# Search results

1. ### Which fluid is more viscous

Homework Statement Which fluid is more viscous at moderate to high Reynolds number if used in this situation? Air or Water? The flow is incompressible and has velocity V and size L. Homework Equations Re = inerial forces/viscous forces = rho*V*L / mu as Re approaches infinity...
2. ### Windmill power generation problem

oh ok. I plugged in the wrong wind velocity for the model. Thanks guys! I appreciate it! I think I got it now.
3. ### Windmill power generation problem

oh ok, how bout this? with D(model) = 1.75ft = 0.5334m Power = 0.603*0.58*(.5334m)^2 *(6.7056)^3 = 30 W rotor tip speed = k*6.7056 = 3.89m/s => if I then divide 3.89m/s by .5334m diameter I get (3.89/.5334) = 7.29s^-1 = 437 rpm thanks again
4. ### Windmill power generation problem

Ok, how does this look... using the equation from http://www.ecolo.org/documents/docum...illFormula.htm [Broken] (P = C*k*D^2 *V^3) where C=constant, k = efficiency = 0.58 and solving for the Constant, C, while assuming the blade diameter = 175ft(53.4m) and the power generated =...
5. ### Windmill power generation problem

I thought 90 mph seemed high too, but I checked the problem again and those were the numbers given, unless the book has a typo. Also if I use the power equation and solve for the radius of the blades I get r = 22.73 meters = 74.6 ft => does this seem too large? For example: 300,000Watts =...
6. ### Windmill power generation problem

Homework Statement A windmill is designed to operate at 20 rpm in a 15 mph wind and produce 300 kW of power. The blades are 1.75 ft in diameter. A model 1.75 ft in diameter is to be tested at 90 mph wind velocity. What rotor speed should be used, and what power should be expected...
7. ### Help solving the Rayleigh Problem in fluid dynamics please

Homework Statement Consider the Rayleigh problem, but allow the plate velocity to be a function of time, V(t). By differentiation show that the shear stress, tau = du/dy*absolute viscosity, obeys the same diffusion equation that the velocity does. Suppose that the plate is moved in such a way...
8. ### Extra force on a train problem

Thanks for the help!
9. ### Extra force on a train problem

I think I get it. If they were both traveling at the same velocity and all other things being the same, the "extra force" would be zero, but since the faster train is moving at 50 ft/s (10 ft/s faster than the slower one), the "extra force" is from this 10ft/s difference. For example, the coal...
10. ### Extra force on a train problem

If that is for the fast train would the force on the slow moving train be the same( for example: the slower train has an "extra force" of +13.3 while the faster train has an "extra force" of -13.3)? Also, when it asked for the force on "each train," would they both be the same but have opposite...
11. ### Extra force on a train problem

So this is what I've come up with... using F = d(mv)/dt = m*dv/dt + v*dm/dt => where dv/dt = 0, so F = v*dm/dt and dm/dt = 4 tons/min F(slower moving train)=d(mv)/dt = (4 tons/min)*(1 min/60sec)*(40ft/sec)/(100 ft of length) = 0.0267 tons-ft/s^2 per unit length = 53.4 lb ft/s^2 per...
12. ### Extra force on a train problem

Homework Statement Two long trains carrying coal are traveling in the same direction side by side on separate tracks. One train is moving at 40 ft/sec and the other at 50 ft/sec. In each coal car a man is shoveling coal and pitching it across to the neighboring train. The rate of coal...

unfortunately I'm still stuck. I'd appreciate any help!

anyone know what to do next?

ok I have this now, the mass of water added to the boat: =rho*A*[W0-W(t)]*t so the total mass of the boat + water scooped up, M(t); M(t) = M0 + rho*A*[W0-W(t)]*t again I'm stumped on how to find find velocity as a function of time, W(t) using the integral momentum equation. Any hints?

ok how about this total mass of water from t = 0 to t: rho * A * S(t=0 to t) W(t) dt wouldn't this just leave me with rho*A*W(t)*t again...unless, the total mass of water is: = rho*A*t*[W0 - W(t)] => would this be the mass of water assuming the velocity stayed at W0 minus the mass...

oops, sorry. The last line would be the mass per time not the mass. I think it would be [W0 - W(t)] instead of [W(t) - W0] also

ok I'll try this again. Integrating the mass of water I come up with this: I'll use "S" as the integration symbol and integrate from W0 to W(t) Mass water as a function of time=density*A*S[from W0 to W(t)] dt = density*A*[W(t)-W0]

here is my understanding of the problem. I may be wrong. The boat is orignally traveling at velocity, M0, and then the motor is shut off. At this point water is scooped onto the boat thus causing it to gain mass and slow down. So, at t = 0, the mass is equal to M0, but when the scoop gathers...

The momentum of the water would be: density*[W(t)^2]*A*t => @t=0 the momentum of the water would be zero...correct? so I think the equation would look like this: M(t)*W(t) = M0*W(t) + density*[W(t)^2]*A*t => @t=0, W(t)=W0...I think. So @ t=0, M(t)*W(t) = M0*W0 does that look right?

the total mass at time t: M(t) = M0 + density * A * W(t) * t the mass of the boat is M0 and the mass of the water is, density*A*W(t)*t the original momentum before the scoop is lowered would just be: M0 * W0 the momentum now would be M(t) * W(t) is this correct? my problem is...

ok, I'm still stuck. Do I need to find an equation for the acceleration(which would be negative) and then use it to find velocity as a function of time. i.e. dW/dt => then seperate variables and integrate from 0 to W(t) => then solve for W(t) Does this sound right? Any hints would be...