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Serhud [2]
4 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]4 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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Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle gra
Vlada [557]

Answer:

Explanation:

Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.

Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.

5 0
4 years ago
Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−
sergeinik [125]

Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

v_{i} = 3.986 m/s

Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

v_{e} = 19.33 m/s

Now, the power output can be calculated from the energy balance eqn:

P_{out} = -\dot{m}W_{s}

P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}

P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW

4 0
3 years ago
We would like to measure the density (p) of an ideal gas. We know the ideal gas law provides p= , where P represents pressure, R
Nostrana [21]

Answer: =

Explanation:

=    P / (R * T) P- Pressure, R=287.058, T- temperature

From the given that

Sample mean(pressure) = 120300 Pa

Standard deviation (pressure) = 6600 Pa

Sample mean(temperature) = 340K

Standard deviation(temperature) = 8K

To calculate the Density;

Maximum pressure = Sample mean(pressure) + standard deviation (pressure) = 120300+6600 = 126900 Pa

Minimum pressure = Sample mean (pressure) - standard deviation (pressure)= 120300-6600 = 113700 Pa

Maximum temperature = Sample mean (temperature) + standard deviation (temperature) = 340+8 = 348K

Minimum temperature = Sample mean (temperature) - standerd deviation (temperature) = 340-8 = 332K

So now to calculate the density:

Maximum Density= Pressure (max)/(R*Temperature (min))= 126900/(287.058*332)= 1.331

Minimum density=Pressure(min)/(R*Temperature (max))= 113700/(287.058*348)= 1.138

Average density= (density (max)+ density (min))/2= (1.331+1.138)/2= 1.2345

cheers i hope this helps

5 0
3 years ago
Two engineers are to solve an actual heat transfer problem in a manufacturing facility. Engineer A makes the necessary simplifyi
deff fn [24]

Answer:

Engineer A results will be more accurate

Explanation:

Analytical method is better than numerical method. Engineer A has used analytical method and therefore his results will be more accurate because he used simplified method. Engineer B has used software to solve the problem related to heat transfer his results will be approximate.

5 0
3 years ago
Bridging are members installed periodically between joists to ensure which of the following?
Pie

Answer:

Distributes a floor load or weight

Explanation:

5 0
3 years ago
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