Stainless steel sphere [k = 16 W/m . OC ] having a diameter of 4 cm is...
Problem Statement: Calculate the steady state temperature for the center of a stainless steel sphere having a diameter of 4 cm, exposed to a convection environment at 20OC with a heat generation rate of 1.0 MW/m3.
Solution:
Given:
Diameter of the sphere, D = 4 cm
Radius of the sphere, r = 2 cm
Thermal conductivity of the sphere, k = 16 W/m . OC
Convection environment temperature, T∞ = 20OC
Heat transfer coefficient, h = 15 W/m2. OC
Rate of heat generation, q'' = 1.0 MW/m3
Assumptions:
The thermal conductivity of the sphere is independent of temperature.
Heat generation is uniform throughout the sphere.
Steady-state conditions prevail.
Negligible radiation heat transfer.
Formulae:
Steady-state heat transfer rate, q = 4πkr2(Tc - T∞)
Rate of heat generation, q'' = Q/V
where Q is the heat generated and V is the volume of the sphere.
Volume of the sphere, V = 4/3 πr3
Critical radius, rC = k/h
Calculation:
Volume of the sphere, V = 4/3 πr3 = 4/3 π(2 cm)3 = 33.51 cm3 = 33.51 x 10^-6 m3
Rate of heat generation, q'' = Q/V = 1.0 MW/m3
Steady-state heat transfer rate, q = q''V = 1.0 x 10^6 W/m3 x 33.51 x 10^-6 m3 = 33.51 W
Critical radius, rC = k/h = 16 W/m . OC / 15 W/m2 . OC = 1.07 m
Since the radius of the sphere, r (2 cm) is less than the critical radius, rC (1.07 m), the heat transfer is dominated by conduction.
Therefore, the steady-state temperature distribution in the sphere is given by:
q/4πk(Tc - T∞)/r = -d/dr(r2 dT/dr)
Integrating and applying boundary conditions, we get:
Tc = T∞ + q''r2/6k = 20OC + 1.0 x 10^6 W/m3 x (2 cm)2 / (6 x 16 W/m . OC) = 20.44OC
Answer: Therefore, the steady-state temperature for the center of the sphere is 20.44OC.
Stainless steel sphere [k = 16 W/m . OC ] having a diameter of 4 cm is...
Is answer 468.6 degreeC?