Problem 45 - University Physics with Modern Physics

Question

20.45. An experimental power plant at the Natural Energy Laboratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water temperatures are \(27^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C}\) , respectively. (a) What is the maximum theoretical effciency of this power plant? (b) If the power plant is to produce 210 \(\mathrm{kW}\) of power, at what rate must heat be extracted from the warm water? At what rate must heat be absorbed by the cold water? Assume the maximum theoretical efficiency. (c) The cold water that enters the plant leaves it at a temperature of \(10^{\circ} \mathrm{C}\) . What must be the flow rate of cold water through the system? Give your answer in \(\mathrm{kg} / \mathrm{h}\) and \(\mathrm{L} / \mathrm{h}\) .

Step-by-Step Solution

Answer

(a) 7.0%. (b) 3000 kW, 2790 kW. (c) 601,860 kg/h or 601,860 L/h.

Step 1: Calculate the Maximum Theoretical Efficiency

To find the maximum theoretical efficiency of the power plant, which functions as a heat engine, we use the Carnot efficiency formula: \[ \eta = 1 - \frac{T_C}{T_H} \]where \(T_C\) is the absolute temperature of the cold reservoir, and \(T_H\) is the absolute temperature of the hot reservoir. The temperatures must be converted to Kelvin:\[ T_H = 27^{\circ} \mathrm{C} + 273.15 = 300.15\, \mathrm{K} \]\[ T_C = 6^{\circ} \mathrm{C} + 273.15 = 279.15\, \mathrm{K} \]Substitute these temperatures into the efficiency formula:\[ \eta = 1 - \frac{279.15}{300.15} \approx 0.070 \text{ or } 7.0\% \]

Step 2: Calculate Heat Extracted from the Warm Water

Now we determine the rate at which heat must be extracted from the warm water. Given the plant generates 210 kW of power, we use the efficiency formula:\[ W = \eta Q_H \]where \(W\) is the power output, \(\eta\) is the efficiency, and \(Q_H\) is the rate of heat extracted from the warm water:\[ 210 = 0.070 Q_H \]Solving for \(Q_H\):\[ Q_H = \frac{210}{0.070} = 3000\, \mathrm{kW} \]

Step 3: Calculate Heat Absorbed by the Cold Water

With the heat extracted from the warm water known, we determine the heat absorbed by the cold water using the energy conservation principle:\[ Q_C = Q_H - W \]Substitute the known values:\[ Q_C = 3000\, \mathrm{kW} - 210\, \mathrm{kW} = 2790\, \mathrm{kW} \]

Step 4: Calculate Flow Rate of Cold Water

To find the flow rate of cold water, we apply the equation:\[ Q_C = m \cdot c \cdot \Delta T \]where \(m\) is the mass flow rate in kg/s, \(c\) is the specific heat capacity of water (approximately 4.18 kJ/kg·°C), and \(\Delta T\) is the temperature change:\[ \Delta T = 10 - 6 = 4\, ^{\circ}C \]Rearranging the formula and solving for \(m\):\[ m = \frac{Q_C}{c \cdot \Delta T} \]Substitute the known values:\[ m = \frac{2790\, \mathrm{kJ/s}}{4.18\, \mathrm{kJ/kg\cdot°C} \cdot 4\, \mathrm{°C}} = \frac{2790}{16.72} = 166.85\, \mathrm{kg/s} \]Convert to kg/h:\[ 166.85\, \mathrm{kg/s} \times 3600 = 601,860\, \mathrm{kg/h} \]To convert the mass flow to a volume flow rate (in L/h) for water, use the fact that 1 kg of water is approximately 1 L:\[ 601,860\, \mathrm{kg/h} \approx 601,860\, \mathrm{L/h} \]

Key Concepts

Carnot Efficiency

The concept of Carnot efficiency is central to understanding the theoretical limits of heat engines. It represents the maximum efficiency that a heat engine can achieve between two reservoirs of different temperatures and is determined by the laws of thermodynamics. Think of it as the ultimate benchmark – no real engine can exceed this efficiency.The Carnot efficiency, \( \eta \), is calculated with the equation:\[ \eta = 1 - \frac{T_C}{T_H} \]where:

• \( T_H \) is the temperature of the hot reservoir (in Kelvin)
• \( T_C \) is the temperature of the cold reservoir (in Kelvin)

In practical applications like the Hawaii power plant example, these temperatures illustrate the ocean's temperature gradient. The surface warm water is hotter than the deep cold water, providing a natural opportunity for energy conversion. In general, a larger temperature difference results in higher Carnot efficiency. However, in this scenario, the temperatures are close, resulting in a theoretical efficiency of just 7%. This highlights the challenge of extracting significant energy from small temperature gradients.

Heat Transfer

Heat transfer is the movement of heat or thermal energy from one object or material to another. In the context of a heat engine, it involves extracting heat from a high-temperature source (hot reservoir) and releasing it into a low-temperature sink (cold reservoir). This flow of energy is what drives the engine. In the Hawaii plant, the warm surface water provides thermal energy, which is transferred to perform work or be converted into electricity. According to the problem: 1. **Heat Extraction**: The plant extracts 3,000 kW of heat from the warm water. 2. **Power Output**: It generates 210 kW of usable power. 3. **Heat Discharge**: The remaining heat (2,790 kW) is discharged into the cold water. This sequence of steps is guided by the principle of energy conservation: energy entering the system must either be converted into useful work or released as waste heat. This basic cycle underlies all heat engines and is essential to understanding their operation and limitations.

Ocean Thermal Energy Conversion

Ocean Thermal Energy Conversion (OTEC) is an innovative method to generate electricity using the temperature difference between warm surface water and cold deep ocean water. As seen in the Hawaii power plant example, OTEC harnesses this natural temperature gradient to operate a heat engine. The process of OTEC involves:

• **Warm Water Intake**: Surface water at a higher temperature is used as the heat source.
• **Cold Water Intake**: Deep ocean water, much colder, acts as the heat sink.
• **Energy Conversion**: The temperature difference is exploited to drive a conventional heat engine, such as a Rankine cycle, generating electricity.
• **Cold Water Discharge**: The cold water exits the system slightly warmer than it entered, influencing calculations for flow rates.

One practical challenge for OTEC is the relatively small temperature differences, which, as discussed, constrain the maximum efficiency. Engineers must optimize the system to maximize energy extraction, considering flow rates and thermal properties. Despite the limitations, OTEC offers a renewable energy option that taps into the vast thermal resources of the world's oceans.