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Question

If turbine efficiency is above 95% in a modern turbine, what is the average overall efficiency including accelerated mass lost in the water as it leaves the bottom of the sluiceway?

Tim Gard's picture
owner Phil-America

I have been disecting machenery forever. I am deeply interested in weaknesses in design. The hydro power industry caught my attention shortly after I got out of college. Later I found the...

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Years ago, an engineer at the Niagara Power plant told me the turbine efficiency was 97% in the Niagara power project. Then I asked him about the overall efficiency including accelerated mass losses. He said in the Niagara river fed generators the efficiency was around 16% optimal, which was not normal, and 14 to 15 % was the average, depending on the river levels and time of year. But when you included their pumped storage system, the average efficiency dropped below 7%. Can anyone here confirm or deny this? In conformation, does that not suggest a possible improvement of efficiency if we drop the turbine designs which insert this loss in its function?

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The power generated during turbine operation is calculated as:

Where (H) is effective head is the head which works effectively for the turbine, and expressed as follows.

H = Hg - Hl

Hg is gross head and for reaction turbine (such as Franacis & kaplan) it is between Dam reservoir level and tailrace tunnel outfall tail water level and in case of impulse (Pelton wheel) it is head between reservoir level and nozzle level at turbine location in power house.

Now Hl is head loss in water conductor system which depend upon various factor such as friction loss in water conductor system, bend loss, entrance loss, exit loss, transition loss, gate loss etc.

η is the product of all of the component efficiencies, which are normally the turbine and generator.

As you pointed out nowadays Francis turbines  has a very high degree of efficiency of more than 96 %  and Kaplan turbines have 95 %.

But in addition to that generator efficiency also contribute in overall efficiency . Typical values for generator efficiency range from 91 to 98%. 

Say if turbine efficiency is 95 % and generator efficiency is 95 % than overall efficiency becomes = 0.95*0.95 = 0.9025 or 90.25 %

The losses in water conductor system varies w.r.t water conductor system and its project specific. For longer WCS the losses may be high.

The total head loss varies and is project specific but for clarification based on my project experience it might be in the range between (4-8) % ( note:- this might varies and mentioned only for clarification) .

So say if for any project head loss is 5% of the maximum gross head at maximum discharge than effective head will be 95 % of gross head.

 

 

Tim Gard's picture
Tim Gard on Jan 5, 2021

I don't seem to be getting a specific answer to a specific question for some reason. Lets try this. What is the amount of energy in mass in motion using a cubic foot of water as it exits the base of the turbine at the Niagara Power project?

I understand how much time I spent doing power loss computations of a triad FM transmission tower installation required I spend many hours with formulas and math calculations was daunting, and how my initial reaction would be to draw from those experiences, but lets try to look at the reality without interjecting math meant for a different purpose. All of the calculations I have seen involves math used for installations advanced computations for turbine science, which as the engineer from Niagara said, was specifically designed for turbine computations.

So, getting back to basics so we can consider if something was missed in the research side, not the engineering side, what is the speed of a cubic foot of water as it exits the lowest level of the sluiceway? Then, does accelerated mass possess energy? Then how much energy does it possess?

Tansel  Yılmaz's picture
Tansel Yılmaz on Jan 6, 2021
  • the speed of water as it exits the lowest level the sluiceway = 1 m/s
  • sluiceway is tailrace
  • a cubic foot of water = 0.028 cubic meters = 28 l/s

= amount of energy in mass in motion using a cubic foot of water as it exits the base of the turbine

= 1/2 m V*V = 140 joule 

m = 28 lİters =28kg = 280 N      V=1 m/s

140 joule is nothing.

 140 joule= 33.5calorie = one cup of milk

Tansel  Yılmaz's picture
Tansel Yılmaz on Jan 7, 2021

When you increase the load on the generator by switching something on, it causes the generator to work harder. Without a governor, it would slow down, lowering both voltage and frequency. Likewise, removing a load by switching something off would cause the generator to speed up, raising voltage and frequency.

With no load whatsoever, the generator would “freewheel,” and run at a very high RPM (possibly causing damage). But by adding progressively higher loads, you would eventually slow the generator until it reached the exact RPM for proper voltage and frequency.

The rotor absorbs power from the grid to speed up and delivers power to the grid in order to slow down. When the machine is running synchronously the frequency of the combined stator and rotor excitation matches the grid frequency, there is no slip and the machine will be synchronised with the grid.

