Finding height of buffer regions and viscous sublayer turbulent flow

[Master1022]

Homework Statement

A fluid flows through a smooth pipe of diameter $150 mm$ with flow rate, density and kinematic viscosity of $180 m^{3} s^{-1}$ and a density $700 kg$ and $0.40 \times 10^{−6} 𝑚^2$ respectively. Calculate the wall shear stress and the height of the viscous sublayer and buffer regions.

Relevant Equations

Reynolds number

Hi,

I was recently attempting a problem about the height of buffer regions and viscous sublayer.

Question:
A fluid flows through a smooth pipe of diameter 150 mm with flow rate, density and kinematic viscosity of 0.180 m^{3} s^{-1} and a density 700 kg and 0.40 \times 10^{−6} 𝑚^2 respectively. Calculate the wall shear stress and the height of the viscous sublayer and buffer regions.

Method:
This was my method:

1. Find the Reynolds number
2. Use a correlation to get the friction factor f from the Reynolds number
3. Find the shear stress \tau from the friction factor f
4. I am not sure how to get from the shear stress to the heights

1. Using the variables above
U = \frac{Q}{A} = \frac{4Q}{\pi D^2} = \frac{4 \times 180}{\pi \times (0.150)^2} = 10.185 m/s
Re = \frac{D U}{\nu} = \frac{0.150 \times 10185.91636}{0.4 \times 10^{-6}} = 3.8197186... \times 10^{6}

This seems like quite a high number to me (maybe even too large), but turbulent nonetheless

2. We are told that we have a smooth pipe and therefore I have used the Blasius correlation for a smooth pipe
f = 0.079 \times (Re)^{-0.25} = (3.8197186... \times 10^{9})^{-0.25} \times 0.079

3. Then we can get \tau from this friction factor
\tau = \frac{1}{2} \times \rho \times U^2 \times f = 64.8812 Pa

I also am not really sure how to make an attempt at the second part of the question at all. Any help or guidance would be greatly appreciated.

 

[haruspex]

You have factors of 1000 that keep coming and going.
0.180 became 180, but you correctly got 10.185. Then that became 10185, yet you then got 3.8... 10⁶ ok. Next, that became 3.8... 10⁹.
Looks like you made two attempts and have posted a hybrid of the working.

The units on the kinematic viscosity should be m²/s.

Your calculation of the linear flow rate is a kind of average. You have effectively assumed it is the same across the flow. Don't you need to consider the velocity gradient? That is what will give rise to a "viscous sublayer", i.e. a non turbulent annulus.

 

[Master1022]

Thank you for your response.

You have factors of 1000 that keep coming and going.
0.180 became 180, but you correctly got 10.185. Then that became 10185, yet you then got 3.8... 10⁶ ok. Next, that became 3.8... 10⁹.
Looks like you made two attempts and have posted a hybrid of the working.

Yes, you are correct. I wrote the problem with converted units. In my original working, I had incorrectly converted the units and forgot to change it throughout the working.

The units on the kinematic viscosity should be m²/s.

That's right. I just blindly copied the question out without much thought about the units.

Your calculation of the linear flow rate is a kind of average. You have effectively assumed it is the same across the flow. Don't you need to consider the velocity gradient? That is what will give rise to a "viscous sublayer", i.e. a non turbulent annulus.

I see, that is a fair point. I don't believe I have really learned about different velocity profiles in pipes, but some searching on the internet yielded a general expression of the form:
u = \frac{1}{k} \sqrt{\frac{\tau_w}{\rho}} ln(y) + c

Does that seem like a reasonable place to start? If it is the right method, I can go to read up more on applying the boundary conditions correctly, etc. The only other velocity profile, which I have learned about, for turbulent flow is the 1/7-th power law. However, I was under the impression that is only applicable to flat-plate flows.

 

[haruspex]

I see, that is a fair point. I don't believe I have really learned about different velocity profiles in pipes, but some searching on the internet yielded a general expression of the form:
u = \frac{1}{k} \sqrt{\frac{\tau_w}{\rho}} ln(y) + c

Does that seem like a reasonable place to start? If it is the right method, I can go to read up more on applying the boundary conditions correctly, etc. The only other velocity profile, which I have learned about, for turbulent flow is the 1/7-th power law. However, I was under the impression that is only applicable to flat-plate flows.

I'm out of my depth here, but I see the exponent in the power law for the velocity distribution depends on the Reynolds number, so it gets tricky: across the width of the pipe the velocity changes, so the Reynolds number changes, so the way the velocity changes changes. Ouch.
1/7 seems to be some commonly observed average.

 

[Master1022]

I think we can assume the flow is fully developed, in which case the velocity distribution will be a constant. Then, I can use a Fanning diagram, instead of the Blasius relation, to obtain the friction factor $f$.