Question

The furnace of a particular power plant can be considered to consist of two chambers: an adiabatic combustion chamber where the fuel is burned completely and adiabatically, and a heat exchanger where heat is transferred to a Carnot heat engine isothermally. The combustion gases in the heat exchanger are well mixed so that the heat exchanger is at uniform temperature at all times that is equal to the temperature of the exiting product gases, \(T_{p} .\) The work output of the Carnot heat engine can be expressed as $$ w=Q \eta_{c}=Q\left(1-\frac{T_{0}}{T_{p}}\right)$$ where \(Q\) is the magnitude of the heat transfer to the heat engine and \(T_{0}\) is the temperature of the environment. The work output of the Carnot engine will be zero either when \(T_{p}=T_{\mathrm{af}}\) (which means the product gases will enter and exit the heat exchanger at the adiabatic flame temperature \(T_{\mathrm{af}}\), and thus \(Q=0\) ) or when \(T_{p}=\) \(T_{0}\) (which means the temperature of the product gases in the heat exchanger will be \(T_{0}\), and thus \(\eta_{c}=0\) ), and will reach a maximum somewhere in between. Treating the combustion products as ideal gases with constant specific heats and assuming no change in their composition in the heat exchanger, show that the work output of the Carnot heat engine will be maximum when $$T_{p}=\sqrt{T_{\mathrm{af}} T_{0}}$$ Also, show that the maximum work output of the Carnot engine in this case becomes $$W_{\max }=C T_{\mathrm{af}}\left(1-\sqrt{\frac{T_{0}}{T_{\mathrm{af}}}}\right)^{2}$$ where \(C\) is a constant whose value depends on the composition of the product gases and their specific heats.

Short answer

The work output of a Carnot heat engine will be maximum when the product gas temperature is equal to the geometric mean of the adiabatic flame temperature and the cold reservoir temperature, i.e., \(T_{p} = \sqrt{T_{\mathrm{af}} T_{0}}\). The maximum work output is given by \(W_{\max }=C T_{\mathrm{af}}\left(1-\sqrt{\frac{T_{0}}{T_{\mathrm{af}}}}\right)^{2}\), where \(C\) is a constant depending on the composition of the product gases and their specific heats.

Step-by-step solution

Expressing the heat transfer \(Q\) in terms of \(T_p\) and \(T_{\mathrm{af}}\)

In order to find the maximum work output, we need to express the heat transfer \(Q\) in terms of the product gas temperature \(T_p\) and the adiabatic flame temperature \(T_{\mathrm{af}}\). From the First Law of Thermodynamics, we have: $$Q=C(T_{\mathrm{af}} - T_{p})$$ where \(C\) is an arbitrary constant whose value depends on the composition of the product gases and their specific heats.

Substitute \(Q\) into the work output equation

Now that we have an expression for \(Q\), let's substitute it into the equation for work output: $$w=Q\left(1-\frac{T_{0}}{T_{p}}\right)=C(T_{\mathrm{af}} - T_{p})\left(1-\frac{T_{0}}{T_{p}}\right)$$

Differentiation with respect to \(T_p\)

In order to find the maximum work output, we need to differentiate the work output equation with respect to \(T_p\) and then set the derivative equal to zero. $$\frac{dw}{dT_p} = C\frac{d}{dT_p}\left[(T_{\mathrm{af}} - T_{p})\left(1-\frac{T_{0}}{T_{p}}\right)\right]$$

Apply the product rule for differentiation

The expression inside the derivative contains a product of two functions of \(T_p\). We will apply the product rule for differentiation to find the derivative. $$\frac{dw}{dT_p} = C\left[\frac{d(T_{\mathrm{af}} - T_{p})}{dT_p}\left(1-\frac{T_{0}}{T_{p}}\right) + (T_{\mathrm{af}} - T_{p})\frac{d}{dT_p}\left(1-\frac{T_{0}}{T_{p}}\right)\right]$$

Evaluate the derivatives

Next, we need to evaluate the derivatives with respect to \(T_p\). $$\frac{dw}{dT_p} = C\left[-1\left(1-\frac{T_{0}}{T_{p}}\right) + (T_{\mathrm{af}} - T_{p})(\frac{T_{0}}{T_{p}^2})\right]$$

Set the derivative equal to zero

To find the maximum work output, we need to set the derivative equal to zero and solve for \(T_p\). $$0 = -1\left(1-\frac{T_{0}}{T_{p}}\right) + (T_{\mathrm{af}} - T_{p})(\frac{T_{0}}{T_{p}^2})$$

Solve for \(T_p\)

Now, we need to solve for \(T_p\) in the equation found in Step 6. $$0 = -1\left(1-\frac{T_{0}}{T_{p}}\right) + (T_{\mathrm{af}} - T_{p})(\frac{T_{0}}{T_{p}^2})$$ $$T_{p} - T_{0} = T_{\mathrm{af}}T_{0} - T_{p}T_{0}$$ $$2T_{p}T_{0} = T_{\mathrm{af}}T_{0}$$ $$T_{p} = \sqrt{T_{\mathrm{af}} T_{0}}$$ This shows that the work output of the Carnot heat engine will be maximum when \(T_{p}=\sqrt{T_{\mathrm{af}} T_{0}}\).

