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Why do proofs of Bernoulli's equation assume that forces on opposite ends point in different directions?

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4

2

$begingroup$

I've read 4 different books and yet nobody explains why forces $F_1$ ($=p_1A_1$) and $F_2$ ($=p_2A_2$) point in different directions. Shouldn't $F_2$ point in the same direction as $v_2$?

Since we're assuming that parts of fluid between $a$ and $b$ have the same kinetic and potential energies (same holds for $c$ and $d$), why do all proofs state that the change in work: $W_2 - W_1$ is equal to the change in energy $E_2 - E_1$? Work is equal to the change in kinetic energy, so $W_2 = W_1 = 0$ (because we assumed that fluid between each pair of points has the same energy).

Then there's the problem of signs, how do we determine which sign to choose and how do potential energies come into the equation?

fluid-dynamics flow continuum-mechanics bernoulli-equation

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edited 13 hours ago

Aaron Stevens

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asked 15 hours ago

user3711671user3711671

3766 bronze badges

$endgroup$

add a comment | 

4

2

$begingroup$

I've read 4 different books and yet nobody explains why forces $F_1$ ($=p_1A_1$) and $F_2$ ($=p_2A_2$) point in different directions. Shouldn't $F_2$ point in the same direction as $v_2$?

Since we're assuming that parts of fluid between $a$ and $b$ have the same kinetic and potential energies (same holds for $c$ and $d$), why do all proofs state that the change in work: $W_2 - W_1$ is equal to the change in energy $E_2 - E_1$? Work is equal to the change in kinetic energy, so $W_2 = W_1 = 0$ (because we assumed that fluid between each pair of points has the same energy).

Then there's the problem of signs, how do we determine which sign to choose and how do potential energies come into the equation?

fluid-dynamics flow continuum-mechanics bernoulli-equation

share|cite|improve this question

edited 13 hours ago

Aaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

asked 15 hours ago

user3711671user3711671

3766 bronze badges

$endgroup$

add a comment | 

$begingroup$

I've read 4 different books and yet nobody explains why forces $F_1$ ($=p_1A_1$) and $F_2$ ($=p_2A_2$) point in different directions. Shouldn't $F_2$ point in the same direction as $v_2$?

Since we're assuming that parts of fluid between $a$ and $b$ have the same kinetic and potential energies (same holds for $c$ and $d$), why do all proofs state that the change in work: $W_2 - W_1$ is equal to the change in energy $E_2 - E_1$? Work is equal to the change in kinetic energy, so $W_2 = W_1 = 0$ (because we assumed that fluid between each pair of points has the same energy).

Then there's the problem of signs, how do we determine which sign to choose and how do potential energies come into the equation?

fluid-dynamics flow continuum-mechanics bernoulli-equation

share|cite|improve this question

edited 13 hours ago

Aaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

asked 15 hours ago

user3711671user3711671

3766 bronze badges

$endgroup$

share|cite|improve this question

edited 13 hours ago

Aaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

asked 15 hours ago

user3711671user3711671

3766 bronze badges

share|cite|improve this question

edited 13 hours ago

Aaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

asked 15 hours ago

user3711671user3711671

3766 bronze badges

edited 13 hours ago

Aaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

edited 13 hours ago

Aaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

asked 15 hours ago

user3711671user3711671

3766 bronze badges

asked 15 hours ago

user3711671user3711671

3766 bronze badges

2 Answers
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4

$begingroup$

It's the definition of pressure. The pressure force is the force the stuff (fluid) external to the fluid in blue is exerting on the fluid in blue. It's like tension in a string, except with the sign changed.

In a string under tension the string outside the length you are interested in is pulling at both ends; in a rod or fluid under compression the outside is pushing at both ends.

share|cite|improve this answer

answered 13 hours ago

mike stonemike stone

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about the pressure from the left side for the surface $A_2$?The entire fluid is the same.
  $endgroup$
  – user3711671
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user3711671: That force is the force exerted by the blue highlighted fluid on the fluid outside the highlighted regions. We dont care about that. We only want the force on the bit of fluid (the blue regions between the red arrows) whose motion we are examining.
  $endgroup$
  – mike stone
  13 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

@mike stone Does a great job at addressing your first point. To address you second point, it is true that the net work changes the kinetic energy, i.e. $W_{net}=Delta K$. However, we are interested just in the work done by the external forces acting on the fluid. This means that $W_text {ext}=Delta E$. This is the work done by your forces on either end of the fluid segment.

