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Serhud [2]
3 years ago
13

Consider a dip-coating process where a very long (assume infinitely long) wire(solid) with radius, ri, is being pulled verticall

y upward with axial velocity, V0, in a co-centric configuration through a very long (assume infinitely long) cylinder with radius, ro . Assume this annular flow is steady state, incompressible, fully-developedand laminar. The liquid density, ????, viscosity, µ, and gravity, g, are also known constants.
a. Simplify the Navier-Stokes equation for this flow (include the body force).
b. Write the boundary conditions for this flow.
c. Determine the velocity profile, uz(r) for the dip-coating process considered here by solving the simplified Navier-Stokes equation in section (a)
d. Determine the shear stress on the surface of the wire (Note: there is no pressure-gradient in this flow, only body force).

Engineering
1 answer:
Gekata [30.6K]3 years ago
6 0

Answer:

See explaination and attachment.

Explanation:

Navier-Stokes equation is to momentum what the continuity equation is to conservation of mass. It simply enforces F=ma in an Eulerian frame.

The starting point of the Navier-Stokes equations is the equilibrium equation.

The first key step is to partition the stress in the equations into hydrostatic (pressure) and deviatoric constituents.

The second step is to relate the deviatoric stress to viscosity in the fluid.

The final step is to impose any special cases of interest, usually incompressibility.

Please kindly check attachment for step by step solution.

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\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

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To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

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The Weibull parameters include:

x_o = 0.02

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Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

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C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

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C_{10} = \dfrac{30962.449}{225}

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