36 Reactive Energy Balances

Consider a propane combustion chamber, where $100 mol/s$ of propane are feed to the chamber at $25^{\circ}C$ and air is fed at $300^{\circ}C$ ($600 mol/s O_{2}(g)$ and $2256 mol/s N_{2}(g)$). The products stream exits at $1000^{\circ}C$ and consists of $100 mol/s O_{2}(g)$, $2256 mol/s N_{2}(g)$, $300 mol/s CO_{2}(g)$, and $400 mol/s H_{2}O(v)$. How much heatis released by this combustion chamber? Assumming atmoshpere pressure, so water boils at 100°C.

$C_{3}H_{8} (g) + 5 O_{2} → 3 CO_{2} (g) + 4 H_{2}O (l)$
$\Delta H^{\circ}_{r} = -2220 kJ$

The following information is provided:

The values listed under each compound are specific enthalpies in kJ/mol

• $C_{P,l(H_{2}O)} = 75.4*10^{-3} \frac{kJ}{molK}$
• $C_{P,v(H_{2}O)} = 33.46*10^{-3}+0.688*10^{-5}*T + 0.7604*10^{-8}*T^{2} - 3.593*10^{-12}*T^{3} (J/molK)$
• $\Delta\hat{H}_{vap}(100^{\circ}C) = 40.66 kJ/mol$

How can we solve the system’s energy balance?

1 – Solve material balances as much as possible

This example already has the material balances solved

2 – Choose reference states for energy calculations. Like we say in the previous figure we want to calculate the changes in energy associated with bringing our reactants to our reference state, calculating the energy of the reaction at the reference state, and then the energy associated with bringing the products to their final state. We will choose reference states based on the information we have to try to make these calculations as easy as possible.

For reacting species:
Since $\Delta H^{\circ}_{r}$ is given, we will assume this is at $25^{\circ}C$ and 1 atm (as there is nothing indicating otherwise), and we will use this as our reference state.

For non-reacting species:
We can use any convenient temperature and pressure as a reference state (inlet temperature, outlet temperature, temperature in enthalpy table). With this example, the enthalpy table given uses a reference state of $25^{\circ}C$, and we will assume everything is at 1 atm (as no other values for pressure are given).

3 – Calculate the extent of reaction for all reactions (in this case we just have the one reaction)

$\xi = \frac{(\dot{n}_{i})_{out}-(\dot{n}_{i})_{in}}{\dot{\nu}_{i}}$

$\xi = \frac{(\dot{n}_{CO_{2}})_{out}-(\dot{n}_{CO_{2}})_{in}}{\dot{\nu}_{CO_{2}}}$

$\xi = \frac{300\frac{mol}{s} - 0 \frac{mol}{s}}{3} = 100 mol/s$

4 – Prepare an inlet-out enthalpy table (this will show what enthalpies we need to calculate associated with energy changes in the reactants or products)

5 – Calculate all the enthalpies. For $\hat{H_{1}}$ to $\hat{H_{6}}$, the specific enthalpies at different temperatures are given as the difference in enthalpy from the reference state (25°C, 1atm, which is the same as what we choose as reference state):

$\hat{H}_{1}=\Delta\hat{H}_{C_{3}H_{8}} (25^{\circ}C → 25^{\circ}C) = 0 kJ/mol$

$\hat{H}_{2}=\Delta\hat{H}_{O_{2}} (25^{\circ}C → 300^{\circ}C) = (8.47 - 0) kJ/mol$

$\hat{H}_{3}=\Delta\hat{H}_{N_{2}} (25^{\circ}C → 300^{\circ}C) = (8.12 - 0) kJ/mol$

$\hat{H}_{4}=\Delta\hat{H}_{O_{2}} (25^{\circ}C → 1000^{\circ}C) = (32.47 - 0) kJ/mol$

$\hat{H}_{5}=\Delta\hat{H}_{N_{2}} (25^{\circ}C → 1000^{\circ}C) = (30.56 - 0) kJ/mol$

$\hat{H}_{6}=\Delta\hat{H}_{CO_{2}} (25^{\circ}C → 1000^{\circ}C) = (48.60 - 0) kJ/mol$

$\hat{H}_{7}=\Delta\hat{H}_{H_{2}O} (l,25^{\circ}C → H_{2}O (v,1000^{\circ}C))$

6 – For $\hat{H_{7}}$, the specific enthalpy for water is calculated. From the reference state of 25°C to the stream temperature of 1000°C, water is first heated to 100°C, which is its natural boiling point at 1atm, then vaporized, then heated in vapor phase to 1000°C. The enthalpy change of each process is calculated separately and added together to get $\hat{H_{7}}$.

