/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 An experimental power plant at t... [FREE SOLUTION] | 91Ó°ÊÓ

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An experimental power plant at the Natural Energy Labo- ratory of Hawaii generates electricity from the temperature gradient of the ocean. The surface and deep-water tempera- tures are \(27^{\circ} \mathrm{C}\) and \(6^{\circ} \mathrm{C}\) , respectively. (a) What is the maximum theoretical efficiency 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 \(\mathrm{L} / \mathrm{h}\) .

Short Answer

Expert verified
(a) 6.99%; (b) 3004 kW (extracted), 2794 kW (absorbed); (c) 167342 kg/h or 167342 L/h.

Step by step solution

01

Maximum Theoretical Efficiency

The maximum theoretical efficiency of a heat engine operating between two temperature reservoirs is given by the Carnot efficiency formula: \[ \eta = 1 - \frac{T_c}{T_h} \] where \( T_h \) is the temperature of the hot reservoir, and \( T_c \) is the temperature of the cold reservoir. These temperatures need to be in Kelvin.Convert each temperature from Celsius to Kelvin:\[ T_h = 27 + 273.15 = 300.15 \ K \]\[ T_c = 6 + 273.15 = 279.15 \ K \]Now compute the efficiency:\[ \eta = 1 - \frac{279.15}{300.15} \approx 0.0699 \text{ or } 6.99\% \]
02

Required Heat Extraction Rate for Power Output

Given the maximum efficiency \( \eta = 0.0699 \) and required power output of \( P = 210 \) kW, we can use the relation:\[ \eta = \frac{W}{Q_h} \]where \( W \) is the work output and \( Q_h \) is the heat input from the hot reservoir.Rearranging gives:\[ Q_h = \frac{W}{\eta} = \frac{210 \text{ kW}}{0.0699} \approx 3004 \text{ kW} \]
03

Heat Absorption Rate by Cold Water

The heat absorbed by the cold water \( Q_c \) can be calculated using the formula:\[ Q_h = W + Q_c \]\[ Q_c = Q_h - W = 3004 \text{ kW} - 210 \text{ kW} = 2794 \text{ kW} \]
04

Cold Water Flow Rate Calculation

The flow rate of the cold water can be calculated using the specific heat capacity formula:\[ Q = mc\Delta T \]where \( Q \) is the heat absorbed, \( m \) the mass flow rate, \( c \) the specific heat capacity of water (approx. 4.18 kJ/kg°C), and \( \Delta T \) the change in temperature of the cold water (\( 10^{\circ}C - 6^{\circ}C = 4^{\circ}C \)). Rearrange for \( m \):\[ m = \frac{Q}{c\Delta T} = \frac{2794 \times 1000}{4.18 \times 4} \approx 167342 \text{ kg/h} \]To convert to L/h, use the density of water (~1000 kg/m³ = 1000 L/m³):\[ \text{Flow rate in L/h} = 167342 \times 1 = 167342 \text{ L/h} \]

