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A hydropower plant utilizes eight Francis turbines. The change in head across the power plant is \(250 \mathrm{~m}\), and the flow through each turbine unit is \(12 \mathrm{~m}^{3} / \mathrm{s}\). The estimated efficiency of each turbine unit is \(95 \%\), and the efficiency of the generator and supporting power delivery systems is \(91 \%\). Estimate the power-generating capacity of the hydropower facility. Assume water at \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The power-generating capacity of the hydropower facility is approximately 20.4 MW.

Step by step solution

01

Calculate the Individual Turbine Power Output

The power output of an individual turbine unit can be determined using the formula: \( P = \eta_t \cdot \rho \cdot g \cdot Q \cdot H \), where \( \eta_t \) is the turbine efficiency, \( \rho \) is the density of water \(1000 \mathrm{~kg/m^{3}}\) at \(20^{\circ} \mathrm{C}\), \( g \) is the acceleration due to gravity \(9.81 \mathrm{~m/s^{2}}\), \( Q \) is the flow rate \(12 \mathrm{~m^{3}/s}\), and \( H \) is the head \(250 \mathrm{~m}\). Substituting these values, we get: \[P = 0.95 \times 1000 \times 9.81 \times 12 \times 250 = 2,798,550 \mathrm{~W}\].The individual turbine power output is approximately \(2.8 \mathrm{~MW}\).
02

Determine the Total Power from All Turbines

The total power generated by all turbines is calculated by multiplying the power output of one turbine by the number of turbines. So, the total power is: \[P_{total} = 8 \times 2,798,550 = 22,388,400 \mathrm{~W}\].Thus, based purely on the turbines, the total power is approximately \(22.4 \mathrm{~MW}\).
03

Account for Generator and System Efficiency

To find the actual power-generating capacity, consider the efficiency of the generator and supporting systems. The effective power output is given by: \[P_{effective} = \eta_g \cdot P_{total} = 0.91 \times 22,388,400 = 20,372,448 \mathrm{~W}\].So, the power-generating capacity of the facility, considering all efficiencies, is approximately \(20.4 \mathrm{~MW}\).

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Key Concepts

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

Francis turbines
Francis turbines are among the most commonly used turbines in the world for hydropower generation, valued for their versatility and efficiency. They are a type of reaction turbine, which means that they generate power from the static pressure of water. Francis turbines are designed to work efficiently with a wide range of heads and flow rates, which makes them suitable for many sites around the globe. These turbines have a radial flow pattern, where water enters the turbine radially and then exits axially. This specific design allows them to handle varying water flows while maintaining high efficiency. The adaptability of Francis turbines makes them ideal for installations where water availability can change through the seasons. Their compact size relative to their power output is another advantage, allowing them to be installed in diverse environments.
turbine efficiency
Turbine efficiency refers to the percentage of the available water energy converted into mechanical energy by the turbine. For Francis turbines, as mentioned, this efficiency can be as high as 95%, which is quite impressive. But what does it really mean? In essence, if a turbine has an efficiency of 95%, it means that 95% of the energy from the water is effectively turned into useful power, while the other 5% is lost, often as heat or noise. This efficiency is crucial for the overall performance of a hydropower plant, as it directly affects how much water power is converted into electricity. To calculate turbine efficiency, you can use the formula: - Efficiency, 畏_t = (Output power) / (Input power from water) Ensuring high turbine efficiency is one reason why regular maintenance and proper design specifications are mandatory for hydropower plants.
generator efficiency
Generator efficiency is the ability of the generator to convert mechanical energy from the turbine into electrical energy. In our example, the generator efficiency is 91%. This percentage reflects the losses that occur typically due to heat generated in the electrical components and the friction of the moving parts. The formula to calculate effective power output considering generator efficiency is: - Effective power output = 畏_g 脳 (Mechanical power input to the generator) In the case of hydropower, efficient generators are key because they ensure that most of the mechanical energy generated by the moving turbine is not wasted. This optimal conversion is critical for maximizing the output of power plants and for reducing operational costs.
hydraulic head
The hydraulic head is a measure of the potential energy available in a body of water. It is crucial in determining the potential energy that can be converted into mechanical energy in a hydropower plant. Essentially, the hydraulic head is the height of the water column, calculated as the difference in elevation between the water source and the turbine. The hydraulic head has a direct impact on the power output of turbines such as Francis turbines. The formula used to calculate theoretical power from water is related to the head, as expressed by: - Power = 蟻 脳 g 脳 Q 脳 H Where: - 蟻 is the water density, - g is the acceleration due to gravity, - Q is the flow rate, - H is the hydraulic head. This value signifies the effectiveness of the hydropower plant's design, affecting how much of the water's energy can be converted into electricity. Adjustments in projects often aim to maximize this head to achieve better energy outputs.

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Most popular questions from this chapter

At a Pelton wheel installation, the water surface elevation in the supply reservoir is \(85 \mathrm{~m}\) above the nozzle; the delivery line has a diameter of \(600 \mathrm{~mm}\), a length of \(300 \mathrm{~m}\), and a roughness height of \(8 \mathrm{~mm}\). The discharge nozzle has a diameter of \(50 \mathrm{~mm}\) and a head loss coefficient of \(0.8 .\) The bucket friction loss coefficient is \(0.5,\) the velocity of water relative to the bucket at the exit from the bucket is \(2 \mathrm{~m} / \mathrm{s},\) and the absolute velocity of the water leaving the bucket is \(6 \mathrm{~m} / \mathrm{s}\). Determine the power that could be derived from the system and the hydraulic efficiency of the turbine.

A pump with a rotary speed of 1725 rpm delivers \(25 \mathrm{~L} / \mathrm{s}\) at its most efficient operating point. Under this condition, the inflow velocity is normal to the inflow surface of the impeller, the component of the velocity normal to the outflow surface of the impeller is \(4 \mathrm{~m} / \mathrm{s}\), and the efficiency of the pump is \(80 \%\). The width of the impeller at the outflow surface is \(15 \mathrm{~mm}\), and the blade angle at the outflow surface is \(50^{\circ}\). (a) Estimate the head added by the pump. (b) Use the affinity laws to estimate the head added and the flow rate delivered by the pump when the rotational speed is changed to \(1140 \mathrm{rpm}\).

A prototype water pump has a specific speed of \(1.2,\) and when operating at its most efficient state, it delivers \(5 \mathrm{~L} / \mathrm{s}\) with an added head of \(10 \mathrm{~m}\). Operation of the pump is to be tested using a \(\frac{1}{5}-\) scale model with various test fluids that have dynamic viscosities in the range of \(5-10\) times that of water. Viscous effects should be accurately accounted for in the model. What range of rotational speeds, flow rates, and heads will be required in model testing? Assess whether accurately accounting for viscous effects is realistic.

Affinity laws are typically used to identify homologous values of flow rate, head, and power between geometrically similar pumps. Develop an affinity law for relating homologous values of torque.

A Pelton wheel is to be designed to harness the available hydropower from a site where the effective head on the turbine will be \(160 \mathrm{~m}\) and the reliable flow rate through the turbine will be \(6 \mathrm{~m}^{3} / \mathrm{s}\). The wheel is to have a rotational speed of \(500 \mathrm{rpm}\), the bucket angle will be \(165^{\circ}\), and the nozzle is expected to have a velocity coefficient of 0.92 . (a) What diameter wheel would maximize the efficiency of the turbine? (b) What power can be expected from the turbine? (c) Assess whether a different type of turbine should be considered for this site. Assume water at \(20^{\circ} \mathrm{C}\).

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