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Ch. 20 - Second Law of Thermodynamics
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 20, Problem 71a

At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750°C and 440°C, and of the second 415°C and 240°C. If the heat of combustion of coal is 2.8 x 10⁷ J/kg, at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the ideal (Carnot) efficiency.

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Step 1: Understand the Carnot efficiency formula. The Carnot efficiency (η_c) is given by the equation: η_c = 1 - (T_c / T_h), where T_h is the high operating temperature and T_c is the low operating temperature. Ensure the temperatures are converted to Kelvin by adding 273.15 to the Celsius values.
Step 2: Calculate the Carnot efficiency for each engine. For the first engine, use T_h = 750°C and T_c = 440°C. For the second engine, use T_h = 415°C and T_c = 240°C. Convert these temperatures to Kelvin and substitute them into the Carnot efficiency formula.
Step 3: Adjust the Carnot efficiency to account for the real-world efficiency of the engines. Since the engines operate at 65% of the ideal Carnot efficiency, multiply the Carnot efficiency of each engine by 0.65 to find the actual efficiency of each engine.
Step 4: Determine the total efficiency of the system. The total efficiency is the product of the efficiencies of the two engines, as the heat output of the first engine becomes the heat input of the second engine.
Step 5: Calculate the rate of coal consumption. Use the formula: Power output = (Efficiency) × (Rate of heat input). Rearrange to find the rate of heat input required. Then, use the heat of combustion of coal (2.8 × 10⁷ J/kg) to find the mass of coal burned per second. Finally, ensure the power output is 950 MW (950 × 10⁶ W) in your calculations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Carnot Efficiency

Carnot efficiency is the maximum possible efficiency of a heat engine operating between two temperature reservoirs. It is defined by the formula η_c = 1 - (T_c / T_h), where T_c is the absolute temperature of the cold reservoir and T_h is the absolute temperature of the hot reservoir. This concept is crucial for understanding the theoretical limits of engine performance and serves as a benchmark for real engines, which can only achieve a fraction of this efficiency.
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Thermal Efficiency

Thermal efficiency measures how well an engine converts heat energy into work. It is calculated as the ratio of the work output to the heat input. In this context, the steam engines operate at 65% of the Carnot efficiency, meaning they convert 65% of the ideal maximum efficiency into useful work, which is essential for determining the actual power output of the engines and the required fuel consumption.
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Power Output and Fuel Consumption

Power output refers to the rate at which work is done or energy is transferred, measured in watts (W). In this scenario, the plant aims to produce 950 MW of power. To achieve this, the rate of coal combustion must be calculated based on the energy content of coal and the thermal efficiency of the engines. Understanding the relationship between power output, energy input, and fuel consumption is vital for determining how much coal needs to be burned to meet the desired power output.
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Related Practice
Textbook Question

Refrigeration units can be rated in “tons.” A 1-ton air conditioning system can remove sufficient energy to freeze 1 ton (2000 pounds = 909 kg) of 0°C water into 0°C ice in one 24-h day. Assume the hot part of a day averages 35°C and the interior of a house is maintained at 22°C by the continuous operation of a 6-ton air conditioning system for 6 hours a day. How much does this cooling cost the homeowner per day, and per month?Assume the work done by the refrigeration unit is powered by electricity that costs \$0.13 per kWh and that the unit’s coefficient of performance is only 18% of an ideal refrigerator. 1 kWh = 3.60 x 10⁶ J .

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Textbook Question

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Textbook Question

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Textbook Question

Suppose a heat pump has a stationary bicycle attachment that allows you to provide the work instead of using an electrical wall outlet. If your heat pump has a coefficient of performance of 2.0 and you can cycle at a racing pace output of about 200 W for a half hour, how much heat can you provide?

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