Flow Rate (Q) of Viscous Flow between Two Plates

[WhiteWolf98]

Homework Statement

Water flows between Horizontal plates 1.5 cm apart and 50 cm wide with a Reynolds number of 1500. Calculate: (a) the flow rate

Relevant Equations

$Q=-\frac {2Wh^3} {3\mu} (\frac {dp} {dx})$
$Re=\frac {\rho u L}{\mu}=\frac {uL}{\nu}$

I know how to derive the formula, but I have no idea how to actually use this. Where it says $50~cm$ wide, I'm not sure if that's in the x-direction or the z-direction. $W$, the width in the equation, is in the z-direction. $\mu$ can be calculated, but I'm not sure what the pressure gradient is.

 

[jackwhirl]

Take it one step at a time. Go ahead and calculate u.

 

[WhiteWolf98]

In order to calculate $u$, I'd have to use the second equation:

$u= \frac {Re\cdot \nu} {L}$

$Re=1500$
$\nu = 1.0 \times 10^{-6} ~m^2/s~(for~ water)$
$L=1.5~cm$

Hence,

$u=\frac {1500 \times (1.0 \times 10^{-6})} {1.5\times 10^{-2}}=0.1~ms^{-1}$

 

[jackwhirl]

$L=1.5~cm$

I don't have a textbook with me, so please double check the characteristic dimension for this regime. (two horizontal plates)

 

[WhiteWolf98]

I don't have a textbook with me, so please double check the characteristic dimension for this regime. (two horizontal plates)

I've tried looking, and I can't seem to find it specifically for two parallel plates. How I decided $L$ was the distance between the plates was knowing that for a pipe, $L$ is the diameter of the pipe. So I assumed it was the equivalent here

 

[jackwhirl]

It's been a few years since I took fluid mechanics, so I hope you'll forgive my hedging. I believe this would qualify as a rectangular wide duct.

 

[WhiteWolf98]

I'm really confused about the dimensions of this duct. What does, '$50~cm~ wide$' mean?

 

[jackwhirl]

What could it mean? A plate generally only has three dimensions: length, width, and thickness. Here, two have been left undefined by the problem statement.

Q: Why not define thickness? A: It doesn't touch the fluid and has no bearing on the problem. Assume it is "thick enough."

Q: Why not define length? Assume it is "long enough." Long enough for what?

 

[haruspex]

I've tried looking, and I can't seem to find it specifically for two parallel plates. How I decided $L$ was the distance between the plates was knowing that for a pipe, $L$ is the diameter of the pipe. So I assumed it was the equivalent here

https://en.m.wikipedia.org/wiki/Reynolds_number#Flow_in_a_wide_duct gives the characteristic dimension for a wide duct as twice the separation.

 

[Chestermiller]

Homework Statement:: Water flows between Horizontal plates 1.5 cm apart and 50 cm wide with a Reynolds number of 1500. Calculate: (a) the flow rate
Relevant Equations:: $Q=-\frac {2Wh^3} {3\mu} (\frac {dp} {dx})$
$Re=\frac {\rho u L}{\mu}=\frac {uL}{\nu}$

I know how to derive the formula, but I have no idea how to actually use this. Where it says $50~cm$ wide, I'm not sure if that's in the x-direction or the z-direction. $W$, the width in the equation, is in the z-direction. $\mu$ can be calculated, but I'm not sure what the pressure gradient is.

You say you know how to derive the first formula. OK. Show us. (I don't think you can, because it does not seem you even understand what W and h represent). Please say in words what you think the physical situation involved here represents.

 

[WhiteWolf98]

The derivation itself is very long, and will take a while to do in LaTex. From my understanding, $W$ is the width of the plate, dz.

