The Crossflow Turbine
Unfortunately bulletin #25 is not a "step-by-step" manual, much to the disappointment of many. When I first saw in in 1978, I found it fragmented, elusive, overly technical missing a few formulas. It' more like the technical ramblings of someone explaining the concept of time and the theory of how a clock works when all you wanted was to know the time. However with some diligent "head scratching" over a suitable period of time you will eventually sort out the relevant pieces of information contained in the bulletin. There are a few point to keep in mind when reading and working on some of the calculations in the bulletin. Keep in mind that it was originally a German document written in the early 1930s. It is very likely that some meaning was lost in OSC translation of the original document. Also bear in mind that some of the formulas in the document do math conventions that we use today ie. addition, subtraction, multiplication then division. You'll have to "play with" the brackets. However taking the bulletin as a whole it follows the same mechanical & hydraulic principles used today in turbines & pumps. Those are a curved blade's reaction to a jet of water (in turbines) and water's reaction to a moving curved blade.
About half of the math in the bulletin deals with the physical relationships between the mechanical turbine elements (blade geometry and runner inside & outside dimensions) and the other half deals with the forces produced on the blades. The forces are represented as "vector diagrams" Vectors diagrams help one to visualize what is going on inside the runner. If you understand them you can analyze changes made to the turbine and or changes made in operating parameter such as head & load. Next to the calculus used in the bulletin, the vectors are probably the most baffling things in the bulletin. The vectors will be explained shortly however, as for the calculus, "I ain't goin' there."
Don't expect to "re-engineer" the crossflow runner. Banki & Mitchell "did their homework" on it. The proportions of the blades to the runner diameters and angles involved are fairly "fixed" and cannot be arbitrarily changes without adversely affecting the power & speed of the runner because these changes affect the efficiency of the turbine. As alluded to in the paragraph above, the runner diameters and blade dimensions are a compromise of mechanical dimensions & mechanical efficiency. Taking all the above into consideration, this article will not be a step-by-step interpretion of bulletin #24 but will be my personal "practical inturpation & explanation" of it. Before getting started I want to clarify something. Bulletin #25 title "The Banki Crossflow Turbine" is somewhat misleading as it only deals with the "runner" aspect of the turbine. After discussion of the "runner" portion of this article I will expound on the turbine design as a whole as well as some abstract theory about the crossflow turbine.
Before calculating much of anything else we need a little understanding of vector diagrams. It will take several stages to illustrate this in it's entirety go get a cup of coffee and come back. In understanding how a jet of water° angle to the plate. In engineering terminology this is called the flat surface held normal to the jet. Go figure, I don't know how or why they come up with this stuff! Anyway, the jet will be forced to make a 90°turn, thusly spreading out over the plane of the plate. In this case the force ( F ) will have no component in the plane of the plate. In other words there will be no forces trying to move the plate "sideways" to the jet. The force ( F ) is computer as F = M * v. Again in engineering terminology it said that "the jet's momentum in it's initial direction is wholly destroyed. This just means as the waters energy was dissipated out radialy 90°into never-never land and that no "work" was done. Remember in high school physics you were taught that for "work" to be done there motion has to be imparted. In our case here we needed the force ( F ) to cause the plate to move for there to have been "work" done. acts on a surface we first use the "Flat Plate Normal to Jet" illustration. In this a flat plate is held at a 90
On the other hand now if we force the water through a smooth 180°turn the force ( F ) is doubled. The force ids doubled because the equation F =( M * v ) + ( -M * -v ). That is F = (Mass * velocity in the initial direction + (Mass * velocity in the opposite directing. Reducing this equation gives us simply F = 2 * M * v. One thing that might be a bit confusing in these two illustrations is the arrows indication the direction of ( F ). It might appear that V is moving in one direction (and it is) and that ( F ) is moving in the opposite direction (which it is not). What the ( F ) directional arrow means is a "resistance" opposite to the direction of ( V ). Again, it's an engineering thing.