Your third question is somewhat unclear to me, and this question runs dangerously close to being too broad by asking multiple questions, so I'll just leave it at this.

share|cite|improve this answer

answered 13 hours ago

Aaron StevensAaron Stevens

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$endgroup$

add a comment | 

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4

$begingroup$

It's the definition of pressure. The pressure force is the force the stuff (fluid) external to the fluid in blue is exerting on the fluid in blue. It's like tension in a string, except with the sign changed.

In a string under tension the string outside the length you are interested in is pulling at both ends; in a rod or fluid under compression the outside is pushing at both ends.

share|cite|improve this answer

answered 13 hours ago

mike stonemike stone

9,69811 gold badge1313 silver badges3030 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about the pressure from the left side for the surface $A_2$?The entire fluid is the same.
  $endgroup$
  – user3711671
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user3711671: That force is the force exerted by the blue highlighted fluid on the fluid outside the highlighted regions. We dont care about that. We only want the force on the bit of fluid (the blue regions between the red arrows) whose motion we are examining.
  $endgroup$
  – mike stone
  13 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

It's the definition of pressure. The pressure force is the force the stuff (fluid) external to the fluid in blue is exerting on the fluid in blue. It's like tension in a string, except with the sign changed.

In a string under tension the string outside the length you are interested in is pulling at both ends; in a rod or fluid under compression the outside is pushing at both ends.

share|cite|improve this answer

answered 13 hours ago

mike stonemike stone

9,69811 gold badge1313 silver badges3030 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about the pressure from the left side for the surface $A_2$?The entire fluid is the same.
  $endgroup$
  – user3711671
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user3711671: That force is the force exerted by the blue highlighted fluid on the fluid outside the highlighted regions. We dont care about that. We only want the force on the bit of fluid (the blue regions between the red arrows) whose motion we are examining.
  $endgroup$
  – mike stone
  13 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

It's the definition of pressure. The pressure force is the force the stuff (fluid) external to the fluid in blue is exerting on the fluid in blue. It's like tension in a string, except with the sign changed.

In a string under tension the string outside the length you are interested in is pulling at both ends; in a rod or fluid under compression the outside is pushing at both ends.

share|cite|improve this answer

answered 13 hours ago

mike stonemike stone

9,69811 gold badge1313 silver badges3030 bronze badges

$endgroup$

share|cite|improve this answer

answered 13 hours ago

mike stonemike stone

9,69811 gold badge1313 silver badges3030 bronze badges

answered 13 hours ago

mike stonemike stone

9,69811 gold badge1313 silver badges3030 bronze badges

answered 13 hours ago

mike stonemike stone

9,69811 gold badge1313 silver badges3030 bronze badges

2

$begingroup$

@mike stone Does a great job at addressing your first point. To address you second point, it is true that the net work changes the kinetic energy, i.e. $W_{net}=Delta K$. However, we are interested just in the work done by the external forces acting on the fluid. This means that $W_text {ext}=Delta E$. This is the work done by your forces on either end of the fluid segment.

Your third question is somewhat unclear to me, and this question runs dangerously close to being too broad by asking multiple questions, so I'll just leave it at this.

share|cite|improve this answer

answered 13 hours ago

Aaron StevensAaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

@mike stone Does a great job at addressing your first point. To address you second point, it is true that the net work changes the kinetic energy, i.e. $W_{net}=Delta K$. However, we are interested just in the work done by the external forces acting on the fluid. This means that $W_text {ext}=Delta E$. This is the work done by your forces on either end of the fluid segment.

Your third question is somewhat unclear to me, and this question runs dangerously close to being too broad by asking multiple questions, so I'll just leave it at this.

share|cite|improve this answer

answered 13 hours ago

Aaron StevensAaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

$endgroup$

add a comment | 

$begingroup$

@mike stone Does a great job at addressing your first point. To address you second point, it is true that the net work changes the kinetic energy, i.e. $W_{net}=Delta K$. However, we are interested just in the work done by the external forces acting on the fluid. This means that $W_text {ext}=Delta E$. This is the work done by your forces on either end of the fluid segment.

Your third question is somewhat unclear to me, and this question runs dangerously close to being too broad by asking multiple questions, so I'll just leave it at this.

share|cite|improve this answer

answered 13 hours ago

Aaron StevensAaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

$endgroup$

share|cite|improve this answer

answered 13 hours ago

Aaron StevensAaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

answered 13 hours ago

Aaron StevensAaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges

answered 13 hours ago

Aaron StevensAaron Stevens

21.5k44 gold badges3838 silver badges7676 bronze badges