The following is written in Physics by Halliday, Resnick and Krane (5th ed.).
$$p + \frac{1}{2} \rho v^2 + \rho g y = \mathrm{constant}$$
Strictly speaking, the points to which we apply Bernoulli's equation should be along the same streamline. However, if the flow is irrotational, the value of the constant is same for all the streamlines in the tube of flow, so Bernoulli's equation can be applied to any two points in the flow.
Here, $p$ is the pressure of the fluid at a point, $\rho$ is the density(assumed constant), $v$ is the velocity of the fluid element, and $y$ is the vertical distance of the element from a fixed reference point. From point to point, $p$, $v$ and $y$ will change.
How can one prove that the constant in Bernoulli's equation does not change along two streamlines for irrotational flow?
The fluid can be assumed to be non-viscous, incompressible and the flow is steady.
Best Answer
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.