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.

Bernouilli is all about conservation of energy. The drop in pressure is necessary so that work can be done on the incompressible fluid - because when it flows faster, it has more kinetic energy and that had to come from somewhere.

There is **no change in temperature**. This is not an ideal gas - you said yourself it is an incompressible fluid. So $PV=nRT$ does not apply.

## Best Answer

Assuming you know the leak rate, you can approximate the flow rate increases by the leak rate, which is though not exact. Knowing the pipe cross section area and fluid density, you can estimate the velocity change. Then you can apply Bernoulli equation to calculate the pressure drop.