It's really helpful to see things explained multiple times, in different ways. Look at the textbook, participate in lecture, watch a video - all 3 will be different, but hopefully in a way that they can complement each other and all fill in the gaps that the others were missing.
When you did Bernoulli between 1 and 2, you did not consider points 3 and 4 as they did not exist! Same for 2 and 4 when you did Bernoulli between 1 & 3 Same for 2 & 3 when you did Bernoulli between 1 &4. Can you explain why?
Consider a single molecule of water that starts at point 1. It has energy. It's energy is stored as pressure, velocity, and gravitational PE. In this branching pipe flow problem, this single molecule could go to either path 2, 3, or 4. The total flow in the entire pipe splits and goes all 3 places. But a single molecule of water can only go down 1 path. So - we follow that 1 molecule, itself. And, no matter which path it takes, conservation of energy still applies. Once that molecule makes it to its branch, either 2 or 3 or 4 ... due to conservation of energy, it will still have the same amount of energy that it had at point 1. It may be in a different form, maybe lower velocity and higher pressure for example. But - it has to have same energy as it started with. When using Bernoulli Equation (or later in the course "Energy Equation"), you follow an individual molecule along a streamline. So any point that this single molecule could reach - they all have to have the same energy. Even though its impossible for a molecule at point 2, to reach point 3 ... since point 1 connects to both 2 and 3, this will make it so that 2 and 3 end up having same amount of energy, even though its not really correct to compare them to each other directly ... you should compare each of them against point 1. I hope this helped. :)
I thought I already posted a comment . First thanks for an informative and easy to understand presentation . I really liked the concept of thinking that the stream line was a series of molecules . But in all the conservation derivation whether heat ,mass , energy they all are based on the principle what goes in comes out . In your analysis it seem what comes out is 3 times more out . Is this a nuance of bernouli equation and not related to conservation laws .
Don't forget the continuity equation, the conservation of mass. The 1 inlet is being split into 3 outlets. Not necessarily exactly 1/3 through each outlet, but for discussion purposes, 1/3 to each outlet. So, if each stream of molecules has the same amount of energy (per molecule, or per unit volume) along the streamlines, you don't end up with 3x energy out, because only 1/3 goes through each path, so 1 in and 1/3 + 1/3 + 1/3 out. Thanks for watching. If you liked this video, I have a full fluid mechanics playlist that you might find valuable.
So for frictionless flow in a manifold with same diameter pipe along manifold the pressure will rise as you cross the lateral since velocity downstream of lateral is slower than upstream of the lateral . Thanks for clarifying
Interesting question, and my first instinct was that there was no change, these points are still along the same streamline and all Bernoulli assumptions are still met, so yes, velocity would be the same , no change - but that's not entirely correct. Your scenario actually kinda breaks the problem. If 3 and 4 are closed - then ALL the flow has to go through 2. That means Q1 = Q2. You can actually solve for velocity as A1V1 = A2V2, just the continuity equation by itself. You get an answer of V2=143 ft/s. This is WAY faster that calculated by Bernoulli Eqn. Why? It's because P2 would now be wrong. If all the flow is going through 2, and none going through 3 or 4, then the given value P2=5.0psi would now be impossible. The better question in this new scenario would be, what would be the pressure at exit 2. You'd use givens at point 1, and continuity equation to find velocity at 2. Then you'd use Bernoulli equation and givens at point 1 and now known velocity at point 2, to find the pressure at point 2 ... which would not be 5psi. Thank you for the question!
in video problem, because mechanical energy is conserved (bernoulli equation), then bernoulli equation must be: mech energy at poin1 = mech energy at point 2 + mech energy at poin 3 + mech energy at poin 4 or (KE + PE + flow energy) at point 1 = (KE + PE + flow energy) at point 2 + (KE + PE + flow energy) at point 3 + (KE + PE + flow energy) at point 4 Why didn't we do that? but in the problem answer, we say: mech energy at point 1 = only mechanical energy at point 2 or (KE + PE + flow energy) at point 1 = (KE + PE + flow energy) at point 2. Isn't it wrong?
