Module 2 Hydro Power PDF

1 Module 2 Hydro Power Introduction This module is all about “Hydro Power” wherein the types of micro-hy...

1 Module 2 Hydro Power Introduction This module is all about “Hydro Power” wherein the types of micro-hydro systems, water resources, theoretical power estimation, and design of micro-hydro systems for AB applications will be discussed. As you read the material, you will be able to learn how to design hydropower conversion systems for agricultural and biosystems applications. You will also learn how to compute basic problems in determining the capacity, the power developed, overall system efficiency, pressure head, etc. Some links to educational videos related to this topic and a journal article on how to design and analyze hydropower conversion systems will be posted in your Moodlecloud platform for you to watch and study as your additional study material. After the content discussion, you are given exercises and activities to work on. All these activities will deepen and strengthen your understanding of the topic presented. Learning Outcome At the end of the module, it is expected that the students will be able to apply the knowledge on how to design a micro-hydro system for Agricultural and Biosystems applications. Learning Objectives The general objective of the module is to impart knowledge about hydro-power and hydro-power collection systems for AB applications. Specifically, this module aimed: to discuss the different types and components of hydro-power conversion systems, hydro resources, and to illustrate the process of solving basic problems in determining the capacity, the power developed, head and other head losses, different types of open channels, overall system efficiency, discharge, slopes, roughness coefficients, etc. required in micro-hydro system design.  2 Content Exploration 2021 Hydropower Status Report (IHA, 2021) The International Hydropower Association (IHA) gives an overview of the global hydropower status. Their mission is to advance sustainable hydropower by building and sharing knowledge on its role in renewable energy systems, responsible freshwater management, and climate change solutions. Hydropower offers clean, affordable, and reliable electricity while meeting our basic needs for water, irrigation, flood, and drought control when delivered responsibly. Figure 1. The Hydropower Capacity by Region in MW (2021) (Image Source: IHA, 2021) Report Summary The report shows that the power sector produced a record 4,370 hours of terawatt (TWh) of clean electricity by 2020 - up from a previous record of 4,306 TWh in 2019. This power is approximately the same as the entire annual electricity consumption of the United States. The total hydropower capacity will reach 1,330 gigawatts (GW) by 2020. This represents an annual increase of 1.6 percent - higher than in 2019 but still significantly lower than the more than 2 percent needed to enable the important hydropower contribution to climate change. By 2020, a total of 21 GW hydropower projects per capacity were installed, in 2019 of 15.6 GW. About two-thirds of this growth came from China, which saw 13.8 GW of new energy. Among other countries introducing new energy by 2020, only Turkey (2.5 GW) added more than 1 GW. The power generation at the pump is up to 1.5 GW of new additions in capacity, up to 304 MW added in 2019. Most of this was in China (1.2 GW), where Israel had approved a 300 MW Mount Gilboa project under a new financial plan. Major projects completed by 2020 include 2.1 GW Lauca in Angola, 1.8 GW Jixi warehouse in China, and Ilisu (1.2 GW), and Lower Kaleköy (0.5 GW) in Turkey. One major increase in capacity was in China, where the Wudongde project included eight out of 12 online units, adding 6.8 GW to the Chinese grid. The remainder is expected to be filed in 2021. China remains the world leader in terms of more than 370 GW of hydropower. Brazil (109 GW), the USA (102 GW), Canada (82 GW), and India (50 GW) make up all the other top five. Japan and Russia are just behind India, followed by Norway (33 GW) and Turkey (31 GW).  