Q2: In a relative non-inertial reference frame, why is fluid velocity zero?

[TSny]

So, does $u_{xyz}$ have different definitions in $\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall$ and $\int_{CS}{u_{xyz}\rho}\vec{V}_{xyz}d\vec{A}$?

The only way it makes sense to me is to take $u_{x, y, z}$ to be the x-component of velocity (relative to the cart) of the mass element $\rho d\forall$. So, $\int_{CV}^{}{u_{xyz}}\rho d\forall$ gives the instantaneous x-component of momentum (relative to the cart frame) of the material inside the CV.

$u_{xyz}$ has the same meaning in $\int_{CS}{u_{xyz}\rho}\vec{V}_{xyz} \cdot d\vec{A}.\,\,\,\,$ $\vec{V}_{xyz}$ is the velocity vector of the element $\rho d\forall$ relative to the cart frame. The integrand gives the flux of x-component of momentum through the surface element $d\vec{A}$ of the control surface.

 

[erobz]

The only way it makes sense to me is to take $u_{x, y, z}$ to be the x-component of velocity (relative to the cart) of the mass element $\rho d\forall$. So, $\int_{CV}^{}{u_{xyz}}\rho d\forall$ gives the instantaneous x-component of momentum (relative to the cart frame) of the material inside the CV.

$u_{xyz}$ has the same meaning in $\int_{CS}{u_{xyz}\rho}\vec{V}_{xyz} \cdot d\vec{A}.\,\,\,\,$ $\vec{V}_{xyz}$ is the velocity vector of the element $\rho d\forall$ relative to the cart frame. The integrand gives the flux of x-component of momentum through the surface element $d\vec{A}$ of the control surface.

Taken directly from my textbook:

 

[TSny]

I always get confused about "why $\frac{\partial}{\partial t}\left( u_{xyz}\cdot M_{_{CV}} \right) =0$ "because $u_{xyz}=u_{xyz}\left( t \right)$.

This expression is not always equal to zero. It might help to consider the case where we choose the CV to include a significant mass of fluid:

Note: The CV remains at rest relative to the cart. So, the CV is a noninertial frame of reference.
[Edited to add this note.]

Let $M$ be the mass of the cart and $m$ be the mass of fluid inside the CV. Both $M$ and $m$ are constant in time. Then $$\frac{\partial}{\partial t}\int u_{xyz}\rho d\forall = m\frac{\partial u_{xyz}}{\partial t} \neq 0.$$ Here, $u_{xyz}$ is the x-component of the velocity of the fluid in the stream relative to the cart, which changes with time.

But, when we chose the CV to exclude any significant mass of fluid (see the figure in post #20), we got zero for this expression since $m = 0$ for that CV. Now, the answer to the problem cannot depend on our choice of CV. So, at first sight, it might seem that something is wrong. However, we are saved by the fact that by changing the CV to include a significant mass of fluid, something changes on the left side of the fundamental equation which fixes things.

On the left side of the fundamental equation, we have the term $$-\int_{CV}a_{rf_x} \rho d\forall$$ where $a_{rf_x}$ is the acceleration of the reference frame of the cart relative to the lab frame. With our new CV, $$-\int_{CV}a_{rf_x} \rho d\forall = -Ma_{rf_x} - ma_{rf_x}.$$ The last term $- ma_{rf_x}$ is what fixes things since it turns out to equal the expression $m\frac{\partial u_{xyz}}{\partial t}$ that occurs on the right side of the fundamental equation. So, these terms cancel. Thus, changing the CV does not affect the final result.

To see the equivalence of$- ma_{rf_x}$ and $m\frac{\partial u_{xyz}}{\partial t}$, we note that $u_{xyz} = -(V+U)$, where $U$ is the instantaneous speed of the cart relative to the lab and $V$ is the speed of the fluid stream relative to the lab. The negative sign in the expression is due to taking positive x direction toward the right. Thus, the x-component of the velocity of the fluid relative to the cart is negative. Since V is a constant, we get $$m\frac{\partial u_{xyz}}{\partial t} = -m\frac{\partial (U+V)}{\partial t} = -m\frac{\partial U}{\partial t} = - ma_{rf_x}$$

 

[TSny]

Taken directly from my textbook:

@tracker890 Source h is using a textbook which develops an equation for the case where the CV is a noninertial frame of reference.

