Tie components

This technique is particularly useful in handling combustion calculations where the nitrogen in the combustion air passes through unreacted and is used as the tie component. This is illustrated in Example 2.8. 

The "other" balance (a tie component) and the CO2 balance are independent equations. We will use mole balances since all of the compositions are in mole fractions. 

Steps 8 and 9 The balances are in kg. The simplest balances to start with are those involving tie components. If selected properly, the equations can be solved sequentially rather than simultaneously. 

Select a basis and a tie component, and write out the relevant equations involved. Select 100 kg mol dry flue gas as the basis. Since nitrogen passes through the system unreacted, select it as the tie component. That is, the other components of the system can be referred to nitrogen as a basis, thus simplifying the calculations. 

Note The flue gas analysis is reported on a dry basis, any water formed having been condensed out. Nitrogen is the tie component. 

The ethanol (EtOH) balance directly relates F and P W is not involved, as there is no ethanol in the W stream. A material or component that goes directly from one stream into another without changing in any respect or having like material added to it or lost from it is called a tie component (element). If a tie component exists in a problem, in effect you can write a material balance that involves only two streams. To detect a tie component, ask yourself the question What component or element passes from one stream to another unchanged with constant mass The answer is the tie component. Frequently, several components pass through a process with continuity so that more than one stream can be connected drectly to its respective companion stream. Sometimes a minor constituent passes through with continuity, but if the... 

Step 9 Notice that because the bone dry cake balance involves only two streams, you can set up a direct ratio of A to B, which is the essence of the tie component ... 

Because we do not have to answer any questions about the water in the exit flue-gas stream, a tie element to relate the test fluid to the dry flue gas and one to relate the air to the dry flue gas would be sufficient to solve the problem. In examining the data to determine whether a tie element exists, we see that the carbon goes directly from the test fluid to the dry flue gas, and nowhere else, so carbon will serve as one tie component. The N2 in the air all shows up in the dry flue gas, so N2 can be used as another tie component. 

We might neglect the C, H, 0, N, and S in the refuse but will include the amounts to show what calculations are necessary if the amounts of the elements are significant. The ash balance is (ash is a tie component)... 

The air and acetone are tie components. (Check to make sure that the equations are independent.)... 

Steps 8 and 9 Solve the S balance for F (inaccuracy in the SOa concentration will cause some error in F, unfortunately) the sulfur is a tie component. Then solve the other four balances simultaneously for G. 

The strategy listed in Table 2.4 is the strategy to be used in solving recycle problems. You can make component and total material balances for each subsystem as discussed in Sec. 2.5, as well as component and total balances for the overall process. Not all of the equations so formulated will be independent, of course. Depending on the information available concerning the amount and composition of each stream, you can determine the amount and composition of the unknowns. If tie components are available, they often simplify the calculations. 

Steps 5 and 6 All the compositions are known and three stream flows, D, IV, and R, are unknown. No tie components are evident in this problem. Two component material balances can be made for the still and two for the condenser. Presumably three of these are independent hence the problem has a unique solution. We can check as we proceed. A balance around either the distillation column or the condenser would involve the stream R. An overall balance would involve D and W but not R. 

Note that other is a tie component. Solution of Eqs. (a) and (c) gives... 

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