this really saved me! love how he goes over the excercise slowly and let us have sometime to gulp down the idea! too bad i didnt find this sooner, i really suffered rereading the books without any help
I am literally passing my class right now because of him. My professor's online lecture is not good enough. Thanks Dr. It really helps a lot when you use many examples, it opens our minds more, and saves us time, it enriches our knowledge so we don't struggle to understand. Instead, we find it interesting and enjoyable.
Thank God we live in the information age. If I was taught fluids by my current professor without internet access, i would have lost interest in engineering...
Thank you so much professor! The midterm is in tomorrow morning. I just watched all those 12 videos from the series and did nothing else. But, I feel really confident. I hope to get an A in my Fluid class.
Thanks to my professor i had started to hate from anything about fluid. But thanks to you i started to understand something and not hate so much any more :) i have exam 2 days later and i am full of confident now.. THANK YOU SO MUCHH
At 38:00, he takes the pressure force P2 as negative! Pressure is a scalar, it has no direction. The area vector is actually pointing in the positive-x direction, so it should have been +P2A2Cos60. Kindly clarify.
+DarthCuber You are correct. He meant the pressure force, which is why he wrote pressure times area in the momentum equation. As you pointed out, pressure does not have a direction. Sometimes we use the term "pressure" instead of "pressure force" by accident (it is easy to make mistakes after lecturing for a long time).
Pressure acts TOWARDS the surface. So the pressure force at that exit can only have components that are downwards and toward the left. That's why it is negative according to the axis selection at the beginning of the problem.
Atakan Okan, but at 42:10 he puts the Pressure’s Y component as positive, looking upwards outwards of the surface. Can you explain why he did it differently, please?
@39:47 Using LME in this problem, what kind of pressure are we really dealing with? Was P1 at gage (it's labeled as "psi") while P2 was absolute (labeled as "psia")
38:30 I think a1 and a2 fields should have been used separately for v1 and v2 cos60 speeds. but we put v1 and v2 cos60 in a single a1 bracket. can you explain
Can someone explain to me the sign convention used during the Turning Vanes problems? How is he assigning the + and - signs when calculating the forces created by pressure at points 1 and 2?
His example at 17:15 for the direction of the force needed to hold the hose stationary doesn't seem to make sense to me. If I hold a pressurized hose and turn it on full blast, I have to push against it to keep it in place, not pull on it. The hose wants to slip back through my hand, rather than shoot forwards like he mentions. So, I think that the force that we are solving for is not the force applied by the person that is holding the hose, but rather the internal force between the nozzle and the hose. i.e. the joint between the hose and the nozzle. Then the direction of the force we're solving for makes sense, because without it, the nozzle would fly off of the hose. Can anyone tell if I'm right or wrong on this? Thanks
Excellent lecture, it was really helpful. However, homework 5.36 through 5.105??? That's 70 problems, or am I misunderstanding the wording. If that's the case, then I am definitely not doing enough practice.
In the Turning Vane Problems why is the Resultant Force orientation different between the two cases? Im sure he is right just a gap in my understanding. *Edit: I now think the diffference is in how the control volume is drawn. By definition the balance of linear momentum is the net external forces acting on the control volume is equal to the change in linear momentum within the CV and across the CS..
For the bracket fitting in the turning vane example, I would think that there be a vertical force on the bracket on the y direction as well caused by the y component of the velocity in the y-dir. Unless the problem explicitly states to find the x-direction only. Kindly clarify?
Great video! just one question: when computing Fr(y) in the "turning vanes"- why is there no contribution of the weight of the water on the left side of the momentum equation?
Over the short period of time the water is in contact with the vane, gravity has a negligible impact on the flow characteristics. If weight were factored in, it wouldn't impact the results significantly.
I think I missed something... why are we saying that the pressure at either ends of the curved turning vane is inward? also how do we know where the restraining force will apply?
I know it's kinda too late but might be useful to others. I think it is because the pressure force is distributed equally in all directions. Meaning that the pressure given is outside of the tube, so from it's point of view, it can only push 'into' the tube.
