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:
mobile phones and internet access
Explanation:
I got it right on my quiz
Answer:
The following statements are true:
A. For flows over a flat plate, in the laminar region, the heat transfer coefficient is decreasing in the flow direction
C. For flows over a flat plate, the transition from laminar to turbulence flow only happens for rough surface
E. In general, turbulence flows have a larger heat transfer coefficient compared to laminar flows 6.
Select ALL statements that are TRUE
B. In the hydrodynamic fully developed region, the mean velocity of the flow becomes constant
D. For internal flows, if Pr>1, the flows become hydrodynamically fully developed before becoming thermally fully developed
Explanation:
Answer:
Relative density = 0.545
Degree of saturation = 24.77%
Explanation:
Data provided in the question:
Water content, w = 5%
Bulk unit weight = 18.0 kN/m³
Void ratio in the densest state, 
= 0.51
Void ratio in the loosest state, 
= 0.87
Now,
Dry density, 
= 17.14 kN/m³
Also,
here, G = Specific gravity = 2.7 for sand
or
e = 0.545
Relative density = 
= 
= 0.902
Also,
Se = wG
here,
S is the degree of saturation
therefore,
S(0.545) = (0.05)()2.7
or
S = 0.2477
or
S = 0.2477 × 100% = 24.77%
