Question 8.3.5: Energy Balance on a Gas Preheater A stream containing 10% CH......
Energy Balance on a Gas Preheater
A stream containing 10% CH_{4} and 90% air by volume is to be heated from 20°C to 300°C. Calculate the required rate of heat input in kilowatts if the flow rate of the gas is 2.00 × 10³ liters (STP)/min.
Basis: Given Flow Rate
Assume ideal-gas behavior
Recall that specifying the flow rate in liters (STP)/min does not imply that the feed gas is at standard temperature and pressure; it is simply an alternative way of giving the molar flow rate.
\dot{n} = \begin{array}{c|c}2000 L (STP)&1 mol \\ \hline min &22.4 L (STP)\end{array} = 89.3 mol/min
The energy balance with kinetic and potential energy changes and shaft work omitted is \dot{Q}=\Delta \dot{H} . The task is to evaluate \Delta \dot{H} = \sum\limits_{out}{\dot{n}_{i}\hat{H}_{i}} – \sum\limits_{in}{\dot{n}_{i}\hat{H}_{i}} . Since each species has only one inlet condition and one outlet condition in the process, two rows are sufficient for the enthalpy table.
The reference condition for methane was chosen so that \hat{H}_{in} could be set equal to zero, and that for air was chosen so that \hat{H}_{in} and \hat{H}_{out} could be determined directly from Table B.8.
The next step is to evaluate all of the unknown specific enthalpies in the table. \hat{H}_{1}, for example, is the specific enthalpy of methane in the outlet gas mixture at 300°C relative to pure methane at its reference temperature of 20°C. In other words, it is the specific enthalpy change for the process
CH_{4}(g, 20°C, 1 atm) \rightarrow CH_{4}(g, 300°C, P in outlet mixture)
We neglect the effect of pressure on enthalpy (i.e., we assume ideal-gas behavior) and we always neglect heats of mixing of gases, so that the enthalpy change is calculated for the heating of pure methane at 1 atm:
\hat{H}_{1}= \int_{20°C}^{300°C}{(C_{p})_{CH_{4}}} dT \\ \left. \Large{\Downarrow} \right. Substitute for C_{p} from Table B.2 \\ = \int_{20°C}^{300°C} (0.03431 + 5.469 \times 10^{-5} T + 0.3661 \times 10^{-8} T^{2} – 11.0\times 10^{-12} T^{3}) dT \\ = 12.07 kJ/mol
Alternatively, in APEx enter the formula =Enthalpy(“CH_{4}”,20,300) into a spreadsheet cell, and the value 12.07 (in kJ/mol) will be returned. The enthalpies of air at the inlet and outlet conditions relative to air at the reference state (\hat{H}_{2} and \hat{H}_{3}, respectively) are determined from Table B.8 as
\hat{H}_{2} = – 0.15 kJ/mol , \hat{H}_{3}= 8.17 kJ/mol
The energy balance now yields
References: CH_{4}(g, 20°C, 1 atm), air(g, 25°C, 1 atm)
| Substance | \dot{n}_{in}
(mol/min) |
\hat{H}_{in}
(kJ/mol) |
\dot{n}_{out}
(mol/min) |
\hat{H}_{out}
(kJ/mol) |
| CH_{4} | 8.93 | 0 | 8.93 | \hat{H}_{1} |
| Air | 80.4 | \hat{H}_{2} | 80.4 | \hat{H}_{3} |
