[Rob Wilk]

I did a Civil Engineering course some years ago and from my textbook I had a question on Turbulent boundary Flow.

In case you are wondering where my calculations are, I am actually blocked on this question and not sure where to start.

May someone please explain to me what things I need to understand, so that I can work towards the answer and get a better understanding on this ?



Take air density = 1.18 kg/m^3 and viscosity = 1.3 * 10^-5 Pa.s

Book Answer 1.31 kg/s


They say calculate mass flow rate entering air intake under zero pressure gradient conditions, how does zero pressure gradient come into this?

From my textbook it says
mass flow rate m = Rho * Uo (Y - boundary Layer Thickness) + Integral between limits y = boundary layer thickness to y = 0 for Rho * u dy

Rho = Density , Uo = free stream velocity u = air velocity

[FMDenaro]

Originally Posted by Rob Wilk

I did a Civil Engineering course some years ago and from my textbook I had a question on Turbulent boundary Flow.

In case you are wondering where my calculations are, I am actually blocked on this question and not sure where to start.

May someone please explain to me what things I need to understand, so that I can work towards the answer and get a better understanding on this ?

https://firebasestorage.googleapis.c...d-f2332b19c09c

[img]Question 9.jpg[/Iimg]

Take air density = 1.18 kg/m^3 and viscosity = 1.3 * 10^-5 Pa.s

Book Answer 1.31 kg/s


They say calculate mass flow rate entering air intake under zero pressure gradient conditions, how does zero pressure gradient come into this?

From my textbook it says
mass flow rate m = Rho * Uo (Y - boundary Layer Thickness) + Integral between limits y = boundary layer thickness to y = 0 for Rho * u dy

Rho = Density , Uo = free stream velocity u = air velocity

Zero pressure gradient is typical of flat plate flow, as illustrated in the classical Blasius solution for laminar flow. Just think of the external inviscid solution where dp/dx=0 from Bernoulli since u is constant.

[Rob Wilk]

Originally Posted by FMDenaro

Zero pressure gradient is typical of flat plate flow, as illustrated in the classical Blasius solution for laminar flow. Just think of the external inviscid solution where dp/dx=0 from Bernoulli since u is constant.

That equation I mentioned from my textbook is that what is used here for this question

mass flow rate m = Rho * Uo (Y - boundary Layer Thickness) + Integral between limits y = boundary layer thickness to y = 0 for Rho * u dy

[FMDenaro]

If you prescribe a velocity profile u=u(y), the mass flow rate across an unlimited vertical section is



rho0* Int[0,+Inf] u(y) dy


but this integral clearly diverges. For that reason one works in terms of mass defect.

If you consider a function delta(x) as sup extrema of the integral it depends on the arbitrary definition of the boundary layer function.

[Rob Wilk]

Originally Posted by FMDenaro

If you prescribe a velocity profile u=u(y), the mass flow rate across an unlimited vertical section is



rho0* Int[0,+Inf] u(y) dy


but this integral clearly diverges. For that reason one works in terms of mass defect.

If you consider a function delta(x) as sup extrema of the integral it depends on the arbitrary definition of the boundary layer function.

Is the boundary layer profile given in the image for my question used for calculating boundary layer thickness ?
This is u/uo = (y / boundary layer thickness)^1/7
and Shear Stress Tau w = 0.025 and the rest of that formula.

It seems to me that you may have to calculate boundary layer thickness first and then the mass flow rate after that.

They do mention in the question that a turbulent boundary layer grows for a distance of 40 m, where does this come into it ?

[FMDenaro]

Originally Posted by Rob Wilk

Is the boundary layer profile given in the image for my question used for calculating boundary layer thickness ?
This is u/uo = (y / boundary layer thickness)^1/7
and Shear Stress Tau w = 0.025 and the rest of that formula.

It seems to me that you may have to calculate boundary layer thickness first and then the mass flow rate after that.

They do mention in the question that a turbulent boundary layer grows for a distance of 40 m, where does this come into it ?

The prescribed power-law is one approximation of the mean velocity profile in the turbulent BL. I strongly suggest to read Sec.7.3 in the textbook of Pope. You will see that the definition of delta(x), as the boundary at which the 99% of the mean velocity is obtained, is arbitrary and different lenghts can be deduced (7.126-127).

Your power-law velocity profile is depicted in Exercise 7.22.

[Rob Wilk]

Originally Posted by FMDenaro

The prescribed power-law is one approximation of the mean velocity profile in the turbulent BL. I strongly suggest to read Sec.7.3 in the textbook of Pope. You will see that the definition of delta(x), as the boundary at which the 99% of the mean velocity is obtained, is arbitrary and different lenghts can be deduced (7.126-127).

Your power-law velocity profile is depicted in Exercise 7.22.

What is the textbook actually called, I might be able to find it online maybe through google books ?

[FMDenaro]

Originally Posted by Rob Wilk

What is the textbook actually called, I might be able to find it online maybe through google books ?

https://www.cambridge.org/core/books...FAC9ED16486B3A

[Rob Wilk]

Originally Posted by FMDenaro

https://www.cambridge.org/core/books...FAC9ED16486B3A

Hi Denaro

I've had a look at that Turbulent Flows book with the section 7 you suggested
on page 299 and 313 in the book.

I'm still confused on how this can be used to work towards the answer.

Are you able to please show me the process of how these calculations are meant to work to work towards the answer for this question ?

[Rob Wilk]

Originally Posted by FMDenaro

https://www.cambridge.org/core/books...FAC9ED16486B3A

Hi Denaro

I've had a look at that Turbulent Flows book with the section 7 you suggested
on page 299 and 313 in the book.

I'm still confused on how this can be used to work towards the answer.

Are you able to please show or explain to me, the process of how these calculations are meant to work to get to the answer for this question ?