Thank u for that explanation , it would be help full, if u elaborate (explain) the 2nd line " Meaning that the pressure given is outside of the tube" .
YOGESH ‘pressure given’ is the pressure he drew on the diagram (P1 & P2). In this case I’d assume it’s Atmospheric pressure. So atmospheric pressure pushes the liquid inside. Let me know if you need more explanation, I’ll try rephrasing.
Саид Гильмутдинов Thank u : but if u look at video 12 @ ( 47:57 ) even though there is no ATM Pressure at both the ends , pr lines where pointing inwards , and I got a reply for CPP saying Pressure lines always point inwards .
Hello sir, Can you please help me solve this? a piping system of diameter 0.4m and length 200 m is leaking . the pipe will be pressurized in an effort to locate the leak. the pipe is pressurized to 55 atmosphere absolute pressure and held for 1hour. after 1 hour , the inside pressure of the pipe was noted to decrease to 50 atmosphere.how much per seconds is leaking from the pipe over this time period ? the compression coefficient k= 4.83*10^-10pa^-1 .( 1 atmosphere pressure =98kn/m^2.
37:50 Shouldnt A1V1 be negative since Area is pointing to the left like in earlier examples? And later P2A2sin60 should be positive since area is pointing outward and vector to the right, not left.
At 37:50 he writes A1P1 and in here Vector A not is in outward direction because we consider Force this time, not momentum (Vector A is negative when we consider only momentum, not force) And this explains your second question as well.
in the turning vane problem, why is v1 = v2 initially? is that something we should always assume when dealing with steady state flows ? also why don't we account for the y direction
Hi, I'm a bit confused about the 60 degree turning vane example. At @, for momentum in x direction, how could you deduct the right part ρV1A1(V2cos60-V1)? Isn't it supposed to be ρV1(-V1A1)+ρV2cos60(V2A2)?
Since mdot = rho V A, U can take them out from the equation. Since we also know mdot1=mdot2, RhoV1A1 = RhoV2A2 Then we end up with mdot*(V2-V1) He repeated this at least 2 times already ever since he first showed this example. So by now it should be clear that you can just do that.
@@maghraouilaila2229 Q1=Q2 because fluid is incompressible. So A1V1 = A2V2. @Zixuan Deng your equation is true but you can write A1V1 instead of A2V2. Thats what our instructor did.
@@sjcan Ah. I see that. Though doesnt that also make the y component negative? So -P2*A2*sin60, because he wrote positive and said "P2 is pointing up" but he just showed later that its pointing down
If anyone is pulling their hair about the results at 22:50 mark. Make sure you use the right units. METERS instead of CENTIMETERS. He got V2 = 9V1 , while calculating in meters it's V2 = 6V1. (Diameter was given in centimeters). I got F = 8571 kN, which I hope is about right.
This professor has the best Fluid lectures ever! Really helped me a lot. Very clear and straight to the point. thank you!
We're glad it helped you.
this really saved me! love how he goes over the excercise slowly and let us have sometime to gulp down the idea! too bad i didnt find this sooner, i really suffered rereading the books without any help
I am literally passing my class right now because of him. My professor's online lecture is not good enough. Thanks Dr. It really helps a lot when you use many examples, it opens our minds more, and saves us time, it enriches our knowledge so we don't struggle to understand. Instead, we find it interesting and enjoyable.
I feel you so much. My professor in fluids isn't even teaching us at all! Left us all alone. Thanks to Doctor Biddle, I'm still learning fluids
Glad to know that man! His lectures really help a lot!
Thank God we live in the information age. If I was taught fluids by my current professor without internet access, i would have lost interest in engineering...
1:02:00 to 1:02:59.. throws in so many great little reminders here and there. Great stuff.
ruclips.net/video/Wpg0BnC7FEU/видео.html
Thank you so much professor! The midterm is in tomorrow morning. I just watched all those 12 videos from the series and did nothing else. But, I feel really confident. I hope to get an A in my Fluid class.
Good luck!
So how was it?