The last term on the left is the net "fictitious" force in the noninertial frame. So, using this equation, you can choose the CV to be at rest relative to the accelerating cart.

 

[tracker890 Source h]

If you are choosing this control volume:

View attachment 336381

Then the governing equation reduces to:

View attachment 336387

This is how you should be reducing the equation for the control volume including the cart.

As you should be able to see, $M_{car} \frac{dU}{dt} \neq 0$?

I understand. $\int_{CV}^{}{\rho d\forall}$ refers to all mass elements (including solid and fluid) inside the control volume, but the fluid appears negligible within the vast control volume, so it can be ignored. However, despite this, the flow remains unsteady because, in reality, $\frac{\partial}{\partial t}$ is not equal to 0.
Thank you for teaching and explaining in such detail and patience.

 

[tracker890 Source h]

This expression is not always equal to zero. It might help to consider the case where we choose the CV to include a significant mass of fluid:
View attachment 336393
Note: The CV remains at rest relative to the cart. So, the CV is a noninertial frame of reference.
[Edited to add this note.]

Let $M$ be the mass of the cart and $m$ be the mass of fluid inside the CV. Both $M$ and $m$ are constant in time. Then $$\frac{\partial}{\partial t}\int u_{xyz}\rho d\forall = m\frac{\partial u_{xyz}}{\partial t} \neq 0.$$ Here, $u_{xyz}$ is the x-component of the velocity of the fluid in the stream relative to the cart, which changes with time.

But, when we chose the CV to exclude any significant mass of fluid (see the figure in post #20), we got zero for this expression since $m = 0$ for that CV. Now, the answer to the problem cannot depend on our choice of CV. So, at first sight, it might seem that something is wrong. However, we are saved by the fact that by changing the CV to include a significant mass of fluid, something changes on the left side of the fundamental equation which fixes things.

On the left side of the fundamental equation, we have the term $$-\int_{CV}a_{rf_x} \rho d\forall$$ where $a_{rf_x}$ is the acceleration of the reference frame of the cart relative to the lab frame. With our new CV, $$-\int_{CV}a_{rf_x} \rho d\forall = -Ma_{rf_x} - ma_{rf_x}.$$ The last term $- ma_{rf_x}$ is what fixes things since it turns out to equal the expression $m\frac{\partial u_{xyz}}{\partial t}$ that occurs on the right side of the fundamental equation. So, these terms cancel. Thus, changing the CV does not affect the final result.

To see the equivalence of$- ma_{rf_x}$ and $m\frac{\partial u_{xyz}}{\partial t}$, we note that $u_{xyz} = -(V+U)$, where $U$ is the instantaneous speed of the cart relative to the lab and $V$ is the speed of the fluid stream relative to the lab. The negative sign in the expression is due to taking positive x direction toward the right. Thus, the x-component of the velocity of the fluid relative to the cart is negative. Since V is a constant, we get $$m\frac{\partial u_{xyz}}{\partial t} = -m\frac{\partial (U+V)}{\partial t} = -m\frac{\partial U}{\partial t} = - ma_{rf_x}$$

View attachment 336395

I got it. Thank you very much for teaching and explaining in such detail and patience.

 

[tracker890 Source h]

Taken directly from my textbook:

View attachment 336391

The content of this book is extremely detailed, it must be a good book. Therefore, I'd like to inquire about the title of this textbook, I hope you don't mind.

 

[erobz]

The content of this book is extremely detailed, it must be a good book. Therefore, I'd like to inquire about the title of this textbook, I hope you don't mind.

Engineering Fluid Mechanics: Crowe, Elger, Williams, Roberson -9th Edition