Bernoulli's equation and the work energy theorem

[Chenkel]

TL;DR Summary

I have a seemingly very simple question which has me stumped. It's related to the link to the following LibreTexts article https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/14%3A_Fluid_Mechanics/14.08%3A_Bernoullis_Equation

Hello physics researchers, teachers and enthusiasts.

I notice one little thing is confusing me in the derivation of Bernoulli's equation in the article, they write:$$dW = dK + dU$$where dW is the work done to the fluid, dK is the change in kinetic energy of the fluid, and dU is the change in gravitational potential energy of the fluid. But I thought with the work energy theorem, the work done to the fluid is equal to the change in kinetic energy:$$dW = dK$$Why is there an additional term of potential energy in their equation? My apologies if this is a little too simple!

 

[kuruman]

Potential energy is gained or lost if there is a change in height of the fluid as it moves.

 

[Chenkel]

Potential energy is gained or lost if there is a change in height of the fluid as it moves.

I agree with that, but the work done to fluid should equal mgh, so I would expect the work energy equation to be something like $$mgh = dK$$The equation they have is different.

Thank you for the response.

 

[kuruman]

I agree with that, but the work done to fluid should equal mgh, so I would expect the work energy equation to be something like $$mgh = dK$$The equation they have is different.

Thank you for the response.

The work energy theorem says $W_{net}=\Delta K$. The term $W_{net}$ is the sum of two works done on the fluid. One work, $dm~ g~\Delta h$ is done by gravity and the other $p\Delta V$ is done by pressure (see equation 14.8.8).

 

[Chenkel]

The work energy theorem says $W_{net}=\Delta K$. The term $W_{net}$ is the sum of two works done on the fluid. One work, $dm~ g~\Delta h$ is done by gravity and the other $p\Delta V$ is done by pressure (see equation 14.8.8).

What happens if the pressure difference is caused by gravity? In that situation would we be summing the work done by gravity twice in a round about way? The mgh term, and the pressure force through a distance term?

 

[kuruman]

Are you asking about the static pressure when the fluid is not moving? Try this at home. Put a straw upright in a glass of water. Plug the top end and lift the straw out of the glass. The water will remain in the straw.
Question 1: What is pressure at the bottom water-air interface? Why?
Now lift your finger that plugs the top end. The water will spill out.
Question 2: As soon as you lift the finger, what is the pressure (a) at the bottom air-water interface and (b) at the top air-water interface? Why?

 

[Chenkel]

Are you asking about the static pressure when the fluid is not moving? Try this at home. Put a straw upright in a glass of water. Plug the top end and lift the straw out of the glass. The water will remain in the straw.
Question 1: What is pressure at the bottom water-air interface? Why?
Now lift your finger that plugs the top end. The water will spill out.
Question 2: As soon as you lift the finger, what is the pressure (a) at the bottom air-water interface and (b) at the top air-water interface? Why?

For question 1: I'm guessing the pressure at the bottom of the straw is 1 atmosphere at sea level, and the pressure at the top of the straw is 1 atmosphere minus $\rho g h$ where h is the height of the straw

For question 2: When you lift your finger I'm guessing the pressure at the bottom is still one atmosphere, and the pressure at the top becomes one atmosphere.

I'm wondering if the pressure in Bernoulli's equation is not pressure from gravity but pressure from some other external force, and I am imagining some moving (non static) liquid. It just doesn't seem to make sense to me to treat the vertical work done by gravity, and the work done by pressure as different work quantities unless the work done by pressure is the result of a non gravitational external force.

 

[erobz]

For question 1: I'm guessing the pressure at the bottom of the straw is 1 atmosphere at sea level, and the pressure at the top of the straw is 1 atmosphere minus $\rho g h$ where h is the height of the straw

For question 2: When you lift your finger I'm guessing the pressure at the bottom is still one atmosphere, and the pressure at the top becomes one atmosphere.

I'm wondering if the pressure in Bernoulli's equation is not pressure from gravity but pressure from some other external force, and I am imagining some moving (non static) liquid. It just doesn't seem to make sense to me to treat the vertical work done by gravity, and the work done by pressure as different work quantities unless the work done by pressure is the result of a non gravitational external force.

