## 650 105

STEP 5. Read to the left to (c) for heat loss Q = 24 Btu/hr ft2.

STEP 6. Read down from the proper surface curve from (d) to (e), which represents ts - ta, to check the surface-temperature assumption. For aluminum, ts - ta (chart) is 21°F, compared with the 120 - 80 = 40°F assumption.

STEP 7. Calculate a new surface temperature 80 + 21 = 101°F; then calculate a new tm, = (300 + 101)/2 =

STEP 3. Check surface temperature assumption by ts = (Q x Rs) + ta using Rs = 0.52. From Table 15.4 for a mastic finish, ts = 50(0.52) + 80 = 106°F

(Note that this in turn changes the ts - ta from 40 to 25, which changes Rs from 0.49 to 0.51, which is insignificant.)

For a graphical solution to this problem, Figure 15.4 is again used. It is simply a matter of reading across the desired Q level and adjusting the ts and R values to reach equilibrium. Thickness is then determined by tk = kR.

### Temperature Drop in a System

The following discussion is quite simplified and is not intended to replace the service of the process design engineer. The material is presented to illustrate how insulation ti es into the process design decision.

Temperature Drop in Stationary Media over Time. The procedure calls for standard heat-flow calculations now tied into the heat content of the fluid. To illustrate, consider the following example.

Example. A water storage tank is calculated to have a surface area of 400 ft2 and a volume of 790 ft3. How much will the temperature drop in a 72-hr period with an ambient temperature of 0°F, assuming that the initial water temperature is 50°F? The tank is insulated with 2-in. fiberglass with a mastic coating.

Before proceeding, realize that the maximum heat transfer will occur when the water is at 50°F. As it drops in temperature, the heat-transfer rate is reduced due to a smaller temperature difference. As a first approximation, it is reasonable to use the maximum heat transfer based on 50°F. Then if the temperature drop is significant, an average water temperature can be used in the second iteration.

STEP 1. Assume a surface temperature, calculate the mean temperature, find the k factor from Table 15.1 or appendix Figure 15A.3, and determine Rs from Table 15.4. With ts = 10°F, m = 30°F, k = 0.22, and Rs = 0.53.

STEP 2. Calculate heat loss with fluid at 50°F. 50-0

= 2080 x 72 = 149,760 Btu. Divide this by the available heat per 1°F drop:

149,760 Btu 49,296 Btu/of

This procedure may also be used for fluid lying stationary in a pipeline. In this case it is easiest to do all the calculations for 1 linear foot rather than for the entire length of pipe.

One conservative aspect of this calculation is that the heat capacity of the metal tank or pipe is not included in the calculation. Since the container will have to decrease in temperature with the fluid, there is actually more heat available than was used above.

Temperature Drop in Flowing Media. There are two common situations in this category, the first involving flue gases and the second involving water or other fluids with a thickening or freezing point. This section discusses the flue-gas problem and the following section, freeze protection.

A problem is encountered with flue gases that have fairly high condensation temperatures. Along the length of a duct run, the temperature will drop, so insulation is added to control the temperature drop. This calculation is actually a heat balance between the mass flow rate of energy input and the heat loss energy outlet.

For a round duct of radius ^ and length L, gas enters at th, and must not drop below tmin (the dew point). The flow rate is M lb/hr and the gas has a specific heat of Cp Btu/lb °F. Therefore, the maximum allowable heat loss in Btu/hr is

STEP 3. Calculate the amount of heat that must be lost for the entire volume of water to drop 1°F.

Available heat per °F

= volume x density x specific heat = 790 ft3 x 62.4 lb/ft3 x 1 Btu/lb F = 49,296 Btu/°F

STEP 4. Calculate the temperature drop in 72 hr by determining the total heat flow over the period: Q

Also, where

(A conservative simplification would be to set th = tin since the higher temperature, tin, will cause a greater heat loss.)

To simplify on large ducts, assume that r1 = r2 (ignore the insulation-thickness addition to the surface area). Therefore, th - ta

2fiT1

0 0