Help with fluid mechanics?
You will need to use Newton's theory of viscosity: http://en.wikipedia.org/wiki/Viscosity#Newton.27s_theory The gradient dU/dy (with U being velocity and y being the axis between the slope and block) is going to be constant since you've been told that the velocity profile is linear. The no-slip condition means that the value of dU/dy is going to be simply the velocity of the block divided by the distance between the block and slope (look at the diagram on the wikipedia article if you don't get what I just said). It appears you are going to be solving for dU/dy, so we'll leave that for a minute. The first step is to calculate the viscous force required for the block to maintain constant velocity. Find the component of the force due to gravity in the direction of the block's motion. The viscous force will be opposite and equal in magnitude to this to this at terminal velocity. The viscous force is given by the shear stress (tau) at the surface of the block times the area of that surface. After calculating tau from the required force and the block contact surface area, you can plug that value into the equation from the wikipedia article along with the viscosity (mu, taken from a table of viscosity values that should be in the back of any textbook). Once you have the velocity gradient you can solve easily for the velocity by multiplying by the gap width. I hope that was all correct. My fluid mechanics is a bit rusty.
Help: Fluid Mechanics........!?
Q1) The density of water is approximately 1 kg per liter. Also, there are 1000 liters in one cubic meter. If the buoy displaces 0.5 m^3 of water, then the buoyant force is: (0.5 m^3) x (1000 L / 1 m^3) = 500 L 500 L x (1 kg / 1 L) = 500 kg ; yes, the vessel will float Q2) First, we need to know the mass of helium in the ballon. Let's start by converting 3 liters to m^3: 3 L x (1 m^3 / 1000 L) = 0.003 m^3 Now, let's find out how much that volume of helium weighs: (0.003 m^3) x (0.179 kg / m^3) = 0.000537 kg helium Calculate the mass of the displaced air: (0.003 m^3) x (1.225 kg / m^3) = 0.003675 kg air Mass of rubber (in kilograms): 2.5 g x (0.001 kg / g) = 0.0025 kg Now, the buoyant force is equal to the total mass of the balloon (the mass of the rubber PLUS the mass of the helium) minus the mass of the displaced air: [(0.0025 kg rubber) + (0.000537 kg helium)] - (0.003675 kg air) = -0.000638 kg (negative sign because the force acts opposite to the balloon's weight) Convert kilograms to grams: 0.000638 kg x (1000 g / kg) = 0.638 grams Q3) Let's start by converting the mass of the boat from tonnes to kilograms: 1.25 tonnes x (1000 kg / 1 tonne) = 1250 kg The boat must displace a volume of water equal in mass to the boat itself. The density of water is (1 kg / liter). So what volume of water must the boat displace? 1250 kg x (1 L / 1 kg) = 1250 L Convert liters to m^3: 1250 L x (1 m^3 / 1000 L) = 1.25 m^3 of water
Fluid mechanics help?
Actually, they don't explain anything. Not a single paper outlining any 'creation science' (read: 'pseudoscience') findings, testable analysis, or hypotheses have been published in any peer reviewed scientific journals or publications. They think that because their explanation is based on the bible, that it should get special treatment, and a free pass, rather than having to be exposed to the rigorous peer review process that ALL scientists must submit their work to in order to have it be taken seriously. Scientists, despite religious people's rants, do not sit around all day trying to prove their work. Instead, they attempt to disprove it. Whatever stands up to testing gets published, at which time everyone else gets a chance to disprove it.
FLUID MECHANICS! please help?
1. Convert the following temperatures to the absolute: 72 degrees Fahrenheit 98.6 degrees Fahrenheit 37 degrees Celsius 18 degrees Celsius 108 degrees Fahrenheit 2. Convert the following gauge pressures to the absolute: 35 PSI 140 KPa 95 PSI 760 KPa 50 PSI 3. A submarine is 630 feet deep in the Atlantic Ocean. What is the external pressure on the hull? 4. There is a hatch on the above submarine that is 3.5 feet in diameter. What is the force being exerted on the hatch? 5. A 5 meter diameter tank is accident filled with two liquids. The specific gravity of one is 1.260 and the specific gravity of the other is 0.911. The total height of fluid in the tank is 6.25 meters and the pressure at the bottom of the tank is 65.9 KPa. What is the volume of each liquid in the tank? Your answer may be in either Meters cubed or Liters.