Tim Gard's picture
Tim Gard on Jan 18, 2021

Excellent! Now we are getting somewhere thanks Tansel! Consider this,

1,420 cubic metres (380,000 US gal) of water per second is diverted from the Niagara River through conduits under the city of Niagara Falls to the Lewiston and Robert Moses power plants Yes, 140 Joules is not much. But what is it when you do that every second of every day?

There are 35.31 cubic feet in a cubic meter. That's 50, 140 cubic feet every second times 140 Joules equals 7,019,628 joules every second of every day there are 86,400mseconds in a day, or 606,495,859,200 Joules every day left untouched and not converted into electricity.  Is that still a small amount?

Satyajeet Sinha's picture
Satyajeet Sinha on Jan 6, 2021

The velocity in concrete lined HRT is around 3.5 m/s and steel lined tunnel is around 7.0 m/s.

You need to calculate head losses in water conductor system. Because the head loss coefficient is different for different condition. So for exact result you need at least some basic calculation.

The formulae that I have shared is the standard formula for power calculation of any hydropower project. The energy can be worked by simply multiplying time.

For example if any pumped storage hydropower is designed for 8 hour than Energy will be computed as E = P (MW) x 8 hr .

Meanwhile I would like to add that at dam spillway might have higher velocity of the order of 30 m/s and proper energy dissipation arrangement is provided to damped this energy. 

Tansel  Yılmaz's picture
Tansel Yılmaz on Jan 6, 2021

Amount of energy in mass in motion using a cubic foot of water as it exits the base of the turbine is asked not big bang.

Tim Gard's picture
Tim Gard on Jan 13, 2021

Thats what I understand. Now, can you calculate the joules of energy inherent in this accelerated mass that flushes into the lower water set without gain by the turbine sets? Here is what the engineer in Niagara was talking about! According to him, seven times greater! Was he wrong?

The writer has used average efficiency in the name of capacity factor “ plant factor”. The capacity factor is a measure of the plant use. It is the ratio of the average load to plant capacity. It may be computed for a day , month year , or any other period of time.

Actually he has confused load factor with capacity factor as well. Load factor and Capacity factor are not same.

If the capacity factor of a hydroelectric plant approaches to 30- 35% then we can call it as a peaking plant.

He says that 14 to 15 was the average. That means the power plant was working 15/100 x 365 x 24 hours = 1314 hours in a year(8760hours) or 3.6 hours in a day all throughout a year. This is not right hydropower engineering. But hydropwer is actually by nature peaking. But there are solutions we can create like using pelton which works with less flows down to 10% of design flow without any problem.

I mean that that 3.6 hours must be minimum 8 hours a day (33%) for that hydro powerplant to be called as a real peaking plant. 4 hours in the morning + 4 hours in the evening.

By selecting a pelton turbine he could have compensated power because pelton turbine can generate same power with a discharge value of 10% -20% of the design flow he used. I gave a real time example below from Hoover Dam with its link. “five wide-head turbines, designed to work efficiently with less flow” are pelton turbines actually. They have not mentioned it there but it is so. You can read my text completely or the yellow coloured ones if you get bored of reading all.

The amount of electricity generated by Hoover Dam has been decreasing along with the falling water level in Lake Mead due to the prolonged drought in the 2010s and high demand for the Colorado River's water. Lake Mead fell to a new record low elevation of 1,071.61 feet (326.63 m) on July 1, 2016 before beginning to rebound slowly.[97] Under its original design, the dam would no longer be able to generate power once the water level fell below 1,050 feet (320 m), which might have occurred in 2017 had water restrictions not been enforced. To lower the minimum power pool elevation from 1,050 to 950 feet (320 to 290 m), five wide-head turbines, designed to work efficiently with less flow, were installed.[98] Due to the low water levels, by 2014 it was providing power only during periods of peak demand.[99] Water levels were maintained at over 1,075 feet (328 m) in 2018 and 2019.[100]

Source: https://en.wikipedia.org/wiki/Hoover_Dam

The load factor is an index of the load characteristics. It is the ratio of the average load over a designated period to the peak load occurring in that period. It may apply to a generating or a consuming station and is usually determined from recording powermeters.

Hydropower plant efficiency η depends on the speed of the turbine, on the head, on the flow processes.

Turbine Power Output

In general, the turbine converts the kinetic energy of the working fluid, in this case water, into rotational motion of the turbine shaft.

Swiss mathematician Leonhard Euler showed in 1754 that the torque on the shaft is equal to the change in angular momentum of the water flow as it is deflected by the turbine blades and the power generated is equal to the torque on the shaft multiplied by the rotational speed of the shaft. See following diagram.

