Bernoulli Equation Problem

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aoifaosf24

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Q) In a production line fluid C is produced. When the bottle comes under the filling heads, fluid A and fluid B are poured into it from separate tanks through two separate tubes.

The final product contains fluid A and fluid B are in a ratio of 1:4 by volume.

Once the bottle is underneath the filling heads, both liquids are filled in exactly the same time so that the bottle can move down the line for sealing and capping and another bottle can be filled up.

a) If the diameter of the fluid A tube is double that of the fluid B
tube, what is the ratio of the velocity of the fluid A: fluid B in their tubes?

b) What is the ratio of the mass flow rate of fluid A: fluid B in their tubes?

heres my working and i think i got the first part right but not sure how to do the second part.

https://pasteboard.co/JBtLiSI.jpg

Eimmanuel

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(Original post by aoifaosf24)
Q) In a production line fluid C is produced. When the bottle comes under the filling heads, fluid A and fluid B are poured into it from separate tanks through two separate tubes.

The final product contains fluid A and fluid B are in a ratio of 1:4 by volume.

Once the bottle is underneath the filling heads, both liquids are filled in exactly the same time so that the bottle can move down the line for sealing and capping and another bottle can be filled up.

a) If the diameter of the fluid A tube is double that of the fluid B
tube, what is the ratio of the velocity of the fluid A: fluid B in their tubes?

b) What is the ratio of the mass flow rate of fluid A: fluid B in their tubes?

heres my working and i think i got the first part right but not sure how to do the second part.

https://pasteboard.co/JBtLiSI.jpg

I think you make some mistakes in your part A.
$D_{\text{A}} = 2D_{\text{B}}$

$A_{\text{B}} = \pi (\dfrac{D_{\text{B}}}{2})^2 = 4\pi (\dfrac{D_{\text{A}}}{2})^2 = 4A_{\text{A}}$

$Q_{\text{B}} = u_{\text{B}} A_{\text{B}} = u_{\text{B}} (4*A_{\text{A}})$

aoifaosf24

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(Original post by Eimmanuel)
I think you make some mistakes in your part A.
$D_{\text{A}} = 2D_{\text{B}}$

$A_{\text{B}} = \pi (\dfrac{D_{\text{B}}}{2})^2 = 4\pi (\dfrac{D_{\text{A}}}{2})^2 = 4A_{\text{A}}$

$Q_{\text{B}} = u_{\text{B}} A_{\text{B}} = u_{\text{B}} (4*A_{\text{A}})$

I am slightly confused then, so what is the ratio of the velocities?

Eimmanuel

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(Original post by aoifaosf24)
I am slightly confused then, so what is the ratio of the velocities?

What are you confused with?

0le

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(Original post by aoifaosf24)
I am slightly confused then, so what is the ratio of the velocities?

Eimmanuel provided you with the working out to get the volumetric flow rate of pipe B. You now need to get the volumetric flow rate of pipe A. If you follow their logic, this should be obvious. Once you have those two equations, the ratio in velocities is expressed by rearranging terms in these two equations:

$u_{A}/u_{B} = \frac{X}{Y}$
Where:
X = Use velocity from volumetric flow rate in pipe A
Y = Use velocity from volumetric flow rate in pipe B

Last edited by 0le; 2 months ago

aoifaosf24

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(Original post by Eimmanuel)
What are you confused with?

Im confused with how you got the second line.

I understand Ab=pi*(Db/2)^2 but if Da=2Db then Db=Da/2 so shouldn't Ab=pi*[(Da/2)/2]^2 => Ab=pi*(Da/4)^2 => Ab=(pi*Da^2)/16

And then as Aa=(pi*Da^2)/4 we can equate the two and get the equation 4Aa=16Ab and therefore Aa=4Ab.

So im just confused with how you got 4Aa=Ab

Eimmanuel

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(Original post by aoifaosf24)
Im confused with how you got the second line.

I understand Ab=pi*(Db/2)^2 but if Da=2Db then Db=Da/2 so shouldn't Ab=pi*[(Da/2)/2]^2 => Ab=pi*(Da/4)^2 => Ab=(pi*Da^2)/16

And then as Aa=(pi*Da^2)/4 we can equate the two and get the equation 4Aa=16Ab and therefore Aa=4Ab.

So im just confused with how you got 4Aa=Ab

You are right that I make a mistake.

It is supposed to be as edited below.

(Original post by Eimmanuel)
I think you make some mistakes in your part A.
$D_{\text{A}} = 2D_{\text{B}}$

$A_{\text{B}} = \pi (\dfrac{D_{\text{B}}}{2})^2 = \dfrac{1}{4}\pi (\dfrac{D_{\text{A}}}{2})^2 = \dfrac{1}{4}A_{\text{A}}$

$Q_{\text{B}} = u_{\text{B}} A_{\text{B}} = u_{\text{B}} (\dfrac{1}{4}A_{\text{A}})$

I was puzzled by your statement in the working:
4Q_A = u_AA_A

Q_B = u_B(4A_B)
which (IMO) do not make sense.

Last edited by Eimmanuel; 2 months ago

aoifaosf24

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(Original post by Eimmanuel)
You are right that I make a mistake.

It is supposed to be as edited below.

I was puzzled by your statement in the working:
4Q_A = u_AA_A

Q_B = u_B(4A_B)
which (IMO) do not make sense.

Sorry about that.

So now i have the equations Qb=Ub(Aa/4) and Qa=UaAa and as 4Qa=Qb [because of the 1:4 (A:B) ratio in the bottle]

So Qb/4=UaAa and Qb=Ub(Aa/4) and therefore Ua=Qb/4Aa and Ub=4Qb/Aa => Ua/Ub=(Qb/4a)/4Qb/Aa) => Ua/Ub=QbAa/16QbAa => Ua/Ub=1/16

so i can conclude Ua : Ub = 1:16

Last edited by aoifaosf24; 2 months ago

Eimmanuel

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(Original post by aoifaosf24)
Sorry about that.

So now i have the equations Qb=Ub(Aa/4) and Qa=UaAa and as 4Qa=Qb [because of the 1:4 (A:B) ratio in the bottle]

So Qb/4=UaAa and Qb=Ub(Aa/4) and therefore Ua=Qb/4Aa and Ub=4Qb/Aa => Ua/Ub=(Qb/4a)/4Qb/Aa) => Ua/Ub=QbAa/16QbAa => Ua/Ub=1/16

so i can conclude Ua : Ub = 1:16

Don't need to apologize.

Your working seems good.

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$\dfrac{Q_{\text{A}}}{Q_{\text{B} }} = \dfrac{1}{4}$

$\dfrac{u_{\text{A}} A_{\text{A}}}{u_{\text{B}} A_{\text{B}}} = \dfrac{1}{4}$

$\dfrac{u_{\text{A}} }{u_{\text{B}} } = \dfrac{1}{4} \dfrac{ A_{\text{B}}}{ A_{\text{A}}}$

$\dfrac{u_{\text{A}} }{u_{\text{B}} } = \dfrac{1}{16}$