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-   -   Point where Shear Stress is 10 Pascals based on sin Velocity Profile (https://www.cfd-online.com/Forums/main/227731-point-where-shear-stress-10-pascals-based-sin-velocity-profile.html)

Rob Wilk June 8, 2020 08:52

Point where Shear Stress is 10 Pascals based on sin Velocity Profile
 
5 Attachment(s)
I did a Civil Engineering course some years ago and from my textbook I had a question on this.

Attachment 78244

I have made an attempt at the calculations for this question based on the attached research below, but I am not getting the correct answer.

Attachment 78245
Attachment 78246
Attachment 78247
Attachment 78248

Where it says Equating the values of τw from Eqs (2) and (3), we obtain
τw = pi * μ * U / (2 * Boundary layer Thickness)

where τw = Shear Stress = 10 Pascals, μ = viscosity = 0.0018 Ns/m^2 and U = 15 m/s

Based on this 10 * 2 = (pi * 0.0018 * 15) / Boundary layer Thickness
so Boundary layer Thickness = 0.00424 m

I have then tried to find x (distance from leading edge) from the boundary layer thickness formula below.

Boundary Layer Thickness = 4.79 * square root [(μ * x) / (ρ * U∞)]

where μ = viscosity, ρ = Density and U∞ = Velocity

Boundary Layer Thickness = 4.79 * square root [(0.0018 * x) / (1000 * 15)]

0.00424 = 4.79 * square root [((0.0018/9.81) * x) / (1000 * 15)]

0.00424 = 4.79 * square root [(0.0018/9.81) / (1000 * 15)] x^0.5

x^0.5 = 0.00424 / [ 4.79 * square root [(0.0018/9.81) / (1000 * 15)]]
x = 0.00424^2 / [ 4.79^2 * [(0.0018/9.81) / (1000 * 15)]]

x = 64.06 m (no way this is far too big and is not the 2.25 mm answer)


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 ?

tas38 June 10, 2020 07:50

Quote:

Originally Posted by Rob Wilk (Post 773778)
I did a Civil Engineering course some years ago and from my textbook I had a question on this.

Attachment 78244

I have made an attempt at the calculations for this question based on the attached research below, but I am not getting the correct answer.

Attachment 78245
Attachment 78246
Attachment 78247
Attachment 78248

Where it says Equating the values of τw from Eqs (2) and (3), we obtain
τw = pi * mu * U / (2 * Boundary layer Thickness)

where τw = Shear Stress = 10 Pascals, mu = viscosity = 0.0018 Ns/m^2 and U = 15 m/s

Based on this 10 * 2 = (pi * 0.0018 * 15) / Boundary layer Thickness
so Boundary layer Thickness = 0.00424 m

I have then tried to find x (distance from leading edge) from the boundary layer thickness formula below.

Boundary Layer Thickness = 4.79 * square root [(μ * x) / (ρ * U∞)]

where μ = viscosity, ρ = Density and U∞ = Velocity

Boundary Layer Thickness = 4.79 * square root [(0.0018 * x) / (1000 * 15)]

0.00424 = 4.79 * square root [((0.0018/9.81) * x) / (1000 * 15)]

0.00424 = 4.79 * square root [(0.0018/9.81) / (1000 * 15)] x^0.5

x^0.5 = 0.00424 / [ 4.79 * square root [(0.0018/9.81) / (1000 * 15)]]
x = 0.00424^2 / [ 4.79^2 * [(0.0018/9.81) / (1000 * 15)]]

x = 64.06 m (no way this is far too big and is not the 2.25 mm answer)


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 ?


Your analysis appears to be correct to me. Using your result for the wall shear stress, I get a location of x = 6.54 [m]. Note that in your calculation at the end, you seem to have the acceleration due to gravity floating around?


Also, repeating the calculation using a parabolic velocity profile in the von Karman integral (to get an expression for wall shear) and using the Blasius expression for boundary layer thickness, I get a location x = 6.7 [m].


What textbook is this? Is there an errata? It appears that the numerical answer provided may not be correct.

Rob Wilk June 10, 2020 10:32

Point where Shear Stress is 10 Pascals based on sin Velocity Profile
 
Quote:

Originally Posted by tas38 (Post 773983)
Your analysis appears to be correct to me. Using your result for the wall shear stress, I get a location of x = 6.54 [m]. Note that in your calculation at the end, you seem to have the acceleration due to gravity floating around?


Also, repeating the calculation using a parabolic velocity profile in the von Karman integral (to get an expression for wall shear) and using the Blasius expression for boundary layer thickness, I get a location x = 6.7 [m].


What textbook is this? Is there an errata? It appears that the numerical answer provided may not be correct.

