Problem 20
Question
A heat engine is to be built to extract energy from the tem- perature gradient in the ocean. If the surface and deepwater temperatures are \(25^{\circ} \mathrm{C}\) and \(8^{\circ} \mathrm{C}\) , respectively, what is the maxi- mum efficiency such an engine can have?
Step-by-Step Solution
Verified Answer
The maximum efficiency is 5.7%.
1Step 1: Convert Temperatures to Kelvin
To calculate efficiency, we need the temperatures in Kelvin. Use the formula: \( K = {}^{ ext{∘C}} + 273.15 \). Surface temperature: \( 25^{ ext{∘C}} + 273.15 = 298.15 \, \text{K} \) Deepwater temperature: \( 8^{ ext{∘C}} + 273.15 = 281.15 \, \text{K} \)
2Step 2: Apply the Carnot Efficiency Formula
The theoretical maximum efficiency of a heat engine operating between two temperatures is given by the Carnot efficiency formula: \[ \eta = 1 - \frac{T_{ ext{cold}}}{T_{ ext{hot}}} \] Plug in the values: \( T_{ ext{cold}} = 281.15 \, \text{K}, \; T_{ ext{hot}} = 298.15 \, \text{K} \).
3Step 3: Calculate the Maximum Efficiency
Substitute the values into the formula: \[ \eta = 1 - \frac{281.15}{298.15} \approx 1 - 0.943 \] \[ \eta \approx 0.057 \text{ or } 5.7 ext{%} \]
4Step 4: Interpret the Result
The maximum efficiency such an engine can achieve, under ideal conditions, is 5.7%. This indicates that only 5.7% of the energy extracted from the temperature difference can be converted into usable work.
Key Concepts
Carnot EfficiencyHeat EngineTemperature ConversionOcean Temperature Gradient
Carnot Efficiency
Carnot Efficiency is a fundamental concept in thermodynamics that establishes the limit on the efficiency any heat engine can achieve. It was introduced by Sadi Carnot, who demonstrated that no machine can be more efficient than a perfect Carnot engine operating between two temperatures. The efficiency is calculated using the formula:
This theoretical maximum highlights that the closer we get to efficiency, the less waste energy we have. However, real-world engines never reach this ideal due to practical inefficiencies and energy losses.
- \( \eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \)
This theoretical maximum highlights that the closer we get to efficiency, the less waste energy we have. However, real-world engines never reach this ideal due to practical inefficiencies and energy losses.
Heat Engine
A Heat Engine is a device that transforms heat energy into mechanical work. It operates on a cycle and typically involves three main steps: absorbing heat from a hot reservoir, converting part of the heat into work, and then expelling the remaining heat to a cooler reservoir.
The efficiency of a heat engine depends largely on the temperature difference between the hot and cold reservoirs. The larger this difference, the more work the engine can perform. For practical cases, the efficiency is always lower than the Carnot efficiency, due to various losses such as friction and heat dissipation.
Two common examples of heat engines are steam engines and internal combustion engines. Understanding their operation helps us appreciate the role of thermodynamics in powering machinery and technology.
The efficiency of a heat engine depends largely on the temperature difference between the hot and cold reservoirs. The larger this difference, the more work the engine can perform. For practical cases, the efficiency is always lower than the Carnot efficiency, due to various losses such as friction and heat dissipation.
Two common examples of heat engines are steam engines and internal combustion engines. Understanding their operation helps us appreciate the role of thermodynamics in powering machinery and technology.
Temperature Conversion
Temperature Conversion is crucial for accurate calculations in thermodynamics. The Kelvin scale is the preferred unit because it is an absolute scale starting from absolute zero, making it ideal for thermodynamic equations.
To convert Celsius to Kelvin, the formula used is:
For example, if we have our temperatures at the ocean’s surface and at depth as \(25^{\circ} \text{C}\) and \(8^{\circ} \text{C}\) respectively, converting to Kelvin becomes \(298.15 \text{K}\) and \(281.15 \text{K}\). This transformation is essential for applying the Carnot efficiency formula.
To convert Celsius to Kelvin, the formula used is:
- \( K = \text{°C} + 273.15 \)
For example, if we have our temperatures at the ocean’s surface and at depth as \(25^{\circ} \text{C}\) and \(8^{\circ} \text{C}\) respectively, converting to Kelvin becomes \(298.15 \text{K}\) and \(281.15 \text{K}\). This transformation is essential for applying the Carnot efficiency formula.
Ocean Temperature Gradient
The Ocean Temperature Gradient refers to the difference in temperature between the surface water and the deeper water in the ocean. This gradient is a potential energy source that can be harnessed by heat engines specifically designed to operate within this thermal difference.
In our example, the surface temperature is \(25^{\circ} \text{C}\) and the deeper, colder water is \(8^{\circ} \text{C}\), creating a gradient. This difference, albeit small, can drive a heat engine to convert some of the heat energy into usable work.
Though the efficiency might seem low – as calculated, only about 5.7% – harnessing the vast and renewable temperature gradient of the ocean could contribute significantly to sustainable energy solutions. Such systems are particularly intriguing for their potential in generating electricity in island and coastal regions.
In our example, the surface temperature is \(25^{\circ} \text{C}\) and the deeper, colder water is \(8^{\circ} \text{C}\), creating a gradient. This difference, albeit small, can drive a heat engine to convert some of the heat energy into usable work.
Though the efficiency might seem low – as calculated, only about 5.7% – harnessing the vast and renewable temperature gradient of the ocean could contribute significantly to sustainable energy solutions. Such systems are particularly intriguing for their potential in generating electricity in island and coastal regions.
Other exercises in this chapter
Problem 18
\(\bullet\) A cooling unit for chilling the water of an aquarium gives specifications of 1\(/ 10\) hp and 1270 \(\mathrm{Btu} / \mathrm{h}\) . Assuming the unit
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A Carnot engine is operated between two heat reservoirs at temperatures of 520 \(\mathrm{K}\) and 300 \(\mathrm{K}\) . (a) If the engine receives 6.45 \(\mathrm
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A Carnot engine has an efficiency of 59\(\%\) and performs \(2.5 \times 10^{4} \mathrm{J}\) of work in each cycle. (a) How much heat does the engine extract fro
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