Answer :
The estimated temperature drop across the wall of the spherical tank is approximately 4.13 °C.
The temperature drop across the wall of the spherical tank can be estimated using the formula for heat conduction through a cylindrical wall. The formula is given by:
ΔT = (Q * r) / (4πkL)
where:
ΔT is the temperature drop in °C,
Q is the heat generation rate per unit volume (168746.9 W m^−3),
r is the radius of the tank wall (average of inner and outer radii) in meters,
k is the thermal conductivity of the steel (18 W m^−1K^−1),
L is the thickness of the tank wall in meters.
To calculate the radius of the tank wall (r):
r = (99.5 cm + 100.2 cm) / 2
= 99.85 cm = 0.9985 m
Assuming the thickness of the tank wall (L) is negligible compared to the radius, we can use this simplified formula:
ΔT = (Q * r) / (4πk)
Substituting the given values into the formula, we have:
ΔT = (168746.9 * 0.9985) / (4π * 18)
Calculating the result:
ΔT = 466.84 / (4π * 18)
≈ 4.13 °C
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