- Can the streamlines of a fluid particle show the position of the particle at a time(using the streamlines)?
- I know that streamlines cannot intersect because at a specific instant the particle reaching the intersection will have two different directions of motion (in the steady flow of particles), but can I say the same thing in turbulent flow for the line of flow of the particles, why or why not?
- Is it necessary that a streamline will be a perfect straight line(to a good approximation)? If no why is it called a line then?
Fluid Dynamics – What Are Streamlines and Line of Flow of Fluid Particles? Understanding Fluid Flow
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This answer is taken from The Feynman Lectures on Physics, Vol. II, Ch. 40: The Flow of Dry Water, Section 40-3: Steady flow—Bernoulli’s theorem.
Imagine a bundle of adjacent streamlines which form a stream tube as sketched in the figure.
From equation of continuity, we can write that $A_1 v_1 = A_2 v_2$. Density of the fluid is constant. Now we calculate the work done by the fluid pressure. The work done on the fluid entering at $A_1$ is $p_1 A_1 v_1 \Delta t$ and th work given up at $A_2$ is $p_1 A_1 v_1 \Delta t$. The net work on the fluid between $A_1$ and $A_2$, is therefore $$p_1 A_1 v_1 \Delta t - p_2 A_2 v_2 \Delta t \tag1$$ which must equal the increase in the energy of a mass $\Delta M$ of fluid in going from $A_1$ to $A_2$. In other words, $$p_1 A_1 v_1 \Delta t - p_2 A_2 v_2 \Delta t = \Delta M (E_2 - E_1) \tag2$$ where $E_1$ is the energy per unit mass of fluid at $A_1$, and $E_2$ is the energy per unit mass at $A_2$. The energy per unit mass of the fluid can be written as $$E = \frac{1}{2}v^2 + \phi + U,$$ where $\frac{1}{2}v^2$ is the kinetic energy per unit mass, $\phi$ is the potential energy per unit mass, and $U$ is an additional term which represents the internal energy per unit mass of fluid. The internal energy might correspond, for example, to the thermal energy in a compressible fluid, or to chemical energy. All these quantities can vary from point to point.
Now, I think that this internal energy can also include the net rotational energy of the individual molecules. So, if the fluid is irrotational, then the contribution of the rotational motion of the individual particles will become zero. If the flow is rotational, we cannot guarantee that the rotational energy of the molecules per unit mass is same everywhere.
Using this form for the energies in $(2)$, we have $$\frac{p_1 v_1 A_1 \Delta t}{\Delta M} - \frac{p_2 v_2 A_2 \Delta t}{\Delta M} = \frac{1}{2}v_2^2 + \phi_2 + U_2 - \frac{1}{2}v_1^2 + \phi_1 + U_1$$ But $\Delta M = \rho A v \Delta t$, so we get $$\frac{p_1}{\rho} + \frac{1}{2}v_1^2 + \phi_1 + U_1 = \frac{p_2}{\rho} + \frac{1}{2}v_2^2 + \phi_1 + U_2,$$ which is the Bernoulli result with an additional term for the internal energy. If the fluid is incompressible and irrotational, the internal energy term is the same on both sides, and we get again that $$p + \frac{1}{2}\rho v^2 + \rho g y$$ holds along any streamline.
Pascal's law says that pressure is isotropic and that, under hydrostatic conditions, this means that it is the same at all horizontal locations within within the fluid. However, if the fluid is flowing, the pressure varies with spatial position to balance any centripetal forces. This is captured in Euler's equation of fluid motion.
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