Now that we are up to speed on velocity, mass & force, lest look at some vector diagrams along with the blade shapes that produced them. In this diagram the jet is being deflected by 70°or so. In applying these momentum theorems or laws as they should be call to turbines is as follows. If a jet of water strikes a curved blade the water is deflected by the angle ?.A force (F) is imparted to the blade in two directions, x & y. These forces are calculated thusly
Fx = M * (v – v cos ° α) or Fx = M * (v – v cos ° α)
&
Fy = M*v * sin α
In this diagram the two velocities are the same but separated by angle α and the triangle is closed by closed by the line ∆ v as dictated by the laws of cosines
∆v = the square root of v1^{2} + v2^{2 }– 2 * v1 * v2 * cos α^{ }
∆v = the square root of 2 * v2 – 2 * v^{2} * cos α
These two forces combined is equal to:
F=M * v * the square root 2 * (1 – cos α)
Here is the general text book vector of a Pelton wheel in motion.
This is the path & vector diagram of a Pelton wheel. It is showing 2 buckets. The bucket on the left is showing the absolute path of the water jet while the right side is showing the relative path if the jet. If the wheel were "locked down", the water path would indeed follow the path as indicated on the left. However in a running machine the wheel is moving in the same direction as V^{1} . The water jet is trying to follow the inside curvature of the bucket but because the bucket is trying to move away from the jet the path is straighter as indicated in blue then had it mage the near U-turn of the absolute path. The vector is compound, in other words its is showing more then just one part of the blade. The left triangle is the vector for the entry of the jet to the bucket & the right side is of course showing the water exit from the bucket. The inset illustration shows what happens when the wheel (or runner) is in over speed. The path flattens out more and you can see in the inset vector that μ^{ 1} is approaching the same length as V^{1}^{ }Very little power is now being produced and aa a matter of fact the power that is being produces is rather in driving the intended load, is being spent on maintaining a high wheel speed and overcoming windage & friction. Notice that three additional notations are included. V^{2,} v^{2} * β. In hydraulics the following notation conventions are used.
Getting a little more complicated visually but still the type vector. Here we have the Francis runner. The actual water path is shown in red. Again the right side of the double vector in showing the entry of water and the exit is shown on the left. Both are shown here on orange. The Green vector is actually the right side entry but for clarity is is duplicated somewhat larger to show the geometry of the lead edger of the blade to the outer periphery of the runner. I think before going on to the crossflow & should as Rickey would say to Lucy, "Le me splain something to you." In hydraulics as in any other engineering field they has it's own set of mathematical notations and also a hierarchy or naming conventions . Most of the confusion in vectors comes from water velocities. If the notation is a big V, then that velocity is an absolute one. If it's a little v, then its a relative velocities. Most turbo machinery only has one in and one out. No so with the crossflow. It's twins! It's got 2 of everything. To keep all the symbols straight I'm not going to "splain" it, but rather illustrate it with a chart. I think the chart and the crossflow illustration can explain this much better then I can. Just so we are clear, the term "quadrants" is mine. The 1st in the entry of water to the at point "A". The 2nd is the exit of water from point "B". The water then crosses the interior of the runner and then re-enters the runner at point "C" in the "minus direction" (remember our discussion above F =( M * v ) + ( -M * -v ). Water then exits the runner in the minus direction at point "D"
| 1 st Quad | 2nd Quad | 3ed Quad | 4th Quad |
Absolute Velocity | V^{1} | V_{2}^{1} | V_{1}^{1} | V^{1} |
Relative Velocity | v^{1} | v_{2}^{1} | v_{1}^{1} | v^{1} |
Blade Angle | β^{1} | β_{2}^{1} | β_{1}^{1} | β^{1} |
Attack Angle | α^{1} | α_{2}^{1} | α_{1}^{1} | α^{1} |
Runner Velocity | μ^{ }^{1} | μ^{ }_{2}^{1} | μ^{ }_{1}^{1} | μ^{ }^{1} |
The problem with the crossflow is just that “crossflow”. Only about 72 % of the waters energy is extracted in the top of the runner leaving only 28% to be extracted from the lower section. The exact ratio is dependant on the actual diameter of the runner, how many blades are being, the length to diameter ratio, the head, bla, bla,bla. Under "ideal circumstances, 50% of the power would be produced in the first section and 50 % from the last section. This will not happen because of the internal crossing of the water in the runner center section. Ideally we should have "laminar flow" all the way through the runner. Laminar flow means that that all the water particles in a given area is flowing parallel to each other and are at the same speed. Think of this like the telescoping antenna on a car where the innermost core has the fastest flow. You will never get true laminar flow due to friction of the surfaces involved. Laminar flow is destroyed by excessive restrictions and abrupt changes in flow area or directions. When it is destroyed friction is the result. When the individual jet filaments cross and interfere with each other that too pretty much "kills ' hell" out of laminar flow. We have a tremendous disruption in flow now plus of air is now being introduced into the water path. By the time the water gets to point "C" it's pretty well diffused to a wide pattern This causes the flow V_{1}^{1} to enter the blade at an attack angle varying widely from the the 16° it should be. That why the 28% of the available power is extracted there. This is illustrated in fig 3 of the bulletin. However there nothing you cab do about it.....or is there? Read on Grasshopper. We'll I think we'll all had enough turbine dynamics for this week so lets move on not to some actual calculations.
