Problem 43 - University Physics with Modern Physics

Question

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\(^\circ\)C and 6\(^\circ\)C, respectively. (a) What is the maximum theoretical efficiency of this power plant? (b) If the power plant is to produce 210 \(kW\) 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\(^\circ\)C. What must be the flow rate of cold water through the system? Give your answer in kg/h and in \(L\)/h.

Step-by-Step Solution

Answer

The maximum theoretical efficiency is 7%. Heat extracted: 3000 kW; absorbed: 2790 kW. Cold water flow rate: 599,400 kg/h or L/h.

Step 1: Understanding the task

We need to calculate the maximum theoretical efficiency of a heat engine working between two temperatures, find the heat exchange rates when producing a certain power, and determine the flow rate of cold water involved in the process.

Step 2: Calculating Maximum Theoretical Efficiency

The maximum theoretical efficiency of a heat engine operating between two reservoirs is given by \[ \eta_{\text{max}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \]where \(T_{\text{hot}}\) and \(T_{\text{cold}}\) are the absolute temperatures in Kelvin. Convert the given temperatures from Celsius to Kelvin:\[ T_{\text{hot}} = 27^\circ C + 273.15 = 300.15 \, K \]\[ T_{\text{cold}} = 6^\circ C + 273.15 = 279.15 \, K \]Substitute into the efficiency equation:\[ \eta_{\text{max}} = 1 - \frac{279.15}{300.15} \approx 0.070 \text{ or } 7\% \]

Step 3: Finding Heat Extraction Rate from Warm Water

The efficiency is also defined as the ratio of the output power \(P\) to the input heat energy \(Q_{\text{in}}\):\[ \eta = \frac{P}{Q_{\text{in}}} \]Rearranging to find \(Q_{\text{in}}\):\[ Q_{\text{in}} = \frac{P}{\eta} \]Substitute \(P = 210 \, kW\) and \(\eta = 0.070\):\[ Q_{\text{in}} = \frac{210}{0.070} \approx 3000 \, kW \]

Step 4: Calculating Heat Absorption Rate by Cold Water

The rate of heat absorption by the cold water is given by:\[ Q_{\text{out}} = Q_{\text{in}} - P \]Substitute the values:\[ Q_{\text{out}} = 3000 \, kW - 210 \, kW = 2790 \, kW \]

Step 5: Determining Flow Rate of Cold Water

The flow rate can be found using:\[ Q = mc\Delta T \]where \( m \) is the mass flow rate, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change. For water, \( c = 4.186 \, \text{kJ/kg°C} \), \( \Delta T = 10^\circ C - 6^\circ C = 4^\circ C \).Convert \( Q_{\text{out}} \) to \( \text{kJ/s} \):\[ 2790 \, kW = 2790 \, \text{kJ/s} \]Rearrange the equation to solve for \( m \):\[ m = \frac{Q}{c\Delta T} = \frac{2790}{4.186 \times 4} \approx 166.5 \, \text{kg/s} \]Convert to \( \,\text{kg/h})\):\[ 166.5 \, \text{kg/s} \times 3600 \, \text{s/h} = 599,400 \, \text{kg/h} \]Convert mass flow to volume flow (using water density \( \approx 1 \, \text{kg/L} \)):599,400 \, \text{L/h}.

Key Concepts

Heat Engine Efficiency

Heat engine efficiency is a crucial concept in understanding how well a power plant or any thermal system converts heat energy into useful work. It is essentially the ratio of useful output energy (work) to the total energy input from the heat source. In our exercise, an experimental power plant is using the thermal gradient of ocean water to generate electricity. The efficiency of such a heat engine is determined by the temperatures of the heat source (warm water) and the heat sink (cold water).

The maximum theoretical efficiency, known as the Carnot efficiency, can be calculated using the formula:

• \( \eta_{\text{max}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \)

The temperatures need to be in Kelvin, an absolute scale, so conversion from Celsius is necessary. In this exercise, the warm surface water is at 27°C, and the deep cold water is at 6°C. Converting these to Kelvin gives us 300.15 K and 279.15 K, respectively. Substituting these into the formula results in a Carnot efficiency of about 7%.

• This theoretical efficiency is the upper limit possible for any heat engine operating between these two temperatures.

Heat Transfer

Heat transfer is the process of energy flowing from a higher temperature to a lower temperature. In the context of our ocean thermal energy plant, heat transfer plays a vital role in the energy conversion mechanism. The plant extracts heat from the warm surface water, which it then converts to electrical energy before transferring the remaining energy to the colder deep water.

To calculate the rate at which heat is extracted from the warm water (\( Q_{\text{in}} \)), we use the relationship between power output and efficiency:

• \( Q_{\text{in}} = \frac{P}{\eta} \)

Where \( P \) is the desired electrical output of 210 kW and \( \eta \) is 0.070 (7%). This results in an extraction rate of approximately 3000 kW from the warm water. The heat absorbed by the cold water (\( Q_{\text{out}} \)) is then calculated as:

• \( Q_{\text{out}} = Q_{\text{in}} - P \)

Which equals 2790 kW in this particular setup, indicating how much heat is transferred to the cold sink.

Energy Conversion

Energy conversion in a heat engine involves transforming thermal energy into mechanical work or electricity, which is what our ocean thermal plant aims to achieve. This conversion process is not 100% efficient due to the second law of thermodynamics, which states that some energy will always be lost to the surroundings.

The plant converts part of the 3000 kW input heat energy into 210 kW of useful electrical power, a process that highlights how the plant's efficiency dictates the portion of input energy successfully transformed into work.

• The remaining portion, which is the difference between the input heat energy and the work done, is transferred to the cold water as waste heat.
• This underscores the importance of maximizing efficiency to gain more work from the same amount of energy input.

Ocean Thermal Energy

Ocean thermal energy takes advantage of the temperature differences in ocean water at varying depths to produce power. The concept relies on the thermodynamic principles that permit heat engines to operate between a hot and a cold reservoir.

In our example from the power plant in Hawaii, the warm surface waters serve as the hot reservoir, while the cooler deep ocean waters act as the cold reservoir. The temperature gradient between these depths is utilized to generate energy. The exercise shows that to maintain efficiency and ensure the sufficiency of the energy conversion process, specific flow rates of cold water are necessary.

Using the formula \( Q = mc\Delta T \), the mass flow rate \( m \) can be calculated. From the problem, \( Q_{\text{out}} = 2790\, \text{kW} \) or \( 2790 \text{kJ/s} \), the specific heat \( c = 4.186 \text{kJ/kg°C} \), and temperature change is 4°C. This yields a mass flow rate of 166.5 kg/s. Converted to hourly rates, this equates to 599,400 kg/h, which is also 599,400 L/h.

• This flow is essential to maintain the cooling process and ensure the continuous operation of the power plant in harvesting ocean thermal energy.