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The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a location where the depth of the water is \(50 \mathrm{m}\). Water is to be supplied at a rate of \(5000 \mathrm{kg} / \mathrm{s}\). If the electric power generated is measured to be \(1862 \mathrm{kW}\) and the generator efficiency is 95 percent, determine \((a)\) the overall efficiency of the turbine- -generator, \((b)\) the mechanical efficiency of the turbine, and ( \(c\) ) the shaft power supplied by he turbine to the generator.

Short Answer

Expert verified
Answer: The overall efficiency of the turbine-generator is 75.88%, the mechanical efficiency of the turbine is 79.88%, and the shaft power supplied by the turbine to the generator is 1960 kW.

Step by step solution

01

Find the power input

Since we know the depth of the water (50 m) and mass flow rate (5000 kg/s), we can find the power input. The formula for power input in a hydraulic system is: Power input = mass flow rate × gravitational acceleration × height where gravitational acceleration (g) is approximately 9.81 m/s². Power input = (5000 kg/s) × (9.81 m/s²) × (50 m) = 2452500 W = 2452.5 kW
02

Find the overall efficiency

Now that we have the power input, we can find the overall efficiency of the turbine-generator using the given power output (1862 kW): Overall efficiency = (Power output) / (Power input) Overall efficiency = (1862 kW) / (2452.5 kW) = 0.7588 or 75.88%
03

Find the shaft power

Since we know the generator efficiency (95%) and the power output (1862 kW), we can find the shaft power supplied by the turbine to the generator using the formula: Power output = Generator efficiency × Shaft power Rearranging to solve for Shaft power: Shaft power = Power output / Generator efficiency = (1862 kW) / 0.95 = 1960 kW
04

Find the mechanical efficiency

Now that we have the shaft power and the power input, we can find the mechanical efficiency of the turbine using the formula: Mechanical efficiency = (Shaft power) / (Power input) Mechanical efficiency = (1960 kW) / (2452.5 kW) = 0.7988 or 79.88% So, the \((a)\) overall efficiency of the turbine-generator is 75.88%, \((b)\) the mechanical efficiency of the turbine is 79.88%, and \((c)\) the shaft power supplied by the turbine to the generator is 1960 kW.

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Key Concepts

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

Power Input Calculation
Understanding how to calculate the power input to a hydraulic turbine is essential for evaluating its performance. In our exercise, we have a large lake and a turbine-generator set. To determine the power input, which is the total potential energy per time unit available for conversion to electricity, we use the relationship:

\[ \text{Power input} = \text{mass flow rate} \times \text{gravitational acceleration} \times \text{height}\]

For our case, the gravitational acceleration, \( g \), remains constant at around \(9.81 \text{ m/s}^2\), a figure provided by standard gravity on Earth's surface. Height, in this context, refers to the water's depth, which translates to potential energy due to gravity. The mass flow rate is the amount of water, by mass, passing through the turbine per second. In practical terms, for our lake's turbine-generator setup, we plugged into the formula the depth of water, the mass flow rate, and the gravitational constant. This provided us with a power input of 2452.5 kW, the baseline against which we measure the system's output efficiency.

When dealing with power input calculations, it's important to ensure the accuracy of the measurements, especially for mass flow rate and height, as these are significant factors that affect the system's potential energy. Moreover, understanding the nature of mass flow rate in relation to the specific hydraulic conditions — like lake size, flow restrictions, and environmental factors — could offer further insights into system optimization.
Mechanical Efficiency of Turbines
Mechanical efficiency is a critical parameter that signifies the ratio of useful mechanical energy output to the total energy input into the system. For turbines, this involves the transformation of kinetic and potential energy of water into mechanical energy. The formula to calculate mechanical efficiency is as follows:

\[ \text{Mechanical efficiency} = \frac{\text{Shaft power}}{\text{Power input}} \]

In our scenario, we computed the mechanical efficiency using the shaft power (the power delivered by the turbine to the generator) and the power input previously calculated. This resulted in a mechanical efficiency of 79.88%. The closer this percentage is to 100%, the better the turbine is at converting the water's energy into mechanical energy without losses.