3 Content Exploration Hydro Power Hydropower or water power is the power derived from the energy of falling water or running water, which may be harnessed for useful purposes. The basic principle of hydropower is that power is generated in the system from the pressure of falling water from higher to lower levels. Most machines that make electricity need some form of mechanical energy to get things started. Mechanical energy spins the generator to make the electricity. In the case of hydropower or hydroelectricity, the mechanical energy comes from large volumes of falling water. For more than 100 years, the Figure 2. The Hydropower Project in Brazil simplest way to produce the volumes of (Image Credit: ENGIE Brazil, 2020) falling water needed to make electricity has been to build a dam. A dam stops the natural flow of a river, building up a deep reservoir behind it. However, large dams and reservoirs are not always appropriate, especially in the more ecologically sensitive areas of the planet. General Principles The basic physical principle of hydropower is that if water can be piped from a certain level to a lower level, the resulting water pressure can be used to do work. If the water pressure is allowed to move a mechanical component then the movement involves the conversion of water energy into mechanical energy. Hydro turbines convert water pressure into mechanical shaft power which can be used to drive an electric generator, a grain mill, and other useful devices. The term “head” refers to the actual height that waterfalls through the system. Power, on the other hand, is the energy converted per second which is the rate of work being done in the system while the energy is the total work done for a certain time. For each cubic meter per second of water falling through a 1-meter head, 9.8 KW Figure 3. The Principles of Hydropower Generation (Image Credit: HowStuffWorks, 2001) of power is generated.  4 Content Exploration Types of Hydro Power Hydropower is used primarily to generate electricity. Broad categories include: Conventional hydroelectric - referring to hydroelectric dams (Figure 2). Run-of-the-river hydroelectricity - which captures the kinetic energy in rivers or streams, without a large reservoir and sometimes without the use of dams. Small hydro projects are 10 megawatts or less and often have no artificial reservoirs. Micro hydro projects provide a few kilowatts to a few hundred kilowatts to isolated homes, villages, or small industries. Watch this video on Figure 3. The Micro-hydropower System Youtube: (Image: ENERGY.GOV) https://www.youtube.com/watch?v=dXBYbRCx 0JQ Conduit hydroelectricity projects utilize water that has already been diverted for use elsewhere; in a municipal water system, for example. Pumped-storage hydroelectricity stores water pumped uphill into reservoirs during periods of low demand to be released for a generation when demand is high or system generation is low. Advantage and Disadvantages of Hydro Power Plants Power is continuously available as long as there is water available. Concentrated energy resource when there is a reasonable head. Energy available is predictable, No imported fuel is required; Low investment and maintenance cost; Minimum technical skills are required to design and install; Manufacturing of the system can be done in small backyard shops; and Long lasting and robust technology Site-specific technology, There is a need for maximum useful power; and Fluctuation on water available limits the power output  5 Content Exploration The basic formula to calculate power on hydro devices 𝑃 = 9, 810 𝐾𝑄𝐻 where: P = Power output, w K = Turbine efficiency, decimals Q = Water flow rate, m3/s H = Head, m Sample Problem Example 1 Compute the power in kW that can be generated by a micro-hydro turbine with water input of 40 lps on a head of 20 m. The turbine efficiency is 70%. Given: Q = 40 lps or 0.04 m3/s H = 20 m K = 0.7 Required: Power Diagram: Figure 4. System Diagram Solution: 𝑃 = 9, 810 𝐾𝑄𝐻 𝑚3 = 9, 810 (0.7) (0.04 ) (20 𝑚) 𝑠 = 54936 𝑊 𝑜𝑟 5.5 𝑘𝑊  6 Content Exploration Water Wheels Waterwheels are traditional devices that can utilize water power. The 4 principal types include the following: Undershoot – where water passes underneath the wheel, for very low head less than 1 m. Overshoot – in which waterfalls