Consider the velocity boundary layer profile for flow over u flat plate to be of the form u = C_1 + C_2 y. Applying appropriate
boundary conditions, obtain an expression for the velocity profile in terms of the boundary layer thickness delta and the free stream velocity Using the integral form of the boundary layer momentum equation (Appendix G). obtain expressions for the boundary layer thickness and the local friction coefficient, expressing your result in terms of the local Reynolds number. Compare your results with those obtained from the exact solution (Section 7.2.11 and the integral solution with a cubic profile (Appendix G).
1 answer:
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Answer:
Explanation:
Given that:
The Inside pressure (p) = 1402 kPa
= 1.402 × 10³ Pa
Force (F) = 13 kN
= 13 × 10³ N
Thickness (t) = 18 mm
= 18 × 10⁻³ m
Radius (r) = 306 mm
= 306 × 10⁻³ m
Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)
Then;
the state of the plane stress can be expressed as follows:
Since d = 2r
Then:
When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.
Thus;
Answer:
4.6 mm
Explanation:
Given data includes:
thin-walled pipe diameter = 100-mm =0.1 m
Temperature of pipe = -15° C = (-15 +273)K =258 K
Temperature of water = 3° C = (3 + 273)K = 276 K
Temperature of ice = 0° C = (0 +273)K =273 K
Thermal conductivity (k) from the ice table = 1.94 W/m.K ; R = 0.05
convection coefficient =2000 W/m².K
The energy balance can be expressed as:
where;
------------- equation (1)
------------ equation(2)
Equating both equation (1) and (2); we have;
Replacing the given data; we have:
r = 0.0454
The thickness (t) of the ice layer can now be calculated as:
t = (R - r)
t = (0.05 - 0.0454)
t = 0.0046 m
t = 4.6 mm