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

[tracker890 Source h]

Homework Statement

The solution provided seems to conflict with the momentum equation.

Relevant Equations

momentum equation

Q： Referring to the solution of this problem, why does the equation (eq) hold true?
$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz}\cdot M \right) =M\left( \frac{\partial}{\partial t}u_{xyz} \right) =M\left( \frac{\partial}{\partial t}\text{(}-\frac{V+U_0}{1+\frac{\rho A\left( V+U_0 \right)}{M}\cdot t}\text{)} \right) =0\cdots \text{(}eq\text{)}
$$
reference

 

[TSny]

$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz}\cdot M \right)
$$

This doesn't look right to me. It will help us if you define the symbol $u_{xyz}$.

 

[erobz]

Homework Statement: The solution provided seems to conflict with the momentum equation.
Relevant Equations: momentum equation

Q： Referring to the solution of this problem, why does the equation (eq) hold true?
$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz}\cdot M \right) =M\left( \frac{\partial}{\partial t}u_{xyz} \right) =M\left( \frac{\partial}{\partial t}\text{(}-\frac{V+U_0}{1+\frac{\rho A\left( V+U_0 \right)}{M}\cdot t}\text{)} \right) =0\cdots \text{(}eq\text{)}
$$
reference
View attachment 336271

How about sharing the diagram of problem 4.154?

 

[tracker890 Source h]

How about sharing the diagram of problem 4.154?

 

[tracker890 Source h]

This doesn't look right to me. It will help us if you define the symbol $u_{xyz}$.

$u_{xyz}$: velocity of the fluid element $d\forall$ relative to the $xyz$ reference.
ref.FLUID DYNAMICS -- Richard H. F. Pao -- 1967
ch2-8 Momentum Equation for a Noninertial Control Volume (84) reference

 

[TSny]

Consider the expression $$\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall$$ As I understand it, this expression is supposed to represent the rate of change of the x-component of the momentum of the material inside the CV. If $u_{xyz}$ is the velocity of a fluid element relative to the $xyz$ coordinate system, then I don't see anything in the integrand of the integral that is picking out the x-component of this velocity. As a fluid element rides up along the vane of the cart, the velocity of the element will not be in the x-direction. Could it be that $u_{xyz}$ is defined as the x-component of the velocity of an element relative to the $xyz$ coordinate system? I don't have access to the textbook. [EDIT: Sorry, I still had in mind the figure of the cart with the vane that was used in your previous thread. In this problem, there is no vane.]

$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz}\cdot M \right) $$

$M$ is the symbol for the mass of the cart. I don't see how $M$ would appear on the right-hand side of the equation above. In the integral on the left, each mass element of the cart would be multiplied by the velocity of the cart relative to the reference frame of the cart. This relative velocity is clearly zero. So, the mass elements of the cart do not contribute to the integral. Only the fluid contributes to the integral.

The expression $\large \frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall$ will not be zero. But you can argue that it can be neglected due to the assumption that the mass of fluid in the CV is much smaller than the mass of the cart. Roughly, the integral is proportional to the mass of the fluid while $\large \frac{\partial u_{xyz}(t)_{\rm fluid}}{\partial t}$ is inversely proportional to the mass of the cart. So, overall, the expression is of order $m/M$ where $m$ is the mass of fluid in the CV.

 

[tracker890 Source h]

Consider the expression $$\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall$$ As I understand it, this expression is supposed to represent the rate of change of the x-component of the momentum of the material inside the CV. If $u_{xyz}$ is the velocity of a fluid element relative to the $xyz$ coordinate system, then I don't see anything in the integrand of the integral that is picking out the x-component of this velocity. As a fluid element rides up along the vane of the cart, the velocity of the element will not be in the x-direction. Could it be that $u_{xyz}$ is defined as the x-component of the velocity of an element relative to the $xyz$ coordinate system? I don't have access to the textbook.$M$ is the symbol for the mass of the cart. I don't see how $M$ would appear on the right-hand side of the equation above. In the integral on the left, each mass element of the cart would be multiplied by the velocity of the cart relative to the reference frame of the cart. This relative velocity is clearly zero. So, the mass elements of the cart do not contribute to the integral. Only the fluid contributes to the integral.

