Given:
Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.
To Find:
a. The distance from the leading edge at which the transition will occur.
b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer
c. Which fluid has a higher heat transfer
Calculation:
The transition from the lamina to turbulent begins when the critical Reynolds
number reaches 



In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.
With the given data we can proceed to calculate the compression stress:



Through Goodman's equations the combined effort by fatigue and compression is expressed as:

Where,
Fatigue limit for comined alternating and mean stress
Fatigue Limit
Mean stress (due to static load)
Ultimate tensile stress
Security Factor
We can replace the values and assume a security factor of 1, then

Re-arrenge for 

We know that the stress is representing as,

Then,
Where
=Max Moment
I= Intertia
The inertia for this object is

Then replacing and re-arrenge for 



Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm
Answer:
a. 51.84Kj
b. 2808.99 W/m^2
c. 11.75%
Explanation:
Amount of heat this resistor dissipates during a 24-hour period
= amount of power dissipated * time
= 0.6 * 24 = 14.4 Watt hour
(Note 3.6Watt hour = 1Kj )
=14.4*3.6 = 51.84Kj
Heat flux = amount of power dissipated/ surface area
surface area = area of the two circular end + area of the curve surface

= 2.136 *10^-4 
Heat flux =
= 2808.99 
fraction of heat dissipated from the top and bottom surface

=11.75%
Answer:
e= 50 J/kg
Explanation:
Given that
Speed ,v= 10 m/s
Diameter of the turbine = 90 m
Density of the air ,ρ = 1.25 kg/m³
We know that mechanical energy given as

That is why mechanical energy per unit mass will be

Now by putting the values in the above equation we get

e= 50 J/kg
That why the mechanical energy unit mass will be 50 J/kg.
Answer:
component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²
magnitude of acceleration is 22.98 m/s²
Explanation:
given data
velocity = 10 m/s
initial time to = 0
distance s = 400 m
time t = 14 s
to find out
components and magnitude of acceleration after the car has travelled 200 m
solution
first we find the radius of circular track that is
we know distance S = 2πR
400 = 2πR
R = 63.66 m
and tangential acceleration is
S = ut + 0.5 ×at²
here u is initial speed and t is time and S is distance
400 = 10 × 14 + 0.5 ×a (14)²
a = 3.37 m/s²
and here tangential acceleration is constant
so velocity at distance 200 m
v² - u² = 2 a S
v² = 10² + 2 ( 3.37) 200
v = 38.05 m/s
so radial acceleration at distance 200 m
ar = 
ar = 
ar = 22.74 m/s²
so magnitude of total acceleration is
A = 
A = 
A = 22.98 m/s²
so magnitude of acceleration is 22.98 m/s²