   

Note that this result does not depend on the turbine configuration or what happens inside the turbine. All that matters is the change in angular momentum of the fluid between the turbine's input and output.

 

Hydroelectric power generation is by far the most efficient method of large scale electric power generation. The overall efficiency can never be 100% however since extracting 100% of the flowing water's kinetic energy means the flow would have to stop.

The conversion efficiency of a hydroelectric power plant depends mainly on the type of water turbine employed and can be as high as 96% for large installations. Smaller plants with output powers less than 5 MW may have efficiencies between 80 and 85 %.

 

Overall efficiency of a HPP  is 85% maximum 86%.

 

The capacity factor for hydroelectric power in the world has been fairly consistent at 40–44% from 1980 to 2008. The capacity factor for hydroelectric power in the United States was 37% in 2008.

Capacity factor has a major influence on the cost of power

 While hydroelectric projects which take their flow directly or indirectly from large reservoirs are suitable for development as peak-load plants , provided that the conduit (headrace tunnel + penstock) required is not too long or too expensive , they are not necessarily peak load plants. (Conduit (headrace tunnel + penstock) is the main artery of a HPP) This is because market conditions may not require all the peak load hydro power available at such a site. Consequently many plants of this type operate on a daily capacity factor of 30 to 100%. Preferably many plants of this type must operate 15 to 20-22 hours a day.

     Although at the time a reservoir plant is installed market conditions may not make it practicable to install capacity on a peak-load basis , the possibility that such conditions may soon change should be considered. If the incremental cost of installation is low , provision should usually be made in the design for the later installation of a large amount of additional capacity.

     In many systems purely peak-load plants prove economical , particularly in connection with storage projects. Assume that , as the load increases in future years , storage is provided on the headwaters of the river system to such an extent that during the minimum December flow and maximum demand week , the plant has available , say , 1,200,000 kw-hr instead of the 438,000 kw-hr which is available without storage. The net effect of the increase in load and this additional energy is to leave at all times the sharp peaks of the load curve projecting above the band that can be served by this plant.

Consequently , these peaks might be served by installing a peak-load hydro plant at the reservoir. In many cases , the additional expense of such a peak load hydro plant need not exceed $70 per kw of installed capacity. This low additional amount results from the fact that the costs of dams and reservoir for storage is already incurred and the additional sum is for intake , conduits (hedrace tunnel + penstock) powerhouse and equipment.

Each hydropower site has unique characteristics. Thus, each hydroelectric project and powerhouse design is different. Each solution must be tailored to the unique characteristics of the site, the transmission and distribution system, and economic and financial resources.

Annual capacity factor at which hydro plants operate is usually limited by the variation in water supply (except in the case of hydro plants like those at Niagara, where installation is less than minimum stream flow and where annual capacity factors may exceed 95%). That means the installed power has been arranged according to be working under less than the minimum stream flow. So in most of the time in a year this plant shall find available water feeding its turbines. But shall not be able to take benefit of a lot of amount of water in most of the time in a year because water will be wasted by overflowing from the spillway. This is not right hydropower engineering.

When the peak load just equals the plant capacity , the capacity factor and load factor are obviously the same. If the maximum demand is less than the plant capacity , the capacity factor may be either greater or less than the load factor depending largely on the load factor itself.

The load for the peak day of the year determines the required generating capacity , while the requirements of the peak week or month (or seasonal or annual) dictate the amount of energy storage required in the form of water. If the scheme is only for power development, then the best use of the water will be by releasing according to the power demand

The average capacity factors are in the typical range for hydropower (≈ 35 to 55%). Capacity factor can be indicative of how hydropower is employed in the energy mix (e.g., peaking vs base-load generation), water availability, or an opportunity for increased generation through equipment upgrades and operation optimization. Potential generation increases achievable by equipment upgrades and operation optimization have generally not been assessed.

The capacity of a power plant is not easily defined. Nameplate capacity or rated capacity of a turbine is usually given in kilowatts or horse power for a given head , discharge and speed at which the best efficiency is obtained. Obviously each of these quantities may vary within definite limits. The rated capacity of a-c generators is usually stated in terms of definite speed, power factor and temperature rise and is usually given in kilovolt-amperes. Each of these quantities may also vary within definite limits.

 

The IEEE definition of generating station capacity is “ the maximum net power output that a generating station can produce without exceeding the operating limit of its component parts.”

The station or plant capacity can therefore be determined for a given station. It may be stated for a peak load over a given period as 15 min. or 1 hr or for a continuous load. It would be higher for short periods than for continuous service if storage regulation exists but is limited by the temperature riğse of the generators. Until the station capacity has been fixed the various factors having to do with capacity can not acquire definite meanings. Where the capacity of a plant has not been fixed , it is customary to take nameplate capacity of the generators as the plant capacity , which is often called installed capacity.