into buckets on the rim of the wheel and the weight of the water carries the wheel around Breast Wheel – essentially a less efficient version of the overshoot wheel but which requires the same effort to build; and Poncelet- a refined version of the undershoot wheel in which the water is forced to a narrower opening at a higher velocity, and the buckets are curved to capture the water-energy more effectively. Figure 5. The Water Wheels (Overshoot and Undershoot) (Image Source: REUK.co.uk) The main reason why waterwheels are rarely used for electricity generation is that they rotate much more slowly than a turbine and therefore need a large and expensive speed-increasing mechanism to drive an alternator at the required 1500 rpm. Components of the hydro system 1. Penstock The penstock is the pipe that conveys water under pressure to the turbine. The penstock often constitutes the major expense in the total cost of the micro-hydro budget. Therefore it is worthwhile to optimize the design of the penstock. The following are the steps in the methodical approach in the design of a penstock: 1. Consider the range of locally available materials and the possible types of joints. Compare maintenance implication cost. Also, list the diameters and wall thicknesses of pipe available on the local market 2. Calculate the expected friction loss for a range of different pipe materials and internal diameters. Use 5% head loss as a starting point 3. Add the extra “turbulence” losses caused by bends, valves, etc. to the friction losses.  7 Content Exploration 4. Predict the likely surge pressure in the case of accidental fast enclosure of the penstock valve, and add it to the static pressure. Calculate suitable wall thicknesses for the range of penstock size preferred 5. Consider the number and type of support, anchors, and joints for the type of penstock preferred 6. Estimate the overall cost of each option and check the availability of components from suppliers 7. Compare the cost-effectiveness of the most promising option 8. Recalculate with 10%, 20%, and 30% head loss to consider the use of smaller and cheaper pipe diameter to be used. Decide if the losses are acceptable and whether the savings in capital cost are sufficient to justify the losses. Note that to help save cost on a high head system, lighter grades of the pipe may be used at the top of the penstock where pressure is relatively smaller. 2. Turbines A turbine converts energy in a form of falling water into rotating shaft power. Its design speed is largely determined by the head under which it operates. Every turbine type has a numerical value associated with it called the specific speed which characterized its performance. Many types of turbines include three major styles suitable for different types of water supplies: impulse turbines, reaction turbines, and submersible propeller turbines. Table 1. Turbine Type Selection Classification of Turbines Impulse turbine – the pressurized water is converted into a high-speed jet by passing it through a nozzle. The jet strikes the specially shaped buckets or blades of the turbine rotor, causing it to rotate. The typical example of impulse turbines is the Pelton wheel and the Turgo wheel. Because an impulse turbine uses jet water in the air, its casing is at atmospheric pressure and serves more as a splash guard than as a container for the water. Impulse turbines are generally used in medium to high head applications. Figure 6. The Impulse Turbine (Image Source: Mechanical Walkins)  8 Content Exploration Reaction turbine – quite different from an impulse turbine in that it runs filled with water. The flow of water through the rotor is deflected in such a way that it creates pressure differences across the blades which cause them to rotate. Propeller turbines, in which rotor shaped like a ship’s propeller is immersed in a flow of water are used at low heads. The more advanced propeller turbines have blades that can be rotated about their point of attachment to the hub so that they cut the water at different angles depending on the Figure 7. The Reaction Turbine flow rate to the turbine, those maximizing the (Image Source: Mechanicalbooster, 2018) efficiency. 