Question

Consider a velocity field \( \vec{V} = 3z \hat{i} + 0 \hat{j} + Cx \hat{k} \), where \( C \) is a constant. If the flow is irrotational, the value of \( C \) is (rounded off to 1 decimal place).

Hint:

To determine if a flow is irrotational, calculate the curl of the velocity field and set it equal to zero. Solving for constants in the velocity components will give the value for the constant.

Answer 1

For a flow to be irrotational, the curl of the velocity field must be zero. The curl of a vector field \( \vec{V} \) is given by: \[ \nabla \times \vec{V} = \left( \frac{\partial V_k}{\partial y} - \frac{\partial V_y}{\partial z} \right) \hat{i} + \left( \frac{\partial V_x}{\partial z} - \frac{\partial V_k}{\partial x} \right) \hat{j} + \left( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right) \hat{k} \] Where \( \vec{V} = 3z \hat{i} + 0 \hat{j} + Cx \hat{k} \), so:
\( V_x = 3z \),
\( V_y = 0 \),
\( V_k = Cx \).
Step 1: Compute the components of the curl
We compute the components of the curl:
\( \frac{\partial V_k}{\partial y} - \frac{\partial V_y}{\partial z} = 0 - 0 = 0 \),
\( \frac{\partial V_x}{\partial z} - \frac{\partial V_k}{\partial x} = 3 - C \),
\( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} = 0 - 0 = 0 \).
Thus, the curl is: \[ \nabla \times \vec{V} = \left( 0 \right) \hat{i} + \left( 3 - C \right) \hat{j} + \left( 0 \right) \hat{k} \] Step 2: Set the curl equal to zero
For the flow to be irrotational, the curl must be zero. Therefore, we set the \( \hat{j} \)-component equal to zero: \[ 3 - C = 0 \] Step 3: Solve for \( C \)
Solving for \( C \): \[ C = 3 \] Step 4: Conclusion
Thus, the value of \( C \) is \( \boxed{3.0} \).