So glad my university uses the same textbook lol. Thank you so much for these lectures!
my exam is in less than 24 hours and you just saved my life. thank you so much you're the best
49 minutes into the video is on point. Great content. Thanks for the time spent creating these videos
ruclips.net/video/Wpg0BnC7FEU/видео.html
I really appreciate first you professor Biddle, then that guy who asks question , they are my questions too 👍
Thanks to my professor i had started to hate from anything about fluid. But thanks to you i started to understand something and not hate so much any more :) i have exam 2 days later and i am full of confident now.. THANK YOU SO MUCHH
At 49:40 why is P2 pointing to the right?
Thanks so much! My teacher really sucks, your lectures are super helpful!!!!
Nice lecture, professor! I'm a brazillian student and I'm loving so much study Fluid Mechanics.
Olá. Glad you are enjoying the videos.
At 38:00, he takes the pressure force P2 as negative! Pressure is a scalar, it has no direction. The area vector is actually pointing in the positive-x direction, so it should have been +P2A2Cos60. Kindly clarify.
+DarthCuber You are correct. He meant the pressure force, which is why he wrote pressure times area in the momentum equation. As you pointed out, pressure does not have a direction. Sometimes we use the term "pressure" instead of "pressure force" by accident (it is easy to make mistakes after lecturing for a long time).
but shouldn't P1A1 be negative since the flow is entering the pipe ?
no, p1a1 is about forces on the system. what you said is applied only during the control surface analysis of the equation.
thank you for question
@@CPPMechEngTutorials thank you for answer
Great class, comments by the proffesor are insightful.
I wish he would teach Thermo and Dynamics so I can go back a review Dynamics with his lectures.
Thank you so much for the nice lecture, professor. The wonderful tips for the seperated figures of forces and velocities is also very useful.
You're welcome!
And the same case at 54:59 in x direction. Besides, at 54:59, the left side shouldn't be -Frx+P1A1-P2A2cosΘ?
Why -P2A2costheta? I know P1A1 shoudnt have the costheta, but why negative
love your lectures sir
@32:49 Why is V2•A2 = +V2A2 ?
Is the 45° not considered?
at 45:00 the for y component of the force is going up, so why isn't P2A2 sin60 positive?
forces direction must toward to CV
Great video!! really helped clear up sign convention for pressure and momentum...thank you good sir!!
ruclips.net/video/Wpg0BnC7FEU/видео.html
In the vane problem, can we solve for V2 using continuity instead of Bernoulli? My numbers do not match the example.
Mee too, i believe that theres something wrong in the given or in the answer key
best prof ever
34:56 pipe vane ;Pin=sumn Pout=0
46:16 piece of pipe network
48:17 NOTE
59:25 width
i appreciate your work sir, very clear teachings..
This is how fluids should be taught.
Thanks. Dr. Biddle has honed his craft for many years.
At 45:08 Why is P2A2sin60 negative ? how does he know which direction its going?
No it shouldnt be toward left, it makes no sense. P2A2sin60 should have been positive as was pointed out by the uploader below.
Pressure acts TOWARDS the surface. So the pressure force at that exit can only have components that are downwards and toward the left. That's why it is negative according to the axis selection at the beginning of the problem.
@Atakan Okan You're a life savor buddy! ^_^
Atakan Okan thanks man ☺️
Atakan Okan, but at 42:10 he puts the Pressure’s Y component as positive, looking upwards outwards of the surface. Can you explain why he did it differently, please?
9:13 and 9:42 I’m lost, why is the velocity 1 positive and dot product v1•A1 negative?
dot product angle between them is 180 area vector is always points outwards
@39:47 Using LME in this problem, what kind of pressure are we really dealing with? Was P1 at gage (it's labeled as "psi") while P2 was absolute (labeled as "psia")
At 50:47 shouldn't the it be P2A2cos(theta)?
nvm someone pointed it out and he fixed it but he multiplied the cos(theta) with P1A1 instead of P2A2.
Bless You
Thank you for your, very comprehensive explanations! I've subbed
Welcome to the channel!
Thank you professor! Save my life! Dammmmn Much better for my own professor!
You're welcome.
