Question

Acetylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) is burned completely during a steady-flow combustion process. The fuel and the air enter the combustion chamber at \(25^{\circ} \mathrm{C},\) and the products leave at \(1500 \mathrm{K} .\) If the enthalpy of the products relative to the standard reference state is \(-404 \mathrm{MJ} / \mathrm{kmol}\) of fuel, the heat transfer from the combustion chamber is \((a) 177 \mathrm{MJ} / \mathrm{kmol}\) (b) \(227 \mathrm{MJ} / \mathrm{kmol}\) \((c) 404 \mathrm{MJ} / \mathrm{kmol}\) \((d) 631 \mathrm{MJ} / \mathrm{kmol}\) \((e) 751 \mathrm{MJ} / \mathrm{kmol}\)

Short answer

Answer: (c) 404 MJ/kmol.

Step-by-step solution

Write down the balanced chemical reaction

First, we need to write down the balanced chemical equation for the complete combustion of acetylene gas. The general equation for acetylene combustion is: $$\mathrm{C}_{2} \mathrm{H}_{2} + \frac{5}{2} \mathrm{O}_{2} \to 2 \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}$$

Apply the first law of thermodynamics for steady flow process

Since the combustion process is steady-flow, we can use the first law of thermodynamics for steady-flow process: $$\dot{Q} - \dot{W} = \dot{m}\left(h_{out} - h_{in}\right)$$ where \(\dot{Q}\) is the heat transfer rate, \(\dot{W}\) is the work done by the system, and \(\dot{m}\) is the mass flow rate. For combustion processes, we assume that there is no work done by the system (\(\dot{W} = 0\)): $$\dot{Q} = \dot{m}\left(h_{out} - h_{in}\right)$$

Calculate the enthalpy change during the process

Given the enthalpy of the products relative to the standard reference state is \(-404\,\text{MJ/kmol}\) of fuel, we can determine the enthalpy change of the process. Since the reactants are at \(25^\circ\text{C}\), we can assume that their enthalpy is equal to the reference enthalpy (i.e., \(0\,\text{MJ/kmol}\)): $$h_{in} = 0\,\text{MJ/kmol}$$ $$h_{out} = -404\,\text{MJ/kmol}$$ $$\Delta h = h_{out} - h_{in} = -404\,\text{MJ/kmol}$$

Calculate the heat transfer per kmol of fuel

Now, we can use the enthalpy change calculated in Step 3 to find the heat transfer per kmol of fuel: $$Q = m\left(h_{out} - h_{in}\right)$$ $$Q = m\left(-404\,\text{MJ/kmol}\right)$$ Let \(m=1\,\text{kmol}\), then we can find the heat transfer using the relationship: $$Q = 1 \times \left(-404\,\text{MJ/kmol}\right) = -404\,\text{MJ}$$ The negative sign indicates that heat is transferred from the combustion chamber, so the heat transfer is: $$Q = 404\,\text{MJ}$$ Based on the given options, the correct option is: \((c) 404\,\text{MJ/kmol}\).

Key concepts

Steady-Flow Combustion

In thermodynamics, steady-flow combustion refers to the process where a continuous supply of fuel and oxidizer is burned, and the resulting products are released at a constant rate. Unlike batch processes, the properties such as pressure, temperature, and flow rates in a steady-flow system remain constant over time.

A common example of a steady-flow combustion process is the operation of an internal combustion engine in a car. Fuel and air enter the engine, combust to release energy, and exhaust gases exit the system continuously. This process is key to keeping engines running smoothly without large fluctuations in power output or efficiency.

A crucial aspect of steady-flow combustion analysis is applying the first law of thermodynamics. This law, which is a principle of energy conservation, allows us to understand how energy enters and exits the system, and how it's converted between different forms, such as thermal energy produced by combustion into mechanical work or output heat.

First Law of Thermodynamics

The first law of thermodynamics is essentially the law of energy conservation, adapted for thermodynamic systems. It states that the energy within a closed system must remain constant over time. Energy within the system can neither be created nor destroyed, only transformed or transferred from one form to another.

Mathematically, the first law for a steady-flow process can be written as: \[\dot{Q} - \dot{W} = \dot{m}(h_{out} - h_{in})\], where \(\dot{Q}\) represents the rate of heat transfer, \(\dot{W}\) is the work done by the system, and \(\dot{m}\) is the mass flow rate. \(h_{out}\) and \(h_{in}\) denote the enthalpy per unit mass of the outflow and inflow, respectively.

In the context of our exercise, since we’re dealing with a combustion chamber with no external work, the work transfer \(\dot{W}\) term can be disregarded. This simplifies the analysis and directly ties the rate of heat transfer to the enthalpy change of the system.

Enthalpy Change

Enthalpy change, represented by \(\Delta h\), is a measure of the total energy change within a system as it undergoes a chemical or physical process, at constant pressure. It accounts for both the internal energy of the system and the work done by the system as it expands or contracts.

For a steady-flow combustion process, the enthalpy change is given by the difference in enthalpy between the outlet and inlet streams: \[\Delta h = h_{out} - h_{in}\]. Returning to our exercise, we consider the enthalpy of the products (\[h_{out} = -404\,\text{MJ/kmol}\]) and we assume that the enthalpy of the reactants at the reference state (\[h_{in} = 0\,\text{MJ/kmol}\]). Therefore, the enthalpy change per kilomole of fuel is \[-404\,\text{MJ/kmol}\].

Understanding this concept is vital as it determines how much energy is released or absorbed during a process, which is a fundamental aspect of thermal systems and energy conversion devices.

Heat Transfer Thermodynamics

Heat transfer in thermodynamics refers to the movement of thermal energy from one place to another driven by temperature differences. It's a cornerstone concept that explains how energy in the form of heat moves between systems and their surroundings.

There are three main modes of heat transfer: conduction, convection, and radiation. In our combustion exercise, we primarily deal with heat released through convection and radiation from the hot combustion products to the surroundings.

The first law of thermodynamics for a steady-flow process simplifies the heat transfer analysis by equating the heat transfer rate to the enthalpy change. By multiplying the mass flow rate by the enthalpy change, we can quantify the total heat transfer per unit of time or fuel mass. In the exercise, we calculate the total heat transfer per kilomole of fuel as \[404\,\text{MJ/kmol}\], indicating that 404 MJ of energy is removed from the combustion chamber via heat transfer for every kilomole of acetylene burned.

Understanding these principles helps us design more efficient energy systems, such as engines or power plants, by managing how heat is utilized and dissipated.