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ELECTRIC DUCTED FAN PHYSICS
********See photos of the sample tested fan at the end of this blog********* Contents: 1. SAMPLE ELECTRIC DUCTED FAN 2. SYMBOLS AND CONVENTIONS 3. WEIGHT AND MASS 4. AIR 5. FORCE=MASS TIMES ACCELERATION 6. ENERGY=ONE HALF THE QUANTITY MASS TIMES VELOCITY SQUARED 7. POWER 8. EDF SIZE 9. EDF THRUST 10. EFFICIENCY 1. SAMPLE ELECTRIC DUCTED FAN Motor, Neu 1527 1.5y 8mm shaft esc, Turnigy 200 amp "monster" Fan, 125mm custom Ducting, inlet ring and split thrust tube with 90% FSA outlet, 14.28 in^2 lipo, 4ea. Zippy 3600 6s 20C in series/parallel (12s 7200) volts, 43.1 current, 146 amps (the meter reads 14.6mv or 146 amps on a 500 amp shunt) watts, 6292 thrust, 20 lb. 1.6 oz. 2. SYMBOLS AND CONVENTIONS v= velocity in feet per second F= force in pounds #= pounds m= weight in pounds divided by g a= acceleration g= acceleration due to gravity or 32ft/sec^2 E= energy P=power; watts, horsepower, or ft lbs/sec. math is done using BASIC symbols in a search window, Google and most others will work. * is multiply; 3*4 = 12 / is divide; 7/2 = 3.5  is subtract + is add ^ raise to the power of....; 3^3 = 27; 5^2 = 25; 16^(1/2) = 4; 27^(1/3) = 3 pi is 3.14...... 3. Weight and mass. Just because a large rock happens to weigh 32# on Earth and ~6# on the moon doesn't pin down its mass. If we divide it by g we get units of mass. 32#/32ft/sec^2 or 1#/1ft/sec^2. If we did this on the moon we would have 6/6 and get the same number. I'm going to jump ahead to show how this works. The 32# rock is in a frictionless sled and we are going to push it with a force of 4# horizontally. How fast does it accelerate? Newton's second law: F = m*a Divide both sides by m: F/m = a or swap a = F/m substitute numbers: a = 4#/1#/ft/sec^2 = 4 ft/sec^2 The units cancellation is confusing in linear format. I included them to show that the units are consistant. When using the search window, only the numbers would be used: 4/1 = 4 So the rock would gain 4 ft per second for each second it is pushed. 4. AIR SLUGS: The density of air is .0748lb. per ft^3 at sea level and 68 degrees F. We need to convert that to mass units and as in section 1, we will divide by g. .0748/32 = 0.0023375 this was cut and pasted from a search window result. In aerodynamics the mass units are called slugs. Density correction for altitude: I am at 900 feet elevation. A correction factor for elevation is 3.16% per 1000 ft. It is accurate enough for our purposes up to 10,000 ft. .0023375900/1000*.0316*.0023375 = 0.0022710215 We will round it to .002271 slugs. Density correction for temperature: The density of air varies inversly with the absolute temperature. My Temperature is 72F To convert to Rankin, 460+72= 532R. The standard air, 460+68=528R. The inverse ratio: 528/532*.002271 = 0.0022539248 rounding to .002254 slugs. This was a small correction of about 1%. Not doing these two corrections would have contributed a 4% error. 5. FORCE=MASS TIMES ACCELERATION F=ma is the cornerstone of the laws of motion. Birds bees planes helicopters all generate lift by accelerating air downward. The induced drag of a wing is proportional to the tangent of the downwash angle. The static thrust of an edf is proportional to the mass and proportional to the velocity of the efflux. I have an edf with a nozzle of 14.28 in^2. What would the thrust be if the efflux was 250 feet per second? Since this is a continuous process, we will take a one second slice. F=ma, we need mass. The cubic feet of air per second would be 250fps*14.28/144. Multiply that by .002254 slugs we previously calculated for the mass per second. We divided the nozzle area by 144 to convert in^2 to ft^2. The mass per second calculation would be: 250*14.28/144*.002254. This mass would accelerate to 250 feet per second in one second, that is our acceleration. we will put into the search window: 250*14.28/144*.002254*250. Here is the result 250*14.28/144*.002254*250 = 13.9701041667. rounding to 14 pounds of thrust. Notice the velocity is in there twice. So we could have squared it but that would obscure what we did. We determined thrust from velocity. My edf has 20.1 pounds of thrust so what is the efflux velocity? We have to solve F=ma for velocity. m = area*v*slugs and a=v/sec so.... F=area*v*slugs*v/sec. divide both sides by area*slugs gives... F/(area*slugs)=v^2. Take the square root of each side. (F/(area*slugs))^(1/2)=v swap sides.... v=(F/(area*slugs))^(1/2). Do the numbers: (20.1/(14.28/144*.002254))^(1/2) = 299.8735874083 Or 300fps. Convert to mph: 60/88*300 = 204.5454545455 rounding to 204mph. 6. ENERGY=ONE HALF THE QUANTITY MASS TIMES VELOCITY SQUARED If we took the 32# rock and threw it at 50 ft/sec, how much kinetic energy would it have? E=mv^2/2 where m= lbs/g E=32lbs/32ft/sec^2*(50 ft/sec)^2/2 The seconds cancel and one ft cancels leaving ft lbs, do the math.... 32/32*50^2/2 = 1,250 ft lbs. A quantity of energy. 7. POWER Power is a continuous flow of energy. If we threw one 32# rock at 50 ft/sec every second that would be 1250 ft lbs/sec. Convert to horsepower: 1250 ft lb*1 hp/550ft lb. The ft lb cancels leaving 2.27 hp. Convert to watts: 2.27 hp*746watts/1 hp. The hp cancels leaving 1695 watts. If we threw a continuous stream of 50 ft/sec tiney rocks at the rate of 32# per second, that would be the same 746 watts. In 3. above, we calculated thrust from mass and velocity using F=ma. so in this case convert # to mass, #/g and 50 ft/sec per one second slice.... F=32#/32ft/sec^2*50ft/sec^2 =50lb TO BE CONTINUED 300^2*.002254*14.28/144*300/2/550 = 5.4864409091 horsepower in efflux 6292/746 = 8.4343163539 watts input converted to horsepower 5.4864409091/8.4343163539 = 0.6504902921 (.65 or 65% is the calculated efficiency of the example fan as a ratio of the watts into the ESC and power in the fan efflux expressed in watts) This will be clarified in sections 8, 9, and 10. ImagesView all Images in thread