Thanks prof. Ive final exam tomorrow morning.
Great lecture!
38:30 I think a1 and a2 fields should have been used separately for v1 and v2 cos60 speeds. but we put v1 and v2 cos60 in a single a1 bracket. can you explain
what's the value for gamma?
Really helpful, thank you!
Can someone explain to me the sign convention used during the Turning Vanes problems? How is he assigning the + and - signs when calculating the forces created by pressure at points 1 and 2?
The 2nd turning vane example... why is P2A2 negative? Thank you for a helpful lecture.
did you get an answer or understanding? @amjadsamra105
@48:30 found my answer
@9:33 why is (A1V1) negative? whats an area vector?
I have the same questions!
at 34:01, how is the velocities the same? and how can you take common for the velocities
His example at 17:15 for the direction of the force needed to hold the hose stationary doesn't seem to make sense to me.
If I hold a pressurized hose and turn it on full blast, I have to push against it to keep it in place, not pull on it. The hose wants to slip back through my hand, rather than shoot forwards like he mentions.
So, I think that the force that we are solving for is not the force applied by the person that is holding the hose, but rather the internal force between the nozzle and the hose. i.e. the joint between the hose and the nozzle. Then the direction of the force we're solving for makes sense, because without it, the nozzle would fly off of the hose.
Can anyone tell if I'm right or wrong on this?
Thanks
Excellent lecture, it was really helpful. However, homework 5.36 through 5.105??? That's 70 problems, or am I misunderstanding the wording. If that's the case, then I am definitely not doing enough practice.
There is a syllabus on www.cpp.edu/meonline/fluid-mechanics.shtml under Dr. Biddle's section.
Do we just assume pressure is constant across the secion of a pipe because the difference between pressure at the top and bottom is negligible?
In the Turning Vane Problems why is the Resultant Force orientation different between the two cases? Im sure he is right just a gap in my understanding.
*Edit: I now think the diffference is in how the control volume is drawn. By definition the balance of linear momentum is the net external forces acting on the control volume is equal to the change in linear momentum within the CV and across the CS..
Do you have any videos dealing with a sluice gate?
Why P2=0??? while we have p(atmosphere)
For the bracket fitting in the turning vane example, I would think that there be a vertical force on the bracket on the y direction as well caused by the y component of the velocity in the y-dir. Unless the problem explicitly states to find the x-direction only. Kindly clarify?
This is in the first example but I see that the second example of turning vane does address this.
Great video! just one question: when computing Fr(y) in the "turning vanes"- why is there no contribution of the weight of the water on the left side of the momentum equation?
Over the short period of time the water is in contact with the vane, gravity has a negligible impact on the flow characteristics. If weight were factored in, it wouldn't impact the results significantly.
@42:14 why is it negative P2A2 for the force in the x but positive P2A2 in y-dir?
@48:30 found my answer
I think I missed something... why are we saying that the pressure at either ends of the curved turning vane is inward? also how do we know where the restraining force will apply?
Can you provide a time stamp for the pressure comment?
I know it's kinda too late but might be useful to others. I think it is because the pressure force is distributed equally in all directions. Meaning that the pressure given is outside of the tube, so from it's point of view, it can only push 'into' the tube.
Thank u for that explanation , it would be help full, if u elaborate (explain) the 2nd line " Meaning that the pressure given is outside of the tube" .
YOGESH ‘pressure given’ is the pressure he drew on the diagram (P1 & P2). In this case I’d assume it’s Atmospheric pressure.
So atmospheric pressure pushes the liquid inside. Let me know if you need more explanation, I’ll try rephrasing.
Саид Гильмутдинов Thank u : but if u look at video 12 @ ( 47:57 ) even though there is no ATM Pressure at both the ends , pr lines where pointing inwards , and I got a reply for CPP saying Pressure lines always point inwards .
