## Kern Unsteady State External Exchanger Isothermal Heating

External 1-2 Exchanger: Heating ln (Ti - t1)/(T1 -12) = (Sw/M)0 (11-35n)

External 1-2 Exchanger: Cooling ln (Ti - ti)/(T2 - t1) = S(wc/MC)9 (11-35o)

The cases of multipass exchangers with liquid continuously added to the tank are covered by Kern, as cited earlier. An alternative method for all multipass-exchanger gases, including those presented as well as cases with two or more shells in series, is as follows:

1. Determine UA for using the applicable equations for counter-flow heat exchangers.

2. Use the initial batch temperature T1 or t1.

3. Calculate the outlet temperature from the exchanger of each fluid. (This will require trial-and-error methods.)

4. Note the FT correction factor for the corrected mean temperature difference. (See Fig. 11-4.)

5. Repeat steps 2, 3, and 4 by using the final batch temperature T2 and t2.

6. Use the average of the two values for F, then increase the required multipass UA as follows:

UA(multipass) = UA(counterflow)/FT

In general, values of FT below 0.8 are uneconomical and should be avoided. FT can be raised by increasing the flow rate of either or both of the flow streams. Increasing flow rates to give values well above 0.8 is a matter of economic justification.

If Ft varies widely from one end of the range to the other, FT should be determined for one or more intermediate points. The average should then be determined for each step which has been established and the average of these taken for use in step 6.

Effect of External Heat Loss or Gain If heat loss or gain through the vessel walls cannot be neglected, equations which include this heat transfer can be developed by using energy balances similar to those used for the derivations of equations given previously. Basically, these equations must be modified by adding a heat-loss or heat-gain term.

A simpler procedure, which is probably acceptable for most practical cases, is to ratio UA or 0 either up or down in accordance with the required modification in total heat load over time 0.

Another procedure, which is more accurate for the external-heat-exchanger cases, is to use an equivalent value for MC (for a vessel being heated) derived from the following energy balance:

Q = (Mc)e(t2 -11) = Mc(t2 -11) + U'A'(MTV )0 (11-35p)

where Q is the total heat transferred over time 0, U'A' is the heat-transfer coefficient for heat loss times the area for heat loss, and MTD' is the mean temperature difference for the heat loss.

A similar energy balance would apply to a vessel being cooled.

Internal Coil or Jacket Plus External Heat Exchanger This case can be most simply handled by treating it as two separate problems. M is divided into two separate masses M1 and (M - M1), and the appropriate equations given earlier are applied to each part of the system. Time 0, of course, must be the same for both parts.

Equivalent-Area Concept The preceding equations for batch operations, particularly Eq. 11-35 can be applied for the calculation of heat loss from tanks which are allowed to cool over an extended period of time. However, different surfaces of a tank, such as the top (which would not be in contact with the tank contents) and the bottom, may have coefficients of heat transfer which are different from those of the vertical tank walls. The simplest way to resolve this difficulty is to use an equivalent area Ae in the appropriate equations where

and the subscripts b, s, and t refer to the bottom, sides, and top respectively. U is usually taken as Us. Table 11-1 lists typical values for Us and expressions for Ae for various tank configurations.

Nonagitated Batches Cases in which vessel contents are vertically stratified, rather than uniform in temperature, have been treated by Kern (op. cit.). These are of little practical importance except for tall, slender vessels heated or cooled with external exchangers. The result is that a smaller exchanger is required than for an equivalent agitated batch system that is uniform.

Storage Tanks The equations for batch operations with agitation may be applied to storage tanks even though the tanks are not agitated. This approach gives conservative results. The important cases (nonsteady state) are:

1. Tanks cool; contents remain liquid. This case is relatively simple and can easily be handled by the equations given earlier.

2. Tanks cool, contents partially freeze, and solids drop to bottom or rise to top. This case requires a two-step calculation. The first step is handled as in case 1. The second step is calculated by assuming an isothermal system at the freezing point. It is possible, given time and a sufficiently low ambient temperature, for tank contents to freeze solid.

3. Tanks cool and partially freeze; solids form a layer of self-insulation. This complex case, which has been known to occur with heavy hydrocarbons and mixtures of hydrocarbons, has been discussed by Stuhlbarg [Pet. Refiner, 38, 143 (Apr. 1, 1959)]. The contents in the center of such tanks have been known to remain warm and liquid even after several years of cooling.

It is very important that a melt-out riser be installed whenever tank contents are expected to freeze on prolonged shutdown. The purpose is to provide a molten chimney through the crust for relief of thermal expansion or cavitation if fluids are to be pumped out or recirculated through an external exchanger. An external heat tracer, properly

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