Deriving formula for sphere displacement vs water flow rate using Bernoulli's principle

[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.

 

[printereater]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$
$$\dot{m}=\frac{T\sin(\beta )}{v_{0}\sin(\theta)}\rightarrow (1)$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=\dot{m}(v_{i}-v_{0}\cos(\theta))$$
$$\dot{m}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}\rightarrow (2)$$

$$(1)=(2)$$
$$\therefore \frac{T\sin(\beta )}{v_{0}\sin(\theta)}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}$$

 

[erobz]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{0}\sin(\theta))$$
$$\dot{m}=\frac{T\sin(\beta )}{v_{0}\sin(\theta)}\rightarrow (1)$$

In y-coordinate direction ,
$$T\cos(\beta )-W=-\dot{m}v_{0}\cos(\theta))-(-mv_{i})$$
$$T\cos(\beta )-W=\dot{m}(v_{i}-v_{0}\cos(\theta))$$
$$\dot{m}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}\rightarrow (2)$$

$$(1)=(2)$$
$$\therefore \frac{T\sin(\beta )}{v_{0}\sin(\theta)}=\frac{T\cos(\beta )-W}{v_{i}-v_{0}\cos(\theta)}$$

You aren't getting what I'm saying ( or I don't understand what making $\dot m$ "the subject " is). Either way, you want each equation to not have any velocity parameters. Just the force terms, mass flowrate $\dot m$, relevant area $A_i,A_o$, and known physical properties (density). Don't join the equations yet.

Think about equating $\dot m = \dot m_i = \dot m_o$ to eliminate each velocity $v_i,v_o$ from each equation.

 

[printereater]

ohh so I need to put $v_{o}$ and $v_{i}$ in terms of $\dot{m}$,$\rho$ and $A$ using $\dot{m}=\rho vA$

 

[erobz]

ohh so I need to put $v_{o}$ and $v_{i}$ in terms of $\dot{m}$,$\rho$ and $A$ using $\dot{m}=\rho vA$

yeah for each location ( inlet/outlet) that holds.

 

[printereater]

Got it. It's about to be 1:30 am now, I will head to bed and continue tomorrow. Thank you:)

 

[erobz]

Got it. It's about to be 1:30 am now, I will head to bed and continue tomorrow. Thank you:)

Get some rest, I'll be unavailable for the weekend. If you want to continue, feel free, someone will help you. If you want to wait till Monday, that is fine too.

 

[printereater]

Oh no. If that's the case, i don't mind sacrificing my sleep HAHAHAH let's continue. Am I close to the final equation

 

[erobz]

You're, going to get tired, and learning abilities are going to decline, and mistakes are going to start to happen in your math. There is certainly no way we can get through the Reynolds Transport Theorem - "The Momentum Equation", with any level of understanding so quick.

Your goal (for the weekend- if waiting for me at least - which is not necessary ) after eliminating the velocities, is then to eliminate the Tension force from the equations by substitution, and then solve the resulting equation containing only parameters $\dot m, A_i, A_o,\rho, \theta$ for the angle the ball makes w.r.t vertical $\beta$.

I strongly suggest you just take a rest, and work at it refreshed.

 

[printereater]

yea you are right, I will just continue tomorrow. Thank you so much, once again

 

[printereater]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{o}\sin(\theta))\rightarrow (1)$$
$$\dot{m}=\rho v_{o}A_{o}$$
$$v_{o}=\frac{\dot{m}}{\rho A_{o}}\rightarrow (2)$$
Substitute (2) into (1) :
$$T\sin(\beta )=\dot{m}(\frac{\dot{m}}{\rho A_{o}})\sin(\theta)$$
$$\therefore T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$

In y-coordinate direction
$$T\cos(\beta )-W=-\dot{m}v_{o}\cos(\theta))+mv_{i}\rightarrow (3)$$
$$v_{i}=\frac{\dot{m}}{\rho A_{i}}\rightarrow (4)$$
Substitute (2) and (4) into (3):
$$T\cos(\beta )-W=-\dot{m}\frac{\dot{m}}{\rho A_{o}}\cos(\theta)+m\frac{\dot{m}}{\rho A_{i}}$$
$$\therefore T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

 

[printereater]