$z$ for "elevation" in Bernoulli's ( work by gravity per unit mass ). It is a different term than $\frac{P}{\gamma}$, they are independent of each other in that derivation. That $P$ is from "something" doing work on the fluid. You are adding the contributions from each together along with a kinetic energy per unit mass $\frac{v^2}{2g}$to get the total energy per unit mass at some point along a streamline.

 

[Chenkel]

$z$ for "elevation" in Bernoulli's ( work by gravity per unit mass ). It is a different term than $\frac{P}{\gamma}$, they are independent of each other in that derivation. That $P$ is from "something" doing work on the fluid. You are adding the contributions from each together along with a kinetic energy per unit mass $\frac{v^2}{2g}$to get the total energy per unit mass at some point along a streamline.

It sounds to me like you are saying the pressure is from something other than gravity in Bernoulli's equation, is it safe to assume in the standard equation that we ignore the effects of gravity on pressure? Also I apologize for being a little new to the terminology, but what is the $\gamma$ in $\frac {P}{\gamma}$ and shouldn't kinetic energy per unit mass be $\frac {v^2}{2}$?

 

[Lnewqban]

Please, see:
https://www.engineeringtoolbox.com/bernouilli-equation-d_183.html

 

[erobz]

It sounds to me like you are saying the pressure is from something other than gravity in Bernoulli's equation, is it safe to assume in the standard equation that we ignore the effects of gravity on pressure?

The pressure from gravity is not ignored, its there. Its just that in the Bernoulli Equation we aren't adding pressures, we are adding energy per unit weight from each contributing source.

and shouldn't kinetic energy per unit mass be $\frac {v^2}{2}$?

Sorry, the terms have dimension energy per unit weight...not mass. My bad.

Also I apologize for being a little new to the terminology, but what is the $\gamma$ in $\frac {P}{\gamma}$

$\gamma$ is the specific weight of the fluid:

$\gamma = \rho g$

Where $\rho$ is the mass density of the fluid

 

[Chenkel]

So after looking at the article again I believe I understand it a little better, the work done to the fluid is equal to the sum of the work done in all dimensions (x, y, and z). So this means we need to consider the vertical work done by gravity, and the lateral work done by pressure, we can relate these two quantities with the work energy theorem.

In the derivation of Bernoulli's equation in the article, the work done by pressure is only lateral (not vertical), I'm wondering if the derivation would get a different result if the work the pressure was doing was in the direction of gravity...

Thank you again for the feedback. I believe I understand much more, but I do wonder about the different derivation from vertical or semi-vertical work from pressure.

 

[erobz]

In the derivation of Bernoulli's equation in the article, the work done by pressure is only lateral (not vertical), I'm wondering if the derivation would get a different result if the work the pressure was doing was in the direction of gravity...

Thank you again for the feedback. I believe I understand much more, but I do wonder about the different derivation from vertical or semi-vertical work from pressure.

The work done by pressure is a path integral. It is the amount of work done on a fluid element as it moves on a path from $a \to b$. You seem to be concerned about what is happening when the path is going "up"?

 

[Chenkel]

The work done by pressure is a path integral. It is the amount of work done on a fluid element as it moves on a path from $a \to b$. You seem to be concerned about what is happening when the path is going "up"?

I was concerned with what happens in the derivation when the velocity is aiming in a vertical or semi vertical angle (against gravity). I believe angle of the initial and final velocity vectors do not matter now that I've analyzed the problem. The main point I want to make that was initially confusing to me, is that there's a difference between net work done on the liquid, and net work done by pressure. I initially thought I could use the work energy theorem to find the work done by pressure, but the work energy theorem tells us the work done to the fluid, and not the work done by pressure.

To illustrate this point, imagine you have an object and you move it vertically, and it maintains the same velocity at beginning and end, the net work done to the body is 0 even considering gravity (using the energy work theorem), but the net work done by you is mgh (definitely not 0). Now imagine you're the "pressure," and the object you're pushing is the "fluid." You will end up doing mgh joules of work, but the work done to the fluid is still 0.

In summary this explains why $dW = dK + dU$ instead of just $dW = dK$.

