you be able to post the remaining parts of this question? I look forward to hear from you. Thank you.
Why not just use the energy form of the Bernoulli equation and use the Velocity head term to find the Velocity? Isn't the Change in height relate to this term?
Mam can you explain why we can't just straight away equate the dynamic pressure i.e. 1/2*d*V^2 (where 'd' is density) to the manometer reading ? Isn't this how it's done in real life where the pitot tube has two outlets - one displays the total pressure, the other static pressure. Both of them are connected to the ends of a manometer and the difference gives the dynamic pressure ?
My treatment starts with a "law of physics" and shows at each step how we make assumptions to get to our answers. Thus, we start with the mechanical energy balance, which is an energy balance that has already been simplified for the following case: single input, single output, steady, no reaction, no temperature or phase change, adiabatic. What you propose is more intuitive, but actually, it does not show what principle of nature makes it correct. So, it may be a good way to remember the solution, but it does not allow us to see how we might analyze a different situation. Starting from the laws of physics (energy is conserved) has this advantage.
Really great video by the way, clear and helpful, thank you very much. Early in the Video you say something along the lines of, "I've chosen a problem from Gene Coplis " I am pretty sure I heard this wrong, what exactly is a 'Gene Coplis' ? I'm guessing that is the writer of a textbook, either way, I would like to know who or what you meant to say, and if there is any book or workbook you're referring to I'd like to know that as well. Thank You.
If the density of the manometer fluid is less than the density of the fluid above it, the layers are unstable and the upper fluid will flow down and exchange places with the lower fluid. Since the lower fluid has higher density, then, we do not get a negative sign under the root.
If the fluid in manometer has less density than the fluid in the flow, then can't we use an inverted U-tube manometer just so that negative root is eliminated.
i have a doubt regarding pitot tube. can you please help me .my doubt is how we can say that velocity at the point 2 is zero and how do we can say that pressure at point 2 is higherthan point1,if it is so how flow will takinplace from low pressure to higher pressure
it is because since the fluid is going through low pressure to high pressure..... it's velocity goes on to decrease and ultimately upon reaching the pitot tube it becomes zero......
The Pitot tube intersects just one streamline of the flow. It's like putting your hand out the car window: you feel a higher pressure on your hand due to the air coming at your hand an your hand stopping the flow. So, just at that location where the Pitot tube intersects the flow, the pressure is high. Out in the tube, for the whole flow, the pressure decreases in the direction of flow (unobstructed).
Yes I just had to sort out the same confusion from my lecturer (don't ask..). The term you need is [delta(v^2)] since it's [(v_2)^2 - (v_1)^2] but I started working with (delta v)^2 because he wrote it the same way it's shown in this video.
The entire time I just kept thinking about amazon wishlisting her any high quality pen or pencil just to make it stop. Everytime i missed simethign and had to rewind I said to myself "is hearing the sharpy really worth getting a better grade? ::sigh::" lol.
Holy shit. Thank you Dr. Morrison! This completely cleared up how to solve these style problems for me. Never even put together that you're really solving two parts.
Your video ends suddenly! But, It's a great explanation!
The density of the working fluid in a manometer must be higher than the fluid above it or the fluids would mix and switch positions. Carbon tetrachloride has a density of 1.5867 g/cc, which is higher than water, density_water=1.0 g/cc.
True; the calculation is completed in the handout. The handout is available at a link that is listed in the video description above.
Thanks alot! Probably the best explanation I could ever get compared to all my lazy shit lecturers in my university
I don't get the static assumption at 5:28, I thought that the venturi effect caused the pressure to drop whenever a flow speeded up, so the pressure between point 1(free fluid) and the surface of the liquid inside the manometer can't be calculated just as a static column of fluid. The only way I can understand that is that the pressure inside the tube, even if the fluid is moving, is constant( the speed doesn't change because the area remains constant) so we can make that static assumption, is that correct? Thank you and sorry for my english.
On 5:20 we have manometer. In left side is it open or close to the pipe? When the diving cylinder is from firm metal and is under water pressure also acts to a gas in bottle. Please answer me. Thanks
Dr. Morrison, Would the approach be different if the pitot static tube were positioned in the center of a venturi meter. Great explanation, thank you.
Ok, does the inlet size of the pitot tube effect the measurement?
Dear Dr. Morrison, I'm currently studying for a final exam in environmental facility design.
Would you be able to post the remaining parts of this question?
I look forward to hear from you.
Thank you.
Thank you Dr Morrison.
I love how you talk and explain things :D
Thanks!
Why not just use the energy form of the Bernoulli equation and use the Velocity head term to find the Velocity? Isn't the Change in height relate to this term?