Before any "design work" is to be done their are a few things that must be known. First you must have a reasonable expectation of the amount of power that might be produced from a given site. For instance, don't expect to supply a full household with electricity produced from a scenic babbling brook running across your back yard. It takes a lot of water to produce electricity. The "head" and "Q" must first be determined. The "head" at least in the US is measured in feet. "Q" is the quantity of water and in "micro-hydro" work it's usually given in CFM (cubic feet per minute) and in larger turbines is given in CFS (cubic feet per second). Be sure when calculation from formulas in other documents you pay attention to & convert units as necessary. In this article I use CFS.
To begin we first determine the power potential of our site. For convenience, (mine) throughout this article I'll be using my own site for the design & evaluation. That is the head ( H ) = 26 feet and the flow ( Q ) = 8 CFS. The formula for determining the potential hydraulic horse power is ( H * Q * 60 ) / 660. This is the raw horse power potential and does not reflect any efficiency or loss's. According to the formula my potential horse power output would be ( 26 * 8 * 60 )/660 or 18.9 HP. Assigning a efficiency figure is difficult. I want to be conservative in this figure so let's use an overall plant efficiency of 75%. Therefore 18.9 * .75 = 14.18 HP. One HP is equivalent to 745.7 watts so 14.18 * 745.7 = 10574 watts or 10.57kW of electrical power. This is enough to "do a house."
Now having that out of the way we can start to design a runner that will accommodate the site. In the bulletin pages 10 through 15 deal with the construction proportions of the runner. The information we need from these pages are: Formula #35 Q=volume of water, Formula #36 L= blade length, Formula #37 ρ=blade radius & Item (E) Central angle on page 15. Once these values ate known you take these figures to a machine shop and have them form the blades from flat stock to conform to “ρ”, machine the blade sections to form the 73.46° arc in item 3, & finely cut the blade to their final length (L). You then pay the man a huge sum of money & prey your calculations were correct. Definitely NOT the way to go. There is a much simpler & cheaper way to arrive at near the same result but first a quick discussion on one aspect of Banki's design. The dimensions and angles in the bulletin represent the "near" optimum dimensions & angles to satisfy mechanical advantage & un-restricted passage of the water. These dimensions are fairly "fixed" and therefore cannot be arbitrary changed without some decrease in efficiency ie. power & speed output. As a result of this there are definite dimensional relation ships between the various components of the runner. I wrote the following formula to determine the runner diameter from the blade radius. D1=2 * ρ /.326.We can use this to great advantage because a supply of blade stock is all around us and is readily available in the form of steel pipe. If you've ever seen some crossflow runners up close before, they mostly look the same, say 12-16 inched in diameter and have an aspect ratio of 1:2, that is 23 -32 inches in length. They also look like the blades were fabricated from 4 inch steel pipe. You're right. But remember what your Momma told you when you were 8. "Just because everybody else is doing it doesn't mean you have to." My point is al lot of these turbine were fabricated from readily available materials and hey, there's nothing wrong with that. However, waiting and searching for that optimal "readily available materials" will save you some money and very likely gain you some efficiency. I'll go through what happens when you design a runner without regard to the project as a whole. Most commonly available is schedule 40 pipe and below are some of it’s specs. Of course there is an almost infinite range pipe sizes in industrial & construction grade so finding a size that will meet your needs should not be a problem. Using this method we supply ρ and let industry supply us with tailor made materials.
