The outer pipe is at 280K. Emissivity of inner and outer surfaces is 0.2 and 0.3 respectively. (i)Determine the radiant heat flow rate over a pipe length of 5m. (ii)If a radiation shield of diameter 12mm and emissivity 0.05 on both sides is placed between the pipes, determine the percentage reduction on heat flow. (iii)What is the equilibrium temperature of the shield?

Case 1

Given,

d1=10mm=0.01m (Diameter of inner pipe)

t1=80K=-193℃ (Temperature of inner pipe)

d2=15mm=0.015m (Diameter of outer pipe)

t2=280K=7℃ (Temperature of outer pipe)

ε1=0.2 (Emissivity of inner pipe surface)

ε2=0.3 (Emissivity of outer pipe surface)

l=5m (Length of both pipes)

Find: Q1-2 (Heat flow rate)

Solution:

Heat Transfer from pipe 1 to pipe 2

Q1-2=Eb1-Eb2Rt

=σ(t14-t24)1-ε1A1ε1+1A1F1-2+1-ε2A2ε2

=A1σ(t14-t24)1-ε1ε1+1F1-2+A1A2(1-ε2ε2) …(1)

As space between 2 pipes is evacuated, so resistance due to space is zero.

∴From (1)

Q1-2=π×0.01×5×5.67×10-8(804-2804)1-0.20.2+0+π×0.01×5π×0.015×5(1-0.30.3)

Q1-2=-9.7882W

Case 2

Given,

d3=12mm=0.012m (Diameter of shield pipe)

ε3=0.2 (Emissivity of shield pipe)

Find: (1) % Reduction in Q

(2) t3 (Temperature of shield pipe)

Solution:

Heat transfer from pipe 1 to pipe 3

Q1-3=σ(t14-t34)(1-ε1A1ε1+1A1F1-3+1-ε3A3ε3)

=A1σ(t14-t34)1-ε1ε1+1F1-3+A1A3(1-ε3ε3) …(2)

As space between the pipes is evacuated, so resistance due to space is zero

∴From (2)

Q1-3=π×0.01×5×5.67×10-8(804-t34)1-0.20.2+0+π×0.01×5π×0.012×5(1-0.050.05)

Q1-3=4.4906×10-10(804-t34) …(3)

Heat transfer from pipe 3 to pipe 2

Q3-2=σ(t34-t24)(1-ε3A3ε3+1A3F3-2+1-ε2A2ε2)

=A3σ(t34-t24)1-ε3ε3+1F3-2+A3A2(1-ε2ε2) …(4)

As space between the pipes is evacuated, so resistance due to space is zero

∴From (4)

Q3-2=π×0.012×5×5.67×10-8(t34-2804)1-0.050.05+0+π×0.012×5π×0.015×5(1-0.20.2)

Q3-2=4.8142×10-10(t34-2804) …(5)

As rate of heat transfer remains same, Equating (3) and (5)

4.4906×10-10(804-t34)=4.8142×10-10(t34-2804)

t3=237.8685K=-35.1614℃

Heat transfer when shield is placed between the two pipes

Q1-2=Q1-3=4.4906×10-10(804-237.86854)

Q1-2=-1.4192W

% Reduction in heat transfer

%↓Q=-9.7882-(-1.4192)-9.7882 ×100

%↓Q=85.5 %