CHEMICAL ENGINEERING

138 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Va l u e s o fTare now assumed and together with values ofhbfrom the given data substituted
into (i) and (ii) until equal values ofqare obtained as follows:

TqD 

693. 3  1. 67 T hb 0. 1074 hb

T 290 0. 317 

T 2901.^25 q

(K) (W/m) W/m^2 K (W/m) (W/m) (W/m)

300 193.3 6.0 6.5 5.7 12.2
320 160.0 6.5 20.9 22.2 43.1
340 126.7 7.1 38.1 42.1 80.2
360 93.3 7.8 58.7 64.2 122.9
380 60.0 8.55 82.7 87.9 170.6
400 26.7 9.55 112.8 113.0 225.8

A balance is obtained whenTD350 KandqD106 W/m.

Case (b)

For conduction through the insulation,xD 0 .006 m and the mean diameter is 15 mm or
0.015 m.

∴ AmD
$ð 0. 015 ð 1. 0 D 0 .0471 m^2 /m

∴ qD
0. 35 ð 0. 0471
420 T/ 0. 006 D
1154  2. 75 TW/m (i)

The outer diameter is now 0.024 m andA 2 D

$ð 0. 024 ð 1. 0 D 0 .0754 m^2 /m

The coefficient due to convection is:

hcD 1 .22[

T 290 / 0 .024]^0.^25 D 3. 10 

T 2900.^25 W/m^2 K

The emissivity is 0.10 and hencehrD 0. 10 hbW/m^2 K

∴ qD[0. 10 hbC 3. 10
T 2900.^25 ]0. 0754
T 290

D 0. 00754 hb

T 290 C 0. 234 

T 2901.^25 W/m (ii)

Making the calculation as before:

TqD 

1154  2. 75 T hb 2 0. 0075 hb

T 290 0. 234 

T 2901.^25 q

(K) (W/m) W/m^2 K (W/m) (W/m) (W/m)

300 329.0 6.0 0.5 4.2 4.7
320 274.0 6.5 1.5 16.4 17.9
340 219.0 7.1 2.7 31.1 33.8
360 164.0 7.8 4.2 47.4 51.6
380 109.0 8.55 5.8 64.9 70.7
400 54.0 9.55 7.9 83.4 91.3

A balance is obtained whenTD390 KandqD81 W/m.