Q: In a relative non-inertial reference frame, the fluid velocity is zero?

[tracker890 Source h]

Homework Statement

It's not apparent why the fluid velocity in the relative non-inertial reference frame is zero.

Relevant Equations

Momentum equation in non-inertial coordinates

Q： Regarding item (4), my understanding aligns with (eq_1), where M is a constant. However, why does $\left( \frac{\partial}{\partial t}u_{xyz} \right)$ in (eq_1) equal 0?

$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz}\cdot M \right) =M\left( \frac{\partial}{\partial t}u_{xyz} \right) =0\cdots \text{(}eq\_1\text{)}
$$
reference

 

[Lnewqban]

I believe that the reason for assumption 4 is to eliminate the reduction of fluid entering velocity as the cart increases its horizontal velocity.

 

[tracker890 Source h]

I believe that the reason for assumption 4 is to eliminate the reduction of fluid entering velocity as the cart increases its horizontal velocity.

Thank you for the explanation. So, for now, let's consider it as an assumption for simplifying the problem.

 

[erobz]

Homework Statement: It's not apparent why the fluid velocity in the relative non-inertial reference frame is zero.
Relevant Equations: Momentum equation in non-inertial coordinates

Q： Regarding item (4), my understanding aligns with (eq_1), where M is a constant. However, why does $\left( \frac{\partial}{\partial t}u_{xyz} \right)$ in (eq_1) equal 0?

$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz}\cdot M \right) =M\left( \frac{\partial}{\partial t}u_{xyz} \right) =0\cdots \text{(}eq\_1\text{)}
$$
reference

What you have written here is incorrect. The integral term on the LHS of the governing equation from the example is what represents the solid mass $M$ momentum accumulation rate (when solid mass is contained inside the control volume - that is accelerating)