 

The average load of a plant or system during a given period of time is a hypothetical constant load over the same period that would produce the same energy output as the actual loading produced (IEEE).

The peak load is a maximum load consumed or produced by a unit or a group of units in a stated period of time. It may be the maximum instantenous load or a maxiumum average load over a designated interval of time.

The maximum average load is generally used. In commercial transactions involving peak load , it is taken as the average load during a time interval of specified duration occurring within a given period of time , that time interval being selected during which average power is greatest (IEEE).

The load factor is an index of the load characteristics. It is the ratio of the average load over a designated period to the peak load occurring in that period. It may apply to a generating or a consuming station and is usually determined from recording powermeters. We may thus have a daily , weekly , monthly  or yearly load factor ; it may apply to a single plan tor to a system. Some plants of a system may be run continuously at a high load factor , whereas variations in load are taken by other plants of the system , either hydro or steam. Hydro plants designed to take such variations must have sufficient regulating storage to enable them to operate on a low load factor. They are often called peak load plants. Operating on a 50 percent load factor , there must be sufficient storage to enable such a plant , in effect , to utilize twice the inflow for half the time : on a 25 percent load factor , the plant should be able to utilize four times the inflow for a quarter of the time , etc. – the lower the load factor , the greater the storage required.

The capacity factor is a measure of the plant use. It is the ratio of the average load to plant capacity. It may be computed for a day , month year , or any other period of time. When the peak load just equals the plant capacity , the capacity factor and load factor are obviously the same. If the maximum demand is less than the plant capacity , the capacity factor may be either greater or less than the load factor depending largely on the load factor itself.

The period usually considered is a month for purposes of billing , although the sale rates of power are often based on the yearly load factor of the consumer.

Firm Power : Power intended to have assured availability to the customer to meet his load requirements.

Primary Energy : Hydroelectric energy which is available from continuous power.

Secondary Energy : All hydroelectric energy other than primary energy.

Surplus System Capacity : The difference between assured capacity and the system peak load for a specified period.

A typical pumped-storage plant (round-trip efficiency now 80%) is a net consumer of energy: it returns approximately 3 kilowatt-hours (kWh) of electricity for each 4 kWh required for pumping. However, it offers the following important benefits:

• The energy generated during peak periods has a higher monetary value than the energy required for pumping during off-peak periods;

• It permits continuous operation of the highest efficiency plants in the utility's system;

• It provides rapid and flexible response to system load changes. Typically, very large load swings can be accommodated; and

• The utility's overall fuel consumption is reduced because the pumped-storage plant's on-peak generation  avoids or displaces generation at the least efficient thermal plants in the system.

Tim Gard's picture
Tim Gard on Dec 25, 2020

This is all true, I agree, but the question was in regard to the accelerated mass, as I referred to. Look at the numbers the engineer from Niagara gave me. Can you give me something showing he was wrong about accelerated mass losses?

 

Pulikkal Ashokan's picture
Pulikkal Ashokan on Apr 8, 2021

Good news

I am an Independent Researcher. Just like a hoppy I started my research to lift water upward without using electricity. Even 1000-1500 feet level, I have invented nearly half-a-dozen methods to lift water upward. I have  forced the water to lift water. I designed a few machines (purely mechanical) to lift water without electricity. These methods can stop the pumped storage activities all over the world.

The upper reservoir can produce electricity 24 hours, and at the same time, the same water can be lifted upward to the Reservoir 24 hours.

Therefore, need not to wait for day or night, and no more pumping storages.  I have made a few working models.

It was my effort to invent a   new technology to generate power from Ocean water and river water etc.  As per these mechanical functions, water can be reached to any height.

I therefore found that instead  of making major type hydro power plants,  if my inventions are applied to lift water upward and generate electricity with lesser height, it will be more profitable due to many factors.  Ocean water, river water, lake water, backwater, etc. can be used freely and  with 24 hours production with abundant volume of power generation is possible once my inventions to lift water upward are applied in the power generation unit.  My efforts are  to install an Independent Power generation Unit (IPGU)   houses, shops, factories, hotels, hospitals, restaurants, malls, cinema theatres, colleges, schools, hills, remote villages, mountains, forests, quarries, agri. fields etc.  The cost for machinery items and operational cost will be trimmed by 50% (machinery)and operational cost will be negligible. This will be 100% automatic. This is about my water project.

