Princeton Ch 8 - Fluids Flashcards
Fluids.
Substances that can flow (ie. gases and liquids).
Difference between liquid and gas.
The molecules in a liquid are able to move around a little more freely than in those of a solid, which molecules typically only vibrate. By contrast, molecules in a gas are not constrained and fly around in a chaotic swarm, with hardly any interaction.
Density.
The amount of mass contained in a unit of volume.
ρ = m/V [kg/m^3]
The density of water.
1000 kg/m^3
1 g/cm^3
1gkg/L, where L stands for liter (a liter is 1000cm3)
Specific gravity.
How dense something is compared to water.
Specific gravity = density of substance/density of water
True or false. For solids, liquids, and gases, density doesn’t change much with changes in pressure or temperature.
Solids and liquids behave the same way under most conditions.
However, the density of gas changes markedly with pressure and temperature. PV =nRT; ρ(gas) = mV = mP/nRT
How does one find the force of gravity acting on a fluid?
ρ = m/V –> m = ρV –> F(grav) = mg = ρVg
If we place an object in a fluid, the fluid exerts a contact force on the object. If we look at how that force is distributed over any small area of the object’s surface, we have the concept of pressure. What is P?
P = Force/Area
Hydrostatic gauge pressure.
The weight of a fluid above a submerged object produces a force that pushes down on a point. This is called hydrostatic because the object is at rest, and gauge because we don’t take the pressure into the atmosphere into account. If we did we/d have to add it.
P(gauge) = ρ(fluid)gD D = depth of the point
If you have an object submeged in water, how do you find the total pressure?
P(total) = P(at surface) + Pguage
P(at the surface) = atmospheric pressure or the layer of gas above the surface of the liquid.
True or false. Density doesn’t increase with depth for liquids.
True. The density of a liquid should remain the same.
True or false. Hydrostatic gauge pressure and the TOTAL pressure is proportional to both the depth and the density of the fluid.
False. Guage pressure = ρ(fluid)gD, is proportional to both depth and density of the fluid.
Total pressure is NOT proportional to either quantity of P(on the surface) isn’t zero.
1 Pascal =?
1N/m^2
If you have a hourglass beaker filled with water, and a regular beaker filled with water, what is the pressure at the bottom right corner?
The pressures are the same if the fluid density and depth are the same. P(guage) = ρfluidgD, regardless of the shape of the container in which the fluid is held. The pressure is the same everywhere along a horizontal dashed line.
Archimedes principle.
The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.
F(buoy) = ρ(fluid)V(sub)g
Buoyant force.
The net upward fluid force.
F(buoy) = ρ(fluid)V(sub)g
For a floating object, __ = __. This is a floating object in equilibrium on the surface.
W(object) = F(buoy)
Vsub/V = ρ(object)/ρ(fluid)
If ρ(object) < ρ fluid, then the object will ___.
The object will float and the fraction of its olume that’s submerged is the same a the ratio of its density to the fluid’s density.
If an object’s density is 3/4 the density of the fluid, then __ of the object will be submerged.
3/4
Vsub/V = ρ(object)/ρ(fluid)
If an object is denser than the fluid, then the object will ___.
The object will sink because the object’s weight is greater than the buoyant force.
For an object that is completely submerged in water, the weight of the object is still the same.
True. The actual weight of the object is still w=ρVg. BUT, the object’s apparent weight is less due to the buoyant force pushing the object upwards.
w(obj)/F(buoy) = ρVg/ρ(fluid)Vg = ρobject/ρfluid
If the fluid was water, note that the ratio of the object weight to the buoyant force is equal to the specific gravity of the object.
A glass sphere has a specific gravity of 2.5. How do you find the density of the glass sphere?
ρ(sphere ) = 2.5*(1000 kg/3^3)
Pascal’s law on fluid pressure.
If you squeeze a container of fluid, the fluid will transmit your squeeze perfectly throughout the container.
For a simple hydraulic jack consisting of two pistons above two cylindrical vessels of fluid, what can we say about force, area, and distance, if we push down on one piston?
F1/A1 = F2/A2 –> F2 = A2F1/A1 –> notice the greater output Force (F2); this is what makes hydraulic jacks useful.
But the volumes have to be the same. A1d1 = A2d2.
So ultimately,
F2d2 = (A2F1/A1)(A1d1)/A2 = F1d1; work you do pushing the left piston equals the work done on the right.
In a simple hydraulic system, let’s say A2 is 5 times bigger than A1. What does this mean regarding the any applied force and the distance?
The increase in F is the same as the decrease in d. If A2 is 5 times larger than A1, then F2 will be 5 times greater than F1, but d2 will only be 1/5 of d1.
Formula for surface tension.
F(surface tension) = 2γL
γ = coefficient of surface tension. 2L = there are two surfaces in which this force acts. The force acts along a line, L.
Flow rate.
The volume of fluid moving through a particular cross sectional area per unit of time.
f = Av = πr²*(v)
Difference between flor rate and flow speed.
Flow rate tells us how MUCH fluid flows per unit time; speed tells us how FAST the fluid moves.
Ex. The hose ejects 4L of water/s versus the water leaves the hose at a speed of 4m/s.
If a pipe carrying an incompressible liquid, then the flow rate must be same everywhere along the pipe. Give the continuity equation for liquids between point1 and point 2.
A1v1 = A2v2
What can A1v1 = A2v2 tell us?
When the tube narrows, the flow speed with the increase, and if the tube widens, the flow speed with decrease. Note that flow speed is inversely proportional to the cross-sectional area. v α 1/A
Bernoulli’s equation only applies to ideal fluids. Define what this is.
1) Fluid is incompressible.
2) Has negligible viscosity ( the force of cohesion between molecules in a fluid - fluid friction)
3) Flow is laminar - the flow is in a “streamline”
4) Flow rate is steady. “f” = Av is constant.
Bernoulli’s equation:
P1 + 1/2ρv(1)² +ρgy1 = P2 + 1/2ρv(2)² +ρgy2
P = total pressure along that point; not guage
P1 - Patm (if 2 was exposed to air) = Pguage at 1
Any fluid exposed to the atmosphere with Bernoulli equations is what pressure?
Atmospheric pressure.
1 atm = 101.3kPa
At any two ponits of equal height (y1=y2), faster fluid flow means (lower/higher) pressure.
At any two points of equal height (y1=y2), faster fluid flow means LOWER pressure. P + 1/2ρv² = constant
For a syringe needle, the needle is smaller, so v is faster than in the plunger. faster fluid means LOWER pressure.
Torricelli’s result.
ρgy1 = 1/2ρv(2)² +ρgy2; P1 = P2 = atm; v1 = 0 because A1>>>>A2
v(2) = √(2gD)
For a large tank exposed to atmosphere and then a small hole poked in the bottom.
Bernoulli (Venturi) effect.
The pressure is lower where the flow speed is greater, if y1 = y2.
Poiseulle’s law (may not have to remember)
ΔP/L = 8nf/(πr^4)
P = pressure gradient across tubing n = viscosity coefficient f = flow rate r = radius of pipe L = length of tubing
Doubling the diameter of a catheter increases the flow rate by 16 fold (r4). The larger the IV catheter the greater the flow.
Increasing viscosity decreases flow through a catheter.
A pipe of constant cross sectional A carries water at a constant flow from the hot water tank in the basement oup to the second floor. What can you say about the speed and pressure at which the water arrive and leaves the water tank.
The water pressure at the second floor must be LOWER than the water pressure at the tank. Since the flow rate and A is constant, the flow speed will be constant. f = Av = constant.
P1 +ρgy1 = P2 +ρgy2
