Question

13.54 A gaseous mixture of butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) and \(80 \%\) excess air at \(25^{\circ} \mathrm{C}, 3\) atm enters a reactor. Complete combustion occurs, and the products exit as a mixture at \(1200 \mathrm{~K}, 3 \mathrm{~atm}\). Coolant enters an outer jacket as a saturated liquid and saturated vapor exits at essentially the same pressure. No significant heat transfer occurs from the outer surface of the jacket, and kinetic and potential energy effects are negligible. Determine for the jacketed reactor (a) the mass flow rate of the coolant, in \(\mathrm{kg}\) per kmol of fuel. (b) the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per kmol of fuel. (c) the rate of exergy destruction, in kJ per kmol of fuel, for \(T_{0}=25^{\circ} \mathrm{C}\) Consider each of two coolants: water at 1 bar and ammonia at 10 bar.

Short answer

Calculate heat released, adjust for excess air, find temperature changes, apply enthalpy changes, determine mass flow from coolant used.

Step-by-step solution

- Write the balanced combustion equation

The combustion of butane \(\text{C}_4\text{H}_{10}\) with theoretically required air (O_2 \( + 3.76 \text{N}_2 \)) is given by: \[ \text{C}_4\text{H}_{10} + 6.5 (\text{O}_2 + 3.76 \text{N}_2) \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} + 24.44 \text{N}_2 \] This ensures complete combustion and accounts for all reactants and products.

- Adjust for 80% excess air

Account for the excess air by multiplying the O_2 term by 1.8: \[ \text{C}_4\text{H}_{10} + 1.8 \times 6.5 (\text{O}_2 + 3.76 \text{N}_2) \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} + 1.8 \times 24.44 \text{N}_2 \]

- Calculate heat released using standard enthalpies

Use the known standard enthalpies of formation at 298 K and calculate the change in enthalpy for the reaction: \[ \text{Reactants heat: } \text{C}_4\text{H}_{10} \text{ + Air} \] \[ \text{Products Heat: } \text{CO}_2, \text{H}_2\text{O}, \text{N}_2} + Excess \] Difference will give the heat added/released.

- Calculate heat capacity changes for temperature

Calculate the enthalpy changes of gases by using their specific heats with the delta T from initial to final temperatures i.e., from 298K to 1200K.

- Calculate enthalpy change for the coolant

Calculate the amount of heat removed by the coolant using its properties for the utilized specific heats. Ensure the coolant changes phase properly and exhibit correct phase transitions - liquid to vapor.

- Determine mass flow rate of the coolant

Link each calculated value to find the precise flow rate in \(\text{kg} \text{per} \text{kmol} \text{of} \text{fuel} \).

- Calculate rate of entropy production

Use the entropy balance equation and the data from previous steps to find the rate of entropy production (in \(\text{kJ} \text{per} \text{kmol} \text{fuel} \)).

- Determine exergy destruction rate

Finally, calculate the exergy destruction rate using the temperature of surroundings \( T_0 = 25^\text{C} \).

Key concepts

Excess Air Calculation

In combustion analysis, 'excess air' refers to the amount of air supplied beyond the theoretical requirement for complete combustion. This is important because providing insufficient air can lead to incomplete combustion, producing pollutants and reducing energy efficiency. For butane \text{C}_4\text{H}_{10}, the theoretical air demand is calculated by balancing the combustion equation.
In our case, the combustion equation for butane is: \[ \text{C}_4\text{H}_{10} + 6.5 (\text{O}_2 + 3.76 \text{N}_2) \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} + 24.44 \text{N}_2 \] Here, the coefficients reflect the stoichiometric amounts. When accounting for 80% excess air, we multiply the oxygen term by 1.8, leading to: \[\text{C}_4\text{H}_{10} + 1.8 \times 6.5 (\text{O}_2 + 3.76 \text{N}_2) \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} + 1.8 \times 24.44 \text{N}_2 \]

Enthalpy Change Calculation

Enthalpy change measures the heat absorbed or released during a reaction at constant pressure. To calculate the enthalpy change during combustion, we use the standard enthalpies of formation at 298 K. The enthalpy of the reactants is compared to the enthalpy of the products.
We break it down like this: the heat provided by butane and air (reactants) is determined, then the heat of the combustion products (CO2, H2O, N2) is calculated. The difference between these values gives the released heat, \(\begin{aligned} \text{Reactants}: & \ \text{Products}: & \end{aligned} \)

Entropy Production

Entropy production signifies the dispersal of energy and the irreversibility of the process. In our combustion system, we determine entropy by considering the entropy changes of the reactants and products, as well as the coolant and surroundings. Using the entropy balance equation: \( \begin{aligned} \frac{\text{d}S_\text{total}}{\text{dt}} = \frac{\text{d}S_\text{system}}{\text{dt}} + \frac{\text{d}S_\text{surroundings}}{\text{dt}} \end{aligned} \)
We can calculate the rate of entropy production.

Exergy Destruction

Exergy destruction represents the loss of potential work due to irreversibilities in the process. It's an essential metric for evaluating energy efficiency. To find exergy destruction, we use the relation involving the environment temperature (T0=25°C): \[E_\text{destroyed} = T_0 \times S_\text{produced} \]
This indicates how much useful work is lost and helps in optimizing the system for improved performance.

Heat Capacity

Heat capacity is a crucial concept in thermodynamics, indicating how much heat a substance can absorb before its temperature changes. It is especially important in combustion analysis as it relates to temperature changes of involved gases and coolant.
\[ Q = mc\triangle T \] where Q is heat absorbed, m is mass, c is specific heat capacity, and \triangle T is temperature change. This helps us determine the phase transition of coolants and the energy balance.