Someone please help me (I give out brainless) This is due very soon

A fluid has a dynamic viscosity of 0.048 Pa.s and a specific gravity of 0.913. For the flow of such a fluid over a flat solid su

sattari [20]

Answer:

Explanation:

First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:

$\tau = \mu \cfrac{\partial v}{\partial y}$

where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.

Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at $y=75 \, mm$ and the velocity at that point is $1.125 \, m/s$

We can put that in mathematical terms as:

$v(y)= A+ By +Cy^2 \\v(0) = 0\\v(75 \, mm) = 1.125 \, m/s\\v'(75 \, mm) = 0\\$

From the no-slip condition, we can deduce that $A=0$ and so we are left with just two terms:

$v(y) = By + C y ^2 \\$

We know that the vertex is at $y= 75 \, mm$ and so we can rewrite the last equation as:

$v(y) = k(y-75 \, mm) ^2+h$

where k and h are constants to be determined. First we check that $v( 75 \, mm) = 1.125 \, m/s$ :

$v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \, m/s$

So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.

$v( 0) = k( -75 \, mm) ^2+ 1.125 \, m/s= 0 \quad (no \, \textendash slip) \\\\k= - \cfrac{ 1.125 \, m/s }{(75 \, mm ) ^2} = - \cfrac{ 1125 \, mm/s }{(75 \, mm ) ^2}\\\\k= - \cfrac{0.2}{mm \times s}$

And thus we find that the final expression for the fluid's velocity is:

$v( y) = 1125- 0.2 ( y -75 ) ^2$

where v is in mm/s and y is in mm.

In SI units it would be:

$v( y) = 1.125- 200 ( y -0.075 ) ^2$

To calculate the shear stress, we need to take the derivative of this expression and multiply by the fluid's viscosity:

$\tau = \mu \cfrac{\partial v}{\partial y}$

$\tau =0.048\, \cdot (-400) ( y-0.075 )$

for $y= 0.050 \, m$ we have:

$\tau =0.048\, \cdot (-400) ( 0.050 -0.075 ) = 0.48\, Pa$

Which is our final result

5 0

4 years ago

Example 1: the two dimensional points P1(0,0) and P2(1,0) and the two tangents P', (1,1) and P2 (0,-1).find the equation of the

Stells [14]

Answer: (0,0)+ (1,0)= 1 lines upwards( suggesting that this is a line graph not saying it is but as an example) an (1,1) and (0,-1) all make a small square ( as this is a 2 dimensional graph that it has a negative side too,(below the positive side)) i hope this helps and is what you are looking for

Explanation:

6 0

3 years ago

A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the

Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})$

Using the Boyle equation we have

$\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})$

Where,

$C_p$ = Specific heat at constant pressure

$T_1$ = Initial temperature of gas

$T_2$ = Final temprature of gas

R = Universal gas constant

$v_1$ = Initial specific Volume of gas

$v_2$ = Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

$T_1 = T_2, v_2=v_1$

The equation would turn out as

$\Delta S = C_p ln1-ln1$

Therefore the entropy change of the ideal gas is 0

Into the surroundings we have that

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

$\Delta S = \frac{230kJ}{(30+273)K}$

$\Delta S = 0.76 kJ/K$

Therefore the increase of entropy into the surroundings is 0.76kJ/K

8 0

3 years ago

What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

Andreyy89

I don’t know but good luck

4 0

3 years ago

Will give brainly

Ksivusya [100]

Answer:

Here you go!!

Explanation:

If I were to write a book I would write my steps down as in figuring out what I what to talk about. Have an Introduction- Using an introduction is needed because you need to Describe your main idea, or what the essay is about, in one sentence. ... Develop a thesis statement, or what you want to say about the main idea. ... List three points or arguments that support your thesis in order of importance (one sentence for each). I would also use a Rising action- This is needed because, To construct the plot, nearly any story can be said to use increasing action. It serves the following purposes: It generates suspense and enhances the sense of urgency around the story's central conflict or problem. Then I would do a Climax- The object of a climax is not to provide the most possible conflict or action. It's also not just about producing the character's greatest reversal of fortunes. Too often we mistake the fortunes of our character with the story's arc, and although fortune is involved, it's not our story arc's key criterion. After that, I would just add finish touches and finish it off before it needs to be shipped off to make copies.

8 0

3 years ago