The expression $\large \frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall$ will not be zero. But you can argue that it can be neglected due to the assumption that the mass of fluid in the CV is much smaller than the mass of the cart. Roughly, the integral is proportional to the mass of the fluid while $\large \frac{\partial u_{xyz}(t)_{\rm fluid}}{\partial t}$ is inversely proportional to the mass of the cart. So, overall, the expression is of order $m/M$ where $m$ is the mass of fluid in the CV.

Thank you for your detailed explanation. Here's my summary based on my understanding of your explanation. Would you mind confirming if it aligns with what you meant?
$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz,fluid\ in\ C\forall}\cdot m_{_{xyz,fluid\ in\ C\forall}} \right) \ +\ \frac{\partial}{\partial t}\left( u_{xyz,C\forall}\cdot M_{_{xyz,car\ in\ C\forall}} \right)
$$
$$
\because m_{_{xyz,fluid\ in\ C\forall}}<<M_{_{xyz,car\ in\ C\forall}}\
$$
$$
u_{xyz,C\forall}=0,\ u_{xyz,fluid\ in\ C\forall}=u_{xyz,fluid\ in\ C\forall}\left( t \right)
$$
$$
m_{_{xyz,fluid\ in\ C\forall}}=cons\tan t\
$$
$$
M_{_{xyz,car\ in\ C\forall}}=cons\tan t\
$$
$$
\therefore \frac{\partial}{\partial t}\left( u_{xyz,fluid\ in\ C\forall}\cdot \cancel{m_{_{xyz,fluid\ in\ C\forall}}} \right) \ +\ \frac{\partial}{\partial t}\left( \cancel{u_{xyz,C\forall}}\cdot \cancel{M_{_{xyz,car\ in\ C\forall}}} \right) =0
$$

 

[erobz]

As far as I can see it's a solid block of mass $M$, there is no fluid accumulating in the control volume over time. Everything that needs to be done to solve this problem will work with the momentum rate efflux integral i.e.

$$ \sum F_{cv} = \int_{i}^{o} \vec{v} \rho \vec{V} \cdot d\vec{A} $$

where $o$ and $i$ refer to control volume outlet and inlet.

 

[tracker890 Source h]

As far as I can see it's a solid block of mass $M$, there is no fluid accumulating in the control volume over time. Everything that needs to be done to solve this problem will work with the momentum rate efflux integral i.e.

$$ \sum F_{cv} = \int_{i}^{o} \vec{v} \rho \vec{V} \cdot d\vec{A} $$

where $o$ and $i$ refer to control volume outlet and inlet.

Could you provide a more mathematically explicit explanation for why (eq) = 0 in my post, to aid in understanding?
Because I think $u_{xyz}$ is a function of t.
Thank you!

 

[erobz]

Draw me the control volume using the picture you appended.

 

[tracker890 Source h]

Draw me the control volume using the picture you appended.

 

[erobz]

View attachment 336319

In the frame of the cart (the pink frame) how does the flow leave the control volume?

I extended your control volume to include the thickness of the fluid jet. I just like it better conceptually.

 

[tracker890 Source h]

In the frame of the cart (the pink frame) how does the flow leave the control volume?

View attachment 336321I extended your control volume to include the thickness of the fluid jet. I just like it better conceptually.

I hope you can explain the reason for (eq) = 0 using mathematics.

 

[erobz]

Ok, so it comes out perpendicular to the direction it went in ( w.r.t the pink frame). Good. And due to the incompressible flow, Bernoulli's, etc... assumptions it comes out at velocity ( again w.r.t. the pink frame)?

 

[tracker890 Source h]

Ok, so it comes out perpendicular to the direction it went in ( w.r.t the pink frame). Good. And due to the incompressible flow, Bernoulli's, etc... assumptions it comes out at velocity ( again w.r.t. the pink frame)?

The assumptions of Bernoulli's equation hold.

 

[erobz]

The assumptions of Bernoulli's equation hold.

Correct. I'm asking you what is the velocity of the outgoing jet in the carts frame due to these assumptions incompressible flow, jet at atmospheric pressure, etc..