 

[kuruman]

OK, let me do a series of applications of the Bernoulli equation to find the difference in pressure at two different points of a pipe containing an incompressible fluid. First the continuity equation says that $$Av=\beta=\text{const.}$$where $\beta$ is the volumetric flow, e.g. gallons/minute. I use the continuity equation to eleiminate speed from Bernoulli's equation to get $$p+\frac{1}{2}\rho\left(\frac{\beta}{A}\right)^2+\rho~g~h=\text{const.}$$The pressure difference between two points in the pipe, $\Delta p=p_2-p_1$, is $$\Delta p=\frac{1}{2}\rho\beta^2\left(\frac{1}{A_2^2}-\frac{1}{A_1^2}\right)+\rho g(h_2-h_1).$$Case I: Horizontal pipe with uniform cross section. WIth $h_2=h_1$ and $A_2=A_1$,$$\Delta p=0.$$ Case II: Horizontal pipe with change in cross section. WIth $h_2=h_1$,$$\Delta p=\frac{1}{2}\rho\beta^2\left(\frac{1}{A_2^2}-\frac{1}{A_1^2}\right).$$If $A_2>A_1$, the pressure drops, else it rises as the fluid flows from 1 to 2.

Case III: Pipe with change in cross section and change in height. $$\Delta p=\frac{1}{2}\rho\beta^2\left(\frac{1}{A_2^2}-\frac{1}{A_1^2}\right)+\rho g(h_2-h_1).$$The change in pressure could be positive, negative or zero depending on the choice of cross-sections and heights.

Case IV: Pipe with change in cross section and change in height but no fluid flow. With $\beta =0$, $$\Delta p=\rho g(h_2-h_1).$$This is the familiar hydrostatic pressure result. Hydrostatic pressure is built in Bernoulli's equation and does not do additional work when the fluid is moving.

I hope this helps.

 

[Lnewqban]

The main point I want to make that was initially confusing to me, is that there's a difference between net work done on the liquid, and net work done by pressure. I initially thought I could use the work energy theorem to find the work done by pressure, but the work energy theorem tells us the work done to the fluid, and not the work done by pressure.

I would like to clarify that the principle is based on an isolated system, with neither work or heat coming into it.
The three forms of energy transform into any of the other, according to the flow characteristics, but the amount of internal energy remains the same.
Somehow, it could be compared to the transformation of kinetic and potential energy for a rollercoaster, depending on the cart position.

Please, see:
https://en.m.wikipedia.org/wiki/Bernoulli's_principle

In order to do work on the fluid, a pump is needed, just like the heart does to make blood circulate up and down the body, overcoming the natural resistance from friction and turbulence.
The heart suplies the delta P between its inlet and outlet, but the internal presion in the circulatory system depends on other factors.

 

[vanhees71]

Once more it's clear that the confusion comes from the avoidance of math ;-). The equation is derivable from Euler's equation for the perfect fluid
$$\rho [\partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v}=-\vec{\nabla} P - \vec{\nabla} V,$$
where $\vec{P}$ is the pressure of the fluid and $V$ the potential of the external force density (e.g., $$V=-\rho \vec{g} \cdot \vec{r}$$
for the potential of the gravitational force.

The only trick is to rewrite the "convection term" by calculating the gradient of $\vec{v}^2$. That's most easily done using the Ricci calculus (for Cartesian components; Einstein summation convention implied),
$$\partial_{j} (v_k v_k)=2 v_k \partial_j v_k = 2 [v_k (\partial_j v_k -\partial_k v_j)+v_k \partial_k v_j] = 2[v_k \epsilon_{jkl}(\vec{\nabla} \times \vec{v})_l +v_k \partial_k v_j] = 2 [\vec{v} \times (\vec{\nabla} \times \vec{v})]_j + 2 (\vec{v} \cdot \vec{\nabla}) v_j,$$
i.e., in vector notation
$$\frac{1}{2} \vec{\nabla} (\vec{v}^2) = \vec{v} \times (\vec{\nabla} \times \vec{v}) +(\vec{v} \cdot \vec{\nabla}) \vec{v}.$$
Plugging this into Euler's equation, you get
$$\rho [\partial_t \vec{v} + \frac{1}{2} \vec{\nabla} (\vec{v}^2) - \vec{v} \times (\vec{\nabla} \times \vec{v})]=-\vec{\nabla}(P+V).$$
Now assuming incompressibility, $\rho=\text{const}$, $\vec{\nabla} \times \vec{v}=0$, and $\partial_t \vec{v}=0$ you get
$$\vec{\nabla} \left (\frac{\rho}{2} \vec{v}^2+P+V \right)=0$$
or
$$\frac{\rho}{2} \vec{v}^2 + P + V =\text{const}.$$
That's Bernoulli's theorem with the clearly stated assumptions when it's valid

