Question 8.3.4: Heat Capacity of a Mixture Calculate the heat required to br......

Heat Capacity of a Mixture
Calculate the heat required to bring 150 mol/h of a stream containing 60% C_{2}H_{6} and 40% C_{3}H_{8} by volume from 0°C to 400°C. Determine a heat capacity for the mixture as part of the problem solution.

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Neglecting changes in potential and kinetic energy and recognizing that there is no shaft work involved in the process, the energy balance becomes \dot{Q} = \Delta \dot{H} = \dot{n} \Delta \hat{H} where \Delta \hat{H} = \int_{0°C}^{400°C}{(C_{p})_{mix}  dT} . The polynomial expressions for the heat capacity of ethane and propane given in Table B.2 are substituted into Equation 8.3-13 to yield

(C_{p})_{mix} (T) = \sum\limits_{\begin{matrix} all\\ mixture\\ components\end{matrix} }{y_{i}C_{pi}(T)}             (8.3-13)

\begin{matrix}(C_{p})_{mix} [kJ/(mol\cdot°C)] = 0.600(0.04937 + 13.92\times 10^{-5}T  –  5.816 \times 10^{-8}T^{2} + 7.280\times 10^{-12} T^{3})\\\\ + 0.400 (0.06803 +  22.59 \times 10^{-5}T  –  13.11 \times10^{-8}T^{2} + 31.71\times 10^{-12} T^{3}) \\\\= \boxed{0.05683 + 17.39 \times 10^{-5}T  –  8.734 \times10^{-8}T^{2} + 17.05\times 10^{-12} T^{3}}\\\\\Delta \hat{H} =  \int_{0°C}^{400°C}{(C_{p})_{mix}  dT} = 34.89  kJ/mol\end{matrix}

If potential and kinetic energy changes and shaft work are neglected, it follows that

\dot{Q} = \Delta \dot{H} = \dot{n} \Delta \hat{H} =\begin{array}{c|c}150  mol &34.89  kJ \\ \hline h&mol\end{array} = \boxed{5230  \frac{kJ}{h}}

As usual, we have assumed that the gases are sufficiently close to ideal for the formulas for C_{p} at 1 atm to be valid.

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