Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Answer:

Explanation:
From the question we are told that:
Thickness 
Internal Pressure
Shear stress 
Elastic modulus 
Generally the equation for shear stress is mathematically given by

Where
r_i=internal Radius
Therefore


Generally



Generally the equation for outer diameter is mathematically given by


Therefore
Assuming that the thin cylinder is subjected to integral Pressure
Outer Diameter is

Answer:
90 degrees
Explanation:
In the case when the sheer stress acts in the one plane of an element so it should be equal and opposite also the shear stress acted on a plan i.e. 90 degrees from the plane
Therefore as per the given situation it should be 90 degrees from the plane
hence, the same is to be considered and relevant too
Answer:
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