A velocity distribution of an incompressible flow field is given by Gravity maybe neglected, but the fluid is not inviscid. a) Find an expression for the pressure gradient. b) Integrate the pressure gradients to find the pressure distribution given that the pressure at point (0,0)is po. c) Find the velocity potential if it exists. If it does not exist, show calculations proving that it does not exist.
The Correct Answer and Explanation is :
To answer the given problem, let’s proceed step by step.
Given Information:
- The flow field is incompressible (divergence-free velocity field).
- Gravity is neglected.
- The fluid is viscous (viscous effects are present).
- Velocity distribution is given (though the exact distribution is not specified in the question).
Let’s assume the velocity distribution is given as:
[
\vec{v} = (u(x,y), v(x,y))
]
where (u(x, y)) is the velocity component in the (x)-direction, and (v(x, y)) is the velocity component in the (y)-direction.
Part a: Find an expression for the pressure gradient
The equation governing the pressure in a viscous fluid flow is given by the Navier-Stokes equation. For an incompressible flow and neglecting external forces like gravity, the Navier-Stokes equation simplifies to:
[
\rho \left( \frac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \nabla \vec{v} \right) = – \nabla p + \mu \nabla^2 \vec{v}
]
where:
- (\rho) is the fluid density,
- (p) is the pressure,
- (\mu) is the dynamic viscosity.
For steady-state flow, (\frac{\partial \vec{v}}{\partial t} = 0), so the equation simplifies to:
[
0 = – \nabla p + \mu \nabla^2 \vec{v}
]
This equation gives us the pressure gradient. Rearranging, we get:
[
\nabla p = \mu \nabla^2 \vec{v}
]
Thus, the pressure gradient depends on the velocity field and the viscosity of the fluid. To get a specific expression, we’d need the exact form of (u(x, y)) and (v(x, y)).
Part b: Integrate the pressure gradients to find the pressure distribution
To integrate the pressure gradient, we first solve for (p) using the equation above. The pressure distribution can be obtained by integrating the pressure gradient. Assuming that the velocity field is known, the integral can be performed as:
[
p(x, y) = p_0 – \int \mu \nabla^2 \vec{v} \, dx
]
where (p_0) is the pressure at the reference point (0, 0). The integral will depend on the specific form of the velocity field (\vec{v}), which is not given here, so we cannot provide a concrete solution without this information.
Part c: Find the velocity potential if it exists
The velocity potential, (\phi), exists if the flow is irrotational, i.e., if the curl of the velocity field is zero:
[
\nabla \times \vec{v} = 0
]
For a two-dimensional incompressible flow, this condition reduces to:
[
\frac{\partial v}{\partial x} – \frac{\partial u}{\partial y} = 0
]
If this condition is satisfied, the velocity field can be expressed as the gradient of a scalar potential:
[
\vec{v} = \nabla \phi
]
In this case, the velocity potential can be found by integrating the components of the velocity field. If the condition is not satisfied, the velocity potential does not exist.
Without knowing the specific form of (u(x, y)) and (v(x, y)), we cannot definitively state whether the potential exists or not. However, if (\frac{\partial v}{\partial x} \neq \frac{\partial u}{\partial y}), the flow is rotational, and thus, a velocity potential does not exist.
Conclusion:
- The expression for the pressure gradient depends on the specific velocity field and viscosity.
- The pressure distribution is obtained by integrating the pressure gradient, but the velocity field is needed to perform the integration.
- The velocity potential exists only if the flow is irrotational. If the curl condition is violated, the potential does not exist.