CHEMICAL ENGINEERING

HEAT TRANSFER 187

PROBLEM 9.55

Derive an expression for the radiant heat transfer rate per unit area between two large
parallel planes of emissivitiese 1 ande 2 and at absolute temperaturesT 1 andT 2 respec-
tively.
Two such planes are situated 2.5 mm apart in air. One has an emissivity of 0.1 and is
at a temperature of 350 K, and the other has an emissivity of 0.05 and is at a temperature
of 300 K. Calculate the percentage change in the total heat transfer rate by coating the
first surface so as to reduce its emissivity to 0.025. Stefan – Boltzmann constantD 5. 67 ð
10 ^8 W/m^2 K^4. Thermal conductivity of airD 0 .026 W/m K.

Solution

The theoretical derivation is laid out in Section 9.5.5 and the heat transfer by radiation is
given by puttingA 1 DA 2 in equation 9.150 to give:

qrD[ e 1 e 2 9/

e 1 Ce 2 e 1 e 2 ]

T^41 T^42

Forconductionbetween the two planes:

qcDkA

T 1 T 2 /x (equation 9.12)

D 0. 026 ð 1. 0

350  300 / 0. 0025 D520 W/m^2

For radiation between the two planes:

qrD[ e 1 e 2 9/

e 1 Ce 2 e 1 e 2 ]

T^41 T^42

D[ 

0. 1 ð 0. 05 ð 5. 67 ð 10 ^8 /

 0. 1 C 0. 05  0. 1 ð 0. 05 ] 

3504  3004

D 13 .5W/m^2

Thus neglecting any convection in the very narrow space, the total heat transferred is
533 .5W/m^2 .Whene 1 D 0 .025, the heat transfer by radiation is:

qrD[ 

0. 025 ð 0. 05 ð 5. 67 ð 10 ^8 /

 0. 025 C 0. 05  0. 025 ð 0. 05 ]

ð 

3504  3004 D 6 .64 W/m^2

and:
qrCqc D 526 .64 W/m^2

Thus, although the heat transferred by radiation is reduced to
100 ð 6. 64 / 13. 5 D
49 .2% of its initial value, the total heat transferred is reduced to
100 ð 526. 64 / 533. 5 D
98 .7%of the initial value.

PROBLEM 9.56

Water flows at 2 m/s through a 2.5 m length of a 25 mm diameter tube. If the tube is at
320 K and the water enters and leaves at 293 and 295 K respectively, what is the value
of the heat transfer coefficient? How would the outlet temperature change if the velocity
was increased by 50%?