Note: The $T$ in the formulas to calculate $C_{P}$ is given in kelvin. This doesn’t make a difference when $C_{P}$ is given as a number, as the scales for one degree of Celcius and kelvin are the same, but the temperature must be converted to kelvin when $T^2$ is higher power of $T$ is used in calculation.

$100°C=373K, 1000°C=1273K$

$$\begin{align*} \hat{H}_{7}& = \int^{100^{\circ}C}_{25^{\circ}C} C_{P,l}dT + \Delta \hat{H}_{vap}(100^{\circ}C) + \int^{1000^{\circ}C}_{100^{\circ}C} C_{P,v}dT\\& = \int^{100^{\circ}C}_{25^{\circ}C} 75.4*10^{-3}dT + \Delta \hat{H}_{vap}(100^{\circ}C) \\& \;\;\;\;+\int^{1273K}_{373K} (33.46*10^{-3}+0.688*10^{-5}*T + 0.7604*10^{-8}*T^{2} – 3.593*10^{-12}*T^{3})dT\\& = 75.4*10^{-3}*(100-25)+ 40.66 \\& \;\;\;\; +(33.46*10^{-3}*T+\frac{1}{2}*0.688*10^{-5}*T^2 + \frac{1}{3}*0.7604*10^{-8}*T^{3} – \frac{1}{4}*3.593*10^{-12}*T^{4})\bigg\vert^{1273K}_{373K}\\& = (5.65 + 40.66 + 35.1)kJ/mol\\& = 81.46 kJ/mol \end{align*}$$

$\hat{H}_{7}= (5.65 + 40.66 + 35.1)kJ/mol = 81.46\frac{kJ}{mol}$

Therefore, we have calculated all the specific enthalpy for the reactants and products:

8 – Finally, solve the energy balance

$\Delta\dot{H} + \Delta\dot{E}_{k} +\Delta\dot{E}_{p} = \dot{Q} + \dot{W}_{s}$

$\Delta\dot{E}_{k}$, $\Delta\dot{E}_{p}$, and $\dot{W}_{s}$ are assumed negligible for this system (as no information is provided on these)

$\dot{Q} = \Delta\dot{H}$

$\dot{Q} = -126 MW$

The combustion chamber releases $126 MW$ of heat to the environment

Let’s consider the propane combustion chamber problem we analyzed before with the heat of reaction method, and let’s see if we can get a similar answer using the heat of formation method.

1 – Solve the material balance as much as possible

This example already has the material balances solved

2 – Choose reference states for energy calculations

For reacting species: elemental species that make up reacting species at standard conditions; we will choose $25^{\circ}C$ at 1 atm with $C(s)$, $H_{2}(g)$, and $O_{2}(g)$.

For non-reacting species (same as $\Delta H_{r}$): Use any convenient temperature (inlet temperature, outlet temperature, temperature in enthalpy table) Here, $25^{\circ}C$ and 1 atm works because of our enthalpy table values

3 – Prepare an inlet-outlet enthalpy table

4 – Calculate all the enthalpies

$\hat{H}_{1} = \Delta\hat{H}^{\circ}_{f,C_{3}H_{8}(g)} = -103.8 kJ/mol$

$\hat{H}_{2}=\Delta\hat{H} (O_{2} (25^{\circ}C → 300^{\circ}C) = (8.47 - 0) kJ/mol$

$\hat{H}_{3}=\Delta\hat{H} (N_{2} (25^{\circ}C → 300^{\circ}C) = (8.12 - 0) kJ/mol$

$\hat{H}_{4}=\Delta\hat{H} (O_{2} (25^{\circ}C → 1000^{\circ}C) = (32.47 - 0) kJ/mol$

$\hat{H}_{5}=\Delta\hat{H} (N_{2} (25^{\circ}C → 1000^{\circ}C) = (30.56 - 0) kJ/mol$

$\hat{H}_{6}=\Delta\hat{H}^{\circ}_{f,CO_{2}(g)} + \int^{1000^{\circ}C}_{25^{\circ}C} C_{p,CO_{2}(g)}dT = -344.9 kJ/mol$

$\hat{H}_{7}=\Delta\hat{H}^{\circ}_{f,H_{2}O(v)} + \int^{1000^{\circ}C}_{25^{\circ}C} C_{p,H_{2}O(v)}dT = -204.1 kJ/mol$

Here, you assume that water forms as vapor directly in the reaction. Therefore, there is no need to account for heat of vaporization.

5 – Calculate $\Delta\dot{H}$ for the reactor

$\Delta\dot{H} = \Sigma\dot{n}_{out}*\hat{H}_{out} - \Sigma\dot{n}_{in}*\hat{H}_{in}=1.26x10^{5} kJ/s$

6 – Finally, solve the energy balance

$\dot{Q} = \Delta\dot{H}$

$\dot{Q} = -126 MW$