I had included acceleration due to gravity because for that part I thought you had to convert the viscosity = μ = 0.0018 Ns/ m^2 to kg/ ms

That formula τw = pi * μ * U / (2 * Boundary layer Thickness)
is for when y = 0 and y is not always equal to 0.


I think that I am meant to be calculating x, the distance along plate.

So I just wonder if I should actually be using the other formula

τw = 0.137 ρ * U^2 * d (that boundary layer symbol) / dx

for working this out, because I think x needs to be calculated.

If it is to be done that way, I wonder how you would rearrange d (that boundary layer symbol) / dx part of the equation, so that x can be solved.

tas38 June 10, 2020 10:55

Quote:

Originally Posted by Rob Wilk (Post 774000)
I had included acceleration due to gravity because for that part I thought you had to convert the viscosity = μ = 0.0018 Ns/ m^2 to kg/ ms

That formula τw = pi * μ * U / (2 * Boundary layer Thickness)
is for when y = 0 and y is not always equal to 0.


I think that I am meant to be calculating x, the distance along plate.

So I just wonder if I should actually be using the other formula

τw = 0.137 ρ * U^2 * d (that boundary layer symbol) / dx

for working this out, because I think x needs to be calculated.

If it is to be done that way, I wonder how you would rearrange d (that boundary layer symbol) / dx part of the equation, so that x can be solved.




You have an expression for \delta \left( x \right). See your last image of your original post. Take the derivative and substitute into the equation you derived for the wall shear. Then insert your known parameters and solve for x.

Rob Wilk June 13, 2020 03:15

Point where Shear Stress is 10 Pascals based on sin Velocity Profile
 
Quote:

Originally Posted by tas38 (Post 774003)
You have an expression for \delta \left( x \right). See your last image of your original post. Take the derivative and substitute into the equation you derived for the wall shear. Then insert your known parameters and solve for x.

Hi Tas38

I have tried to calculate this based on what you have said.

Firstly I am finding an expression for boundary layer thickness

Boundary Layer Thickness = 4.79 * sqrt (μ0 / (ρ * U0)) * x^1/2

Now we need to find the derivative for this boundary layer thickness formula
So differentiate boundary layer thickness

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 x^1/2 - 1

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 x^-1/2

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 / x^1/2


Now I will put d (boundary layer thickness) / dx into the shear stress formula

We will use the following shear stress formula:

τw = 0.137 ρ * U^2 * d (that boundary layer thickness) / dx

τw = 0.137 ρ * U^2 * 4.79 * sqrt (μ0 / (ρ * U0)) * (1/2 / x^1/2)

Now we can put in the values and solve for x

τw = 0.137 * 1000 * 15^2 * 4.79 * sqrt (0.0018 / (1000 * 15)) * (1/2 / x^1/2)

τw = 147651.75 * 0.00034641 * (1/2 / x^1/2)
τw = 25.574 / x^1/2

The question has given Shear Stress τw = 10 Pa or 10 N/ m^2

so 10 = 25.574 / x^1/2

10 x^1/2 = 25.574

x^1/2 = 25.574/ 10
x^1/2 = 2.5574
x = 2.5574^2 = 6.54 m

I am not there yet why is this, I believe I have done my differentiation correct with boundary layer thickness.

How do you get this answer of 2.25 mm or is it not possible ?

tas38 June 13, 2020 06:35

Quote:

Originally Posted by Rob Wilk (Post 774233)
Hi Tas38

I have tried to calculate this based on what you have said.

Firstly I am finding an expression for boundary layer thickness

Boundary Layer Thickness = 4.79 * sqrt (μ0 / (ρ * U0)) * x^1/2

Now we need to find the derivative for this boundary layer thickness formula
So differentiate boundary layer thickness

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 x^1/2 - 1

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 x^-1/2

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 / x^1/2


Now I will put d (boundary layer thickness) / dx into the shear stress formula

We will use the following shear stress formula:

τw = 0.137 ρ * U^2 * d (that boundary layer thickness) / dx

τw = 0.137 ρ * U^2 * 4.79 * sqrt (μ0 / (ρ * U0)) * (1/2 / x^1/2)

Now we can put in the values and solve for x

τw = 0.137 * 1000 * 50^2 * 4.79 * sqrt (0.0018 / (1000 * 15)) * (1/2 / x^1/2)

τw = 1640575 * 0.00034641 * (1/2 / x^1/2)
τw = 568.31 / x^1/2

The question has given Shear Stress τw = 10 Pa or 10 N/ m^2

so 10 = 568.31 / x^1/2

10 x^1/2 = 568.31

x^1/2 = 568.31/ 10
x^1/2 = 56.831
x = 56.831^2 = 3229.76 m (no way that is way too much)

I am not there yet why is this, I believe I have done my differentiation correct with boundary layer thickness.

