Derivation of Bernoulli's equation via Newton's second law

  • Thread starter Shinaolord
  • Start date
  • #1
92
4
In the derivation on Wikipedia, it says the following
## \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v = \frac{d}{dx}[\frac{v^2}{2}] ##

How do they go from the second to last to last equation? I've been trying to understand, but I think I'm just looking over something incredibly simple.
 
Last edited:

Answers and Replies

  • #2
94
10
Hi,

first they replace ##\frac{dx}{dt}## with ##v##, by definition. Then they use the chain rule of differentiation 'backwards'.
 
  • #3
92
4
Yes I know where the v comes from, I just don't see how they get the 1/2

Could you perhaps show what you mean mathematically? The chain rule backwards, that reminds me of integration by parts, but I don't really seem to understand what you mean.
Thank you for replying though!

Specifically I don't understand how ##\dfrac{dv}{dx}v =\frac{d}{dx}\frac{v^2}{2} ##
 
  • #4
94
10
Its because ##\frac{d}{dx} v^2 = 2 v\cdot \frac{dv}{dx} ##. So you have to take care of the 2.
 
  • #5
92
4
Oh wow that is easy. I knew it was something simple, thank you!
 
  • #7
vanhees71
Science Advisor
Insights Author
Gold Member
16,990
8,141
Well, it's a bit inconsistent, because on the one hand they use ##\vec{x}## in the sense of Euler coordinates of fluid dynamics and on the other they use it as the trajectory ##\vec{x}(t)## of the material fluid element. The correct definition of the "material time derivative" of the velocity (and of any other quantity in the Euler notation)
$$\mathrm{D}_t \vec{v}=\partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v}.$$
So the correct derivation is as follows
$$\mathrm{D}_t (\vec{v}^2)=2 \vec{v} \cdot \mathrm{D}_t \vec{v}.$$
For an incompressible fluid you have
$$\rho \mathrm{D}_t \vec{v}=-\vec{\nabla} p.$$
Multiplying this equation by ##\vec{v}## gives
$$\frac{\rho}{2} \mathrm{D}_t (\vec{v}^2)=-\vec{v} \cdot \vec{\nabla} p.$$
If now you are in a static situation, where the pressure is time-independent you get
$$\frac{\rho}{2} \mathrm{D}_t (\vec{v}^2)=-\mathrm{D}_t p.$$
Since ##\rho=\text{const}## for an incompressible fluid, you finally get by integration over time
$$\frac{\rho}{2} \vec{v}^2 + p=\text{const},$$
which is Bernoulli's Law in the most simple case.

The most general extension of this law is the energy balance of a fluid, which expresses the energy-conservation law.
 
  • Like
Likes dextercioby

Related Threads on Derivation of Bernoulli's equation via Newton's second law

Replies
20
Views
2K
Replies
42
Views
11K
Replies
1
Views
27K
Replies
6
Views
406
Replies
12
Views
3K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
16
Views
978
  • Last Post
Replies
16
Views
1K
Replies
5
Views
2K
  • Last Post
Replies
3
Views
1K
Top