Recent content by W R-P

I'd say assume no frictional losses, then solve in terms of bend losses etc. You can then obtain V1 in terms of V2,eg. V1= sqrt(k2/k1)V2 (Where k is the sum of the loss coefficients ,the Ks will vary for each pipe.). Then plug that into Q=A1V1 + A2V2 to get the value of V2, and subsequently Q2...

Hey everyone. I've been working on a axial flux permanent magnet DC generator to use on a 100W wind turbine(small I know), to charge 12v batteries in low wind speed areas. Now that it's complete, I'd like to determine the full power curve at different rpms. So far I've been running it on NO...

The angles show the direction each force is acting,ANTICLOCKWISE relative to the positive x-axis ----> 150 N acts at 120 degrees, starting from ---> go anticlockwise 90 degrees,then add 30 degrees = 120 degrees 50 N acts at 250 degrees,not 290 go 180 degrees anticlockwise then add 70 degrees...

Well i those two equations are not that far off.lol Just a quick idea, you have your output speed in rpm. 1) using the speed-torque graph, find the torque at that speed. 2) using the torque, check the power-torque graph and find the mech power the motor is producing in Watts 3) Convert the...

A more pragmatic approach is to isolate the generator and : 1) perform a Prony brake test or 2) couple it to a motor(with a motor performance chart provided) to find the minimum torque(provided by the motor) required to start the generator. Hope this helps

Well we've got a vertical outlet, moving the fluid upwards against gravity by a certain height, H(the head). SO we can say the pump is doing work against gravity ie [W][/pump]= Force x distance = weight of fluid x head = m g H power is the rate of...

Your equations look correct but YOU FORGOT TO ADD THE LOSSES.lol Have a look http://postimage.org/image/2ydg1exwk/ V1 is the velocity JUST before it enters the pipe.In the pipe the flow takes a parabolic profile when it is fully developed,due to boundary layer conditions,so velocity is maximum...

If you really want to go into lava flow, http://www.itg.cam.ac.uk/people/heh/Paper221.pdf The top layer solidifies due to cooling,creating a tube like surface for the rest of the lava to flow,that's why the top layer(dark) moves slower than the rest(orange).

I think your diagram is similar to fig. 6.6 http://www.creatis.insa-lyon.fr/~dsarrut/bib/others/phys/www.mas.ncl.ac.uk/%257Esbrooks/book/nish.mit.edu/2006/Textbook/Nodes/chap06/node12.html The layer of fluid DIRECTLY in contact with the surface is at rest( due to boundary layer...

I'm sorry to say that calculation is wrong. Some airfoils(blade profile) will give you these forces as a function of the radius eg. L=0.902(r^2).But since this isn't given find the lift and drag forces like this: Lift is given as L= Cl(0.5x air density x WIND SPEED^2), where Cl is the lift...

I've written down what I was trying to explain (I assumed the pipe is horizontal,so z1=z2): http://postimage.org/image/2ydg1exwk/ From the description you've given me,it looks like you're discharging to the atmosphere..if you are, then it will be easy to find the flow velocity.P2 will go to...

this can be done with some simple construction..DRAW EVERY THING TO SCALE Draw an arc showing the motion of the rod fixed to B. Then divide that arc into maybe 5 pts.Measure the length the 150 rod. From each of those 5 pts, draw an arc of radius 150(the 150 rod length).This shows the...

I thought u were looking for the velocity. If you take the inlet as point1 and outlet(orifice) as point2 1) The average velocity can be taken as 0.99x the inlet velocity (V= 0.99 x V1). 2) The velocity a the orifice will be a function of the average flow velocity ie (pipe area/orifice...

Well your pipe is quite short so I think you can neglect friction losses. I think you should apply Bernoulli's equation between the inlet and outlet(since the pressures are known), however the losses in the system will be due to the bends. The general form of the pipe fitting losses are ...

looks ok to me