(a) ideal fluid
(b) incompressibility
(c) stationary flow
(d) irrotational flow

 

[Lnewqban]

Please, also see:
https://www.engineeringtoolbox.com/pumps-power-d_505.html

https://www.engineeringtoolbox.com/bernouilli-equation-d_183.html

https://www.engineeringtoolbox.com/static-pressure-head-d_610.html

https://www.engineeringtoolbox.com/velocity-head-d_916.html

 

[erobz]

Just as a heads up.

There is a "sticky" situation here. Bernoulli's is derived for inviscid flow. It's just an approximation to real flow over short distances.

The conceptual trouble always comes in when we try to use it to solve for the $\Delta P$ of straight/uniform horizontal section of pipe. Something that is completely not a problem in real viscous flow suddenly disagrees (dare I say even seems a bit paradoxical) with what we know about flow in pipes. However, it should be expected if we remember the assumptions.

 

[erobz]

Once more it's clear that the confusion comes from the avoidance of math ;-). The equation is derivable from Euler's equation for the perfect fluid
$$\rho [\partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v}=-\vec{\nabla} P - \vec{\nabla} V,$$
where $\vec{P}$ is the pressure of the fluid and $V$ the potential of the external force density (e.g., $$V=-\rho \vec{g} \cdot \vec{r}$$
for the potential of the gravitational force.

The only trick is to rewrite the "convection term" by calculating the gradient of $\vec{v}^2$. That's most easily done using the Ricci calculus (for Cartesian components; Einstein summation convention implied),
$$\partial_{j} (v_k v_k)=2 v_k \partial_j v_k = 2 [v_k (\partial_j v_k -\partial_k v_j)+v_k \partial_k v_j] = 2[v_k \epsilon_{jkl}(\vec{\nabla} \times \vec{v})_l +v_k \partial_k v_j] = 2 [\vec{v} \times (\vec{\nabla} \times \vec{v})]_j + 2 (\vec{v} \cdot \vec{\nabla}) v_j,$$
i.e., in vector notation
$$\frac{1}{2} \vec{\nabla} (\vec{v}^2) = \vec{v} \times (\vec{\nabla} \times \vec{v}) +(\vec{v} \cdot \vec{\nabla}) \vec{v}.$$
Plugging this into Euler's equation, you get
$$\rho [\partial_t \vec{v} + \frac{1}{2} \vec{\nabla} (\vec{v}^2) - \vec{v} \times (\vec{\nabla} \times \vec{v})]=-\vec{\nabla}(P+V).$$
Now assuming incompressibility, $\rho=\text{const}$, $\vec{\nabla} \times \vec{v}=0$, and $\partial_t \vec{v}=0$ you get
$$\vec{\nabla} \left (\frac{\rho}{2} \vec{v}^2+P+V \right)=0$$
or
$$\frac{\rho}{2} \vec{v}^2 + P + V =\text{const}.$$
That's Bernoulli's theorem with the clearly stated assumptions when it's valid

(a) ideal fluid
(b) incompressibility
(c) stationary flow
(d) irrotational flow

Thats notation is a little intimidating though!

Eulers Formula:

$$ -\frac{ \partial }{ \partial l} \left( p + \gamma z \right) = \rho \left( V \frac{ \partial V}{ \partial l} + \frac{\partial V}{ \partial t} \right) $$

(but I even think this is intimidating)

 

[vanhees71]

I don't understand your notation. What's $l$? Also note that the velocity is a vector!

 

[erobz]

I don't understand your notation. What's $l$? Also note that the velocity is a vector!

According to the textbook I copied it from $l$ is a pathline in the flow. When they derive Bernoulli’s from it, the assumption is the flow is steady and the “local acceleration” $\frac{\partial V}{\partial t}= 0$. Then they say that with this assumption the “pathline” becomes a “streamline” $s$ and the derivatives become normal w.r.t. $s$.