 

[erobz]

Correct. I'm asking you what is the velocity of the outgoing jet in the carts frame due to these assumptions incompressible flow, jet at atmospheric pressure, etc..

Never mind, I see you have it written in green $v+u$. Ok, now switch to the inertial frame on the ground. What is the velocity of the outgoing jet in that frame (the black frame)?

Spoiler: Hint

You on the cart are seeing it exit the control volume purely vertical with magnitude $v+u$. Someone on the ground sees it also moving along with the cart at velocity $u$ in the $x$-direction.

 

[erobz]

I hope you will cut me some slack, but I think I want the mass of the cart to be excluded from the control volume moving forward:

 

[erobz]

I think I want to clarify the symbols used in the notation I suggest in post #8 because they are not the symbols of the problem (and that could be confusing). With the control volume in post #18 replacing the cart with an external force $\vec{F}_c$ acting on the $cv$ (I hope the color coding is clear):

$$ \sum \vec{F} = \int_{cs} \vec{v} \rho \vec{V} \cdot d \vec{A} \tag{1} $$

$\vec{V}$ (at each station) is the velocity of the flow measured with respect to the control surface.

I represent it as exiting upward, but it could be exiting in all directions perpendicular to the cart face it's impacting. The important part for this analysis in the $x$ - direction is that relative to the control surface it comes in parallel to $\hat x$ and leaves $\perp \hat x$.

$\vec{v}$ is (must be) the velocity of the flow crossing the control surface w.r.t. an inertial frame.

To evaluate (1) integrate over each area of each outflow and subtract the integration over each area of each inflow.

So can you simplify the RHS of (1) based on these definitions and the assumptions?

 

[TSny]

Thank you for your detailed explanation. Here's my summary based on my understanding of your explanation. Would you mind confirming if it aligns with what you meant?
$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz,fluid\ in\ C\forall}\cdot m_{_{xyz,fluid\ in\ C\forall}} \right) \ +\ \frac{\partial}{\partial t}\left( u_{xyz,C\forall}\cdot M_{_{xyz,car\ in\ C\forall}} \right)
$$
$$
\because m_{_{xyz,fluid\ in\ C\forall}}<<M_{_{xyz,car\ in\ C\forall}}\
$$

Yes. In my last post I had not seen the link to your reference that gives the diagram for this problem. So, I was still visualizing the cart with the vane that was used in your other thread. For this problem, as you note, we can choose the CV so that negligible mass of fluid is within the CV.

So, $m_{\rm xyz \, fluid \, in \, CV} \approx 0$.

$$u_{xyz,C\forall}=0,\ u_{xyz,fluid\ in\ C\forall}=u_{xyz,fluid\ in\ C\forall}\left( t \right)
$$
$$
m_{_{xyz,fluid\ in\ C\forall}}=cons\tan t\
$$
$$
M_{_{xyz,car\ in\ C\forall}}=cons\tan t\
$$
$$
\therefore \frac{\partial}{\partial t}\left( u_{xyz,fluid\ in\ C\forall}\cdot \cancel{m_{_{xyz,fluid\ in\ C\forall}}} \right) \ +\ \frac{\partial}{\partial t}\left( \cancel{u_{xyz,C\forall}}\cdot \cancel{M_{_{xyz,car\ in\ C\forall}}} \right) =0$$

This looks good to me.

 

[erobz]

If the control volume is (or contains) the mass $M$, then momentum is accumulating in the control volume. The accumulation term is not zero for that selection of control volume. Its negative.

With the mass $M$ enclosed there are no external forces acting on the $cv$ (in the direction of motion) $\sum \vec{F} = 0$.

$$ 0 = \overbrace{M \frac{d \vec{U}}{dt}}^{\text{accumulation}} + \overbrace{\lambda(\vec{U}) }^{ \text{net efflux}}$$

where the function $\lambda ( \vec{U}) > 0$ (given by the net momentum efflux integral ) until the velocity of the cart is equal to the velocity of the jet (the cart has completely turned around, and is heading away from the jet ), making $\lambda = 0$, in your problem the cart is only just coming to a stop.

