Question

Let \(\rho(x, y, z, t)\) and \(\mathbf{u}(x, y, z, t)\) represent density and velocity, respectively, at a point \((x, y, z)\) and time \(t\). Assume \(\frac{\partial \rho}{\partial t}\) is continuous. Let \(V\) be an arbitrary volume in space enclosed by the closed surface \(S\), and \(\mathbf{\hat{n}}\) be the outward unit normal of \(S\). Which of the following equations is/are equivalent to: \[ \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{u}) = 0? \]

• A. \(\int_V \frac{\partial \rho}{\partial t} dv = - \int_S \rho \mathbf{u} \cdot \mathbf{\hat{n}} ds\)
• B. \(\int_V \frac{\partial \rho}{\partial t} dv = \int_S \rho \mathbf{u} \cdot \mathbf{\hat{n}} ds\)
• C. \(\int_V \frac{\partial \rho}{\partial t} dv = - \int_V \nabla \cdot (\rho \mathbf{u}) dv\)
• D. \(\int_V \frac{\partial \rho}{\partial t} dv = \int_V \nabla \cdot (\rho \mathbf{u}) dv\)

Hint:

The divergence theorem is key to converting volume integrals into surface integrals, particularly when analyzing fluid dynamics and applying the continuity equation.

Answer 1

The Correct Option is A

Step 1: Utilize the divergence theorem.
The divergence theorem states that the volume integral of \(\nabla \cdot (\rho \mathbf{u})\) over a region \(V\) is equal to the flux of \(\rho \mathbf{u}\) through the surface \(S\) enclosing \(V\): \[ \int_V \nabla \cdot (\rho \mathbf{u}) \, dv = \int_S \rho \mathbf{u} \cdot \mathbf{\hat{n}} \, ds. \] Step 2: Relate to the continuity equation.
The continuity equation is given by: \[ \int_V \frac{\partial \rho}{\partial t} \, dv = - \int_V \nabla \cdot (\rho \mathbf{u}) \, dv. \] Using the divergence theorem, this becomes: \[ \int_V \frac{\partial \rho}{\partial t} \, dv = - \int_S \rho \mathbf{u} \cdot \mathbf{\hat{n}} \, ds. \] Step 3: Verify the options.
By comparing the expressions derived from the continuity equation: - Options (1) and (3) align with the derived equations and satisfy the continuity condition.
- Options (2) and (4) do not hold. Final Answer: \[ \boxed{{(1), (3)}} \]