Question

13.35 The energy required to vaporize the working fluid passing through the boiler of a simple vapor power plant is provided by the complete combustion of methane with \(110 \%\) of theoretical air. The fuel and air enter in separate streams at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Products of combustion exit the stack at \(150^{\circ} \mathrm{C}\), \(1 \mathrm{~atm}\). Plot the mass flow rate of fuel required, in \(\mathrm{kg} / \mathrm{h}\) per MW of power developed by the plant versus the plant thermal efficiency, \(\eta\). Consider \(\eta\) in the range \(30-40 \%\). Kinetic and potential energy effects are negligible.

Short answer

Calculate \( m_{fuel} = \frac{3600 \, \text{kg/h}}{\eta \cdot 50,020} \), then plot \( m_{fuel} \ against \( \eta \) from 30 to 40%.

Step-by-step solution

- Understand the Problem

The task is to determine the mass flow rate of methane required to generate power in a vapor power plant, while considering the thermal efficiency of the plant and complete combustion with 110% theoretical air.

- Define Parameters and Equations

Let the thermal efficiency of the plant be \( \eta \). The energy output of the plant per MW of power developed is given by \[ Q_{out} = \frac{1 \, \text{MW}}{\eta} \]. The energy input \( Q_{in} \) required from methane combustion is \[ Q_{in} = \frac{Q_{out}}{\eta} \].

- Combustion Equation

Write the balanced chemical equation for the complete combustion of methane (CH_4) with the given excess air: \[ CH_4 + 2.2(O_2 + 3.76N_2) \rightarrow CO_2 + 2H_2O + 8.272N_2 \]

- Determine Heat of Combustion

Calculate the heat released during the combustion of 1 kmol of methane using standard enthalpy values or reference tables, noting that combustions typically release about \( 802.3 \text{ kJ/mol} \). Thus, per kilogram, this energy can be expressed as \[ \Delta H_{combustion} = 50,020 \text{ kJ/kg} \].

- Calculate Mass Flow Rate

The energy provided by combustion for a given mass flow rate of fuel ( \( \dot{m}_{fuel} \), in \( kg/h \)) can be expressed as \[ Q_{combustion} = \dot{m}_{fuel} \cdot \Delta H_{combustion} \]. Combining the equations from previous steps, we get \[ \dot{m}_{fuel} = \frac{Q_{in} \cdot 3600}{\Delta H_{combustion}} = \frac{1 \, \text{MW} \cdot 3600}{\eta \cdot 50,020 \text{ kJ/kg}} \].

- Plot the Relation

To plot the mass flow rate of methane versus plant thermal efficiency, vary \( \eta \) between 0.30 and 0.40 (or 30% and 40%) and calculate \( \dot{m}_{fuel} \) for each value of \( \eta \). This can be plotted to visually represent the relationship.

Key concepts

Mass flow rate calculation

In a vapor power plant, the mass flow rate of fuel is a critical parameter. It's essential to know how much fuel is needed to produce the required power output. Mass flow rate is calculated by dividing the total energy required by the energy produced per kilogram of fuel.
To find the mass flow rate of methane in a vapor power plant, we have:
1. Determine the power output needed per hour.
2. Find the thermal efficiency of the plant, denoted as \( \eta \).
3. The energy output can be expressed as \( Q_{out} = \frac{1 \hspace{0.1cm} \text{MW}}{\eta}\ \).
4. The energy input required, taking into account the efficiency, is \( Q_{in} = \frac{Q_{out}}{\eta}\ \).
5. Adjust for the total energy input within the given efficiency range to find the mass flow rate:
\( \dot{m}_{fuel} = \frac{Q_{in} \cdot 3600}{\Delta H_{combustion}}\ \).
This completes the calculations to determine how much methane is needed per hour for each MW of power produced, and how the amount changes with varying efficiencies.

Methane combustion

Methane (CH_4) is a common fuel for vapor power plants because it burns cleanly and is highly efficient. In the combustion process, methane reacts with oxygen (O_2) to produce carbon dioxide (CO_2) and water (H_2O), releasing energy.
The complete chemical reaction for combustion with 110% theoretical air is given by:
\( CH_4 + 2.2(O_2 + 3.76N_2) \rightarrow CO_2 + 2H_2O + 8.272N_2 \).
This equation shows that methane combines with oxygen and nitrogen in the air, resulting in carbon dioxide, water, and nitrogen. The energy released during this reaction forms the basis for the power generated in the plant.
The key here is the amount of energy released per kilogram of methane. This energy is known as the heat of combustion, noted as \( \Delta H_{combustion}\ \), typically around \( 50,020 \text{kJ/kg} \). By knowing this value, one can calculate how much methane is required to produce a certain amount of energy.

Thermal efficiency

Thermal efficiency (\( \eta \)) is a measure of how well a vapor power plant converts the heat from the fuel into useful work or electrical power. It is a crucial factor for the overall performance and cost-effectiveness of the power plant.
Thermal efficiency is defined as:
\( \eta = \frac{Q_{out}}{Q_{in}} \times 100 \%\ \).
Here, \( Q_{out}\ \) is the energy output of the plant, and \( Q_{in}\ \) is the energy input from the fuel. Efficiency values for many vapor power plants typically range between 30-40%.
To calculate the impact of thermal efficiency on the necessary mass flow rate of methane, vary the efficiency within the given range and observe how the mass flow rate changes:
\( \dot{m}_{fuel} = \frac{1 \hspace{0.1cm} \text{MW} \cdot 3600}{\eta \cdot 50,020 \text{kJ/kg}}\ \).
By plotting the efficiency against the mass flow rate, you can visualize how more efficient plants require less fuel to produce the same amount of power. This graphical representation helps highlight the importance of improving plant efficiency.