Answer:
h = 375 KW/m^2K
Explanation:
Given:
Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm
steel thermal conductivity k = 15 W / mK
Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C
Air Temp T_∞ = 100 C
Assuming there are no other energy sources, energy balance equation is:
E_in = E_out
q"_cond = q"_conv
Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2
q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)
=15KW/m^2
Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:
q"_conv = h * ( T_∞ - T_s ) ----> h = q"_conv / ( T_∞ - T_s )
h = 15000 W / (100 - 60 ) C = 375 KW/m^2K
It would be 72cm bc u need to add up all the line in the back to
Answer:
not sure if this helps but
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
