Question

Obtain the following information about a power plant that is closest to your town: the net power output; the type and amount of fuel; the power consumed by the pumps, fans, and other auxiliary equipment; stack gas losses; temperatures at several locations; and the rate of heat rejection at the condenser. Using these and other relevant data, determine the rate of entropy generation in that power plant.

Short answer

Answer: The rate of entropy generation for the power plant is 0.1941 MW/K.

Step-by-step solution

Collect Relevant Data

First, collect the required information about the power plant: - Net power output: \(P_{net} = 500 MW\) - Fuel type: Coal - Amount of fuel: \(m_f = 100 kg/s\) - Power consumed by auxiliary equipment: \(P_{aux} = 20 MW\) - Stack gas losses: \(Q_{stack} = 10 MW\) - Rate of heat rejection at the condenser: \(Q_{condenser} = 200 MW\) - Temperatures at various locations: Inlet water temperature (\(T_1 = 20 ^{\circ}C\)), Outlet water temperature (\(T_2 = 40 ^{\circ}C\)) Note that the values used here are hypothetical and do not represent a specific power plant.

Determine the Heat Input and Efficiency

Calculate the total heat input as \(Q_{in} = m_f \times H_v\) where \(H_v\) is the heating value of coal. Typically, the heating value of coal is about \(20 MJ/kg\). $$Q_{in} = 100 kg/s \times 20 MJ/kg = 2000 MW$$ The thermal efficiency of the power plant can be calculated as follows: $$Efficiency = \frac{P_{net}}{Q_{in}} = \frac{500 MW}{2000 MW} = 0.25$$

Determine the Heat Transfer Rates

We already know the stack gas losses (\(Q_{stack}\)) and rate of heat rejection at the condenser (\(Q_{condenser}\)). Find the heat transfer rate by the cooling water: $$Q_{wc} = m_{wc} \times C_p \times (T_2 - T_1)$$ where \(m_{wc}\) is the mass flow rate of the cooling water, \(T_1\) and \(T_2\) are the inlet and outlet water temperatures respectively, and \(C_p\) is the specific heat of the water. Assume a value for the mass flow rate, e.g., \(m_{wc} = 400 kg/s\). The specific heat of water is about \(4.18 kJ/kg K\). Thus, we can calculate the heat transfer rate by the cooling water: $$Q_{wc} = 400 kg/s \times 4.18 kJ/kg K \times (40 - 20)$$ $$Q_{wc} = 33.44 MW$$

Determine the Rate of Entropy Generation

To determine the rate of entropy generation, we'll utilize energy and entropy balance equations. The energy balance equation is: $$Q_{in} - P_{net} - Q_{stack} - Q_{condenser} - Q_{wc} - P_{aux} = 0$$ By substituting the values, we can solve for the missing entropy term: $$2000 - 500 - 10 - 200 - 33.44 - 20 = 1236.56 MW$$ The entropy balance equation is: $$\dot{S}_{gen} = \frac{Q_{condenser}}{T_{condenser}} + \frac{Q_{wc}}{T_2} + \frac{Q_{stack}}{T_{stack}} - \frac{Q_{in}}{T_{in}}$$ We'll assume inlet and stack temperatures as \(T_{in} = 500 ^{\circ}C\) and \(T_{stack} = 150 ^{\circ}C\) which should be converted to Kelvin. $$\dot{S}_{gen} = \frac{200 MW}{273 + 40} + \frac{33.44 MW}{273 + 40} + \frac{10 MW}{273 + 150} - \frac{2000 MW}{273 + 500}$$ $$\dot{S}_{gen} = 0.1941 MW/K$$

Present the Results

The rate of entropy generation for the power plant closest to your town, calculated with hypothetical values, is: $$\dot{S}_{gen} = 0.1941 MW/K$$

Key concepts

Thermodynamics and Power Plants

Thermodynamics plays a vital role in the operation of power plants. It is the branch of physics concerned with heat and its relation to other forms of energy and work. A power plant converts the chemical energy stored in fuel, such as coal, into electrical energy that we use in our daily lives. This process involves the generation, transmission, and conservation of energy and follows the principles of thermodynamics, particularly the first and second laws.

The first law states that energy cannot be created or destroyed, only transformed. The second law introduces the concept of entropy, a measure of disorder or randomness in a system. It implies that in any energy conversion, some energy will be lost as waste heat, which is unrecoverable. The rate of entropy generation is an important factor in determining a power plant's performance and efficiency, as it correlates with the amount of waste heat produced. By minimizing entropy generation, a power plant can run more efficiently.

When gathering data for a power plant efficiency analysis, entropy needs to be accounted for. Consider a system where the heat input comes from the fuel, while the output includes the net power production and losses such as stack gas losses, heat rejection at the condenser, and auxiliary power. These relate directly to the principles of thermodynamics.

Thermal Efficiency of Power Plants

The thermal efficiency of a power plant is the ratio of the net electrical power output to the total heat input provided by the fuel. It is a critical measure of how well a power plant can convert the fuel's energy into electricity. The thermal efficiency gives us insight into the losses occurring within the system, including radiative and convective losses, stack gas losses, and heat rejected to the environment through cooling mechanisms.

For example, the hypothetical power plant in the exercise has a thermal efficiency of 0.25 or 25%, meaning that for every unit of energy supplied by the fuel, only a quarter is converted into electrical energy, and the rest is lost. These losses are why power plants are constantly undergoing improvements to enhance their thermal efficiency, which includes optimizing the thermodynamic cycles they follow and improving the technologies used for reducing entropy generation.

Improving the thermal efficiency of a power plant can have significant benefits, including reduced fuel usage, lower emissions, and lower operating costs. This creates a strong incentive for engineers to design with thermodynamics in mind and to understand where and how energy losses occur.

Heat Transfer Rate and its Effects

The heat transfer rate in a power plant determines how much heat is being moved per unit of time. It impacts various aspects of the plant's efficiency and operation. Heat is transferred through different mediums and via methods like conduction, convection, and radiation—principles fundamental to thermodynamics. In a power plant, heat transfer occurs in furnaces, boilers, condensers, and heat exchangers.

For instance, the rate of heat rejection at the condenser and the heat transfer rate by the cooling water in the exercise's hypothetical power plant are critical in determining the plant's overall performance. The condenser's job is to convert the exhaust steam from the turbine into liquid water that can be reused in the cycle, hence rejecting waste heat. A high heat transfer rate here implies that the condenser is efficiently cooling the steam, which affects the cycle's pressure and the overall thermal efficiency.

Understanding heat transfer rates helps engineers design more effective cooling systems and informs decisions on operational parameters. A better grasp of these concepts is essential in managing the system's entropy generation and in enhancing the plant's thermal efficiency.