Question

A spherical ball has 3cm internal diameter and its inner surface temperature is 250˚C and the outer surface temperature is 30˚C. Calculate the outer diameter of the ball if the heat loss from the ball is 1600W and has 2.75W/m.˚C thermal conductivity.

Answer 1

Answer:

The outer diameter of the ball is 6.2138 cm

Explanation:

The formula to apply is ;

Heat loss ,

$Q/t=kA*\frac{( T_1-T_2)}{d}$

where ;

Q/t=total heat loss from the ball = 1600 w

k=coefficient of heat transmission through the ball= 2.75 W/m.˚C

A=area in m² of the ball with the coefficient of heat transmission

T₁=Hot temperature

T₂=Cold temperatures

d=thickness of the ball

Area of spherical ball using internal diameter, 3cm= 0.03 m will be

Radius = half the diameter = 0.03/2 = 0.015

Area = 4 *π*r²

Area = 4*π*0.015² = 0.002827 m²

Apply the formula for heat loss to get the thickness as:

1600 = {2.75 * 0.002827 *(250-30 ) }/d

1600 =1.711/d

1600d = 1.711

d=1.711/1600 = 0.001069 m

d= 0.1069

Using internal radius and the thickness to get outer radius as;

3 + 0.1069 = 3.1069 cm

Outer diameter will be twice the outer radius

2*3.1069 = 6.2138 cm