Hello sir,
Can you please help me solve this?
a piping system of diameter 0.4m and length 200 m is leaking . the pipe will be pressurized in an effort to locate the leak. the pipe is pressurized to 55 atmosphere absolute pressure and held for 1hour. after 1 hour , the inside pressure of the pipe was noted to decrease to 50 atmosphere.how much per seconds is leaking from the pipe over this time period ? the compression coefficient k= 4.83*10^-10pa^-1 .( 1 atmosphere pressure =98kn/m^2.
really so effective lesson thnx lot..
Our pleasure. :)
19:57 don't you multiply by V1 first before you subtract the velocities?
23:49 why doesn't the force have a y-component?
thank you so much
+kevin shepard There will be more videos from the ME department in the near future too. :)
+CPPMechEngTutorials this chanel is very useful for my lesson so I subscribed .one more thing... soundtruck is awesome.:-)
+BEN KADİR Very cheerful.
what book is the professor checking?
Can you provide a time where he checks the book?
25:10
The class textbook: Munson et al., "Fundamentals of Fluid Mechanics", 7th ed.
01:01:54 what happened to "mü" kinematic viscosity, was it?
37:50 Shouldnt A1V1 be negative since Area is pointing to the left like in earlier examples? And later P2A2sin60 should be positive since area is pointing outward and vector to the right, not left.
At 37:50 he writes A1P1 and in here Vector A not is in outward direction because we consider Force this time, not momentum (Vector A is negative when we consider only momentum, not force) And this explains your second question as well.
Can anybody tell me at which video they teach vortex motion.
Wait how did he get V2=9V1 didn’t he say it’s 9/3^2 which is 1
thank u
@ 33:54 does the professor assume that V1=V2? is this because the areas are equal where the fluid enters and leaves the cv?
Yup, you're right. Since areas are equal, therefore from continuity equation, V1=V2.
in the turning vane problem, why is v1 = v2 initially?
is that something we should always assume when dealing with steady state flows ?
also why don't we account for the y direction
Contunity equation va1=va2 area is the same so v1=v2
Hi, I'm a bit confused about the 60 degree turning vane example. At @, for momentum in x direction, how could you deduct the right part ρV1A1(V2cos60-V1)? Isn't it supposed to be ρV1(-V1A1)+ρV2cos60(V2A2)?
Since mdot = rho V A,
U can take them out from the equation.
Since we also know mdot1=mdot2,
RhoV1A1 = RhoV2A2
Then we end up with mdot*(V2-V1)
He repeated this at least 2 times already ever since he first showed this example. So by now it should be clear that you can just do that.
@@guanhan898 can you replay , me too i didn't understand nothing
@@maghraouilaila2229 Q1=Q2 because fluid is incompressible. So A1V1 = A2V2. @Zixuan Deng your equation is true but you can write A1V1 instead of A2V2. Thats what our instructor did.
For the pipe fitting where does the minus sign for “-P2A2cos60” come from
The pressure force at the right side acts toward the cv. So the pressure force is acting left and down. Therefore the x-component would be negative.
@@sjcan Ah. I see that. Though doesnt that also make the y component negative? So -P2*A2*sin60, because he wrote positive and said "P2 is pointing up" but he just showed later that its pointing down
@@XplosiveAction He switches it later on.
Which textbook is used in this course? Can I find a copy online?
Munson et al. "Fundamentals of Fluid Mechanics (7th ed)"
Sir book name please🙏
@47.57 why is P2 drawn inwards ? Fluid is flowing outwards .
Pressure forces always point inwards.
CPPMechEngTutorials : Thank u
wheres angular momentum?
I do not see how Fy ended up being 1685 if we ad up the term in the right to the P2A2*sin(60)?
I am stuck with the same problem. Can someone explain please?
Mee too, i think theres something wrong
If anyone is pulling their hair about the results at 22:50 mark. Make sure you use the right units. METERS instead of CENTIMETERS. He got V2 = 9V1 , while calculating in meters it's V2 = 6V1. (Diameter was given in centimeters). I got F = 8571 kN, which I hope is about right.
I dont think v2=6v1
23:18
What the fuck happened to A2
When Los problemas de valencia son mas hards v:
as
We
Wow
Okay q
THE PROFESSOR WAS ALSO CONFUSED WHAT HE HAD DOING