A fluid could be incredibly viscous but if there is no shear stress in the fluid, that high viscosity doesn't matter and the fluid will still behave as an inviscid fluid. That's why as long as you are outside of the boundary layer, for example on an airplane wing, you can still use Bernoulli's equation to get the pressure distribution.
As for the streamlines question, how familiar are you with vector calculus? It would be hard to explain without familiarity with calculus. Basically, without starting too early in the process, the motion of an inviscid fluid is governed in the fullest sense by what is called Euler's equation, and one possible method of writing Euler's equation gives you
\rho\left( \dfrac{\partial\vec{V}}{\partial t} + \vec{V}\cdot \nabla\vec{V} \right) = \rho \vec{f} - \nabla p.
Through an identity, this is also equal to
\rho\left( \dfrac{\partial\vec{V}}{\partial t} + \nabla\dfrac{V^2}{2} - \vec{V}\times\nabla\times\vec{V} \right) = \rho \vec{f} - \nabla p,
where \vec{f} is the body force (e.g. gravity). If you assume the flow is irrotational and thus \nabla \times \vec{V} = 0 and the velocity can therefore be described in terms of a potential function \vec{V} = \nabla \Phi, then Euler's equation simplifies to the following
\rho\left( \nabla\dfrac{\partial\Phi}{\partial t} + \nabla\dfrac{V^2}{2} \right) = \rho \vec{f} - \nabla p,
or
\left( \nabla\dfrac{\partial\Phi}{\partial t} + \nabla\dfrac{V^2}{2} \right) = \vec{f} - \nabla \dfrac{p}{\rho}.
Now, body forces are conservative and also governed by a potential function, \vec{f} = \nabla\Phi_f, so
\left( \nabla\dfrac{\partial\Phi}{\partial t} + \nabla\dfrac{V^2}{2} \right) = \nabla\Phi_f - \nabla \dfrac{p}{\rho}. We can gather all the gradient terms together then and take the gradient of the entire expression
\left( \nabla\dfrac{\partial\Phi}{\partial t} + \nabla\dfrac{V^2}{2} - \nabla\Phi_f + \nabla \dfrac{p}{\rho}\right) = 0.
Or,
\nabla\left( \dfrac{\partial\Phi}{\partial t} + \dfrac{V^2}{2} - \Phi_f + \dfrac{p}{\rho}\right) = 0.
From calculus, then, you know that this is the equivalent of saying
\dfrac{\partial\Phi}{\partial t} + \dfrac{V^2}{2} - \Phi_f + \dfrac{p}{\rho} = \text{constant}.
If the gravity is the only body force, then \Phi_f is simply the gravitational potential and is -gh, which gives you the a modified Bernoulli's equation that can handle unsteady flows,
\dfrac{\partial\Phi}{\partial t} + \dfrac{V^2}{2} + \dfrac{p}{\rho} + gh= \text{constant}.
Or, in a steady-state flow, you get the classical Bernoulli's equation:
\dfrac{V^2}{2} + \dfrac{p}{\rho} + gh= \text{constant}.
This required the following assumptions to derive:
- Inviscid flow
- Irrotational flow
- Incompressible flow
- Steady flow (if you want the classic form)
So then what happens if you have a rotational flow and \nabla \times \vec{V} \neq 0? We can start again with
\left( \dfrac{\partial\vec{V}}{\partial t} + \nabla\dfrac{V^2}{2} - \vec{V}\times\nabla\times\vec{V} \right) = \vec{f} - \nabla \dfrac{p}{\rho}.
Here, we can't assume a potential function exists since the flow field is not irrotational. What we can do is use the concept of a streamline (the direction of the velocity vector, \vec{V}). To make things easier, let's introduce the vorticity, \nabla\times\vec{V} = \vec{\zeta} so
\left( \dfrac{\partial\vec{V}}{\partial t} + \nabla\dfrac{V^2}{2} - \vec{V}\times\vec{\zeta} \right) = \vec{f} - \nabla \dfrac{p}{\rho}.
We can then assume that the entire thing only applies along a streamline, which is the equivalent of taking the dot product with the velocity vector \vec{V}. Looking at that vorticity term in the streamline direction
\vec{V}\cdot\left(\vec{V}\times\vec{\zeta}\right) = \vec{\zeta}\cdot\left(\vec{V}\times\vec{V}\right) = 0.
So,
along a streamline, we then have, analogous to before,
\left[\dfrac{\partial\vec{V}}{\partial t} + \nabla\left(\dfrac{V^2}{2} + \dfrac{p}{\rho} + gh \right)\right]_{s} = 0
where the s subscript means the whole expression was taken in the streamline direction. If you then assume a steady flow once again, you get
\left(\dfrac{V^2}{2} + \dfrac{p}{\rho} + gh \right)_{s} = \text{constant}.
This holds for any streamline, but the value of the constant will change from one streamline to another. The assumptions (and thus limitations) here were slightly different, and were
- Inviscid flow
- Incompressible flow
- Steady flow (a hard requirement this time)