In x-coordinate direction,
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}}{\rho A_{i}A_{o}}-\frac{\dot{m}^{2}A_{i}\cos(\theta)}{\rho A_{i}A_{o}}+\frac{W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}\rightarrow (6)$$

(5)=(6),
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{ A_{i}\cos(\beta )}$$

 

[erobz]

In x-coordinate direction,
$$ T\sin(\beta )=\dot{m}v_{o}\sin(\theta))\rightarrow (1)$$
$$\dot{m}=\rho v_{o}A_{o}$$
$$v_{o}=\frac{\dot{m}}{\rho A_{o}}\rightarrow (2)$$
Substitute (2) into (1) :
$$T\sin(\beta )=\dot{m}(\frac{\dot{m}}{\rho A_{o}})\sin(\theta)$$
$$\therefore T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$

In y-coordinate direction
$$T\cos(\beta )-W=-\dot{m}v_{o}\cos(\theta))+mv_{i}\rightarrow (3)$$
$$v_{i}=\frac{\dot{m}}{\rho A_{i}}\rightarrow (4)$$
Substitute (2) and (4) into (3):
$$T\cos(\beta )-W=-\dot{m}\frac{\dot{m}}{\rho A_{o}}\cos(\theta)+m\frac{\dot{m}}{\rho A_{i}}$$
$$\therefore T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

Keep factoring the y coordinate equation to clean it up. Otherwise, ok. Then keep going.

 

[erobz]

In x-coordinate direction,
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}}{\rho A_{i}A_{o}}-\frac{\dot{m}^{2}A_{i}\cos(\theta)}{\rho A_{i}A_{o}}+\frac{W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T\cos(\beta )=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}}$$
$$T=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}\rightarrow (6)$$

(5)=(6),
$$\frac{\dot{m}^{2}\sin\(\theta)}{\rho A_{o}\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{\rho A_{i}A_{o}\cos(\beta )}$$
$$\frac{\dot{m}^{2}\sin\(\theta)}{\sin(\beta )}=\frac{\dot{m}^{2}A_{o}-\dot{m}^{2}A_{i}\cos(\theta)+W\rho A_{i}A_{o}}{ A_{i}\cos(\beta )}$$

Your algebra looks unnecessarily complicated here. Just use the x direction equation to eliminate $T$ in the y direction.

 

[printereater]

Oh alright

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

 

[erobz]

$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}})$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$

factor the RHS a bit more and then proceed to solve for $\beta$, and you are done with that part.

 

[printereater]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})+W$$
$$\frac{\rho A_{o}\tan(\beta )}{\dot{m}^{2}\sin(\theta)}=\frac{\rho}{\dot{m}^{2}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{1}{W}$$
$$\tan(\beta )=\frac{\rho\dot{m}^{2}\sin(\theta)}{\dot{m}^{2}\rho A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}$$
$$\tan(\beta )=\frac{\sin(\theta)}{A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}$$
$$
\tan(\beta )=\arctan[\frac{\sin(\theta)}{A_{o}}( A_{i}-\frac{A_{o}}{ \cos(\theta)})+\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}W}]$$

 

[printereater]

@erobz @Delta2 Please take a look at it when you are free and let me know if I am on the right track.

 

[erobz]

In x-coordinate direction
$$T\sin(\beta )=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}}$$
$$T=\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\sin(\beta )}\rightarrow (5)$$

In y-coordinate direction,
$$T\cos(\beta )-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}\rightarrow (6)$$

Substitute (5) into (6),
$$\frac{\dot{m}^{2}\sin(\theta)\cos(\beta )}{\rho A_{o}\sin(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho A_{i}}-\frac{\dot{m}^{2}\cos(\theta)}{\rho A_{o}}$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}-W=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})$$
$$\frac{\dot{m}^{2}\sin(\theta)}{\rho A_{o}\tan(\beta )}=\frac{\dot{m}^{2}}{\rho}(\frac{1}{ A_{i}}-\frac{\cos(\theta)}{ A_{o}})+W$$

You go off the rails on the next step. In general, $\frac{1}{a+b} \neq \frac{1}{a}+\frac{1}{b}$.

 

[printereater]

You go off the rails on the next step. In general, $\frac{1}{a+b} \neq \frac{1}{a}+\frac{1}{b}$.

Sorry, I just saw your message. I will try again today. The goal is it to factorise the right hand side and make $\beta$ the subject of the equation right