Consider the blending system shown schematically in Fig. 8.10. Two identical liquids of different temperatures T_{1} and T_{2}, flowing with different flow rates Q_{1} and Q_{2}, are perfectly mixed in a blender of volume V. The mixture of liquids is also heated in the blender by an electric heater supplying heat at a constant rate, Q_{hgen}. There are heat losses in the system, and the coefficient of heat transfer between the blender and ambient air of temperature T_{a} is h_{c}. Although the mixing in the tank is assumed to be perfect, the work done by the mixer is negligible and the kinetic energies of the flows Q_{1} ,Q_{2}, and Q_{3} are very small. Derive a mathematical model of the blending process.
Question 8.2: Consider the blending system shown schematically in Fig. 8.1...
The unsteady-flow energy-balance equation includes the following terms:
\left ( \begin{matrix} rate \\ of \\ enthalpy \end{matrix} \right ) _{1} +\left ( \begin{matrix} rate \\ of \\ enthalpy \end{matrix} \right ) _{2}+\left ( \begin{matrix} rate \\ of \\ heat \end{matrix} \right ) _{gen}=\left ( \begin{matrix} rate \\ of \\ enthalpy \end{matrix} \right ) _{3} +\left ( \begin{matrix} rate \\ of \\ heat \end{matrix} \right )_{loss}+\left ( \begin{matrix} rate \\of \\ change \\ of \\ energy \end{matrix} \right ) _{sto}.The first two terms in this equation represent the rates of enthalpy supplied with the two incoming streams of liquids given by
Q_{h1} =\rho c_{p} Q_{1} \left(T_{1} -T_{a} \right) ,Q_{h2} =\rho c_{p} Q_{2} \left(T_{2} -T_{a} \right) .
The next term in the energy-balance equation represents the heat generated by the heater, Q_{hgen}. The heat carried away from the tank is represented by the first two terms on the right-hand side of the energy-balance equation. The rate of enthalpy carried with the outgoing stream of the mixture of the two input liquids is
Q_{h3} =\rho c_{p} Q_{3} \left(T_{3} -T_{a} \right)=\rho c_{p} \left(Q_{1} +Q_{2} \right)\left(T_{3} -T_{a} \right)The rate at which heat is lost by the liquid through the sides of the tank to the
ambient air is
Finally, the rate of change of energy stored in the liquid contained in the tank is
\frac{d\xi}{dt} =\rho c_{p} V\frac{dT_{3} }{dt}Substituting detailed mathematical expressions for the heat and enthalpy rates into the energy-balance equation yields
\rho c_{p}Q_{1} \left(T_{1} -T_{a} \right)+\rho c_{p}Q_{2} \left(T_{2} -T_{a} \right)+Q_{hgen} =\rho c_{p}\left(Q_{1} +Q_{2} \right)\left(T_{3} -T_{a} \right)+h_{c} A\left(T_{3} -T_{a} \right)+\rho c_{p}V\frac{dT_{3} }{dt} (8.36)
Equation (8.36) can be rearranged into the simpler form
\frac{dT_{3} }{dt} =-\frac{1}{V} \left(\frac{h_{c} A}{\rho c_{p} }+Q_{1} +Q_{2} \right) T_{3} +\frac{Q_{1} }{V} T_{1} +\frac{Q_{2} }{V} T_{2}+\frac{1}{\rho c_{p}V } Q_{hgen}+\frac{h_{c} A }{\rho c_{p}V } T_{a} (8.37)
Equation (8.37) represents a first-order multidimensional model with six potential
input signals, Q_{1} ,T_{1} ,Q_{2},T_{2},Q_{hgen}, and T_{a}. The system time constant is
The model can be further simplified if the blender is assumed to be perfectly insulated, h_{c} =0, and if there is no heat generation in the system Q_{hgen}=0. Under such conditions the system state equation becomes
\frac{dT_{3} }{dt}=\left(-\frac{1}{C} \right) \left(\frac{1}{R_{1} }+\frac{1}{R_{2} } \right) T_{3} +\left(\frac{1}{R_{1}C } \right) T_{1}+\left(\frac{1}{R_{2}C } \right) T_{2},
where the lumped-model parameters C, R_{1}, and R_{2}, are defined as follows:
C=\rho c_{p} V,
R_{1} =\frac{1}{\rho c_{p} Q_{1} },
R_{2} =\frac{1}{\rho c_{p} Q_{2} }.
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