@wateressam1889 It depends on whether you are considering the perspective of a single water molecule or the total flow. When using the Bernoulli Equation, you should be thinking in terms of a single unit of water, not the entire flow. Think of a single molecule of water. That molecule has its own amount of energy, split in KE, PE, and Pressure. As it moves, it's own energy can change forms. That's the Bernoulli Equation. The Bernoulli Equation follows a single molecule of water along its path and looks at how that single molecule's energy changes forms. The Bernoulli Equation does NOT follow ALL of the water. It's not a summation of all the energy in the entire stream. That sort of analysis IS possible though. Probably the next chapter of your textbook will start to introduce you to "Control Volumes" and the "Reynolds Transport Theorem". That is when Energy1 = Energy2 + Energy3 + Energy4. This is essentially done by combining both the Bernoulli Equation and the Continuity Equation into a single equation (that single equation is Reynolds Transport Theorem) that includes both energy and mass flow rate. To sum up - for right now, consider Bernoulli Equation as only following a single molecule of water as it completes its own individual journey. Then later, you will learn a new solution technique involving control volumes, where you analyze the sum of all flow into a junction vs all the flow out of a junction, and this will combine mass flow rate and energy (or mass flow rate and momentum , since this technique of using control volumes is especially good for solving momentum problems).
ok, it's a single particle, but we can treat/choose this single particle as a thermodynamic system. so, the question is: can we say that "the total energy of this thermodynamic system at exit 2 does not change whether we close exit 3 and exit 4 together or we don't" ??? can we say that "the total energy of this thermodynamic system at exit 2 equals to the total energy of this thermodynamic system at inlet 1 whether we close exit 3 and exit 4 together or we don't" ???
we can also choose the branched pipe system (single energy input and three energy outputs at 1,2,3 and 4 respectively) as a thermodynamic system (control volume system) and apply conservation of energy princible (first law of thermodynamics) on it. Cann't we do that????
I found a question on the same example saying that bernoulli between point 1 and any outlet point will get us an inaccurate result . so the question was what should we introduce to bernoulli to get accurate result ? can u help me with this question?
It sounds like they are hinting towards Friction (ie Head Loss). One of the assumptions of the Bernoulli Equation is that there is zero head loss. This is useful for general understanding, but to get a more accurate answer, you would need to use Head Loss, which is friction removing energy (head) from the system. You'll calculate this friction head loss using the Darcy Weisbach Equation, and within that equation, you'll use the Moody Diagram to find the "f" term in the Darcy Weisbach Equation. There are also "minor losses", which are energy losses due to things like corners and bends and changes in pipe diameter. But overall, "head loss" is probably what you are looking for.
Thank you so much for this. I was lost all week in lecture, great explanation and clean layout!!
It's really helpful to see things explained multiple times, in different ways. Look at the textbook, participate in lecture, watch a video - all 3 will be different, but hopefully in a way that they can complement each other and all fill in the gaps that the others were missing.
U make it very clear and easy tnx a lot bruh keep going
I'm happy to help and glad you like it. I'll keep at it.
When you did Bernoulli between 1 and 2, you did not consider points 3 and 4 as they did not exist!
Same for 2 and 4 when you did Bernoulli between 1 & 3
Same for 2 & 3 when you did Bernoulli between 1 &4.
Can you explain why?
Consider a single molecule of water that starts at point 1. It has energy. It's energy is stored as pressure, velocity, and gravitational PE. In this branching pipe flow problem, this single molecule could go to either path 2, 3, or 4. The total flow in the entire pipe splits and goes all 3 places. But a single molecule of water can only go down 1 path. So - we follow that 1 molecule, itself. And, no matter which path it takes, conservation of energy still applies. Once that molecule makes it to its branch, either 2 or 3 or 4 ... due to conservation of energy, it will still have the same amount of energy that it had at point 1. It may be in a different form, maybe lower velocity and higher pressure for example. But - it has to have same energy as it started with. When using Bernoulli Equation (or later in the course "Energy Equation"), you follow an individual molecule along a streamline. So any point that this single molecule could reach - they all have to have the same energy. Even though its impossible for a molecule at point 2, to reach point 3 ... since point 1 connects to both 2 and 3, this will make it so that 2 and 3 end up having same amount of energy, even though its not really correct to compare them to each other directly ... you should compare each of them against point 1. I hope this helped. :)
I thought I already posted a comment . First thanks for an informative and easy to understand presentation . I really liked the concept of thinking that the stream line was a series of molecules . But in all the conservation derivation whether heat ,mass , energy they all are based on the principle what goes in comes out . In your analysis it seem what comes out is 3 times more out . Is this a nuance of bernouli equation and not related to conservation laws .
Don't forget the continuity equation, the conservation of mass. The 1 inlet is being split into 3 outlets. Not necessarily exactly 1/3 through each outlet, but for discussion purposes, 1/3 to each outlet. So, if each stream of molecules has the same amount of energy (per molecule, or per unit volume) along the streamlines, you don't end up with 3x energy out, because only 1/3 goes through each path, so 1 in and 1/3 + 1/3 + 1/3 out. Thanks for watching. If you liked this video, I have a full fluid mechanics playlist that you might find valuable.