Mam can you explain why we can't just straight away equate the dynamic pressure i.e. 1/2*d*V^2 (where 'd' is density) to the manometer reading ? Isn't this how it's done in real life where the pitot tube has two outlets - one displays the total pressure, the other static pressure. Both of them are connected to the ends of a manometer and the difference gives the dynamic pressure ?
My treatment starts with a "law of physics" and shows at each step how we make assumptions to get to our answers. Thus, we start with the mechanical energy balance, which is an energy balance that has already been simplified for the following case: single input, single output, steady, no reaction, no temperature or phase change, adiabatic. What you propose is more intuitive, but actually, it does not show what principle of nature makes it correct. So, it may be a good way to remember the solution, but it does not allow us to see how we might analyze a different situation. Starting from the laws of physics (energy is conserved) has this advantage.
Nice video, I might be absent when my proff explained this topic, hahahah. Thanks for this video, it is really helpful =)
answer is negative?
density of Pc is lower than Pw
i a bit confuse can u explain?
Fantastic Explanation !!AWesome
hi teacher this is very good presentation for pitot tube thank you
Really great video by the way, clear and helpful, thank you very much.
Early in the Video you say something along the lines of, "I've chosen a problem from Gene Coplis "
I am pretty sure I heard this wrong, what exactly is a 'Gene Coplis' ? I'm guessing that is the writer of a textbook, either way, I would like to know who or what you meant to say, and if there is any book or workbook you're referring to I'd like to know that as well. Thank You.
I'm referring to the textbook by Christie J. Geankoplis, Transport Processes and Unit Operations, 4th Edition, Prentice Hall, New York (2003).
Superb explanation.. awesome
Dat Ending
If the density of the manometer fluid is less than the density of the fluid above it, the layers are unstable and the upper fluid will flow down and exchange places with the lower fluid. Since the lower fluid has higher density, then, we do not get a negative sign under the root.
If the fluid in manometer has less density than the fluid in the flow, then can't we use an inverted U-tube manometer just so that negative root is eliminated.
Yeah, sorry about that. The solution is completed in the handout, which is available at the link listed in the video description above.
awesome and excellent explanation
thank you very much.your video is great
in real life where are they applied? eg which machine and how does it work?
They're on every airplane to measure air speed.
+DrMorrisonMTU I laughed out loud when someone called it a lightning rod
luv luv the explanations....very clear
Why am I getting negative under my root? Density of working fluid is less than density of water....manometer turn up or down shouldn't matter
Thank you so much 💕 You are great.
this is very helpful to instrumentation technician students
Thanks!
Thanks for the great video.
thank you....but the video is incomplete...
One of the greatest mysteries explained ( to me at least ) :-)
Great! Glad you found it of value. I'll make some more.
i have a doubt regarding pitot tube. can you please help me .my doubt is how we can say that velocity at the point 2 is zero and how do we can say that pressure at point 2 is higherthan point1,if it is so how flow will takinplace from low pressure to higher pressure
it is because since the fluid is going through low pressure to high pressure..... it's velocity goes on to decrease and ultimately upon reaching the pitot tube it becomes zero......
The Pitot tube intersects just one streamline of the flow. It's like putting your hand out the car window: you feel a higher pressure on your hand due to the air coming at your hand an your hand stopping the flow. So, just at that location where the Pitot tube intersects the flow, the pressure is high. Out in the tube, for the whole flow, the pressure decreases in the direction of flow (unobstructed).
@@DrMorrisonMTU Thank you for clarification
cool video ..the problem i couldn't handle the calligrapher sound
I got the handout, thank you for your valuable response
See the handout for final solution!
Thank you!
Delta V squared is NOT V1 squared - V2 squared.
Its just lazy notation. The equation is (v_2)^2 - (v_1)^2. Idk why she wrote (delta v)^2
Yes I just had to sort out the same confusion from my lecturer (don't ask..). The term you need is [delta(v^2)] since it's [(v_2)^2 - (v_1)^2] but I started working with (delta v)^2 because he wrote it the same way it's shown in this video.
@obaidCarkey turn your speakers down :)
Thank you so much
thanks a lot.
good job!
best explanation
Wow, that's very impressive... But the end of the video got cut off and there's no final solution!
RhoC is 1.58 so RhoC-RhoH20 is 0.58! No negative sign there!
very useful
thank u very much.......
nice one
thanks!
wow thxxx a lot.
Sopranos ripped off your ending
Incomplete vedio.. And focus on actual question
Good explanation but good god please stop writing in sharpie! I cringe everytime I hear that nails-on-a-chalkboard like squeal come from the paper.
The entire time I just kept thinking about amazon wishlisting her any high quality pen or pencil just to make it stop. Everytime i missed simethign and had to rewind I said to myself "is hearing the sharpy really worth getting a better grade? ::sigh::" lol.
Get me a panadol and scotch quick
please don't use this pen again because the noise made me stop the video and write the comment.thanks
Thank you!
thank you!