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Chapter 7: Problem 52

Water at \(24^{\circ} \mathrm{C}, 1\) bar is drawn from a reservoir \(1.25 \mathrm{~km}\) above a valley and allowed to flow through a hydraulic turbine- generator into a lake on the valley floor. For operation at steady state, determine the maximum theoretical rate at which electricity is generated, in MW, for a mass flow rate of \(110 \mathrm{~kg} / \mathrm{s}\). Let \(T_{0}=24^{\circ} \mathrm{C}, p_{0}=1\) bar and ignore the effects of motion.

Short Answer

Expert verified
The maximum theoretical rate at which electricity is generated is 1.351 MW.

Step by step solution

01

- Identify the Potential Energy Conversion

When water flows from a higher elevation to a lower elevation, its potential energy converts into kinetic energy. In this scenario, water falling from the reservoir (1.25 km above the valley) into the lake at the valley floor can be used to generate electricity using a hydraulic turbine-generator.
02

- Calculate the Potential Energy

The potential energy per unit mass is given by \[ PE = m \times g \times h \] where: -\( m \)is the mass flow rate (\( 110 \text{ kg/s} \)), -\( g \)is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \)), -\( h \)is the height difference (\( 1.25 \text{ km} = 1250 \text{ m} \)).
03

- Perform the Calculation

Substitute the known values into the potential energy formula: \[ PE = 110 \text{ kg/s} \times 9.81 \text{ m/s}^2 \times 1250 \text{ m} = 1.351 \times 10^6 \text{ J/s} = 1.351 \text{ MW} \] Hence, the maximum theoretical power generated is 1.351 MW.

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Key Concepts

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

potential energy
Potential energy is the energy stored in an object due to its position. In this exercise, water flows from a high elevation (the reservoir) to a lower elevation (the valley). The potential energy of the water at the reservoir is converted into kinetic energy as it falls. This energy is harnessed by a hydraulic turbine to generate electricity. The potential energy per unit mass is calculated using the formula: \( PE = m \times g \times h \) - \( m \) is the mass flow rate: 110 kg/s - \( g \) is the acceleration due to gravity: 9.81 m/s虏 - \( h \) is the height difference: 1.25 km or 1250 m Consuming potential energy to produce electricity efficiently is the core of hydraulic energy conversion.
mass flow rate
Mass flow rate indicates the amount of mass flowing through a point per unit of time. For water flowing into the turbine, the mass flow rate is provided as 110 kg/s. This high mass flow rate is crucial for generating significant power. A larger mass flow rate means more potential energy is converted into kinetic energy, which can then be converted into electricity. By optimizing the mass flow rate, the efficiency of the hydraulic energy conversion process can be maximized. Measuring mass flow rate helps engineers design systems that can handle the required load and produce the needed output.
gravity
Gravity plays a key role in hydraulic energy conversion. It provides the force that pulls water down from the reservoir to the valley. The acceleration due to gravity, represented by \( g \), is approximately 9.81 m/s虏. This constant influences the potential energy calculation, as expressed in the formula: \( PE = m \times g \times h \) Because gravity is consistent and predictable, it can be relied upon to calculate the potential energy accurately. Understanding the effects of gravity helps in predicting the behavior of fluids within hydraulic systems, ensuring efficient energy conversion.
electricity generation
Electricity generation via hydraulic energy conversion involves transforming the potential energy of water into electrical energy. As water moves from a high elevation to a low elevation, it passes through turbines which spin due to the kinetic energy of falling water. These turbines are connected to generators that convert mechanical energy into electricity. In this exercise, the theoretical maximum power generated is 1.351 MW. This is derived from the potential energy calculation for the given mass flow rate and height difference. Effective electricity generation from hydraulic energy also depends on factors like turbine efficiency and system design.

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Most popular questions from this chapter

The sun shines on a \(300-\mathrm{ft}^{2}\) south-facing wall, maintaining that surface at \(98^{\circ} \mathrm{F}\). Temperature varies linearly through the wall and is \(77^{\circ} \mathrm{F}\) at its other surface. The wall thickness is 6 inches and its thermal conductivity is \(0.04 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{R}\). Assuming steady state, determine the rate of exergy destruction within the wall, in Btu/h. Let \(T_{0}=77^{\circ} \mathrm{F}\).

An ideal gas is stored in a closed vessel at pressure \(p\) and temperature \(T\). (a) If \(T=T_{0}\), derive an expression for the specific exergy in terms of \(p, p_{0}, T_{0}\), and the gas constant \(R\). (b) If \(p=p_{0}\), derive an expression for the specific exergy in terms of \(T, T_{0}\), and the specific heat \(c_{p}\), which can be taken as constant. Ignore the effects of motion and gravity.

Steam at 30 bar and \(700^{\circ} \mathrm{C}\) is available at one location in an industrial plant. At another location, steam at 20 bar and \(400^{\circ} \mathrm{C}\) is required for use in a certain process. An engineer suggests that steam at this condition can be provided by allowing the higher-pressure steam to expand through a valve to 20 bar and then cool to \(400^{\circ} \mathrm{C}\) through a heat exchanger with heat transfer to the surroundings, which are at \(20^{\circ} \mathrm{C}\). (a) Evaluate this suggestion by determining the associated exergy destruction rate per mass flow rate of steam \((\mathrm{kJ} / \mathrm{kg})\) for the valve and heat exchanger. Discuss. (b) Evaluating exergy at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\) and assuming continuous operation at steady state, determine the total annual cost, in \(\$$, of the exergy destruction for a mass flow rate of \)1 \mathrm{~kg} / \mathrm{s}\(. (c) Suggest an alternative method for obtaining steam at the desired condition that would by preferable thermodynamically, and determine the total annual cost, in \)\$$, of the exergy destruction for a mass flow rate of \(1 \mathrm{~kg} / \mathrm{s}\). Let \(T_{0}=20^{\circ} \mathrm{C}\), \(p_{0}=1 \mathrm{~atm}\).

Steam enters a turbine operating at steady state at \(4 \mathrm{MPa}\), \(500^{\circ} \mathrm{C}\) with a mass flow rate of \(50 \mathrm{~kg} / \mathrm{s}\). Saturated vapor exits at \(10 \mathrm{kPa}\) and the corresponding power developed is \(42 \mathrm{MW}\). The effects of motion and gravity are negligible. (a) For a control volume enclosing the turbine, determine the rate of heat transfer, in MW, from the turbine to its surroundings Assuming an average turbine outer surface temperature of \(50^{\circ} \mathrm{C}\), determine the rate of exergy destruction, in MW. (b) If the turbine is located in a facility where the ambient temperature is \(27^{\circ} \mathrm{C}\), determine the rate of exergy destruction for an enlarged control volume including the turbine and its immediate surroundings so heat transfer takes place at the ambient temperature. Explain why the exergy destruction values of parts (a) and (b) differ. Let \(T_{0}=300 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\).

An electric water heater having a \(200-L\) capacity heats water from 23 to \(55^{\circ} \mathrm{C}\). Heat transfer from the outside of the water heater is negligible, and the states of the electrical heating element and the tank holding the water do not change significantly. Perform a full exergy accounting, in kJ, of the electricity supplied to the water heater. Model the water as incompressible with a specific heat \(c=4.18 \mathrm{~kJ} / \mathrm{kg}+\mathrm{K}\). Let \(T_{0}=23^{\circ} \mathrm{C}\).
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