A spherical tank 1m in diameter is maintained at a temperature of $120^0C$ and exposed to a convection environment with $h= 25W/m^2-K$ and temperature of ambient air is $15^0C$. What thickness of urethane foam $(k = 20×10^{-3} W/m-K)$ should be added to ensure that the outer temperature of the insulation does not exceed $40^0C$? What percentage reduction in heat loss results from installing this insulation?

Case 1

Given,

d = 1m (Diameter of spherical tank)

t1 = 120℃ = 393K (Temperature of sphere)

t2 = 15℃ = 288K (Temperature of surrounding environment)

h = 25 W/m2K (Convective heat transfer coefficient of surrounding environment)

Find: (1) Q (Heat transfer rate)

Solution:

Heat transfer rate

Q1 = dtR = t1 - t21h. As = 393 - 288125 × 4π × 0.52

Q1 = 8246.68 W

Case 2

Given,

d = 1m (Diameter of spherical tank)

t1 = 120℃ = 393K (Temperature of sphere)

t3 = 15℃ = 288K (Temperature of surrounding environment)

h = 25 W/m2K (Convective heat transfer coefficient of surrounding environment)

K = 20 × 10 - 3W/mK (Thermal conductivity of urethane)

t2 = 40℃ = 313K (Outer temperature of urethane)

Find:

(1) T (Thickness of urethane)

(2) % Reduction in Q (After applying the insulation)

Solution:

Heat transfer rate

Q2 = dtR1 = t1 - t2r2 - r14π.K.r2r1 = 393 - 313r2 - 0.54π × 20 × 10 - 3 × r2 × 0.5

Q2 = 80(r2 - 0.5)0.125r2 …(1)

also,

Q2 = dtR2 = t2 - t31h. As = t2 - t31h × 4π × r22 = 313 - 288125 × 4π × r22

Q2 = 251100π × r22 …(2)

Equating (1) and (2)

80(r2 - 0.5)0.125r2 = 251100π × r22

80100π × r22 = 25(r2 - 0.5)0.125r2

10 = 7853.981r22 - 3926.99r2

r2 = 0.502m

Thickness of urethane

T = r2 - r1 = 0.502 - 0.5

T = 0.002m = 2mm

∴Q2 = 251100π × 0.5022

Q2 = 1979.234 W

% Reduction in Q

%↓Q = Q1 - Q2Q1 × 100 = 8246.68 - 1979.2348246.68 × 100

%↓Q = 75.99 %