3. Drive System The Drive System connects the generator to the turbine. On the one hand, it allows the turbine to spin at the most efficient speed possible. It runs the generator at the correct speed to create the correct voltage and frequency on the other end (frequency is only for AC systems). Direct, one-to- one coupling between the turbine and the generator is the most efficient and dependable drive system. For many sites, this is doable, but not for all head and flow combinations. In many cases, particularly with AC systems, the transfer ratio must be adjusted so that both the turbine and the generator run at their optimal (but different) speeds. These types of drive systems can use gears, chains, or belts, each of which introduces additional efficiency losses into the system. 4. Generator Alternators with brushes work well and are still used for their lower cost. The major drawback is the alternator’s brushes need regular replacement. These days, brushless permanent magnet alternators are available and are the better choice, since they eliminate the need for brush replacement. In addition, brushless permanent magnet alternators perform higher efficiencies, increasing the hydro system’s output. 4.1. Alternator Configuration Utilizing different field configurations (configurable by a qualified technician), the alternator can produce 12V, 24V, 48V, or 120V (3-phase AC). Standard Configuration – Extra Low Voltage (12V, 24V, or 48V) Externally Rectified - Extra Low Voltage (12V, 24V, or 48V) Long Transmission - Low Voltage (120V, 3-phase)  9 Content Exploration 5. Controls 5.1. AC Controls Pure AC systems had no batteries or inverter. AC is used by loads directly from the generator, and surplus electricity is burned off in dump loads (usually resistant heaters). Governors and other controls help ensure that an AC generator constantly spins at its correct speed. The most common types of governors for small hydro systems accomplish this by managing the load on the generator (New, 2004). With no load, the generator would “freewheel” and run at a very high speed. By adding progressively higher loads, the generator slows down until it reaches the exact speed for proper AC voltage and frequency. As long as the level of the designed load is maintained, electrical output will be correct. A governor performs this action automatically. 5.2. DC Controls DC hydro systems are not the same as AC hydro systems. The batteries are charged by the generator output. Excess energy is diverted to a dump load using a diversion controller. Smaller streams with a potential of less than 3 kW benefit from DC systems. AC systems are limited to a peak load that is equal to the generator's output. Even if the producing capacity is lower, DC systems with a battery bank and a large inverter can offer a high peak load from the batteries. Figure 8. Diagram of a Typical Battery-based Hydro Power System (Source: EA Energy Alternatives Ltd.) Since the generator cannot run without a load, series charge controllers, such as those used with solar electric systems, are not employed with hydro systems (open circuits). Over-speeding can potentially harm the alternator windings and bearings. A diversion (or shunt) controller must be utilized instead. These often redirect energy from a battery to a resistive heater (air or water) to maintain a consistent load on the generator while keeping the battery voltage at the correct level.  10 Content Exploration 6. Settling Basin and Forebay Tanks The water drawn from the river and fed to the turbine will usually carry a suspension of small particles. This sediment will be composed of hard abrasive materials such as sand which can cause expensive damage and rapid wear to turbine runners. To remove this material, the water flow must be slowed down by a settling basin so that silt particles will settle on the basin floor. This deposit formed is then periodically flushed away. Figure 9. Parts of a Micro Hydro System (Image Source: T. Wijenayake ) 7. Spillways Spillways are designed to permit controlled overflow at certain points along the channel. Flood flows through the intake can be twice the normal channel flow, so the spillway must be large enough to remove this excess flow. It can be combined with control gates to provide a means of emptying the channel. 