Understanding the Incompressibility Condition

For a two-dimensional, incompressible flow, the continuity equation must be satisfied. This implies that the divergence of the velocity field is zero:

$ \nabla \cdot \vec{V} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 $

The given velocity field is $\vec{V} = u \hat{\mathbf{i}} + v \hat{\mathbf{j}}$, where $u = 2 \sinh x$. We need to find the unknown component $v(x, y)$ and evaluate it at a specific point.

Calculating the Velocity Component Derivative

First, find the partial derivative of the known velocity component $u$ with respect to $x$:

$ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (2 \sinh x) $

The derivative of $\sinh x$ is $\cosh x$. Therefore:

$ \frac{\partial u}{\partial x} = 2 \cosh x $

Determining the Unknown Velocity Component $v(x, y)$

Substitute the calculated $\frac{\partial u}{\partial x}$ into the continuity equation:

$ 2 \cosh x + \frac{\partial v}{\partial y} = 0 $

Isolate $\frac{\partial v}{\partial y}$:

$ \frac{\partial v}{\partial y} = -2 \cosh x $

Integrate this expression with respect to $y$ to find $v(x, y)$. Treat $x$ as a constant during this integration. The integration constant will be a function of $x$, let's call it $f(x)$:

$ v(x, y) = \int (-2 \cosh x) dy = (-2 \cosh x) y + f(x) $

Applying the Boundary Condition

Use the given boundary condition $v(x, 0) = \cosh x$. Substitute $y=0$ into the expression for $v(x, y)$:

$ v(x, 0) = (-2 \cosh x)(0) + f(x) $

This simplifies to:

$ \cosh x = f(x) $

So, $f(x) = \cosh x$. Substitute this back into the equation for $v(x, y)$:

$ v(x, y) = -2y \cosh x + \cosh x $

Calculating the Specific Velocity Value

The final step is to calculate $v(0, -1)$ by substituting $x=0$ and $y=-1$ into the derived expression for $v(x, y)$:

$ v(0, -1) = -2(-1) \cosh(0) + \cosh(0) $

Recall that $\cosh(0) = 1$. Plugging this value in:

$ v(0, -1) = -2(-1)(1) + 1 $

$ v(0, -1) = 2 + 1 $

$ v(0, -1) = 3 $