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.
Re = inerial forces/viscous forces = rho*V*L / mu
as Re approaches infinity...
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
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 =...
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 =...
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...
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...
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...
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...
ok I have this now,
the mass of water added to the boat:
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...
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...
ok, I've come up with this to describe the mass of the boat plus the scooped up water as a function of time:
M(t) = d/dt (density of water)*Volume + Mo
M(t) = [(density of water)*(velocity)*(area of Scoop))* (time)] + M0
I'm still not quite sure what to do to find the velocity of the...
Oh ok. I think I completly missed the point of the question. I was thinking that the scoop was just acting as a "brake," but it makes more sense that it is collecting water and adding weight to the boat.
I'm thinking I'll need the continuity equation to show the flow onto the boat as a...
I posted this in the advanced forum by mistake so I'm am posting it here as I think it may be more appropriate in this forum.
A motor boat is speeding at velocity W0 when the motor is turned off and a scoop is lowered into a still lake. The scoop captures flow with a cross...
I believe the solution to this is probably easy, but for some reason I just can't understand it. Any help would be GREATLY appreciated!
Consider a point at x2 = h/2. Find the rate of closure of the angle between two material lines in the x1 and x2 directions. Find the rate...