Understanding the Linear Momentum Equation

The question asks for the linear momentum equation for steady flow with a fixed control volume. This equation is derived from Newton's second law of motion, \(\sum F = ma\), applied to a continuous fluid flow within a defined region called a control volume.

The general form of the linear momentum equation for a control volume is:

\(\sum F = \frac{\partial}{\partial t} \mathop \smallint \nolimits_{cv} \rho V d\mathcal{V} + \mathop \smallint \nolimits_{cs} \rho V(V.\hat n)\;dA\)

Where:

• \(\sum F\) represents the sum of all external forces acting on the control volume.
• \(\frac{\partial}{\partial t} \mathop \smallint \nolimits_{cv} \rho V d\mathcal{V}\) is the time rate of change of momentum within the control volume.
• \(\mathop \smallint \nolimits_{cs} \rho V(V.\hat n)\;dA\) is the net rate of momentum efflux (flow out minus flow in) across the control surface (\(cs\)).
• \(\rho\) is the fluid density.
• \(V\) is the velocity vector of the fluid.
• \(d\mathcal{V}\) is an infinitesimal volume element within the control volume (\(cv\)).
• \(\hat n\) is the outward normal unit vector to the differential area \(dA\) on the control surface (\(cs\)).
• \(dA\) is an infinitesimal area element on the control surface.

Applying Conditions for Steady Flow and Fixed Control Volume

We are given two specific conditions:

1. Steady Flow: Steady flow means that fluid properties at any point within the control volume do not change with time. Mathematically, this means \(\frac{\partial}{\partial t} (\text{property}) = 0\). For the momentum equation, the time rate of change of momentum within the control volume is zero:
   \(\frac{\partial}{\partial t} \mathop \smallint \nolimits_{cv} \rho V d\mathcal{V} = 0\)
2. Fixed Control Volume: A fixed control volume means the boundaries of the control volume do not move or deform. This simplifies the analysis of the flow crossing the control surface, but the convective term (the second integral) structure remains the same for a fixed, non-deforming control volume, with \(V\) representing the fluid velocity relative to the fixed control surface.

Simplifying the Momentum Equation

Applying the steady flow condition to the general linear momentum equation:

\(\sum F = 0 + \mathop \smallint \nolimits_{cs} \rho V(V.\hat n)\;dA\)

Thus, for steady flow with a fixed control volume, the linear momentum equation simplifies to:

\(\sum F = \mathop \smallint \nolimits_{cs} \rho V(V.\hat n)\;dA\)

Analyzing the Options

Let's compare our simplified equation with the given options:

1. \(\sum F = \mathop \smallint \nolimits_{cs} \rho V(V.\hat n)\;dA\)

2. \(\sum F = \mathop \smallint \nolimits_{cs} \rho V(V \times \hat n)dA\)

3. \(\sum F = \mathop \smallint \nolimits_{cs} \rho \left( {V \times V} \right) \cdot \hat ndA\)

4. \(\sum F = \mathop \smallint \nolimits_{cs} \rho (V \cdot \hat n)dA\)

Option 1 exactly matches the derived equation for steady flow with a fixed control volume. Option 2 includes a cross product \((V \times \hat n)\) which is incorrect for the momentum flux term. Option 3 has a cross product of \(V\) with itself, which is always zero \((V \times V = 0)\), making the entire term zero, which is incorrect. Option 4 represents the mass flow rate per unit area \(\rho (V \cdot \hat n)\), not the momentum flow rate.

Conclusion

Based on the derivation from the general linear momentum equation under the conditions of steady flow and a fixed control volume, the correct form of the equation is \(\sum F = \mathop \smallint \nolimits_{cs} \rho V(V.\hat n)\;dA\). This equation represents the balance between the forces acting on the control volume and the net rate of momentum flowing out across the control surface.