The explanation for this assumption is the same for most assumptions: because it makes the problem easier. This equation is (generally) applied to a single streamline due to the assumptions that were made when the equation was derived. Many mistakenly ascribe Bernoulli's Law to the principle of conservation of energy, when in reality it is a direct consequence of Newton's linear momentum equation. From the relatively straightforward force analysis of a differential fluid mass, it can be shown that

$$-\frac{\partial p}{\partial s}=\rho a_s=\rho v\frac{\partial v}{\partial s},
\tag{i}$$

$$+\frac{\partial p}{\partial n}=\rho a_n=\rho\frac{v^2}{R}, \tag{ii}$$

where $p$ is the static pressure, $\rho$ is the fluid density, $a$ is the local acceleration, $v$ is velocity, $R$ is the local radius of curvature, and $s$ and $n$ are curvilinear coodinates along and normal to the streamline, respectively. A partial differential is used because the pressure and velocity (in general) change in both the $n$ and $s$ directions. Now, if we limit our analysis to changes only *along* the streamline, we can replace the original partial differentials in (1) with exact differentials. Rearranging, this gives us

$$\frac{dp}{ds}+\rho V\frac{dV}{ds}=0, \tag{iii}$$

which can be simplified further into the classic differential Bernoulli equation:

$$\frac{dp}{\rho}+VdV=0. \tag{iv}$$

It is *this* version of the equation (with it's inherent assumptions) that is then integrated to give the classic textbook version of Bernoulli's Eqn. mentioned earlier.

$$p+\frac{1}{2}\rho V^2=p_0$$

Why would we do this? Well, there are several flow situations in which it is approximately valid (e.g. irrotational flows), where the stagnation pressure is uniform everywhere and needs to be calculated only once. For viscous flows, the equation can still be used to determine the stagnation pressure at a given location in the flow, but there should be no expectation that the stagnation pressures will be equal between streamlines.

This is a correct way to solve exercises involving viscosity (within certain constraints, e.g., constant viscosity). Eqn. 1 is a version of the Bernoulli equation, modified to include a frictional head loss, and is definitely valid, provided the velocities used are the average velocities. Eqn. 1 without the $h_L$ is valid along a streamline, even for a viscous flow. If Eqn. 1 is being used for a *laminar* viscous flow (say in a tube of slowly varying cross section), the kinetic energy terms should not have a 2 in the denominator. See Bird, Stewart, and Lightfoot, Transport Phenomena for details. Here they show that, if the 2 is to be included, then instead of using the average value of v squared, one should use the average value of $v^3$ divided by the average value of v. For laminar flow in a tube, this reduces to twice the square of the average velocity, so, if the average velocity squared is being used in the kinetic energy term, the 2 should not be included in the denominator if the flow is laminar.

## Best Answer

Laminar flow occurs when the liquid layers move parallel. Otherwise stated, the motion of the particles of the fluid is very orderly with all particles moving in straight lines parallel to the pipe walls. The condition in which Bernoulli's principle has been derived is that the viscosity of the fluid should be negligible. Hence we can assume that the given fluid has got nearly zero viscosity. In such a case, the velocity vector will be steady and has no rotation at all and is a constant at every points.

In such a case, you may write laminar flow corresponds to irrotational flow. However, in reality the effects of viscous force are not that negligible. Hence no real flows are irrotational. Hence in general, if you cannot demand that a laminar flow should correspond to an irrotational flow. In reality, due to the effects of viscosity, the increase in both its dynamic pressure and kinetic energy does not occurs with a simultaneous decrease in (the sum of) its static pressure, potential energy and internal energy. There will be a drag or a delaying effect.

However, in deriving the Bernoulli's equation from the work energy principle, you should neglect that the liquid has got negligible viscosity and can assume that the fluid element has to be along the flow. So, in the present context, $rot\vec{v}=0$ corresponds to a stream line flow, as long as the assumptions holds good.