A steel sphere is of inner diameter 40
cm and
outer diameter 45 cm is used to store liquid oxygen (B.P is minus
183^{o}C). The sphere is covered with one layer of insulation, of thickness 50 mm whose K is 0.35 W/m.K and another insulation, of thickness 50 mm whose K is 0.098 W/m.K. The sphere is exposed to atmosphere of 25^{o}C. Find out the rate of oxygen becoming vapor every minute. Latent heat of oxygen is 370 kJ/kg. Thermal conductivity of steel = 20 W/m.K. Heat transfer coefficient of ambient air = 80 W/m^{2}.K

Calculations:

For composite concentric sphere, the rate of heat transfer from outside to the sphere Q, is given by,

Q = (T_{a} - T_{b})/R

T_{a} = 25^{o}C

T_{b} = -183^{o}C

Where, R = R_{s} + R_{ins,1} + R_{ins,2} + R_{a}

R_{a} = air film resistance = 1 / (4pr_{a}^{2}h_{a})

R_{ins,2} = Resistance offered by the outer insulation = (r_{a} - r_{1}) / (4pk_{ins,2}r_{a}r_{1})

R_{ins,1} = Resistance offered by the inner insulation = (r_{1} - r_{2}) / (4pk_{ins,1}r_{1}r_{2})

R_{s} = Resistance offered by the metal wall (steel) =
(r_{2} -
r_{b}) / (4pk_{s}r_{2}r_{b})

Given Data:

Diameter of sphere with insulation = 45 + 2 x 5 + 2 x 5 = 65 cm

r_{a} = 32.5 cm = 0.325 m

r_{1} = 32.5 - 5 = 27.5 cm = 0.275 m

r_{2} = 27.5 - 5 = 22.5 cm = 0.225 m

r_{b} = 20 cm = 0.2 m

Substituting these data in the above equations,

R_{a} = 1/(4 x 3.142 x 0.325^{2} x 80) = 9.4175 x 10^{-3} ^{o}C/W

R_{ins,2} = (0.325 - 0.275)/(4 x 3.142 x 0.098 x 0.325 x 0.275) = 0.4543 ^{o}C/W

R_{ins,1} = (0.275 - 0.225)/(4 x 3.142 x 0.35 x 0.275 x 0.225) = 0.1837 ^{o}C/W

R_{s} = (0.225 - 0.2)/(4 x 3.142 x 20 x 0.225 x 0.2) = 2.2105 x 10^{-3} ^{o}C/W

R = 9.4175 x 10^{-3} + 0.4543 + 0.1837 + 2.2105 x 10^{-3} = 0.6496 ^{o}C/W

Q = (25 + 183)/0.6496 = 320.2 W

Rate of heat in per second = 320.2 J

Rate of heat in per minute = 320.2 x 60 = 19212 J = 19.212 kJ

370 kJ of heat is needed to vaporize 1 kg of liquid oxygen. Therefore, for the heat input of 19.212 kJ/min,

rate of oxygen becoming vapor = 19.212/370 = 0.0519 kg/min