Another invention is based on Artificial wind creation and generate energy 24 hours. No electric power or diesel power etc.  will be  used.  Here too, all are mechanical functions and highly cost effective.  Here also, Independent Power Generation Unit is  is possible.

One more simple technology I have invented for energy generation. Will explain to you later.

With these inventions, what I have observed that the energy that is required for one year in a country, as my per inventions, the total required energy can be produced in two months itself, and the cost will be very less.  Also, maintenance cost will be very less.

My inventions are profit offering and there can be high level industrial growth as  power is available very cheaply, and anybody can produce energy  self.

Companies will have major role , as they can manufacture machineries as per these simple methods, and start generating power abundantly.  The can do business all over the world.

If anybody is interested: Please write to me to

power.ashoka@gmail.com ph. 9074575071,  Kochi - Kerala,

, India.

 

Tansel  Yılmaz's picture
Tansel Yılmaz on Dec 25, 2020

Accelerated mass lost in the water

 

The dynamic behaviour at part load has been a major problem for low head and medium head Francis turbines. The main reason for this has been inter blade separation and unstable swirl flow in the draft tube. Hydraulic design of X-BLADE runners obtain stable operation on the whole range of operation by reducing the cross flow.

 

It is important to design a Francis runner with pressure balanced blades especially for high specific speed runners. A further study of the pressure waves from the blade passing through the guide vanes is important for low specific speed runners in order to reduce pressure pulsations and avoid noise and blade cracking.

 

High frequency pressure pulsation is created by the wakes from the guide vanes. There is possibility of interaction caused by travelling pressure shocks and the blades passing of the wakes for a certain number of blades versus number of guide vanes. Unstable flow with  unacceptable low frequency pressure pulsations in the draft tube  causes blade cracking,

Flow in the runner blade channels regarding the cross flow from hub to band is possible. It is possible to reduce this unfavourable flow by a negative blade lean at the inlet and further balance the pressure by adjustment of the blade lean all the way towards the outlet of the blades. The blade lean angle of a runner blade is defined by the angle Θ between the blade and a meridian plane, measured perpendicular to the stream surface. For a pressure balanced runner, the pressure gradient dh/dy = 0 when Θ is adjusted correctly. For the X-blade runner , the pressure gradient dh/dy ≈ 0 is obtained by adjusting Θ all the way from inlet to outlet for the preliminary blade shaping.

Reference to the design on low and medium specific speed runners is “X-BLADE” runner designed for Three Gorges Power Plant in China.

A negative blade lean at the inlet of the runner is necessary to reduce the cross flow on the pressure side of the blade. The X-BLADE runner for the Three Gorges Power Plant provided this in 1996 and proved a very stable performance over the whole range of flow.

A  traditionally designed runner installed in the turbine at the power plant has a higher specific speed than the X-BLADE runner but traditionally designed runner installed in the turbines  with high specific speed have proven that stable operation and a high efficiency have been achieved with X-BLADE runner design which fulfilled the guaranteed limit of pressure surges and the efficiency and strain gauge measurements on the blade outlets proved a low level of dynamic stress amplitudes in the blades.

With the  traditionally designed runner the turbine can not be operated continuously below 60% load without blade cracking problems.

 

 

High frequency pressure pulsations from blade passing at high head Francis turbines.

The frequency of the dominating sound of a high head Francis turbine, without cavitation problems, is the blade passing frequency. The sound is caused by the runner blades when passing the guide vane wakes which is creating rotating pressure waves travelling in the water between the runner and guide vane cascade. The frequency of these blade passing pressure shocks is the runner speed multiplied by the number of runner blades at the runner inlet i.e. including the splitter blades, if the runner is furnished with splitter blades. However, inside the runner in the runner channels, the frequency of the pressure shocks will be the speed of the runner multiplied by the number of guide vanes.

 

Several strain gauge measurements and pressure measurements of the stresses in the runner blades of various high head Francis turbines, have proven the described pressure pulsations in the blade channels. In some cases too high stress amplitudes have caused fatigue cracking of runner blades. The reason for this has normally been the shape and thickness of the blades versus the magnitudes of the pressure pulsations.

In special cases , very high pressure pulsations caused by the runner blade passing of the guide vane wakes may occur for a certain combination of the number of runner blades versus the number of guide vanes.

If the shock propagation wave with speed in water reaches the blade in front of the regarded blade simultaneously with the passing of this blade through the next guide vane wake , then an interaction and amplification of the high frequency pressure pulsations occurs.

 

In the turbine at power plant Hemsil I in Norway the shock propagating speed in the water was calculated to be around a = 900 m/sec taking into account the flexibility of the head cover and bottom cover.