There seems to be a lot of 4 inch steel pipe around. Let's see if we can design a runner around that size. We start by finding the jet thickness which started at item 4 on page 17 of the bulletin. Area of Jet = Q/V
L = 210.6 * Q / D1 * H^^{.5}
L = 210.6 * 8 / 12.27 * 5.099 = 25.5 inches.
Oh by the way, H^^{.5}^{ }is the same thing as^{ } "the square root of "H". It took me a while to figure out that one. Anyway with the initial blade length calculated as 25.5 inches, divide the "Jet area" of 28.8 square inches by 25.5 to get the S_{o} which in this case = 1.13 inches.
The only thing we need to factor in now is the nozzle efficiency and adjust the length accordingly. If you follow good hydraulic principle and design a good gate/nozzle you should be able to achieve an nozzle coefficient of .98.That’s only a 2% off peak which would be 1.356 which translate to a .223 increase in runner width to compensate. This would bring the runner length to 25.72 inches. If it were me I’d bring the runner on out to 26 to28 inches just for grins and a little more error margin. Now we’re looking at a D:L ratio of 1:2.08 Not terribly bad but! A 28 inch wide small diameter turbine is going to be a machining & welding nightmare. Building the runner itself it not too bad but I would add in 2 extra center support disk for rigidity. Of course you’ll want to extend the runner length again to compensate for the width of the extra center supports. We’ll we’re now out to 28.5 inches. Do I here 30?
My Daddy used to tell me, “You’ve got to use some horse sense”. Although I was never very good with math, I do have to ability to “visualize” how things function & anticipate problem areas. Having some “horse since” also helps. Here comes the first major problem in designing a turbine. Let’s tentatively select 4-inch pipe to make out blade sections from. We might select it because it look good & stout and because it is relatively easy to come by. That will make us a runner 12.27 inches in diameter. At this point my horse is telling me “there aint no way”, you’re getting 8 cubic feet of water a second through a 12 inch runner without major difficulty. It’s not impossible just not practical and here’s why. Anyway fabricating all the flat stock for the gate & nozzle assembly will really be the difficult part. That’s an awfully wide gate assembly. At a 26 foot head you only have a shade over 11 PSI at the lower end of the system but think about it. You have a 28.5 inch wide gate perhaps transitioning back several feet to a round penstock. That might present top panel behind the gate valve of 28.5 x 36 inches. Multiple that times 11.25 PSI and you’ll have in the neighborhood of 11,550 pounds of force acting just on the top panel of the gate. Even if your welds held, the things going to bulge out & distort like a balloon. Personally I’d give it around a 100% failure rate within 10 minutes.
What's a fellow do do? We'll before I through the preverbal monkey wrench into the mix, lets fix this problem first. To get a narrower runner we need to make it'd diameter larger. This is done make it larger by choosing a blade with a larger radius. This time we'll step up to blades made from 6 inch pipe. Building a 18 gate & nozzle would be child’s play compared to a 28 or 30 inch gate. The mechanical stresses by water pressure would be reduced almost 70 %. The machine will cost more to build mainly due to the heaver drive components required because of the slower speed & greater torque when compared to the smaller machines of equal horsepower. But here you’re getting into a serous machine of much higher durability and a much greater potential for increasing the efficiency beyond the apparent fixed limit or 87%.I’ll comment on this a little later. The runner built from 8 inch pipe is even better with some qualifications. Again, the cost will be higher because of even larger bearings and shafting required. However building a nozzle, gate & transition 13 inches would really be a piece of cake. The biggest concern with a runner this large in diameter is the the loss in head due to the higher inlet. It's only a couple of feet in this example but may be a consideration.
Alright, as promised, I'm throwing a monkey wrench into the works. The problem comes when calculating how long the runner needs to be. Notice in the calculations & illustrations in the bulletin all the math used an infinitely thin blade. If this is not realized it will cause you to calculate the runner too short. You might not notice this until you go to full load and “it just aint makin’ it”. Refer to page 9 of the bulletin to figure 5 for the spacing used. Using our 12.27 turbine as an example, if we multiply it’s diameter by pi we have a circumference of 38.52 inches. This gives a blade spacing( t )in the outer periphery of the runner of 2.14 inches. The illustration to the right shows the problem very well using a exaggerated blade thickness in blue. The original S_{1} value "A" that should be is 1.25 inches had been reduces to "B" 1.04 inches. Our jet thickness S_{o} "C" has dropped from .85 inched to "D" .64 inches. That's a 21% decrease in jet area from the original calculated value. This means to keep the efficiency as high as possible the runner length will have to be increased 21%. You don't need any fancy math or trig. to figure out just subtract the blade thickness from the calculated value of S_{o}. Calculate the percent difference & multiply you original blade length by that percentage as we've done above.
