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An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately \(300 \mathrm{~K}\), while the temperature at reasonable depths is approximately \(280 \mathrm{~K}\). The warmer water is used as a heat source to evaporate the working fluid, while the colder water is used as a heat sink for condensation of the fluid. Consider a power plant that is to generate \(2 \mathrm{MW}\) of electricity at an efficiency (electric power output per heat input) of \(3 \%\). The evaporator is a heat exchanger consisting of a single shell with many tubes executing two passes. If the working fluid is evaporated at its phase change temperature of \(290 \mathrm{~K}\), with ocean water entering at \(300 \mathrm{~K}\) and leaving at \(292 \mathrm{~K}\), what is the heat exchanger area required for the evaporator? What flow rate must be maintained for the water passing through the evaporator? The overall heat transfer coefficient may be approximated as \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Step by step solution

01

Calculate the heat input

Given that the power plant generates 2 MW of electricity at an efficiency of 3%, we can calculate the heat input (Q_in) as follows: Efficiency = (Electric Power Output) / (Heat Input) => Heat Input = (Electric Power Output) / Efficiency \[Q_{in} = \frac{2 \times 10^6 \mathrm{~W}}{0.03} = 6.67 \times 10^7 \mathrm{~W}\]

02

Calculate the heat transfer rate in the evaporator

The heat transfer rate (Q_evaporator) can be calculated as the difference between the heat input and the electric power output: \[Q_{evaporator} = Q_{in} - P_{output} \] \[Q_{evaporator} = 6.67 \times 10^7 \mathrm{~W} - 2 \times 10^6 \mathrm{~W} = 6.47 \times 10^7 \mathrm{~W}\]

03

Calculate the required heat exchanger area

We are given the overall heat transfer coefficient (U) as 1200 W/m虏路K. To find the heat exchanger area (A), we can use the following equation: \[Q_{evaporator} = U \cdot A \cdot \Delta T_{lm}\] where \(\Delta T_{lm}\) is the log mean temperature difference. We can calculate \(\Delta T_{lm}\) as follows: \[\Delta T_{lm} = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\frac{\Delta T_1}{\Delta T_2})}\] We know that ocean water enters at 300 K and leaves at 292 K, and the working fluid is evaporated at 290 K. So, we have: \[\Delta T_1 = T_{in} - T_{evaporation} = 300 K - 290 K = 10 K\] \[\Delta T_2 = T_{out} - T_{evaporation} = 292 K - 290 K = 2 K\] Now we can calculate \(\Delta T_{lm}\): \[\Delta T_{lm} = \frac{(10 K - 2 K)}{\ln(\frac{10 K}{2 K})} \approx 6.37 K\] Then we can find the heat exchanger area (A): \[A = \frac{Q_{evaporator}}{U \cdot \Delta T_{lm}}\] \[A = \frac{6.47 \times 10^7 \mathrm{~W}}{1200 \mathrm{~W/m^2 \cdot K} \cdot 6.37 \mathrm{~K}} \approx 8516 \mathrm{~m^2}\]

04

Calculate the flow rate for the water

The flow rate (峁�) can be calculated using the specific heat capacity (c_p) of the water and the temperature difference. \[Q_{evaporator} = \dot{m} \cdot c_p \cdot \Delta T\] We know that the specific heat capacity of water is approximately 4.18 kJ/kg路K. Therefore, we have: \[6.47 \times 10^7 \mathrm{~W} = \dot{m} \cdot 4.18 \times 10^3 \mathrm{~J/kg \cdot K} \cdot (300 K - 292 K)\] Now we can solve for the flow rate (峁�): \[\dot{m} = \frac{6.47 \times 10^7 \mathrm{~W}}{4.18 \times 10^3 \mathrm{~J/kg \cdot K} \cdot 8 K} \approx 1945 \mathrm{~kg/s}\] So, the heat exchanger area required for the evaporator is approximately 8516 m虏 and the flow rate for the water passing through the evaporator is approximately 1945 kg/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ocean Thermal Energy Conversion (OTEC)

Ocean Thermal Energy Conversion (OTEC) is a fascinating process that uses the natural temperature difference between warm surface seawater and cold deep seawater to generate electricity. This method is particularly effective in tropical regions where the surface ocean temperatures are consistently warmer than deeper water levels. In an OTEC system, warm surface water is used to heat and vaporize a working fluid, which then drives a turbine to produce electricity. After passing through the turbine, the vaporized fluid is cooled by deep ocean water, causing it to condense back into a liquid, ready to be used again.

The key advantage of OTEC is that it provides a continuous source of renewable energy, assuming the temperature gradient between surface and deep waters is sufficient. However, successful OTEC systems require careful engineering design, especially in the heat exchanger components, to ensure efficient and effective energy transfer. The goal is to maximize energy extraction while maintaining environmental sustainability.

Power Cycle Efficiency

Power cycle efficiency is a measure of how effectively a power cycle converts heat energy into electric energy. It is defined as the ratio of the electric power output produced by the cycle to the heat energy input into the system. In our exercise, the OTEC plant operates with a power cycle efficiency of 3%, meaning that only 3% of the heat energy from the warm seawater is converted into electricity.

This might seem low compared to other power generation methods, but it reflects the inherent challenges of working with such small temperature differences in OTEC systems. The efficiency is limited by the second law of thermodynamics, which restricts efficiency based on the available temperature range. Improving efficiency could involve optimizing the working fluid properties, enhancing the heat exchanger design, and using advanced materials to minimize energy losses.

• The formula for efficiency is: \[ \text{Efficiency} = \frac{\text{Electric Power Output}}{\text{Heat Input}} \]
• Efficiency factors include fluid choice, turbine effectiveness, and heat transfer optimization.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient is a crucial parameter in the design and operation of a heat exchanger. It represents the heat transfer capability of the exchanger and incorporates the effects of conduction and convection on both the inside and outside surfaces of the exchanger. A higher overall heat transfer coefficient signifies more efficient heat transfer between the working fluid and seawater in the OTEC system.

In this exercise, an approximate value of 1200 W/m虏路K is provided for the heat exchanger in the evaporator. This value influences the design, particularly the surface area required to achieve the necessary heat transfer to evaporate the working fluid. The formula used in our solution, \( Q = U \cdot A \cdot \Delta T_{lm} \), connects the transfer rate \( Q \) with the heat transfer coefficient \( U \), heat exchanger area \( A \), and the logarithmic mean temperature difference \( \Delta T_{lm} \). Keeping this value optimal involves selecting appropriate materials and maintaining cleanliness to avoid fouling.