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).
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).
1 answer:
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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Answer:
Option D
Explanation:
Given information
Bulk unit weight of 107.0 lb/cf
Water content of 7.3%,=0.073
Specific gravity of the soil solids is 2.62
Specifications
Dry unit weight is 113 lb/cf
Water content is 6%.
Volume of embankment is 440,000-cy
Borrow material
Embankment
Considering that the volume of embankment is inversely proportional to the dry unit weight
Therefore,
Therefore, volume of borrow material is 498594-cy
(b)
The weight of water in embankment is found by multiplying the moisture content and dry unit weight.
Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf
Since
The embankment requires water of 6.78*27*440000= 80546400 lb
Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf
Borrow material requires water of 7.27959*27*498594=97998120 lb
Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb
1 gallon is approximately hence
That's approximately 4.2 gallons
Answer:
The head loss in Psi is 0.390625 psi.
Explanation:
Fluid looses energy in the form of head loss. Fluid looses energy in the form of head loss when passes through the valve as well.
Given:
Factor cv is 48.
Flow rate of water is 30 GPM.
GPM means gallon per minute.
Calculation:
Step1
Expression for head loss for the water is given as follows:
Here, cv is valve coefficient, Q is flow rate in GPM and h is head loss is psi.
Step2
Substitute 48 for cv and 30 for Q in above equation as follows:
h = 0.390625 psi.
Thus, the head loss in Psi is 0.390625 psi.
Answer:
0.5m^2/Vs and 0.14m^2/Vs
Explanation:
To calculate the mobility of electron and mobility of hole for gallium antimonide we have,
(S)
Where
e= charge of electron
n= number of electrons
p= number of holes
mobility of electron
mobility of holes
electrical conductivity
Making the substitution in (S)
Mobility of electron
Mobility of hole in (S)
Then, solving the equation:
(1)
(2)
We have,
Mobility of electron
Mobility of hole is
