Chapter 7: Problem 1
Pareto Trade-Off between Economy and Ecology in Design
Biomass (B) is converted to a hydrocarbon fuel product (P) and carbon dioxide
\(\left(\mathrm{CO}_{2}\right)\) in a well-mixed continuous-flow reactor,
operating at steady state. The nominal chemical composition of
\(\mathrm{B}=\left[-\mathrm{C}_{1} \mathrm{O}_{0.6}
\mathrm{H}_{1.4}-\right]_{n}\) and that of \(\mathrm{P}=\left[-\mathrm{C}_{1}
\mathrm{H}_{2}-\right]_{n}\). The corresponding mass-based reaction
stoichiometry is (in close approximation)
\(1 \mathrm{~kg} \mathrm{~B} \stackrel{\mathrm{k}}{\rightarrow} 0.425
\mathrm{~kg} \mathrm{P}+0.575 \mathrm{~kg} \mathrm{CO}_{2}\) with
\(\mathrm{k}=10^{-3}\left(\mathrm{~s}^{-1}\right)\)
This reaction is the first order in the biomass amount; the symbol
\(\mathrm{k}\) represents the associated first-order reaction rate coefficient.
The biomass is only partially converted as full conversion would require a
reactor of infinite size and cost. The degree of conversion of the biomass
\((X)\) is defined as
\(X=\frac{F_{\mathrm{B}}^{(\mathrm{in})}-F_{\mathrm{B}}^{(\text {out
})}}{F_{\mathrm{B}}^{(\mathrm{in})}}\) with \(0 \leq X<1\)
The biomass intake rate \(\left(F_{\mathrm{B}}^{(\mathrm{in})}\right)\) and the
volume flow \(\left(\varphi_{V}\right)\) are kept constant:
$$
F_{\mathrm{B}}^{(\text {in })}=1\left(\mathrm{~kg} \cdot
\mathrm{s}^{-1}\right) ; \varphi_{V}=4.0 \times 10^{-3}\left(\mathrm{~m}^{3}
\cdot \mathrm{s}^{-1}\right)
$$
The degree of conversion is an independent design variable. The dependent
process variables are
$$
\begin{array}{lrl}
\text { Outflow of } \mathrm{B}\left(\mathrm{kg} \cdot \mathrm{s}^{-1}\right):
& F_{\mathrm{B}}^{(\text {out })} & =(1-X) F_{\mathrm{B}}^{(\text {in })} \\
\text { Outflow of } \mathrm{P}\left(\mathrm{kg} \cdot \mathrm{s}^{-1}\right):
& F_{\mathrm{P}}^{(\text {out })} & =0.425 X F_{\mathrm{B}}^{(\text {in })}
\\\
\text { Outflow of } \mathrm{CO}_{2}\left(\mathrm{~kg} \cdot
\mathrm{s}^{-1}\right): & F_{\mathrm{CO}_{2}}^{(\text {out })} & =0.575 X
F_{\mathrm{B}}^{(\text {in })} \\
\text { Reactor volume }\left(\mathrm{m}^{3}\right): & V_{\mathrm{R}} &
=\frac{\varphi_{V} X}{k(1-X)}
\end{array}
$$ The process performance is evaluated by means of two different indicators.
The ecological indicator is the carbon atom efficiency
\(\left(\varepsilon_{\mathrm{C}}\right)\). This efficiency is the fraction of
carbon atoms in biomass B ending up in product P; carbon left in the
unconverted biomass and in \(\mathrm{CO}_{2}\) is "lost":
$$
\varepsilon_{\mathrm{C}}=\frac{w_{\mathrm{c}, \mathrm{p}}
F_{\mathrm{p}}^{(\text {out })}}{w_{\mathrm{c}, \mathrm{B}}
F_{\mathrm{B}}^{(\text {in })}} \quad \text { with } w_{\mathrm{c},
\mathrm{P}}=\frac{12}{12+2 \cdot 1} ; w_{\mathrm{c},
\mathrm{B}}=\frac{12}{12+0.6 \cdot 16+1.4 \cdot 1}
$$
The economic performance is expressed by means of a profit function \((P)\). The
profit function is made up of earnings from product sales minus the costs of
feed and wastes and the capital charge (cost of capital investment per unit of
time). There are trade-offs in this design. A higher conversion increases
product output as well as reactor size and cost. The carbon efficiency
benefits from a high conversion. The company for which you design this reactor
aims for a better balance between ecological and economic performances.
The profit per hour is given by
$$
P=\left(F_{\mathrm{P}}^{(\text {out })}
p_{\mathrm{P}}-F_{\mathrm{B}}^{(\mathrm{in})}
c_{\mathrm{B}}-F_{\mathrm{B}}^{(\text {out })}
c_{\mathrm{W}}-F_{\mathrm{CO}_{2}}^{(\text {out })} c_{\mathrm{CO}_{2}}\right)
3600-C_{\text {cap, h }}\left(\$ \mathrm{~h}^{-1}\right)
$$
in which the capital charge, \(C_{\text {cap, }, \mathrm{h}}\), is given by
$$
C_{\text {cap, } \mathrm{h}}=\frac{c c f}{\mathrm{t}_{\mathrm{prod}}} I_{\text
{cap }}\left(\$ \mathrm{~h}^{-1}\right)
$$
with the capital investment, \(I_{\text {cap, }}\) given by
$$
I_{\text {cap }}=I_{0}\left(\frac{V}{V_{0}}\right)^{q}
$$
Data are given in Table \(7.8\).
Evaluate performance functions \(\varepsilon_{\mathrm{C}}\) and EP over the
conversion interval \(0.3
Short Answer
Step by step solution
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