How do you get this answer of 2.25 mm or is it not possible ?

1.) Your approach appears to be correct. But I believe you made a numerical error (it appears that 50 m/s was inserted for velocity).

2.) I highly recommend that you manipulate/derive equations symbolically and only enter numerical values at the very end. This is far less error prone. Additionally, you are much more likely to solicit helpful responses on this forum if the expressions are presented symbolically.

3.) Regarding my previous post, I got the result x = 6.54 [m]. I did not get the numerical result provided in the text and question its validity.

Rob Wilk June 15, 2020 11:29

Point where Shear Stress is 10 Pascals based on sin Velocity Profile
 
Quote:

Originally Posted by tas38 (Post 774243)
1.) Your approach appears to be correct. But I believe you made a numerical error (it appears that 50 m/s was inserted for velocity).

2.) I highly recommend that you manipulate/derive equations symbolically and only enter numerical values at the very end. This is far less error prone. Additionally, you are much more likely to solicit helpful responses on this forum if the expressions are presented symbolically.

3.) Regarding my previous post, I got the result x = 6.54 [m]. I did not get the numerical result provided in the text and question its validity.

I have changed the velocity from 50 m/s to 15 m/ s and put in the calculations with , so I have now got x = 26.16 m.

I have tried to calculate this based on what you have said.

Firstly I am finding an expression for boundary layer thickness

Boundary Layer Thickness = 4.79 * sqrt (μ0 / (ρ * U0)) * x^1/2

Now we need to find the derivative for this boundary layer thickness formula
So differentiate boundary layer thickness

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 x^1/2 - 1

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 x^-1/2

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 / x^1/2


Now I will put d (boundary layer thickness) / dx into the shear stress formula

We will use the following shear stress formula:

τw = 0.137 ρ * U^2 * d (that boundary layer thickness) / dx

τw = 0.137 ρ * U^2 * 4.79 * sqrt (μ0 / (ρ * U0)) * (1/2 / x^1/2)

Now we can put in the values and solve for x

τw = 0.137 * 1000 * 15^2 * 4.79 * sqrt (0.0018 / (1000 * 15)) * (1/2 / x^1/2)

τw = 147651.75 * 0.00034641 * (1/2 / x^1/2)
τw = 51.148 / x^1/2

The question has given Shear Stress τw = 10 Pa or 10 N/ m^2

so 10 = 51.148 / x^1/2

10 x^1/2 = 51.148

x^1/2 = 51.148/ 10
x^1/2 = 5.1148
x = 5.1148^2 = 26.16 m


Just wondering how you got x = 6.54 m, what formulas did you use for that ?

It seems that answer of 2.25 mm is impossbile to get.

tas38 June 15, 2020 13:57

Quote:

Originally Posted by Rob Wilk (Post 774562)
I have changed the velocity from 50 m/s to 15 m/ s and put in the calculations with , so I have now got x = 26.16 m.

I have tried to calculate this based on what you have said.

Firstly I am finding an expression for boundary layer thickness

Boundary Layer Thickness = 4.79 * sqrt (μ0 / (ρ * U0)) * x^1/2

Now we need to find the derivative for this boundary layer thickness formula
So differentiate boundary layer thickness

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 x^1/2 - 1

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 x^-1/2

d (boundary layer thickness) / dx = 4.79 * sqrt (μ0 / (ρ * U0)) * 1/2 / x^1/2


Now I will put d (boundary layer thickness) / dx into the shear stress formula

We will use the following shear stress formula:

τw = 0.137 ρ * U^2 * d (that boundary layer thickness) / dx

τw = 0.137 ρ * U^2 * 4.79 * sqrt (μ0 / (ρ * U0)) * (1/2 / x^1/2)

Now we can put in the values and solve for x

τw = 0.137 * 1000 * 15^2 * 4.79 * sqrt (0.0018 / (1000 * 15)) * (1/2 / x^1/2)

τw = 147651.75 * 0.00034641 * (1/2 / x^1/2)
τw = 51.148 / x^1/2

The question has given Shear Stress τw = 10 Pa or 10 N/ m^2

so 10 = 51.148 / x^1/2

10 x^1/2 = 51.148

x^1/2 = 51.148/ 10
x^1/2 = 5.1148
x = 5.1148^2 = 26.16 m


Just wondering how you got x = 6.54 m, what formulas did you use for that ?

It seems that answer of 2.25 mm is impossbile to get.

The methodology is fine.

You made an error going from (1) to (2) below...

(1) τw = 147651.75 * 0.00034641 * (1/2 / x^1/2)
(2) τw = 51.148 / x^1/2

Again, I cannot emphasize enough to manipulate the equation symbolically and only insert numerical values at the very end. Your numerical mistake is a testament to how error-prone your evaluation of the answer is.


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