So for frictionless flow in a manifold with same diameter pipe along manifold the pressure will rise as you cross the lateral since velocity downstream of lateral is slower than upstream of the lateral . Thanks for clarifying
if we closed the exit at 3 and 4 together, is there any change in bernoulli equation from 1 to 2? if there, what is it?
Interesting question, and my first instinct was that there was no change, these points are still along the same streamline and all Bernoulli assumptions are still met, so yes, velocity would be the same , no change - but that's not entirely correct. Your scenario actually kinda breaks the problem. If 3 and 4 are closed - then ALL the flow has to go through 2. That means Q1 = Q2. You can actually solve for velocity as A1V1 = A2V2, just the continuity equation by itself. You get an answer of V2=143 ft/s. This is WAY faster that calculated by Bernoulli Eqn. Why? It's because P2 would now be wrong. If all the flow is going through 2, and none going through 3 or 4, then the given value P2=5.0psi would now be impossible. The better question in this new scenario would be, what would be the pressure at exit 2. You'd use givens at point 1, and continuity equation to find velocity at 2. Then you'd use Bernoulli equation and givens at point 1 and now known velocity at point 2, to find the pressure at point 2 ... which would not be 5psi. Thank you for the question!
in video problem, because mechanical energy is conserved (bernoulli equation), then bernoulli equation must be:
mech energy at poin1 =
mech energy at point 2 +
mech energy at poin 3 +
mech energy at poin 4
or
(KE + PE + flow energy) at point 1 =
(KE + PE + flow energy) at point 2 +
(KE + PE + flow energy) at point 3 +
(KE + PE + flow energy) at point 4
Why didn't we do that?
but in the problem answer, we say:
mech energy at point 1 = only mechanical energy at point 2
or
(KE + PE + flow energy) at point 1 =
(KE + PE + flow energy) at point 2.
Isn't it wrong?
@wateressam1889 It depends on whether you are considering the perspective of a single water molecule or the total flow. When using the Bernoulli Equation, you should be thinking in terms of a single unit of water, not the entire flow. Think of a single molecule of water. That molecule has its own amount of energy, split in KE, PE, and Pressure. As it moves, it's own energy can change forms. That's the Bernoulli Equation. The Bernoulli Equation follows a single molecule of water along its path and looks at how that single molecule's energy changes forms. The Bernoulli Equation does NOT follow ALL of the water. It's not a summation of all the energy in the entire stream. That sort of analysis IS possible though. Probably the next chapter of your textbook will start to introduce you to "Control Volumes" and the "Reynolds Transport Theorem". That is when Energy1 = Energy2 + Energy3 + Energy4. This is essentially done by combining both the Bernoulli Equation and the Continuity Equation into a single equation (that single equation is Reynolds Transport Theorem) that includes both energy and mass flow rate. To sum up - for right now, consider Bernoulli Equation as only following a single molecule of water as it completes its own individual journey. Then later, you will learn a new solution technique involving control volumes, where you analyze the sum of all flow into a junction vs all the flow out of a junction, and this will combine mass flow rate and energy (or mass flow rate and momentum , since this technique of using control volumes is especially good for solving momentum problems).
ok, it's a single particle, but we can treat/choose this single particle as a thermodynamic system. so, the question is: can we say that "the total energy of this thermodynamic system at exit 2 does not change whether we close exit 3 and exit 4 together or we don't" ???
can we say that "the total energy of this thermodynamic system at exit 2 equals to the total energy of this thermodynamic system at inlet 1 whether we close exit 3 and exit 4 together or we don't" ???
we can also choose the branched pipe system (single energy input and three energy outputs at 1,2,3 and 4 respectively) as a thermodynamic system (control volume system) and apply conservation of energy princible (first law of thermodynamics) on it. Cann't we do that????
I found a question on the same example saying that bernoulli between point 1 and any outlet point will get us an inaccurate result . so the question was what should we introduce to bernoulli to get accurate result ? can u help me with this question?
It sounds like they are hinting towards Friction (ie Head Loss). One of the assumptions of the Bernoulli Equation is that there is zero head loss. This is useful for general understanding, but to get a more accurate answer, you would need to use Head Loss, which is friction removing energy (head) from the system. You'll calculate this friction head loss using the Darcy Weisbach Equation, and within that equation, you'll use the Moody Diagram to find the "f" term in the Darcy Weisbach Equation. There are also "minor losses", which are energy losses due to things like corners and bends and changes in pipe diameter. But overall, "head loss" is probably what you are looking for.
@@BrianBernardEngineering thank you
Only in an imperial system can "Head" ever equal "Feet" smh
😂