Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
hat of the surrounding air, i.e. 20°C. Its width is 5.0 cm; thickness is 1.0 mm; thermal conductivity is 200 W/m·K; and base temperature is 40°C. The heat transfer coefficient is 20 W/m2 ·K. Estimate the fin temperature at a distance of 5.0 cm from the base and the rate of heat loss from the entire fin.
1 answer:
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Answer:
(b)False
Explanation:
Given:
Prandtl number(Pr) =1000.
We know that
Where is the molecular diffusivity of momentum
is the molecular diffusivity of heat.
Prandtl number(Pr) can also be defined as
Where is the hydrodynamic boundary layer thickness and
is the thermal boundary layer thickness.
So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
In given question Pr>1 so hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.
So hydrodynamic layer will be thicker than the thermal boundary layer.
Answer:
T = 167 ° C
Explanation:
To solve the question we have the following known variables
Type of surface = plane wall ,
Thermal conductivity k = 25.0 W/m·K,
Thickness L = 0.1 m,
Heat generation rate q' = 0.300 MW/m³,
Heat transfer coefficient hc = 400 W/m² ·K,
Ambient temperature T∞ = 32.0 °C
We are to determine the maximum temperature in the wall
Assumptions for the calculation are as follows
- Negligible heat loss through the insulation
- Steady state system
- One dimensional conduction across the wall
Therefore by the one dimensional conduction equation we have
During steady state
= 0 which gives
From which we have
Considering the boundary condition at x =0 where there is no heat loss
= 0 also at the other end of the plane wall we have
hc (T - T∞) at point x = L
Integrating the equation we have
from which C₁ is evaluated from the first boundary condition thus
0 = from which C₁ = 0
From the second integration we have
From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows
→ C₂ =
T(x) = and T(x) = T∞ +
∴ Tmax → when x = 0 = T∞ +
Substituting the values we get
T = 167 ° C