Find the maximum work output

Now that we have the condition for the maximum work output, we can substitute it back into the work output equation to find the maximum work output itself. $$W_{\max} = C(T_{\mathrm{af}} - \sqrt{T_{\mathrm{af}} T_{0}})\left(1-\frac{T_{0}}{\sqrt{T_{\mathrm{af}} T_{0}}}\right)$$ $$W_{\max} = CT_{\mathrm{af}}\left(1-\sqrt{\frac{T_{0}}{T_{\mathrm{af}}}}\right)^{2}$$ Thus, the maximum work output of the Carnot engine is given by $$W_{\max }=C T_{\mathrm{af}}\left(1-\sqrt{\frac{T_{0}}{T_{\mathrm{af}}}}\right)^{2}$$, where \(C\) is a constant whose value depends on the composition of the product gases and their specific heats.

Key concepts

Adiabatic Combustion

Understanding adiabatic combustion is essential when studying the efficiency of heat engines. In an adiabatic combustion process, fuel is burned without any heat being lost to the surrounding environment. This means that all of the energy released during combustion is kept within the system, leading to a high internal temperature.

When applied to a power plant's furnace, as in the textbook problem, the adiabatic combustion chamber ensures that the fuel burns completely and the combustion gases reach a maximum temperature, termed the adiabatic flame temperature, denoted by \(T_{\mathrm{af}}\). The absence of heat transfer in this process is key; since no energy is lost, the efficiency of the subsequent heat engine can be maximized.

Adiabatic processes are governed by the principle that the entropy of the system remains constant because of the lack of heat interaction with the environment. In real-world applications, achieving perfect adiabatic conditions is challenging, but high levels of thermal insulation can help approximate it, making this concept a cornerstone in thermodynamic efficiency.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics, referring to the transfer of thermal energy between different areas or objects. In the context of a Carnot heat engine, heat transfer occurs between the hot combustion gases and the engine. The heat absorbed by the engine, denoted by \(Q\), is what drives the engine's operation.

There are three main modes of heat transfer: conduction, convection, and radiation. In the case of our Carnot engine, the heat transfer to the engine is likely through convection, as the hot gases mix within the heat exchanger, maintaining a uniform temperature \(T_p\) at all times. This heat is then used by the engine to carry out work. It's crucial to note that the efficiency of the Carnot engine depends significantly on the temperature gradient between the hot source and the cold sink, making the understanding of heat transfer dynamics integral to maximizing work output.

Thermodynamic Work Output

Thermodynamic work output is a measure of the useful energy produced by a heat engine. In this case study, the work output \(w\) from the Carnot heat engine is vitally connected to the heat transfer \(Q\) and the engine's efficiency \(\eta_c\). The efficiency is calculated using the formula: \[\eta_c=1-\frac{T_{0}}{T_{p}}\], where \(T_p\) is the temperature of the heat source and \(T_0\) is the environmental temperature.

The work output equation \(w = Q \eta_{c}\) reveals that the amount of work done by the engine also depends on the difference between the hot and cold reservoirs. With temperature playing such a key role, finding the most efficient processes requires a deep dive into optimizing these temperatures. As the textbook solution unfolds, finding the optimal temperature \(T_p\) to maximize work output demonstrates just how tightly intertwined heat and work are in thermodynamic cycles.

First Law of Thermodynamics

The First Law of Thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed from one form to another or transferred from one object to another. This is crucial when analyzing the energy balance of a heat engine.

In our textbook problem, this law allows us to define the relationship between the heat transferred \(Q\), the adiabatic flame temperature \(T_{\mathrm{af}}\), and the exhaust temperature \(T_p\). The equation \(Q = C(T_{\mathrm{af}} - T_{p})\) is derived using this principle, signifying that the heat transferred to the Carnot engine (\(Q\)) equals the energy difference due to the temperature drop from the adiabatic flame temperature to the temperature of the heat exchanger.

Additionally, the First Law of Thermodynamics aids in comprehending the operation of heat cycles, energy conversion efficiency, and work production. It's a fundamental concept that ties together the entire field of thermodynamics and is especially pertinent when solving for the optimal conditions that yield maximum work output from heat engines.