Cold streams

The analysis of the heat exchanger network first identifies sources of heat (termed hot streams) and sinks (termed cold streams) from the material and energy balance. Consider first a very simple problem with just one hot stream (heat source) and one cold stream (heat sink). The initial temperature (termed supply temperature), final temperature (termed target temperature), and enthalpy change of both streams are given in Table 6.1. 

Consider the simple flowsheet shown in Fig. 6.2. Flow rates, temperatures, and heat duties for each stream are shown. Two of the streams in Fig. 6.2 are sources of heat (hot streams) and two are sinks for heat (cold streams). Assuming that heat capacities are constant, the hot and cold streams can be extracted as given in Table 6.2. Note that the heat capacities CP are total heat capacities and... 

Figure 6.1 A simple heat recovery problem with one hot stream and one cold stream.

Figure 6.2 A simple flowsheet with two hot streams and two cold streams.

Figure 6.4 The cold streams be combined to obtain a composite cold stream.

Consider now the possibility of transferring heat between these two systems (see Fig. 6.76). Figure 6.76 shows that it is possible to transfer heat from hot streams above the pinch to cold streams below. The pinch temperature for hot streams for the problem is 150°C, and that for cold streams is 140°C. Transfer of heat from above the pinch to below as shown in Fig. 6.76 transfers heat from hot streams with a temperature of 150°C or greater into cold streams with a temperature of 140°C or less. This is clearly possible. By contrast. Fig. 6.7c shows that transfer from hot streams below the pinch to cold streams above is not possible. Such transfer requires heat being transferred from hot streams with a temperature of 150°C or less into cold streams with a temperature of 140°C or greater. This is clearly not possible (without violating the ATmin constraint). 

An alternative inappropriate use of utilities involves heating of some of the cold streams below the pinch by steam. Below the pinch, cooling water is needed to satisfy the enthalpy imbalance. Figure... 

First, determine the shifted temperature intervals T from actual supply and target temperatures. Hot streams are shifted down in temperature by and cold streams up by AT J2, as detailed... 

Fig. 6.16 set to ATmin/2 below hot stream temperatures and ATn,in/2 above cold stream temperatures. 

Now cascade any surplus heat down the temperature scale from interval to interval. This is possible because any excess heat available from the hot streams in an interval is hot enough to supply a deficit in the cold streams in the next interval down. Figure 6.18 shows the cascade for the problem. First, assume that no heat is supplied to the first interval from a hot utility (Fig. 6.18a). The first interval has a surplus of 1.5 MW, which is cascaded to the next interval. This second interval has a deficit of 6 MW, which reduces the heat cascaded from this interval to -4.5 MW. In the third interval the process has a surplus of 1 MW, which leaves -3.5 MW to be cascaded to the next interval, and so on. 

More than 7.5 MW could be added from a hot utility to the first interval, but the objective is to find the minimum hot and cold utility. Thus from Fig. 6.186, QHmin = 7.5MW and Qcmm = 10MW. This corresponds with the values obtained from the composite curves in Fig. 6.5a. One further important piece of information can be deduced from the cascade in Fig. 6.186. The point where the heat flow goes to zero at T = 145°C corresponds to the pinch. Thus the actual hot and cold stream pinch temperatures are 150 and 140°C. Again, this agrees with the result from the composite curves in Fig. 6.5a. 

Next, calculate the shifted interval temperatures. Hot streams are shifted down by 2.5°C, and cold streams are shifted up by 2.5°C (Table 6.5). 

So far it has been assumed that any hot stream could, in principle, be matched with any cold stream, providing there is feasible temperature diflerence between the two. Often, however, practical constraints prevent this. For example, it might be the case that if two... 

The grand composite curve is obtained by plotting the problem table cascade. A typical grand composite curve is shown in Fig. 6.24. It shows the heat flow through the process against temperature. It should be noted that the temperature plotted here is shifted temperature T and not actual temperature. Hot streams are represented ATn,in/2 colder and cold streams AT iJ2 hotter than they are in practice. Thus an allowance for ATj in is built into the construction. 

I = total numher of hot streams in enthalpy interval k J = total numher of cold streams in enthalpy interval k K = total number of enthalpy intervals... 

By constrast, Fig. 7.46 shows a diflFerent arrangement. Hot stream A with a low coefficient is matched with cold stream D, which also has a low coefficient but uses temperature diflferences greater than vertical separation. Hot stream B is matched with cold stream C, both with high heat transfer coefficients but with temperature differences less than vertical. This arrangement requires 1250 m of area overall, less than the vertical arrangement. 

Figure 7.6 now shows the stream population for each enthalpy interval together with the hot and cold stream temperatures. Now set up a table to compute Eq. (7.6). This is shown in Table 7.2. Thus the network area target for this problem for = 10°C is 7410 m. ... 

The final temperature of the hot stream is higher than the final temperature of the cold stream, as illustrated in Fig. 7.8a. This is called a temperature approach. This situation is straightforward to design for, because it can always be accommodated in a single 1-2 shell. 

Another way to relate these principles is to remember that heat integration will always benefit by keeping hot streams hot and cold streams cold. ... 

Adiabatic operation. If adiabatic operation leads to an acceptable temperature rise for exothermic reactors or an acceptable fall for endothermic reactors, then this is the option normally chosen. If this is the case, then the feed stream to the reactor requires heating and the efiluent stream requires cooling. The heat integration characteristics are thus a cold stream (the reactor feed) and a hot stream (the reactor efiluent). The heat of reaction appears as elevated temperature of the efiluent stream in the case of exothermic reaction or reduced temperature in the case of endothermic reaction. 

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