Specific Speed
Specific speeds is a dimensionless number. In broadcast engineering they call this term "normalizing", if any of you are familiar with Smith Charts. The term is used to “level the playing field” if you will, so that all types of runners can be evaluated under the same conditions .As a result the term via it’s number define the shape of the runner. I remember from a long time ago one hydraulic document described it this way. If a model of any given turbine were build with a 1 foot diameter and operated with a 1 foot head, then the specific speed is the speed that the runner would turn to produce 1 horsepower. I guess that about sums it up for a level playing field.
What it all means is that a turbine with a high specific speed will while running a full load, be turning faster then one with a low specific speed. An extreme example are the Pelton wheel which has a very low specific speed. It is usually thought of as a high-head machine. However it can be very efficient a low heads. It’s just that it turns so slow at low heads that the cost the equipment needed to increase the shaft to something usable by a alternator may cause the whole project scrapped or re-dome with a turbine of a higher specific speed. Also a low specific speed is also thought of as a low volume unit. This really makes it an ideal selection for mountainous terrains where large quantities may not be available.
On the other hand you have the Kaplan Turbine which is an axial flow (propeller) turbine. It has a large specific speed and is used mainly on large dammed hydro sites where then the is somewhat low but the quantities of water available are staggering. These turn relatively fast rate when compared to the crossflow, Francis & Pelton. They would not be suitable for medium to hi-head as they would turn much to fast to be practical. When every thing above is considered the crossflow would be an excellent choose for low to medium head operations. However it’s not a weekend project and must be engineered properly if it’s going to be efficient and last. If I were King of the world I’d make all crossflow builder applicants take the following test. Can you draft? Can you weld? Can you run a milling machine and a lath? What is the square root of 2? Convert 1 PSIG to Head. If you’re carrying all the feathers you can carry, can you carry one more? You had better be able to rattle off without blinking, “yes, yes, yes, 1.414, 2.31 and no. ”The point is this is a real engineering project and is not for the typical do-it-yourselfers.
Nozzles
I believe that bigger is better up to a point. In the case of selecting a runner diameter, using a larger & therefore narrower runner not only saves money and add durability but does offer up a few extra chances at increasing the efficiency of the crossflow. Take a second look at the Horse power formulas #2 & #6 on page 7 of the bulletin. Remember the lows of cosines? The 16 ° α^{1} is usually chosen as a compromise between hydraulic efficiency and mechanical clearances in adjusting S_{o}. Therefore if a1 is reduced the efficiency & power output will increase. With a large diameter runner this is much easier then in a small turbine. You could lay that angle down to maybe as low as 8 °.Of course you would want to lengthen the runner to compensate for the smaller S_{o}.
General Layout of Flow In Nozzle
The nozzle diagram above is meant to show some general proportions. For maximum efficiency the runner should be designed for single blade operation. However in the interest of construction difficulties in building a wider runner, a double nozzle - blade arrangement may be used at some loss in efficiency. The proportions are general. For instance I chose the radius of the nozzle curvature arbitrarily at 2 times the runner diameter. The exact radius in not important so long at it gives a nice long gentle sweep into the blades. The arc of the nozzle is also an arbitrary figure. I placed this one at 73 degrees “just because”. That long sweep & mechanical clearance is all that matters. You could go “straight in” as the folks a OSC did when they built their turbine using the freshly translated document from Banki’s original papers. By the way, does anyone know how to get or has a copy of the “original Banki papers? However they had some pretty horrible efficiency numbers with their turbine. They may have been “Jim Dandy” mathematicians but would have made a few changes on their nozzle design & transition assembly. Probably the thing that hurt them the most was the nozzle. It was a sliding gate that opens & closed “laterally”, that is across the runner face. What happened when you pot your thumb over the end of a garden hose? Using their arrangement that’s exactly what happened in their turbine. My guess is that at small gate openings the water might have been disbursed by 10-20 degrees. Another thing that hurt a little was not having a smooth transition between their supply pipe and the nozzle. It was a blunt sharp edge transition. That hurt them more at full power then anything else. I’m not trying to belittle any of the people involved I’m just trying to make you aware of “design flaws in engineering.” Left click on the illustration above to save it to your computer. It's actual size is about twice what you are seeing here.
