Page 139 - Chemical process engineering design and economics

Process Circuit Analysis 123

After counting all the equations and variables in Tables 3.4.1 and 3.4.2, we
find that we now have zero degrees of freedom. Thus, we have defined the prob-
lem, and we can now outline the solution procedure. The twenty-two equations
are decoupled, i.e., it is not necessary to solve all them simultaneously. By inspec-
tion we find that we can solve the mole balance equations independently of the
energy balance. This frequently occurs, usually when the temperatures in some of
the lines are known. Furthermore, in this case, we do require an iterative calcula-
tion procedure. We again obtained a solution procedure by inspection, which is
given in Table 3.4.3.
Frequently, we do not analyze simple process problems by the approach
given in Tables 3.4.1 to 3.4.3. Instead, from the beginning, we assume that a solu-
tion is possible, and we carry out the calculations, introducing equations as
needed. With experience one can recognize that certain problems have solutions,
however, in most cases it is not evident that there is enough information to solve a
problem, particularly when the solution contains many equations. In this problem,
we will calculate the mole balance quickly without a formal analysis, once we
know that the degrees of freedom are zero.
Because there is 1000 kmol/h (2204 Ib mol/h) of methanol in line 3, there
will be 200 kmol/h (440.8 Ib mol/h) of methanol and 800 kmol/h of formaldehyde
in line 4 because the conversion is 80 %. If 80 % of the stoichiometric quantity of
oxygen is required, there will be 0.8 (1/2) (1000) = 400 kmol/h (881.6 Ib mol/h) of
oxygen at lines 2 and 3 and zero at line 4. The nitrogen flow rate in lines 2, 3 and 4
is (0.79/0.21) (400) = 1505 kmol/h (3317 Ib mol/h). It is good practice to tabulate

Table 3.4.2 Summary of Equations for a Mixer_______________

Mole Balances

m I' = y3, 1m 3 (3.4.23)

J2.2 m 2 = y 3, 2 m 3 (3.4.24)
y2, 5'm 2= y 3, 5m 3 (3.5.25)

Variables

Additional Variable is m 2.

Degrees of Freedom

F = 22 - 22 = 0

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