Tom, can you share some of your construction methods you used to attach your blades to the machined hubs you made for the 6 and 12 blade fans?
Also, I'd love to ask you a question regarding prefan stators. Do you have any experience with pre fan stators? In turbojet engines, often there are pre compressor fixed blade stators that get the air rotating in the same direction axially as the compressor rotates. I'm wondering, in a layman's way, if this could potentially improve fan efficiency in some way. Really why I'm interested in this is for airplanes without optimum ducting. I'm looking for ways to improve ducting issues in some of these cheap foamies with long ducts  the Hobby Top Gun F100 comes to mind immediately as its one of my models in the production line up. The ducting is a little nightmarish though at least it is ducting  some don't even have that. But I was considering running the fan without a spinner on it and instead fitting a short rocket tube of the same diameter of the spinner in front of the fan and then it would have a static rocket nose cone in front of it to fair into the fan. The tube would be held in the duct by stators, and those stators would not just be flat but would introduce some small amount of swirl to the air to get it twisting in the same direction as the fan. Not too much, but just a little. I'm being a little bit of a copycat here but I'm wondering of some of those other engines have something in these pre stators. Also, the thought of not having a spinner on the fan is another way the fan can be more easily balanced. The spinners tend to create balance problems in all but the best fans that are dynamically balanced. So if the pre stators can improve blade efficiency by offering optimized air flow at the fan face and the fan can be more easily balanced with less mass via no spinner then aside from the setup being more of a hassle, it may be an improvement? 



Also Tom, I was interested in a comment you made in another thread but thought I'd ask the question here as it's directed toward your fan knowledge and not so much the thread it came up in  here it is:
********* Phil, The formula you used would only be correct if there were no least common denominator between the rotor and stator count. My fan has 12 and 5 or 60 individual passages per rev so at 30,000 rpm you get 30 khz. If it had 12 and 6 then there would be only 12 simultaneous passages of 6 blades so at 30,000 rpm you would get a powerful 6 khz. That is how a siren works! ********* I'm not sure how this goes, I can see 12 blades, 5 stators being 12 blades passing 5 stators in one revolution, or 60 blade/stator passes (12*5=60) Is the fact that the fan blades pass the stators nearly simultaneously on the even blade count/even stator count the key for khz sound? I'm thinking it is  it's typically nice to have something other than a whole number when dividing blade count by stator count, right? For example, I am refanning and motoring one of my older models. I have some very old Robbe Rojet fans in a 4 engine BAe146. I'd like to maintain a nice sound that it currently has or even improve it along with a power upgrade it will get with more powerful brushless motors. It has 4 blade rotors and 3 blade stator shrouds in it's current stock configuration at about 200 watts per motor. I'm thinking of using a 5 blade rotor, or a 6 bladed rotor or a 9 bladed rotor at a little less than 300 watts per motor. What concerns me now is the stator count. Am I assuming correctly that I should hear some loud fan noise from the 6 or 9 bladed rotors due to the stator count? My next experiment would be to fashion some stator blades to add to the shroud. This would not only increase strength of the motor mount but if it could fine tune the sound on say a 9 bladed fan it may be worth the hassle. What do you think? Feel free to answer on PM and erase these posts if you believe this will clutter up your blog thread on fan design. Thanks again and thanks for posting your thoughts on fan design. It's a great subject these days with so many copycat fans running around with interesting features but questionable combinations of features. 



That does it! I'll message you when I do.
My wife is finnish. tom 
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