 

The technical specification for this turbine is Hn = 510 m, Pn = 35.7 MW and speed n = 750 rpm.

 

The number of guide vanes in the turbine is 28 and the first installed runner had 30 blades at the inlet. (15 splitter blades and 15 full length blades) and then an interaction of the pressure waves and blade passing could be proven.

 

This interaction of the pressure waves caused a serious noise problem with a sound level of 120 dB in the surrounding of the turbine. In addition blade cracking occurred in the runner. Because of the high noise level and blade cracking, the pressure pulsations were measured between the runner band and bottom cover with amplitudes of 45 m peak to peak and with a frequency of 375 Hz which was the number of runner blades of 30 multiplied by the speed of 750 rpm (=12.5Hz).

 

The runner with 30 blades was then exchanged with a new runner with identical geometry, but with 16+16 = 32 blades and the noise level dropped to around 80 dB which is very good for a high head turbine operating at 510 m net head.

 

Later several measurements of stress amplitudes on the blades at runner outlets of high head turbines have been made, mainly because of blade cracking caused by weld defects and residual stresses in the stainless weld deposits between blade and band or hub.

 

The number of runner blades and guide vanes for Francis turbines operating at 400-600 m net head must be carefully chosen to avoid interference as described. An alternative to 32 runner blades and 28 guide vanes would be 30 runner blades and 24 guide vanes which also have proven a smooth running. For runners operating at heads below 300 m, 30 blades and 28 guide vanes may be chosen because of a lower rim speed so interaction is avoided.

It should also be mentioned that resonance with the natural frequencies of the runner have not been proven to be the reason for the problems of noise and or blade cracking of low specific speed high head Francis turbines. This is proven by the fact that the measured frequency of pressure pulsations is different on the outside and inside of the runner.

 

In high specific speed Francis runners there are normally not observed similar dominating high frequency pressure pulsations from the blade passing as for high head runners. However, resonant problems and stress amplitudes driven by low frequency draft tube problems have been observed

The reason for the absence or weak influence from the blade passing frequency is the larger distance from the guide vane outlets to the runner blade inlets at the crown compared to the band. The natural frequency of the low head runners may also in some cases be in resonance with the low frequency pressure surges in the draft tube

 

Low frequency load on Francis turbine runners caused by draft tube surging.

Low frequency rotating voids in draft tubes which cause pressure surges and the part load “swirling rope” which often has caused a major problem for low head and medium head Francis turbines, and from time to time blade cracking has been reported. Also high head turbines may have part load problems caused by draft tube surges. The unstable swirl flow rope is caused by reversed flow in the centre of the runner and cross flow from crown towards the band on the pressure side of the runner blades. Swirling rope at part load will always occur in a Francis turbine, but no damage with blade cracking caused by unstable flow, will occur if the runner is well designed.

Wirling rope caused by reversed flow towards the centre of the runner and cross flow at the blade outlets at the hub at part load, may cause unstable conditions and blade cracking  and the reversed rotating centre void at overload may also in some cases cause problems

It is important that the pressure on the pressure side of the blades is balanced with strongly reduced cross flow all the way from the blade inlet towards the blade outlet by adjustment of the blade lean angle. This procedure must be carried out as the first step during the creation of a new runner.

 

The most famous runners, designed in 1996 were the X-Blade runners for Three Gorges Power Plant in China. These runners have so far been operated from 61 m to 110 m net head with a non - restricted range of operation. No serious problems have occurred for these runners according to the reports from the power station.

 

 

 

.

Tim Gard's picture
Tim Gard on Dec 28, 2020

Because I was simply a curious communications engineer, the question of accelerated mass losses the engineer at the Niagara power plant spoke of sparked my curiosity. Because it suggested there was a huge prize at the end of the tunnel. So I began studying and bringing all the science into focus everyone here has been writing about. And I agree with every syllable all of you have written. 100%. But we have yet addressed the fact that could increase our electrical power generation worldwide by 100 fold. What are the accelerated masses, and is there a way to generate electricity with elevated water without , or with much less accelerated mass. Forget without. It was a written word that is impossible, but I left it in to make sure you understand I am not suggesting over unity. Unity is an unbreakable lay in the physical realm, and I know what the limitations are. But if my electrical engineer friend was right, we are missing a golden opportunity. I am hoping I can get a bunch of us together to see if we can find something that will have such a major impact on the Earth it would be spoken of for many thousands of years! Anyone here want to try to see what I am seeing? First point. If this problem can be solved, we might be able to harness energy from very low mass and low elevation sources that were never considered as energy sources with turbine designs. Imagine the power. I believe I have found a solution that my Dad taught me when I was a child in an unrelated field. This is fascinating when you begin to see it!

