Recent content by UFeng

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...

oh ok. I plugged in the wrong wind velocity for the model. Thanks guys! I appreciate it! I think I got it now.

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

Ok, how does this look... using the equation from http://www.ecolo.org/documents/docum...illFormula.htm (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 = 300kW, I get...

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 =...

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...

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...

Thanks for the help!

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...

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...

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...

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?