Formula derivation connecting vertical water flowrate & horizontal distance moved by a suspended sphere

[printereater]

I am not really sure how I can label these equations, Can I write $\sum F_{x} = ...$

 

[erobz]

I am not really sure how I can label these equations, Can I write $\sum F_{x} = ...$

That means "add up the forces on this side of the equation". Then you equate that to the momentum rate change ( the other side of the equation).

 

[printereater]

That means "add up the forces on this side of the equation".

so $F_{x}$ and $F_{y}$ will suffice?

 

[erobz]

so $F_{x}$ and $F_{y}$ will suffice?

No, the forces(external) go on the left side. There are two of them acting on the stuff inside the control volume. You have written them down...

 

[printereater]

OHHH so the external forces are being balanced by the internal forces?

 

[printereater]

But they seem to be acting in the same direction though

 

[erobz]

Don't think about it like an internal force. Its really not. Think about it as Newton Second Law. But instead of the $\frac{dp}{dt} = 0$ when solid mass is in equilibrium, now we have fluid mass that is undergoing a change in momentum inside the control that must be accounted for, even though the control volume itself is stationary hence $\frac{dp}{dt} \neq 0$ here.

So you are adding up the external forces, and equating it to the rate of change of momentum (the flowing fluid mass momentum change within the control volume)

 

[printereater]

Don't think about it like an internal force. Its really not. Think about it as Newton Second Law. But instead of the $\frac{dp}{dt} = 0$ when solid mass is in equilibrium, now we have fluid mass that is undergoing a change in momentum inside the control that must be accounted for, even though the control volume itself is stationary hence $\frac{dp}{dt} \neq 0$ here.

Oh, alright. So,
In x-coordinate direction,
$$T\sin(\beta )=\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$T\cos(\beta )+W=\dot{m}v_{i}+\dot{m}v_{0}\cos\left ( \theta \right )$$

 

[erobz]

Oh, alright. So,
In x-coordinate direction,
$$T\sin(\beta )=\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$T\cos(\beta )+W=\dot{m}v_{i}+\dot{m}v_{0}\cos\left ( \theta \right )$$

Step in the right direction, but still not there. What is your convention for positive directions - are you using my diagram or making up your own?

 

[printereater]

Step in the right direction, but still not there. What is your convention for positive directions - are you using my diagram or making up your own?

At first, i was using mine but after a while i started using yours. currently I am taking -ve x and -ve y directions as positive

 

[erobz]

At first, i was using mine but after a while i started using yours

in y direction you have T and W acting in the same direction on the left hand side of the equation? And remember its outflow minus inflow on the right hand side.

 

[printereater]

in y direction you have T and W acting in the same direction on the left hand side of the equation? And remember its outflow minus inflow on the right hand side.

ohhh sorry I just realised that. For the right hand side, I shouldn't be looking at the direction of the force? (for the outflow and inflow part)

 

[erobz]

ohhh sorry I just realised that. For the right hand side, I shouldn't be looking at the direction of the force? (for the outflow and inflow part)

for that side you look at the direction of the flow velocity $\vec{v}$ w.r.t. your positive convention. But it is still outflow minus inflow.

 

[printereater]

in y direction you have T and W acting in the same direction on the left hand side of the equation? And remember its outflow minus inflow on the right hand side.

In x-coordinate direction,
$$T\sin(\beta )=-\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$W-T\cos(\beta )=\dot{m}v_{0}\cos\left ( \theta \right)- \dot{m}v_{i}$$

 

[printereater]

for that side you look at the direction of the flow velocity $\vec{v}$ w.r.t. your positive convention. But it is still outflow minus inflow.

wouldn't it just be $\dot{m}v_{i}+\dot{m}v_{0}\cos\left ( \theta \right)$ then because both vectors are facing downwards?

 

[erobz]

In x-coordinate direction,
$$T\sin(\beta )=\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$W-T\cos(\beta )=\dot{m}v_{0}\cos\left ( \theta \right)- \dot{m}v_{i}$$

Good.

Just to practice this whole convention issue you were(hopfully) having rewrite your equations for "up is positive" and "right is positive".

 

[printereater]

Good.

Just to practice this whole convention issue you were(hopfully) having rewrite your equations for "up is positive" and "right is positive".

HAHAHA yea. btw for my x-coordinate direction it should be $-\dot{m}v_{0}\sin\left ( \theta \right )$ and not $\dot{m}v_{0}\sin\left ( \theta \right )$ right

 

[erobz]

wouldn't it just be $\dot{m}v_{i}+\dot{m}v_{0}\cos\left ( \theta \right)$ then because both vectors are facing downwards?

no, becuase it is outflow minus inflow vectorally

$$ \sum \vec{F} = \dot m \vec {v_o} - \dot m \vec {v_i}$$

 

[printereater]

Oh so, I am not literally adding up the vectors, I am using the formula that was derived from Reynold's transport theorem?

 

[erobz]

Oh so, I am not literally adding up the vectors, I am using the formula that was derived from Reynold's transport theorem?

yes, it is the simplified formula of RTT. The minus sign comes from the dot products in the full treatment.