Tansel  Yılmaz's picture
Tansel Yılmaz on Dec 29, 2020

For determining the exploitation benefits of a hydropower plant also the outfall curve is important. This curve represents the time of outfall of the specific amount of water on a specific place on the river. The more these outfall rates are balanced the more advantageous the exploitation of hydro energy is.

Efficiency of a solution, should be judged with respect to feasibility, reliability, ease of construction, quality control and cost .

How to check reliability of Ins. Pow.-disharge in penstocks, Peak Powerhouse Capacity factor by Specific Water Consumption for Power :

Specific Water Consumption for Power :

The amount of work done by 1 ton (1 m3) of water as in mgh= 9. 8 m/s2   x  1000 kg  x  100m.  is 980 000 joule.

For head= 100 m.                          

The amount of work done by 1 ton (1 m3) of water

 shall be

980 000 / 3 600 000 = 0.27 kwh ,   as 1kWh equals to 3 600 000 joule.

That 100m at           

(9. 8 m/s2   x  1000 kg  x  100m.  = 980 000)

can be revised to be your project’s net head ! )

If we read it from the opposite side in term of mathematics ;

It requires to consumpt 3 600 000 / 980 000 = 3.67 m3 water in order to obtain 1kWh energy.

Amount of water consumpted for 1 kWh is called Specific Water Consumption for Power which is subject to net head. Specific Water Consumption for Power rises as the water level at dam reservoir head pond (or dam reservoir)  drops down which means that more water is consumpted for obtaining the same energy.

Let us assume that head pond (or dam reservoir)  shall supply with a capacity factor (plant factor) of 0.50 of .

( 365 x 24 = 8760 hours is the number of hours in one year)

10 400 kW x 8760 hours /2

=45 552 000 kWh in one year

which requires

45 552 000 kWh x 3.67 m3 water

= 167 175 840 m3 water = 167.2 million m3 water.

Now what is the amount of water turbined into powerhouse via penstocks with a discharge of say 23.33 m3/s in one year when 3 Kaplan units x 23.33 m3 /s =70 m3/s working?

= 31 536 000 seconds  in one year  x 70 m3/s / 2

=  1 103 760 000m3 water  (3 Kaplan units  x 23.33 m3 /s =70 m3/s)

This means that we need 167.2 million m3 water to generate 10.4 MW in one year but we discharge 1 103 760 000m3 water from the 3 turbine outlets. This shows discrepancy.Because Specific Water Consumption for Power was For head= 100 m. and Kaplan does not work above 26-27 m heads. This was to show the method to check reliability of Ins. Pow.-disharge in penstocks, Peak Powerhouse Capacity factor by Specific Water Consumption for Power.

 

Mass further out from the center of rotation generates more momentum of inertia, which is why ice skaters spin faster when they align their arms along the center of rotation. In general, you will see that reaction turbines have shorter blades that allow for higher spinning.

Pelton turbine generates power by converting the energy of falling water to kinetic energy by using a nozzle and striking the turbine by thus formed jet. Further  the jet strikes, more moment the exerts on bucket arm thus effective harnessing of power (it is easier to open doors with effort applied farther from hinge) . However we cannot make very large runners just to ease power harnessing as it will have very high inertia. Now, for geometrically similar turbine, the diameter of runner will be much larger compared to other counterparts thus lowering the speed of runner hence lower specific of turbine.

Turbine speed is not wholly dependent on water velocity; the turbine will turn at a constant speed because it is directly coupled to the generator, where a Governor is maintaining stable RPM by controlling the load. Load vary inversely with the RPM. Turbine efficiency varies with load(electricity).  But as the disparity between actual and optimum water velocity grows, less of the energy from the water is transferred to the turbine

 

For power benefits, the energy output will vary in accordance with the inflow, outflow, and change in storage and corresponding head, tailwater elevation, turbine capacity and plant efficiency. If the plant is a part of a system, the output may be subject to varying demands of the system load curve and whether the plant is to be used as a base load plant or a peaking plant.

The change of angular momentum of the fluid between runner entrance and runner exit causes a torque acting on the rotating blades. This torque is proportional to the angular momentum change per unit mass and to the mass flow rate passing through the runner blade row.

