Respuesta :
Answer:
The convection coefficient is [tex]15456.48\ W/m^{2}K[/tex]
Solution:
Mass flow rate, [tex]\dot{m} = 0.4\ kg[/tex]
Inner diameter of the tube, d = 0.014 m
Fluid density, [tex]\rho_{f} = 990\ kg/m^{3}[/tex]
Specific Heat, C = 3845 J/K
Thermal Conductivity, K = 0.74
Prandtl Number, [tex]P_{r} = 8.6[/tex]
Heat flux, [tex]\dot{q} = 71,297\ W/m^{2}[/tex]
Viscosity, [tex]\mu = 0.00079\ Ns/m^{2}[/tex]
Now,
To calculate the convection heat coefficient, h:
Determine the cross sectional area of the circular tube:
[tex]A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}[/tex]
Determine the velocity of the fluid inside the tube by mass flow rate:
[tex]\dot{m} = \rho_{f}A_{c}v[/tex]
[tex]0.4 = 990\times 1.54\time 10^{- 4}v[/tex]
v = 2.624 m/s
Determine the Reynold's Number, [tex]R_{e}[/tex]:
[tex]R_{e} = \frac{\rho_{f}dv}{\mu}[/tex]
[tex]R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253[/tex]
Thus it is clear that [tex]R_{e}[/tex] > 10,000 hence flow is turbulent.
Now,
Determine the Nusselt Number:
[tex]N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}[/tex]
[tex]N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42[/tex]
Also,
[tex]N_{u} = \frac{dh}{K}[/tex]
where
h = convection coefficient
Now,
[tex]292.42 = \frac{0.014\times h}{0.74}[/tex]
[tex]h = 15456.48\ W/m^{2}K[/tex]