8. Channels The different types of channel for the hydro system includes the following: 1. Simple earth channel 2. Earth channel lined with stone and mortar 3. Earth channel sealed with clay or cement slurry 4. Open concrete channel 5. Cupped concrete channel, shielded against debris 6. Supported steel sheet channel 7. Closed pipe. 8.1. Open Channel – An open channel is a water conveyance structure in which the stream is not completely enclosed by solid boundaries and therefore has a free surface subjected only to atmospheric pressure. The flow in such a channel is caused by some external head, but rather by the gravity component along the slope of the channel. In an open channel flow, the hydraulic grade line is coincident with the stream surface since the pressure at the surface is atmospheric. The flow in an open channel may Figure 9. Types of Open Channels (Image Source: GoldSim ) either be uniform or non-uniform.  11 Content Exploration 8.1.1. Most Efficient Cross Sections Also known as most economical sections, these are sections which, for a given slope S, channel cross-sectional area A, and roughness n, the rate of discharge Q is at maximum. From Manning Formula, 𝟏 𝟐/𝟑 𝟏/𝟐 𝑸=𝑨 𝑹 𝑺 𝒏 It can be seen that with A, n, and S constant, Q is maximum when the hydraulic radius R is maximum, and since R = A/P, then R is maximum if P is minimum. Therefore, requires the least cost of grading and lining, which makes it the most economical. *Note: Of all the canal shapes, the semi-circular open channel is the most efficient. 8.1.2. Proportions of Most Efficient Sections To derive the proportions for most efficient sections, minimize the perimeter with the cross-sectional area constant. Rectangular Section Base b = 2d Hydraulic Radius R = d/2 Trapezoidal Section Top width x = 2y Hydraulic Radius R = d/2 Angle Ө = 300 Wetted Perimeter, P P = b + 2y The most efficient trapezoidal section (including the rectangle) has its top width (x) equal to the sum of the sides (2y), which is the proportion for half-hexagon. This shows that the most efficient trapezoidal section is the half-regular hexagon (all sides are equal)  12 Content Exploration Triangular Section Area A = d2 Angle Ө = 90o The most efficient triangle section is the 900 V-notch Manning’s Formula To find the velocity, v, (S.I. Units) 1 2/3 1/2 𝑣= 𝑅 𝑆 𝑛 To find the discharge, Q, (S.I. Units) 1 2/3 1/2 𝑄=𝐴 𝑅 𝑆 𝑛 where: n = rougness coefficient (Table 2) R = hydraulic radius S = slope of energy gradeline A = Area of Open Channel Table 2. Values of Rougness Coefficient (n) for Manning’s Equation (Source: Gillesania, 2015)  13 Sample Problem 2. Determine the uniform flow through a trapezoidal concrete-lined canal (Cement mortar surfaces) of a hydropower system having a side slope of 3H to 4V and bottom width of 2 m if the depth of flow is 2 m. The channel is laid on a slope of 3 m per 2 km. (Use the average n- value) Solution: Find the values of 𝑦 and 𝑥1 : (using the similar triangles method) 5 𝑦 = ( ) 2 = 2.5 𝑚 4 2 𝑥1 = ( ) 3 = 1.5 𝑚 4 Find the Area of Trapezoid, 𝐴: 𝑥+𝑏 5+2 𝐴= 2 (d) = 2 (2) = 7 𝑚2 Find the wetted perimeter, 𝑃 𝑃 = 2 + 2.5(2) = 7 𝑚 Find the hydraulic radius, 𝑅 𝐴 7 𝑚2 𝑅= = = 1𝑚 𝑃 7𝑚 Find the value of 𝑛 0.011 + 0.015 𝑛= = 0.013 2 Find the slope, S 3𝑚 𝑆= = 0.0015 2000 𝑚  14 Sample Problem Find the discharge, 𝑄 1 2/3 1/2 𝑄=𝐴 𝑅 𝑆 𝑛 1 𝑄 = 7 𝑚2 (1 𝑚) 2/3 (0.0015) 1/2 0.013 𝑄 = 20.85 𝑚3 /𝑠 3. Given the computed water discharge in sample problem no. 2. If a small hydropower station has a gross head of 8.2 meters and a head loss in the water conductor system of 0.5 meters, calculate the power generated by the system at a turbine efficiency of 83 %. Solution: Compute the total Head, 𝐻 𝐻 = 𝐺𝑟𝑜𝑠𝑠 𝐻𝑒𝑎𝑑 − 𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑠𝑒𝑠 𝐻 = 8.2 𝑚 − 0.5 𝑚 𝐻 = 7.7 𝑚 Calculate the power generated, 𝑃𝑔 𝑃𝑔 = 9, 810 𝐾𝑄𝐻 𝑚3 ( ) 𝑃𝑔 = 9, 810 0.83 (20.85 )(7.7 𝑚) 𝑠 𝑃𝑔 = 1.31 𝑀𝑊  15 Sample Problem 4. A triangle with the most efficient proportion discharges water at a rate of 2 m 3/s. Assuming n = 0.018 and S = 0.0021, calculate the normal depth of flow in meters. =? y= Solution: For most