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).
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Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 × = 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × / 3
so d = 0.0133 m
so diameter is 14 mm
<h2>Answer:</h2>
7532V
<h2>Explanation:</h2>For a given transformer, the ratio of the number of turns in its primary coil () to the number of turns in its secondary coil (
) is equal to the ratio of the input voltage (
) to the output voltage (
) of the transformer. i.e
=
----------------(i)
<em>From the question;</em>
= number of turns in the primary coil = 8 turns
= number of turns in the secondary coil = 515 turns
= input voltage = 117V
<em>Substitute these values into equation (i) as follows;</em>
=
<em>Solve for </em><em>;</em>
= 117 x 515 / 8
= 7532V
Therefore, the output voltage (in V) of the transformer is 7532