 

[tracker890 Source h]

Yes. In my last post I had not seen the link to your reference that gives the diagram for this problem. So, I was still visualizing the cart with the vane that was used in your other thread. For this problem, as you note, we can choose the CV so that negligible mass of fluid is within the CV.
View attachment 336334
So, $m_{\rm xyz \, fluid \, in \, CV} \approx 0$.This looks good to

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

 

[tracker890 Source h]

If the control volume is (or contains) the mass $M$, then momentum is accumulating in the control volume. The accumulation term is not zero for that selection of control volume. Its negative.

With the mass $M$ enclosed there are no external forces acting on the $cv$ (in the direction of motion) $\sum \vec{F} = 0$.

$$ 0 = \overbrace{M \frac{d \vec{U}}{dt}}^{\text{accumulation}} + \overbrace{\lambda(\vec{U}) }^{ \text{net efflux}}$$

where the function $\lambda ( \vec{U}) > 0$ (given by the net momentum efflux integral ) until the velocity of the cart is equal to the velocity of the jet (the cart has completely turned around, and is heading away from the jet ), making $\lambda = 0$, in your problem the cart is only just coming to a stop.

I modified the original equation as follows, keeping the same problem: why is $\frac{\partial}{\partial t}\left( u_{xyz}\left( t \right) \right)$ = 0?
The flow $u_{xyz}=u_{xyz}\left( t \right)$, thus it's an unsteady flow, so $\frac{\partial}{\partial t}$ shouldn't be equal to 0. Is this understanding incorrect?
$$
\frac{\partial}{\partial t}\int_{CV}^{}{u_{xyz}}\rho d\forall =\frac{\partial}{\partial t}\left( u_{xyz}\cdot M_{_{CV}} \right) =\frac{\partial}{\partial t}\left[ u_{xyz}\cdot \left( M_{car}+\cancel{m_{fluid}} \right) \right] =\frac{\partial}{\partial t}\left( u_{xyz}\cdot M_{car} \right) =M_{car}\frac{\partial}{\partial t}\left( u_{xyz}\left( t \right) \right) =0
$$

 

[tracker890 Source h]

I think I want to clarify the symbols used in the notation I suggest in post #8 because they are not the symbols of the problem (and that could be confusing). With the control volume in post #18 replacing the cart with an external force $\vec{F}_c$ acting on the $cv$ (I hope the color coding is clear):

View attachment 336331$$ \sum \vec{F} = \int_{cs} \vec{v} \rho \vec{V} \cdot d \vec{A} \tag{1} $$

$\vec{V}$ (at each station) is the velocity of the flow measured with respect to the control surface.

I represent it as exiting upward, but it could be exiting in all directions perpendicular to the cart face it's impacting. The important part for this analysis in the $x$ - direction is that relative to the control surface it comes in parallel to $\hat x$ and leaves $\perp \hat x$.

$\vec{v}$ is (must be) the velocity of the flow crossing the control surface w.r.t. an inertial frame.

To evaluate (1) integrate over each area of each outflow and subtract the integration over each area of each inflow.

So can you simplify the RHS of (1) based on these definitions and the assumptions?

The content you expressed implies a steady flow, but my understanding is that it's an unsteady flow, as shown in the diagram below. Please correct me if I'm mistaken.

 

[erobz]

The content you expressed implies a steady flow, but my understanding is that it's an unsteady flow, as shown in the diagram below. Please correct me if I'm mistaken.
View attachment 336375

1) In this control volume selection that does not include the cart mass the term $\int ~a ~\rho ~d~V\llap{-} = 0$.

2) The momentum accumulation of the portion of the fluid jet that is inside the control volume is taken to be very small so $\frac{ \partial}{\partial t} \int u \rho ~d~V\llap{-} \approx 0$. Imagine the jet dispersing evenly in all directions in a very thin disk that will have negligible mass in comparison to the cart. When I extended it the $cv$, it was to check that we know fluid has to enter/exit "the control volume" for this to work.

3) The integral $\int u \rho V \cdot dA$ does not evaluate to $\rho ( V+U)A$. Try that again and show step by step reasoning.