When evaluating a turbine's performance, it is vital to distinguish mechanical efficiency from overall efficiency. The former looks exclusively at the turbine, disregarding any downstream components like generators. It is therefore a good indication of the turbine's performance on its own. This can help engineers and operators identify potential areas of energy loss within the turbine itself, such as friction or turbulence, and make the necessary adjustments.
Shaft Power Output
The shaft power output, which is a direct factor of mechanical efficiency, reflects the amount of power that is transmitted through the turbine's shaft to the generator. It is a measure of a turbine's capability to do work, which in turn is converted to electricity. In essence, this metric can be visualized as the turbine's muscle power.

To calculate the shaft power output, we base our calculations on the power that the generator produces which is then back-calculated, using the generator efficiency, to find out how much power the turbine delivered to the generator. The formula used is:

\[ \text{Shaft power} = \frac{\text{Power output}}{\text{Generator efficiency}} \]

In the textbook example, given the electrical power output and generator efficiency, we determined the shaft power to be 1960 kW. One key insight from this calculation is that the generator efficiency — how well the generator converts mechanical energy to electrical energy — plays a pivotal role in understanding the energy flow through the turbine-generator system.

Paying attention to shaft power output helps in understanding the distribution of energy losses between the hydraulic turbine and the generator. It also guides technicians in maintenance practices and can signal when a turbine or generator is not performing optimally. Increased energy losses in this aspect of turbine operation can point to mechanical issues that, when addressed, can lead to better turbine efficiency.

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Most popular questions from this chapter

A water pump that consumes \(2 \mathrm{kW}\) of electric power when operating is claimed to take in water from a lake and pump it to a pool whose free surface is \(30 \mathrm{m}\) above the free surface of the lake at a rate of \(50 \mathrm{L} / \mathrm{s}\). Determine if this claim is reasonable.

Large wind turbines with blade span diameters of over \(100 \mathrm{m}\) are available for electric power generation. Consider a wind turbine with a blade span diameter of \(100 \mathrm{m}\) installed at a site subjected to steady winds at \(8 \mathrm{m} / \mathrm{s}\). Taking the overall efficiency of the wind turbine to be 32 percent and the air density to be \(1.25 \mathrm{kg} / \mathrm{m}^{3},\) determine the electric power generated by this wind turbine. Also, assuming steady winds of \(8 \mathrm{m} / \mathrm{s}\) during a 24 -hour period, determine the amount of electric energy and the revenue generated per day for a unit price of \(\$ 0.09 / \mathrm{kWh}\) for electricity.

For heat transfer purposes, a standing man can be modeled as a 30 -cm diameter, 175 -cm long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of \(34^{\circ} \mathrm{C} .\) For a convection heat transfer coefficient of \(10 \mathrm{W} / \mathrm{m}^{2} \cdot^{\circ} \mathrm{C},\) determine the rate of heat loss from this man by convection in an environment at \(20^{\circ} \mathrm{C}\).

Consider a \(24-\mathrm{kW}\) hooded electric open burner in an area where the unit costs of electricity and natural gas are \(\$ 0.10 / \mathrm{kWh}\) and \(\$ 1.20 /\) therm \((1 \text { therm }=105,500 \mathrm{kJ}),\) respectively. The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners. Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners.

An aluminum pan whose thermal conductivity is \(237 \mathrm{W} / \mathrm{m} \cdot^{\circ} \mathrm{C}\) has a flat bottom whose diameter is \(20 \mathrm{cm}\) and thickness \(0.6 \mathrm{cm} .\) Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 700 W. If the inner surface of the bottom of the pan is \(105^{\circ} \mathrm{C}\), determine the temperature of the outer surface of the bottom of the pan.

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