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3 years ago

What is differentiation

Engineering

1 answer:

oksian1 [2.3K]3 years ago

7 0

Answer:

the action or process of differentiating or distinguishing between two or more things or people.

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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

A closed, 5-m-tall tank is filled with water to a depth of 4 m. The top portion of the tank is filled with air which, as indicat

Brums [2.3K]

Answer:

The pressure that the water exerts on the bottom of the tank is 59.2 kPa

Explanation:

Given;

height of tank, h = 5m

height of water in the tank, $h_w$ = 4m

pressure at the top of the tank, $P_{top}$ = 20 kPa

The pressure exerted by water at the bottom of the tank is the sum of pressure on water surface and pressure due to water column.

$P_{bottom} = \gamma h + P_{top}\\\\P_{bottom} = (9.8*10^3*4 \ \ + \ 20*10^3)Pa\\\\P_{bottom} = 59200 \ Pa\\\\P_{bottom} = 59.2 \ kPa$

Therefore, the pressure that the water exerts on the bottom of the tank is 59.2 kPa

6 0

3 years ago

What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-

Hitman42 [59]

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

$\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}$

Where $\sigma_{m}$ is the magnitude of he maximum stress at the tip of the crack, $\sigma_{p}$ is the magnitude of the tensile stress, $l_{c}$ is $1/2$ the length of the internal crack, and $r_{c}$ is the radius of curvature of the crack.