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Key Concepts

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

Carnot Efficiency
The concept of Carnot efficiency is a cornerstone of thermodynamics and plays a crucial role in understanding how heat engines like the Ocean Thermal Energy Conversion (OTEC) plant operate. The Carnot efficiency, denoted as \( \eta \), defines the maximum possible efficiency any heat engine can achieve when operating between two temperature reservoirs. This maximum efficiency is theoretically achievable under ideal conditions, where no energy is lost to friction or other imperfection.
To calculate the Carnot efficiency, use the formula:
  • \[ \eta = 1 - \frac{T_c}{T_h} \]
where:
  • \( T_h \) is the absolute temperature of the hot reservoir,
  • \( T_c \) is the absolute temperature of the cold reservoir.
These temperatures should always be converted to Kelvin by adding 273.15 to the Celsius values. This formula shows that the greater the temperature difference (or gradient) between the hot and cold reservoirs, the higher the potential efficiency of the heat engine. However, no real-world heat engine can reach this efficiency due to inevitable energy losses.
Heat Engine
A heat engine is a system that converts heat or thermal energy into mechanical work. In the context of the OTEC plant, the heat engine uses the temperature difference between the ocean's warm surface water and the colder deep water to generate electricity. Here's how it operates:
  • Heat is extracted from the warm surface water (the hot reservoir).
  • This heat is used to perform work, such as turning a turbine, producing electrical energy.
  • The excess or unused heat is then expelled to the colder deep ocean water (the cold reservoir).
The efficiency of a heat engine like this is determined by its ability to convert the extracted heat from the hot reservoir into useful work. The key aspects of understanding a heat engine include:
  • The greater the temperature gradient between the warm and cold reservoirs, the more efficient the engine can potentially become.
  • Heat engines must comply with the Second Law of Thermodynamics, meaning not all heat can be converted into work – some is always expelled.
Understanding this concept helps explain why extracting energy from natural temperature gradients, like those in the ocean, can be a viable way to generate electricity.
Specific Heat Capacity
Specific heat capacity, often simply called specific heat, is an essential property of materials that indicates how much heat is needed to change the temperature of a given mass. For water, the specific heat capacity is about 4.18 kJ/kg°C.
The OTEC plant relies on this property to calculate how much cold deep water needs to flow through the system to absorb the waste heat expelled from the heat engine. When calculating the flow rate, the formula used is:
  • \[ Q = mc\Delta T \]
where:
  • \( Q \) is the total heat absorbed (in kJ),
  • \( m \) is the mass flow rate of the water (in kg),
  • \( c \) is the specific heat capacity of water (approximately 4.18 kJ/kg°C),
  • \( \Delta T \) is the change in temperature (in °C).
This equation allows engineers to predict how much water must be circulated to effectively carry away unwanted heat, ensuring the system runs efficiently.
Temperature Gradient
Temperature gradient refers to the difference in temperature between two points. In the case of the OTEC, the gradient occurs between the warm surface water and the colder deep water of the ocean.
Temperature gradients are vital in determining the operational efficiency of the OTEC process. The larger the temperature gradient, the more energy can theoretically be harnessed from the ocean.
  • A high gradient allows for more efficient energy conversion, as shown by the Carnot efficiency formula.
  • This principle is applied in various thermal power plants beyond OTEC, where engineers strive to maximize the temperature difference between reservoirs to improve output efficiency.
The concept of temperature gradients also emphasizes the importance of using renewable energy sources effectively, particularly in locations like Hawaii, where natural thermal gradients in the ocean are readily available. This understanding encourages innovations in sustainable energy production, contributing to less reliance on fossil fuels.

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Most popular questions from this chapter

\bullet A cylinder contains oxygen gas \(\left(\mathrm{O}_{2}\right)\) at a pressure of 2.00 atm. The volume is \(4.00 \mathrm{L},\) and the temperature is 300 \(\mathrm{K}\) . Assume that the oxygen may be treated as an ideal gas. The oxygen is carried through the following processes: (i) Heated at constant pressure from the initial state (state 1)to state \(2,\) which has \(T=450 \mathrm{K}\) . (ii) Cooled at constant volume to 250 \(\mathrm{K}\) (state 3\()\) . (iii) Compressed at constant temperature to a volume of 4.00 \(\mathrm{L}\) (state \(4 ) .\) (iv) Heated at constant volume to 300 \(\mathrm{K}\) , which takes the sys- tem back to state I. (a) Show these four processes in a \(p V\) diagram, giving the numerical values of \(p\) and \(V\) in each of the four states. (b) Cal- culate \(Q\) and \(W\) for each of the four processes. (c) Calculate the net work done by the oxygen. (d) What is the efficiency of this device as a heat engine? How does this efficiency compare with that of a Carnot-cycle engine operating between the same minimum and maximum temperatures of 250 \(\mathrm{K}\) and 450 \(\mathrm{K}\) ?

\(\cdot\) Fach cycle, a certain heat engine expels 250 \(\mathrm{J}\) of heat when you put in 325 \(\mathrm{J}\) of heat. Find the efficiency of this engine and the amount of work you get out of the 325 \(\mathrm{J}\) heat input.

\(\bullet\) 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 produces its 1\(/ 10\) hp at 70.0\(\%\) efficiency, calculate its perform- ance coefficient.

\(\cdot\) An aircraft engine takes in 9000 \(\mathrm{J}\) of heat and discards 6400 \(\mathrm{J}\) each cycle. (a) What is the mechanical work output of the engine during one cycle? (b) What is the thermal efficiency of the engine?

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?

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