efficient cross section of a triangular canal, A = d2 y = d sec 450 Compute the depth of flow (d) using Manning’s Equation 1 2/3 1/2 𝑄=𝐴 𝑅 𝑆 𝑛 2/3 𝑚3 2 1 𝑑2 2 = (𝑑 ) ( ) (0.0021)1/2 𝑠 0.018 2 (𝑑 sec 45𝑜 ) 𝑑 = 1.18 𝑚 *sec x = 1/cos x  16 Sample Problem 5. A rectangular concrete channel (cement mortar surfaces), 15 m wide is to carry water at the rate of 22 m3/s. If the channel slope is 0.00025, determine the normal depth of flow. (Use the average n- value) Solution: Compute the average value of 𝑛 (Table 2) 0.011+0.015 𝑛= = 0.013 2 Compute the normal depth of flow using Manning’s Equation 1 2/3 1/2 𝑄=𝐴 𝑅 𝑆 𝑛 1 15𝑑 2/3 22 𝑚3 /s = (15𝑑) ( ) (0.00025)1/2 0.013 15 + 2𝑑 𝑑 = 1.187 𝑚 Exercises 1. A rectangular channel with the most efficient cross-section made of concrete (monolithic) discharges water at a rate of 20 m3/s. Calculate the normal depth of flow in meters at a slope of 0.0021. (Use the average n-value). 2. If the same channel with the same water discharge in the item above will be used in a hydropower system with a penstock head of 40 m and other head losses of 1.3 m, calculate the power that can be generated in MW by the system at turbine efficiency of 82 %. 3. A trapezoidal canal section having a side slope of 2H to 3V has a total depth of 1.5 m. For the most efficient proportion, what is the required bottom width in meters? 4. A triangular channel with the most efficient proportion discharges water at a rate of 1m3/s. Assuming n = 0.018 and S = 0.0021, calculate the normal depth of flow in meters. 5. Given the same water discharge in exercise item no. 4. If a micro hydropower station generated power of 0.25 MW by the system at a turbine efficiency of 85 %, calculate the total head in meters.  17 Activities Answer the following: 1. Type of hydroelectric power that typically produces from 5 kW to 100 kW of electricity using the natural flow of water. 2. A rotary mechanical device that extracts energy from a fluid flow and converts it into useful work. 3. A refined version of the undershoot wheel. 4. It is the power derived from the energy of falling water or running water, which may be harnessed for useful purposes. 5. A type of hydropower that utilizes water that has already been diverted for use elsewhere. 6. Refers to the actual height that water falls through the system. 7. A turbine where the pressurized water is converted into a high-speed jet by passing it through a nozzle. 8. Traditional devices that can utilize water power. 9. A man-made barrier across the river that was built to keep the water level constant with the turbine. 10. A turbine that runs completely filled with water. Problem Solving 1. What are the best dimensions (b x d) for a rectangular brick channel (n = 0.015) designed to carry 5 m3/s of water in uniform flow with S = 0.001? 2. What are the best dimensions (b x d) for a triangular brick channel with the same S, n, and Q values in the 1st item? 3. Determine the uniform flow through a trapezoidal open channel of a hydropower system with a neat cement surface having a slope of 1.75 m/km. The side slope of the channel is 2H to 3V with a bottom width of 2.5 m and the depth of flow is 2 m. The channel is laid on a slope of 1 m per km. (Use the average n-value) 4. Given the same water discharge in item no. 3. If a hydropower station has a penstock height of 30 m and other head losses of 2.1 m, calculate the power in MW that can be developed by the system at turbine efficiency of 80 %. 5. A small hydropower plant with a 50m head can generate a power of 9 megawatts (MW). Find the water flow rate of the power plant if the turbine has 80% efficiency.  18 References 1. Belonio, A.T. (1998). Hydropower: In Farm Power. Central Philippine University, Iloilo City. 2. Gillessania, D.I. (2015). Fluid Mechanics and Hydraulics. Cebu City, Philippines: Cebu DGPrint, inc. 3. International Hydropower Association (IHA). Hydropower Status Report (2021). Retrieved from https://assets-global.website- files.com/5f749e4b9399c80b5e421384/60c37321987070812596e26a_IHA20212405-status- report-02_LR.pdf 4. Thapar, O. D. (2008). Teachers Manual HYDROPOWER ENGINEERING FOR DIPLOMA LEVEL COURSES. Uttarakhand, India. Compiled by: ELBERT M. GARCIA, ABE, MSAE