What is important is the angle the water hits the blade at. This is generally taken at 16 deg. However, that is "relative: to the blade angle B, which itself is relative to the periphery of the runner. Through out the bulletin you see constant reference to a1. This is an extremely important angle, for it more then any other factor, determines the power output of the runner. However I’d have to say that 16 degrees is the maximum angle that one should use. Us it as a design figure then see if you can go smaller. Getting small requires “laying the nozzle down” closes to the runner. If you use a large enough radius and a long enough arc for the nozzle, you could get a1 on down to the 8 to 12 degree range. Any smaller though, and you’ll have to start thing about lengthen the runner. Going to excess on this could get you a nozzle with a low coefficient because of excess friction because of excess length.
There are several nozzle arrangements that may be used. Most of the commercial crossflow turbines built in Europe use mutable blade inlets. In all of these the nozzle in intrigal with the turbine housing. If you are makingare also nylon A deviation of the "sharp-edge orifice" is used to help eliminate the spreading of the jet as it leaves the nozzle The actual length of the nozzle is a bit longer then shown and the nozzle will have an adjustable pivot mount at the end of the assembly where it meets the runner housing. The other end its attached to the transition/diffuser assembly witch is mounted at the other end of the turbine sub frame assembly. There are two critical considerations when mounting a nozzle like this. Because the runner, nozzle and transition/diffuser are mounted together on the same frame, the alignment to the penstock is critical to that undue stress is not imposed and ether assembly. Ideally a flexible coupling would be the ticket but a commercial coupling would be rather expensive for a 12 inch penstock. Later I'll be adding to this section about flexible coupling and alternatives. A flexible coupling does three things. It does allow for some mis-alignment, It isolated the penstock from the turbine from mechanical vibration, it allows for expansion & contraction of the penstock due to temperature and it will allow for some movement should the penstock try to settle of shift due to waterhammer.
commercial turbines that's a very good idea because it saves manufacturing cost and makes an extremely ridged turbine assembly. This method is a little impractical for us little guys because not too many of us have casting facilities in our back yards or want to shell out major bucks for some "real machine shop work" Besides if the nozzle is cast in with the housing we can't adjusts the attack angle now can we? I took my design from one of the old Ossburger design. In stead of the water following the runner housing after it leaves the gate it follows a curved guide which is the top of the nozzle assembly. In mine I'll use a nylon or duron spring loaded back seal on the gate shaft. The side seals and shaft bearing are not show in my illustration but they are mounted externally on the nozzle housing. The sealing method on edges of the gate plate are not shown but they
Something Radical
A large diameter narrow turbines lends itself well to a radical departure from standard design. Since all the blades are fixed and have a fixed relationships no part of any blade can be moved by itself. In other words likeβ^{1} & β^{2}^{ } and β ^{1}_{1} & β^{2}_{1}. However it is possible to in effect change β^{2}_{1}. While β^{2}_{1} itself does not actually change, you can change the angle at which the water enters quadrant 3 an angle β^{2}_{1 }by using an inside guide within the runner. This would necessitate having an “open faced” on one side. I can already here some folks now in that condescending nasally voices. "Well if you do that then all the water will run out! Then what are you going to do?" Not really. Anyway what do I care what they think . Besides, these are the people who failed my test miserably!