3) Edit for 3. I see the squared sign now. you have probably evaluated it correctly. Just note that $A_1 = A_2 = A$

The resulting equation for the $cv$ is?

And using Newtons Third Law (for $F_c$) the resulting equation for the cart of mass $M$ is?

 

[erobz]

It looks like you've tried to solve the correct ODE for $U$, but it's not quite what I get - I think you have an algebra mistake.

Anyhow, what's all this business about $M \frac{dU}{dt} = 0$...? It's not.

$$ M\frac{dU}{dt} = -\rho A \left( V+U\right)^2 $$

The RHS is clearly $< 0$ for all times when $U \geq 0$ ?

 

[tracker890 Source h]

It looks like you've tried to solve the correct ODE for $U$, but it's not quite what I get - I think you have an algebra mistake.

Anyhow, what's all this business about $M \frac{dU}{dt} = 0$...? It's not.

$$ M\frac{dU}{dt} = -\rho A \left( V+U\right)^2 $$

The RHS is clearly $< 0$ for all times when $U \geq 0$ ?

The diagram has been modified as follows. Ans I always get confused about "why $\frac{\partial}{\partial t}\left( u_{xyz}\cdot M_{_{CV}} \right) =0$ "because $u_{xyz}=u_{xyz}\left( t \right)$.

 

[erobz]

The diagram has been modified as follows.

View attachment 336376

Ok, not including the cart mass $M$ the sign is wrong on the last term. $F_c$ is the normal force reaction between the cart and our control volume in front of it. The cart is pushing the control volume into the jet with force $F_c$.

So strictly for the control volume chosen without the mass, the equation is:

$$ F_c = \rho A \left( V+U \right)^2 $$

Then, to examine the motion of the cart (by Newtons Third Law) the control volume is pushing on the cart of mass $M$ with force $-F_c$, giving:

$$ M \frac{dU}{dt} = -F_c = - \rho A \left( V+ U\right)^2 $$

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

As you have re-written it:

$\frac{\partial}{\partial t}\left( u_{xyz}\cdot M_{_{CV}} \right) \neq 0$

I don't know why you keep writing that... The equation is:

$\frac{\partial}{\partial t}\left( U\cdot M_{_{CV}} \right) = - \rho A ( V+U ) ^2$

Where $M_{CV} = M_{car} + m_{fluid}$. (the control volume encloses the cart in this interpretation)

Are you looking at what is on the RHS of the eq you obviously derived? $- \rho A ( V+U ) ^2$. It's clearly negative (non-zero) for all positive $U$.

As for the mass in the fluid mass in the control volume, we don't have enough information to determine how much fluid mass $m$ is in the control volume. Its constant in time, its speed is changing...sure. but we don't know how much there is without more information regarding the incoming jet. Furthermore, we expect it to be small in comparison to $M$, the mass of the cart. For example, if we had the radius of the jet as a "given", or even "its a cylindrical jet - most likely", I believe we could account for the momentum acc. of the fluid mass in the control volume. However, for these set of assumptions doing so is not going to fundamentally alter the resulting ODE characteristics.

Also, when you write $M$ to mean $M_{cv}$ and the carts mass is $M$ in the problem you are overusing the symbol and not helping us untangle what you are asking. To me it just seemed like you were very confused about it, and I don't think you really are given you can solve the problem.

Speaking of which did you find a mistake in the algebra with $U(t)$?

 

[erobz]

Lets try to be quantitative: Lets assume a cylindrical jet. This is effectively the control volume, and the fluid mass inside is that red cylinder:

We have $A = 0.01 \rm{m^2}$ and $\rho = 999 \rm{ \frac{kg}{m^3}}$

The is no accumulation of mass inside the control volume, but there is accumulation of momentum because the $cv$ is moving speed $U$ (which is changing in time).

By Bernoulli's and mass conservation we know the outflow area ( dark red cylindrical shell ) must be equal to the inflow area ( circular cross-section $A$ ). What is the mass of the fluid in the control volume in terms of $A, \rho$?

 

[erobz]

If you are choosing this control volume:

Then the governing equation reduces to:

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

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

 

[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.