 

[printereater]

Got it. Alright, I need to go and sleep now. Thank you so much for your help, I really appreciate it so much. Can we continue tomorrow please

 

[erobz]

Got it. Alright, I need to go and sleep now. Thank you so much for your help, I really appreciate it so much. Can we continue tomorrow please

Sounds good.

 

[printereater]

I suspect it's not a trivial problem to solve theoretically. But experimentally, you should be able to find some functional relationship between the outflow angle and the mass flowrate. you have to measure ( control) incoming flowrate, and measure the angle of deflection.

You are going to be working with a control volume that is something like this:

View attachment 339090

And you apply Reynolds Transport Theorem - "The Momentum Equation" in fluid mechanics.

Which says for steady flow with uniform velocity distribution (one inlet-one outlet):

$$ \sum \vec{F} = \dot m \vec{v_o} - \dot m \vec{v_i} $$

Also, can you please show me how you simplify Reynold's transport theorem to this formula please or provide a website that shows it. I tried to look for it but I could not find anything

 

[erobz]

Also, can you please show me how you simplify Reynold's transport theorem to this formula please or provide a website that shows it. I tried to look for it but I could not find anything

We can go over that tomorrow. But tomorrow as a favor to both of us(you and I) you can open with rewriting the equations for "up positive, right positive" just to be sure you aren't still confused(please). I'm hoping to see the step where you take the sign of $v$ relative to coordinate into consideration before simplifying. It's always trivial until we get into uncharted territory, then everyone forgets which way is up, and why that minus sign is there...

 

[erobz]

In x-coordinate direction,
$$T\sin(\beta )=-\dot{m}v_{0}\sin\left ( \theta \right )$$

In y-coordinate direction,
$$W-T\cos(\beta )=\dot{m}v_{0}\cos\left ( \theta \right)- \dot{m}v_{i}$$

Change the sign back in the $x$ direction when you wake up...its not negative on the RHS.

 

[printereater]

We can go over that tomorrow. But tomorrow as a favor to both of us(you and I) you can open with rewriting the equations for "up positive, right positive" just to be sure you aren't still confused(please). I'm hoping to see the step where you take the sign of $v$ relative to coordinate into consideration before simplifying. It's always trivial until we get into uncharted territory, then everyone forgets which way is up, and why that minus sign is there...

Hello! Yes, definitely. I am used to taking up and right as positive all the time but I took left and down as positive to avoid dealing with negative values. i ended up confusing my self so much though.

 

[printereater]

In x-coordinate direction,
$$-T\sin(\beta )=-\dot{m}v_{0}\sin(\theta))$$
$$\therefore T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))+mv_{i}$$

 

[erobz]

In x-coordinate direction,
$$-T\sin(\beta )=-\dot{m}v_{0}\sin(\theta))$$
$$\therefore T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))+mv_{i}$$

Alright, it seems like you get it.

The next step is to use continuity to eliminate the inflow/outflow velocities in terms of variables you plan to control $\dot m$, and variables you can measure/estimate $A_i,A_o$.

 

[printereater]

oh, how do I do that?

 

[erobz]

oh, how do I do that?

Mass flowrate for incompressible flow into control volume must equal mass flowrate out for steady flow ( i.e. fluid mass is not accumulating inside the control volume over time). Velocities $v_i,v_o$ assumed to be uniformly distributed over the cross-sectional inlet/outlet areas. Are you familiar with the expression that describes this?

 

[printereater]

oh so,
In x-coordinate direction,
$$ T\sin(\beta )=\rho v_{o}A_{o} v_{o}\sin(\theta))$$
$$ T\sin(\beta )=\rho (v_{o})^{2}A_{o} \sin(\theta))$$

In y-coordinate direction ,
$$ T\cos(\beta )-W=-\rho v_{o}A_{o}v_{o} \cos(\theta)+\rho v_{i}A_{i}v_{i} $$
$$ T\cos(\beta )-W=-\rho (v_{o})^{2}A_{o} \cos(\theta)+\rho (v_{i})^{2}A_{i} $$

 

[erobz]

oh so,
In x-coordinate direction,
$$ T\sin(\beta )=\rho v_{o}A_{o} v_{o}\sin(\theta))$$
$$ T\sin(\beta )=\rho (v_{o})^{2}A_{o} \sin(\theta))$$

In y-coordinate direction ,
$$ T\cos(\beta )-W=-\rho v_{o}A_{o}v_{o} \cos(\theta)+\rho v_{i}A_{i}v_{i} $$
$$ T\cos(\beta )-W=-\rho (v_{o})^{2}A_{o} \cos(\theta)+\rho (v_{i})^{2}A_{i} $$

you need to eliminate $v_i$ and $v_o$, and keep $\dot m$

 

[printereater]

ohhh nevermind, I am supposed to make $\dot{m}$ the subject of the equation and equate them

 

[printereater]

Hello! Yes, definitely. I am used to taking up and right as positive all the time but I took left and down as positive to avoid dealing with negative values. i ended up confusing my self so much though.

idk why but the same message keeps popping up on my reply box. I accidentally ended up sending it again just now

 

[erobz]

idk why but the same message keeps popping up on my reply box. I accidentally ended up sending it again just now

It happens sometimes. I don't know why.