Using just one blade set water leaves the blade between A & B. It re-enters the runner for it's 3ed quadrant of operation at D through E. At full gate operation using 2 blade sets, water leaves quadrant 3 from A to C & then re-inters at D through F. In the illustration shown here the water path for single blade operation is the blue lines. Adding a 2nd blade is shown by the green lines. This illustrated the spread of the water upon entering the 3edred. The pink area shows the water path for single blade operation. The light blue shows the path should 2 blades be used. In single blade operation point E is about as far back to your right as the jet will reach at normal speeds.. What I've attempted to do here is re-direct the water back open more blade set and inter at 16 degrees at that point. A similar situation occurs when the 2nd blade section is added. The water in confined to a course that does not vary with speed and it is always forces to enter quadrant 3 at 16 deg. quadrant of operation. I plotted the water path graphically and came up with the curve necessary for an internal guide shown in
The bottom surface of the guide would be either the center red portion above or the left red portion depending on weather you were using one or two blades in operation. The right red outline shows the top edge of the side walls of the guide. The illustration here is the concept of what the guide might look like. It's only a Photoshop assisted freehand drawing but I think you can see what I'm getting at here. A guide like this presents me with some interesting possibilities. However it is mounted, the mounting mechanism should be extremely ridged. There would of course be a standard fixed mount that would during installation but what about an "on the fly adjustment?" One way of doing this would have the lower edge be the pivot point so the quadrant 3 entry angle could be varied to suit flow & load conditions. Another possible mount scenario would be to "hang" the guide from the runner shaft using pillow block bearings. These bearing would of course be of the "double sealed" type suitable for such as wet environment. The guide would have a lever attached between it and the activation mechanism
As mentioned above in order to use this type of guide the runner would have to be open faced. This does present a small engineering problem. With this runner it is not necessary that the shaft extend thru the runner. It only need to can if the guide were under hung from it using pillow block bearings. However a more likely mounting configuration is what's called " an overhung load. An overhung load is defined as the radial load on the output shaft extension produced by a pulley, chain sprocket, gear, crank arm, cam or other similar device. It necessitates using a larger diameter shaft & beadings. In addition it requires a larger mounting boss to the runner end plate. More to come as this page expands.
APPENDIX I
Notes on the "Efficiency Formula" in Bulletin #25
The efficient formula is every bit a complicated as it looks. I really have very little though on how it works. Due to my lack of understanding or motion mechanics I’m forced to take Banki’s word on this one. However I can tell you what is going on in the formula. In the formula “C” is the nozzle coefficient. He’s accounting for that in the first part of the formula in dealing with ?^{ 1} & V^{1}.In the middle of the formula the term y is the factor describing the loss of energy caused by the separate jets crossing each other between the 3ed & 4^{th} quadrants. I believe that the loss of power is also represented in ? due to “shock” loss. Shock loss is when the relative velocities v^{2}_{1} & v^{1}_{1}are not parallel (in the vector diagram). This can be seen on page 11 of the bulletin on the left side of fig 7.The relative velocity of v^{2}_{1} suggest that the inside diameter of the runner at point “B” in the drawing above is turning at a different speed to the corresponding point “C”These actually turning at different speeds id clearly impossible since they the same physical surface. The last part of the formula is the velocity difference between V^{1} & ?^{ 1}to extract power.
APPENDIX II
Notes on the "Horse Power Formula" in Bulletin #25
The Horse power formula is not as complicated as it might seem. The formula can be divided into parts.
(W * Q * ?^{ 1} / g)the momentum part.g=gravity constant at 32.3 In checking my Excel and Q-Basic programs that calculate this I found a discrepancy withW.As stated earlier W the weight of water per cubic foot of at sea level is 62.2 ft. However to make the formula produce the correct answer in Excel a value of .13 has to be used. At the moment I don’t have time to fix it so I use the correction multiplier of .00291 in the horse power equations. In Q-Basic the value of 62.6 works just fine. I think they mean "mass" rather then weight.
- ((V1 * cos(a1) – u1) takes account of the laws of cosines & subtracts the wheel peripheral speed from V1 to that power is produced.
- (1+y) * (cos b1 / cos b2) the lump sum factor or runner coefficients taking into account the cosine blade angles of the 1^{st} quadrant and the 4^{th} quadrant. Their ratio would always be 1:1 because they are physically the same piece of steel.
This page will be an ongoing document. It will be up-dates and expanded as I have time. Eventually I hope to cover every aspect of building a crossflow turbine. I welcome comments on this page. Please let me know of any mistakes you find or clarifications that nee to be made. I will also entertain contributions to this page. Just email me and let talk about it.
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hi there,
pictures are missing. please upload.
This is Hassan from GIK institute. Currently working on micro hydro CFT. Please image do not show up. Kindly mail me to hasanifte@gmail.com
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