Help with Fluid Mechanics Problem?
Water flows from left to right in this parallel pipe system. Which pipe has the greatest velocity? Pipe A: L=6000ft Diameter=1.5ft and friction coefficient=0.012 Pipe B: L=2000ft Diameter=0.5ft and friction coefficient=0.02 Pipe C: L=5000ft Diameter=1.0ft and friction coefficient=0.015 As it said above, the pipes are in parallel series like in a current division problem in Circuits. My professor mentioned using something like Current Division, but I just don't see it. This problem consists of head loss (hence the givens) and that's about all I can figure out about it. I was thinking maybe the relation was that the mass doesn't change therefore the flow rate doesn't change??? Plz, any help would be appreciated. Thanks
Physics: Fluid Mechanics?
you're given the objective quantity, and the time to fill it. Use this to locate the volumetric flow fee: V_dot = V/t Geometric relation between velocity and volumetric flow fee: V_dot = v*A area for a around go area: A = Pi*d^2/4 replace and simplify: V_dot = v*Pi*d^2/4 V/t = v*Pi*d^2/4 remedy for v: v = 4*V/(Pi*d^2*t) information: V:=20600 cm^3; d:=a million.21 cm; t := sixty 8.4 sec; result: v = 261.9 cm/sec or v= 2.619 meters/sec
Need Help with a Fluids Mechanics Problem?
a) you draw a administration volume around the nozzle and notice the momentum equation to calculate the unknown stress that's the reaction on the nozzle, as a results of alter of momentum of the flow. you additionally can notice continuity and Bernoulli's equation interior the pipe. nonetheless, i can't see a thank you to do this with the concepts you have offered! I seem lacking an upstream speed or a community. additionally, i'm undecided related to the 'ability of the jet' - i assume it is the fee of KE of the flow 0.5.mdot.v^2 b) is comparable - draw a administration volume around a blade. stress on the blade is the fee of momentum replace of the flow, in step with replace of absolute flow speed (so which you desire inlet and outlet absolute flow velocities interior the x-course (inlet flow course)). you recognize completely the inlet speed, rotor speed and as a result relative inlet speed, yet are no longer given any outlet situations including relative go out speed or perspective, from which to calculate absolute go out speed - so back i do no longer see the way you do it? Does everyone else recognize? thank you
Fluid Mechanics Problem?
Hello Wzr, Surprised that this one was not snatched right up within the first ten minutes, it is not complicated. Pressure is a Force spread out over an Area. Pressure = Force / Area P = F / A Here the force is the man's weight, and the Area is the bottom of his shoe pushing on the ground. F = 200 lb Lets call the Area of the bottom of ONE shoe (Aone) = 72 in^2 (one shoe supporting all 200 lb) Lets call the Area of the bottom of TWO shoe (Atwo) = 144 in^2 (two shoes supporting 200 lb) a)If he stands on both shoes: P = F/A = F/Atwo = 200 lb / 144 in^2 P = 1.39 lb/in^2 (psi) b) If he stands on one shoe: P = F/A = F/Aone = 200 lb / 72 in^2 P = 2.78 lb/in^2 (psi) I hope that this helped a little Good Luck Wzr
[Physics][Fluid Mechanics][Problem Help]?
You would need to know the air pressure inside the room. Specifically you need the pressure difference. Force = ΔP * A This assumes that there is no air velocity impacting the window, in which case that value would have to be corrected.
Fluid mechanic question. Please help?
When the canoe is about to sink it will be displacing its own volume of water. Volume displaced = 3 x 3 x 6 x (0.3048)³ = 1.52911 m³ Weight of displaced water = 1529.11 kg Part of the displacement is caused by the man and the canoe's own weight = 90 + 25 = 115 kg. The rest of the displacement = 1529.11 - 115 = 1414.11 kg must be due to the water that is in the canoe. The fraction of the canoe volume that is filled with water when the canoe is about to sink is thus 1414.11/1529.11 = 92.48%