Efficiency of turbines is also related with how much power is generated and and how much of this generated power is extracted from the powerhouse by consumers. Balance between installed power and consumed power by the consumers. Powerhouse being close or far to the consumers also effects the efficiency of the turbine.
Turbines have 3 efficiencies hydraulic +mechanical +volumetric
Hydraulic turbines are machines that develop torque from the dynamic and pressure action of wate

Relation between torque and power is rotational speed.The amount of torque the engine can exert usually varies with rpm.
 

HP = TORQUE x rpm / 5252 Power is dependent on torque and rpm

 

A characteristic feature of turbines is that the torque decreases with increasing speed. A turbine has its maximum torque at standstill, whereas the torque becomes zero at the so-called runaway speed. It is one of the tasks of the speed governor to prevent the runaway of a turbine. Today water turbines are usually controlled by digital electronic speed governors, the electric signals of which are transformed to the hydraulic actuating system e.g. by means of proportional valves.

In practice, the head of a turbine is not a constant but varies within certain limits. A similar consideration of the velocity triangles shows that both shock-less entry and zero angular momentum at the runner exit cannot be achieved with another head in case of a Francis turbine. Hence efficiency will drop when operating apart from the rated head.

Low heads are difficult to exploit economically, as the physical dimension of the turbine increases when the head is reduced. Therefore, heads below 2.5 to 3 m are rarely exploited on the basis of the conventional turbine technique.

Turbine power is mass flow multiplied by specific enthalpy drop. Thus low enthalpy drop means high mass flow and high volumetric flow, which makes the turbine design stage easier (larger relative blade height).

Thanks for your question.  Both NYPA's Robert Moses and OPG's Sir Adam Beck are flagship power stations.  They've been studied and optimized for decades, so it is not surprising that their turbines can operate near 97% efficiency. The economics behind the scheduling and marketing of energy are complex, and I am sure NYPA strives to dispatch the plants to produce at peak efficiency.  But there are doubtless heavy load hours every year when it's beneficial to run turbines below their peak.  And, perhaps, one or more of their turbine runners are optimized to produce additional power, at a sacrifice of some efficiency, whereas others are optimized for peak efficiency, at a sacrifice of some power.

The overall plant efficiency includes unavoidable losses which come about from moving water through pipes toward the turbine (friction and momentum changes), conveying water away from the turbine downstream, electrical resistance losses in the generator, and stepping up electricity as needed to connect it to the grid.  I suspect the overall efficiency of the Robert Moses plant is more like 85% when all these factors are included.  But the figures 15% and lower must be referring to something other than overall plant efficiency.

And well designed pumped storage systems like these usually have a round-trip efficiency above 75%.  (My information is a little dated and I'd welcome a correction!)  By round-trip, think of a blob of water, then consider the electric energy consumed to pump it uphill then subtract the electric energy produced when it spins the turbine going downhill.

Anyway, in summary 85% efficiency is not uncommon from a hydro power station. The low-percentage figures you remember hearing must be referring to some other aspects.

Tim Gard's picture
Tim Gard on Dec 24, 2020

The engineer I spoke to referred to the speed and mass of the water exiting the system. The mass 32,500 cubic feet per second is the same as it leaves to plant as it was went it went in. Which is logical as water can not be compressed. If a cubic foot of water weighs 62.3 pounds, and there are 32,500 cubic feet passing through the American turbines per second, that means there are 2,024,750 pounds of water entering and leaving these turbines every second. I know mass in motion is energy, and that is one huge mass moving very quickly! I asked him why they didn't simply load the turbines more to generate more electricity, and he said if they loaded the generators that much it would cause the turbines to slow too much until they would not turn fast enough to generate any at all. This is how he explained the very low efficiency numbers he gave me. And being he was the lead engineer at the Niagara Power Plant, I took him at his word. But I can not find anyone to verify this, and his explanation and experience is something I can not take lightly! Was he right?

Tansel  Yılmaz's picture
Tansel Yılmaz on Dec 26, 2020

 

The efficiency of generators is generally greater if the unit is large , but it too depends upon the load carried :

Inside the scroll case there are stay vanes called the wicket Gates. These control how (and howmuch) water flows into the turbine. The wicket Gates are controlled by an external governor system. This opens and closes them and respond to conditions on the national grid. A high load on the system slows the turbine. The governor system reacts by opening the Gates further to speed it back up. Conversely it will narrow the wicket Gates from response to no load on the system.

 

The efficiency of transformers increases rapidly with installed capacity and load within certain limits  whereas that of transmission lines increases with installed capacity but degreases with load as electrical energy regarding the electrical losses are equal to the square of electrical current (Ampere) times Resistance (Ohm).

 

I wish you all a Happy New Year.

 

 

 

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