Insulation Calculations

Insulation and refractory performance may be evaluated using the following procedure. The heat loss from a surface is given by

05 q . = . 5 67 10 × ×∈× ( 273 + t

c )( − 273 + t a )  + . 0 00195 × ( t c − t a ) ××  ( V + 21 ) /  21   (6.5)

∈ , the emissivity of casing, may be taken as 0.9 for oxidized steel, 0.05 for polished aluminum, and 0.15 for oxidized aluminum.

q = Ktt/ ( − c )   { ( d + 2 L / ln d ) 2 } { ( + 2 L /d ) }  = K t t /L ( – c ) e (6.12)

where t a ,t c , t are the ambient, casing, and hot surface temperatures, °C

V is the wind velocity, m/min K is the thermal conductivity of insulation, W/m K

d is the pipe outer diameter, m L, L e are thickness of insulation and its equivalent thickness, m (see Table 6.2)

Example 6.5

Determine the thickness of insulation to limit the casing temperature to 93°C when ambient is 27°C, wind velocity V = 80 m/min, casing emissivity = 0.15. K = thermal conductivity of insulation = 0.05 W/m K. The pipe diameter is 0.3 m and is at 426°C.

4 4 1 .. 25 05 q= . . 5 67 10 × × . 0 15 ×

= . 0 8837 kW/m 2 = 883 7 . W/m 2

883.7 = 0.05(426 − 93)/L e or L e = 0.019 m (0.74 in.). One may use look-up tables to find out the thickness-based d and L. For large diameters, L e will be nearly same as L. Here, L = 0.01905 m.

Example 6.6

What is the casing temperature when 0.025 m thick insulation is used in the earlier case?

L e = {(0.3 + 2 × 0.025)}/2 ln[(0.3 + 2 × 0.025)/0.3] = 0.0269 m Let t c = 65°C. Use a K = 0.0486 W/m K for the revised average insulation temperature.

LHS = . 5 67 10 × − 11 ×

2 kW/m W/m RHS = 0.0486 × (426 − 65)/0.0269 = 652 W/m 2 . Since there is a mismatch, try 77°C.

LHS = 0.6273 kW/m 2 and RHS = 0.630 kW/m 2 . Hence, casing temperature = 77°C. A computer program is ideal for these calculations particularly if multilayer insulation is involved.

Miscellaneous Boiler Calculations 321

TABLE 6.2

Equivalent Thickness of Insulation

Thickness of Insulation, in. (mm)

Tube OD, in. (mm)

A horizontal flat surface is at −12.2°C. Ambient temperature is 27°C and relative humidity is 80%. Determine the thickness of fibrous insulation that will prevent condensation of water vapor on the surface. Use K = 0.04 W/m K. Wind velocity may be neglected. Use emissivity of 0.9 for casing.

The surface must be above the water dew point to prevent the condensation of water vapor. From steam tables, the saturated vapor pressure is 0.51 psia. At 60% relative humidity, the vapor pressure = 0.6 × 0.51 = 0.408 psia. This corresponds to a saturation temperature of 73°F from steam tables. So we must select the thickness of insulation so that it is above 23°C (73°F).

− 11  4 4 1 q= .225 . 5 67 10 × × 09 . × 

= . 0 0326 kW/m 2 = . 32 6 W/m 2 q = 0.04 × (23 + 12.2)/L = 32.6 or L = 0.043 m = 43 mm. Use slightly higher thickness

considering variations in ambient conditions.

Example 6.8

A 1.5 in. Sch 40 pipe (1.9 in. OD, 1.61 in. ID), 300 m long carries hot water at 150°C. What is the heat loss from its surface if not insulated (case 1) or has 25, 50, and 75 mm thick insulation (case 2)? Assume K of insulation as 0.036 W/m K. Ambient temperature is 27°C and wind velocity is zero.

Case 1: For bare pipe, × − 11 × × 

 ( 273 150 )( 273 27 )  . 0 00195 × ( 150 27 − = 20 . kW/m ) 2 . Case 2: Using a computer program, the heat loss was determined for various cases,

q= 11 25 . 5 67 10 09 .

which is shown later. It can be shown that the heat losses are, respectively, 0.085, 0.044, 0.0284 kW/m 2 for the 25, 50, and 76 mm thick insulated pipes. If the 50 mm thick insulation pipe is considered,

322 Steam Generators and Waste Heat Boilers: For Process and Plant Engineers

the total heat loss = 3.14 × .0.044 × 0.148 × 300 = 6.134 kW. If the specific heat of water is taken as 4.3 kJ/kg K and flow as 1 kg/s, the decrease in temperature of water is about 6.134/4.3 = 1.5°C.

Superheated steam pipes are large in diameter, and also, the specific heat of steam is lower. Hence, the temperature drop of steam can be as much as 5°C–7°C depending on insulation thickness and pipe sizes and length. Hence, such a calculation may be per- formed to estimate the drop in steam temperatures as lower temperature at steam tur- bine inlet reflects loss in power production. The pipe length can be split up into segments, and the effect of variations of K with temperature can be considered, and hence, more accurate results can be obtained. One may also arrive at the optimum thickness by evalu- ating the loss in energy in dollar terms and cost of installation for various thicknesses.

Insulation Calculations