Find the attachment for complete solution
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:
γ =0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ = displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ ------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V = V
= 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v = 0.001057 m³/kg
v = 1.0037 m³/kg
u = 486.82 kJ/kg
u 2524.5 kJ/kg
h = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V/v
+ V
/v
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m = m₁ - m₂
m = 1.89414 - 0.003985
m = 1.890155 kg
so, Initial internal energy will be;
U₁ = mu
+ m
u
U₁ = (V/v
)u
+ (V
/v
)u
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E - E
= ΔE
QΔt - mh
= m₂u₂ - U₁
QΔt - mh
= m₂u
- U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW