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).
To find the value of \(C\) that makes the velocity field \(\vec{V} = 3z\hat{i} + 0\hat{j} + Cx\hat{k}\) irrotational, we must ensure the curl of \(\vec{V}\) is zero. The curl of a vector field \(\vec{V} = V_x\hat{i} + V_y\hat{j} + V_z\hat{k}\) is computed as: \[ \nabla \times \vec{V} = \left( \frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z} \right)\hat{i} + \left( \frac{\partial V_x}{\partial z} - \frac{\partial V_z}{\partial x} \right)\hat{j} + \left( \frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y} \right)\hat{k} \] Given: \(V_x = 3z\), \(V_y = 0\), \(V_z = Cx\). Calculating the components: \( \frac{\partial V_z}{\partial y} = 0 \) \(\frac{\partial V_y}{\partial z} = 0 \) \( \frac{\partial V_x}{\partial z} = 3 \) \( \frac{\partial V_z}{\partial x} = C \) \( \frac{\partial V_y}{\partial x} = 0 \) \(\frac{\partial V_x}{\partial y} = 0 \) The curl expression gives: \(\nabla \times \vec{V} = (0-0)\hat{i} + (3-C)\hat{j} + (0-0)\hat{k} = 0. \) For the flow to be irrotational, \(3-C = 0\), solving gives \(C = 3\). The value \(C = 3.0\) satisfies the given range \([2.9, 3.1]\).
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