To derive flow rate from the local velocity profile:

$u=\frac 1 {2\mu} (\frac {dp} {dx})(y^2-h^2)$

The formula for flow rate is used:

$dQ=u \cdot dA$

Now, $dA$ is the area of the cross-section (of the control volume), $dydz$. As the width $dz$ remains constant, it can be replaced with $W$, making the formula:

$dA=u \cdot Wdy$

Substituting in $u$:

$dQ= \frac 1 {2\mu} (\frac {dp} {dx})(y^2-h^2)Wdy$

Integrating this between the height of the duct ($h$ and $-h$), yields the flow rate:

$Q=-\frac {2W} {3\mu} (\frac {dP} {dx}) h^3$

The only way I can think to find this pressure gradient is finding $u$ using the formula for Reynolds number, and then using the local velocity profile equation at $y=0$.

 

[Chestermiller]

https://en.m.wikipedia.org/wiki/Reynolds_number#Flow_in_a_wide_duct gives the characteristic dimension for a wide duct as twice the separation.

I realize that the hydraulic diameter is twice the separation of the plates 4h, but I think most people would just use 2h in defining Re for this.

 

[Chestermiller]

The derivation itself is very long, and will take a while to do in LaTex. From my understanding, $W$ is the width of the plate, dz.

To derive flow rate from the local velocity profile:

$u=\frac 1 {2\mu} (\frac {dp} {dx})(y^2-h^2)$

The formula for flow rate is used:

$dQ=u \cdot dA$

Now, $dA$ is the area of the cross-section (of the control volume), $dydz$. As the width $dz$ remains constant, it can be replaced with $W$, making the formula:

$dA=u \cdot Wdy$

Substituting in $u$:

$dQ= \frac 1 {2\mu} (\frac {dp} {dx})(y^2-h^2)Wdy$

Integrating this between the height of the duct ($h$ and $-h$), yields the flow rate:

$Q=-\frac {2W} {3\mu} (\frac {dP} {dx}) h^3$

The only way I can think to find this pressure gradient is finding $u$ using the formula for Reynolds number, and then using the local velocity profile equation at $y=0$.

How is Q related to u in this system?

 

[WhiteWolf98]

For the TOTAL flow rate:

$Q=\bar u A$

Either way, $\bar u$ still requires this pressure gradient to calculate

 

[Chestermiller]

For the TOTAL flow rate:

$Q=\bar u A$

Either way, $\bar u$ still requires this pressure gradient to calculate

What do you get if you solve this for u, and then substitute the result into the equation for Re, given that A = 2hW?

 

[WhiteWolf98]

I'm getting confused between $u$ and $\bar u$. The equation for $Re$ is:

$Re=\frac {\rho u L} {\mu}$

Is this the same as:

$Re=\frac {\rho \bar u L} {\mu}$?

 

[Chestermiller]

I'm getting confused between $u$ and $\bar u$. The equation for $Re$ is:

$Re=\frac {\rho u L} {\mu}$

Is this the same as:

$Re=\frac {\rho \bar u L} {\mu}$?

Yes

 

[WhiteWolf98]

Yes

In that case, the question is solved. For the sake of completion, I shall write out the full solution here. From above:

$Re=\frac {\rho \bar u L} {\mu}$

The formula for flow rate is:

$Q=\bar u A \Rightarrow \bar u = \frac Q A$

Hence,

$Re=\frac {\rho Q L}{\mu A} \Rightarrow Q= \frac {Re \mu A}{\rho L}$

$Re=1500~\mu=1 \times 10^{-3}~kg~m^{-1}~s^{-1}$
$\rho=1000~kg~m^{-3}~L=1.5\times 10^{-2}~m$

$A=(1.5\times 10^{-2})(50\times 10^{-2})$
$=7.5\times 10^{-3}~m^2$

$Q=\frac {1500(1 \times 10^{-3})~kg~m^{-1}~s^{-1}~(7.5\times 10^{-3})~m^2} {1000~kg~m^{-3}~(1.5\times 10^{-2})~m}$

$Q=7.5\times 10^{-4}~m^3/s~(or~0.75~l/s)$

Thanks to everyone for their help! It was very much appreciated.