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
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Explanation: See above.
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Answer:
Training is a way for employers to provide tools to enable employees to protect themselves and others from injuries.
Explanation:
Answer:
Explanation:
Given that:
diameter = 100 mm
initial temperature = 500 ° C
Conventional coefficient = 500 W/m^2 K
length = 1 m
We obtain the following data from the tables A-1;
For the stainless steel of the rod 
Here, we can't apply the lumped capacitance method, since Bi > 0.1
However, on a single rod, the energy extracted is:
Hence, for centerline temperature at 50 °C;
The surface temperature is:
Answer:
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