[FMDenaro]

Originally Posted by Rob Wilk

Hi Denaro

I've had a look at that Turbulent Flows book with the section 7 you suggested
on page 299 and 313 in the book.

I'm still confused on how this can be used to work towards the answer.

Are you able to please show or explain to me, the process of how these calculations are meant to work to get to the answer for this question ?

Use the law delta(x) for turbulent BL you can find in many textbooks (for example 0.38*x/Re_x^1/5 and integrate u(y)/U0 from zero to delta(x).

[Rob Wilk]

Originally Posted by FMDenaro

Use the law delta(x) for turbulent BL you can find in many textbooks (for example 0.38*x/Re_x^1/5 and integrate u(y)/U0 from zero to delta(x).

What is law delta(x) ?

[FMDenaro]

The approximate expression for the boundary layer thickness along x

[Rob Wilk]

Originally Posted by FMDenaro

Use the law delta(x) for turbulent BL you can find in many textbooks (for example 0.38*x/Re_x^1/5 and integrate u(y)/U0 from zero to delta(x).

I found this Boundary Layer 01 youtube video https://www.youtube.com/watch?v=35qN3jjyIL0.
5 minute in it talks about mass flow rate.

It looks like it does the integration u(y)/U0 from zero to delta(x)

From the velocity profile and shear stress distribution I get boundary layer thickness= 0.37*x/Re_x^1/5 for this question.

x maximum for this question is 40 m isn't it.
in the video they talk about b, that seems like it is the same as x

[FMDenaro]

Originally Posted by Rob Wilk

I found this Boundary Layer 01 youtube video https://www.youtube.com/watch?v=35qN3jjyIL0.
5 minute in it talks about mass flow rate.

It looks like it does the integration u(y)/U0 from zero to delta(x)

From the velocity profile and shear stress distribution I get boundary layer thickness= 0.37*x/Re_x^1/5 for this question.

x maximum for this question is 40 m isn't it.
in the video they talk about b, that seems like it is the same as x

For each position x you compute the thickness delta(x), then the flow rate through the section y=0 .. delta(x). Change the position to a different x and repeat. Of course you have a flow rate also across delta from two different positions.

That's all.

[Rob Wilk]

Originally Posted by FMDenaro

For each position x you compute the thickness delta(x), then the flow rate through the section y=0 .. delta(x). Change the position to a different x and repeat. Of course you have a flow rate also across delta from two different positions.

That's all.

I think I am on the right track with getting closer to solving this, just want to check a few things.

When you say for each position x, do you mean first for position x = 0 at the start of the air intake and for position x = 40 m (end of boundary layer)

I know that flow rate = area * velocity and we are given 300mm * 300mm
and for mass flow rate you have to take into account density as well.

From velocity profile u / uo = (y / delta(x))^1/7

We can say that the velocity u = uo * (y / delta(x))^1/7

so from this equation mass flow rate I think needs to be integrated for y between limits of 0 and maximum boundary layer thickness as taken from the video I showed you.

so mass flow rate = Integral between limits of 0 to boundary layer thickness for uo * (y / delta(x))^1/7 * area * dy

and y gets integrated between the limits I mentioned

Is this the final equation ?

[FMDenaro]

Yes, at each position between x=0 and x=end you compute the BL thickness and integrate in y. The assumption is for 2d flow so that the third dimension is simple assumed to be unity in lenght (1 meter dimensionally). However for the mass flow rate you need to multiply for the constant density otherwise it is the volumetric flow rate.

[Rob Wilk]

Originally Posted by FMDenaro

Yes, at each position between x=0 and x=end you compute the BL thickness and integrate in y. The assumption is for 2d flow so that the third dimension is simple assumed to be unity in lenght (1 meter dimensionally). However for the mass flow rate you need to multiply for the constant density otherwise it is the volumetric flow rate.

It looks like i am on the right track then with the
equation mass flow rate = Integral between limits of 0 to boundary layer thickness for Rho * uo * (y / delta(x))^1/7 * area * dy

I added Rho because you have to take density into account as well.

In the video https://www.youtube.com/watch?v=35qN3jjyIL0 5 minutes 55 seconds in, they have used elemental area of A = b * dy.
For this question though, they say that the intake is square and 300 mm * 300 mm, so how do I deal with this ?

When I get the chance I'll go through the calculations and see if I can get the answer of 1.31 kg/s

[Rob Wilk]

Originally Posted by FMDenaro

Yes, at each position between x=0 and x=end you compute the BL thickness and integrate in y. The assumption is for 2d flow so that the third dimension is simple assumed to be unity in lenght (1 meter dimensionally). However for the mass flow rate you need to multiply for the constant density otherwise it is the volumetric flow rate.

I'm nearly there with this, I think.
I have one important part to clarify.

In the Boundary Layer 01 video https://www.youtube.com/watch?v=35qN3jjyIL0 5 minutes 55 seconds in, they have used elemental area of A = b * dy.

For this question though, they have given both the width and height of 300 mm * 300 mm for the square air intake.

So how do I deal with the elemental area for this question or do I not need to, can I take the 300m * 300mm as the area ?

[FMDenaro]

Originally Posted by Rob Wilk

I'm nearly there with this, I think.
I have one important part to clarify.

In the Boundary Layer 01 video https://www.youtube.com/watch?v=35qN3jjyIL0 5 minutes 55 seconds in, they have used elemental area of A = b * dy.

For this question though, they have given both the width and height of 300 mm * 300 mm for the square air intake.

So how do I deal with the elemental area for this question or do I not need to, can I take the 300m * 300mm as the area ?

You need of a vertical surface that is delta(x)*(spanwise length). In the 2D theory, the spanwise lenght is assumed to be 1