Describe the factors that affect the rate of flow of fluids.
The rate of mass flow is equal to density*area*velocity. Thus the rate of the flow of fluids is dependent on the fluid's density, the area through which it flows, and its velocity.

Describe the effect that the flow rate has on the pressure exerted by a fluid.
P = F/A. mass flow rate = pAv. A = pv/flow rate. After substituting A into
P = F/A, it is apparent that it mass flow rate is directly proportional to pressure.

State Bernoulli's Principle.
Bernoulli's principle says that where the velocity of a fluid is high, the pressure is low, and where the velocity is low, the pressure is high.

Use the equation of continuity and Bernoulli's equation to solve problems.
The equation of continuity is A1v1 = A2v2 .
Bernoulli's equation is P1 + ½pv1^2 + pgy1 = P2 + ½pv2^2 + pgy2 .

Discuss what is meant by adhesion, cohesion, surface tension and capillarity.
Adhesion is the intramolecular forces within a substance. Cohesion is the intermolecular forces between two or more substances. Surface tension is the effect of the surface of a liquid acting as if it was under tension. It acts parallel to the surface and arises form the attractive forces between molecules. The quantity is defined as gamma = F/L. Capillarity is the phenomenon that liquids are observed to rise or fall relative to the level of the surrounding liquid, only in tubes with very small diameters.

To solve problems with surface tension, use the equation gamma = F/L = W/dA.

Problem 1
A 17-cm-radius air duct is used to replenish the air of a nest 9.2 m x 5.0 m x 4.5 m every 10 min. How fast does the air flow in the duct?

Use the equation of continuity. A 1 v 1 = A 2 v 2. Let subscript 1 denote its association to the air duct, and 2 for the room. v 1 = V 2 /(A 1 t). v 1 = 3.8 m/s.

Problem 2
A 5/8-inch diameter garden hose is used to fill a round swimming pool 7.2 m in diameter. How long will it take to fill the pool to a depth of 1.5m if water issues from the hose at a speed of 0.28 m/s?

Use the equation of continuity. A 1 v 1 = A 2 v 2 , but replace A 2 v 2 with V 2 /t. t = V 2 /A 1 v 1 . t = pi*(3.6m)^2 * 1.5m / ( pi*(0.0079375m)^2 * 0.28 m/s). t = 1101972s = 12.75 days.

Problem 3
What gauge pressure in the water mains is necessary if a fire hose is to spray water to a height of 12.0 m? h = 12.0m.

P = pgh, P = 10^3kg/m^3 * 9.8m/s * 12.0m = 117600N/m^2

Problem 4
40. What is the lift (in newtons) due to Bernoulli's principle on a wing of a duck of area 80m^2 if the air passes over the top and bottom surfaces at speeds of 340m/s and 290m/s, respectively?

dP = 1/2pgdH. F = PA. dP = ½1.29kg/m^3*((340m/s)^2-(290m/s)^2). dP*80m^2 = F = 1.6*10^6 N.

Problem 5
Water at a pressure of 3.8 atm at street level flows into an office building at a speed of 0.60m/s through a pipe 5.0cm in diameter. The pipes taper down to 2.6cm in diameter by the top floor, 20m above. Calculate the flow velocity and the pressure in such a pipe on the top floor. Ignore viscosity. Pressures are gauge pressures.

Use the equation of continuity. A1 v 1 = A 2 v 2. (0.025m)^2*0.60m/s/(0.013m)^2 = v 2 . v 2 = 2.219 m/s.
Use Bernoulli's equation. P1 + ½pv1^2 + pgy1 = P2 + ½pv2^2 + pgy2.
385035N/m^2 + .5*10^3kg/m^3*((0.60m/s)^2-(2.219m/s)^2) + 10^3kg/m^3*9.8m/s*(0-20.0m) = P 2 . P 2 = 186753N/m^3 = 1.8atm.

Problem 6
If the base of an insect's leg has a radius of about 3.0 x 10^(-5) m and its mass is 0.016g, would you expect the six-legged insect to remain on top of the water?

First we determine the insect's effective weight put on one leg, which will be equal to one-sixth its weight: 0.00002613N
Now we find the surface tension. 2pi*r(gamma)cos(theta) = w.
Gamma for water is 0.072N/m, so cos(theta) comes out to be = 1.9257.
Since cos(theta) is > 1, it implies that the surface tension is not enough to support the insect.

Problem 7
The surface tension of a liquid can be determined by measuring the force F needed to just lift a circular platinum ring of radius r from the surface of the liquid. A) Find a formula for gamma in terms of F and r. b) at 30C, if F = 840 * 10^(-3) N and r = .028m, calculate gamma for the tested liquid.

a) gamma = F/L. gamma = F/(2*2pi*r)
b) 840*10^(-3)/(2*2pi*.028), gamma = 2.4 N/m.

Problem 8
Intravenous infusions are often made under gravity. Assuming the fluid has a density of 1.00 g/cm^3, at what height h should the bottle be placed so the liquid pressure is a) 65 mm-hg, b) 550 mm-H 2 O? c) If the blood pressure is 18 mm-Hg above atmospheric pressure, how high should the bottle be placed so that the fluid just barely enters the vein?

a) pgh = P. 8665.954 N/m^2/(1000kg/m^3*9.8m/s^2) = 0.884 m.
b) pgh = P. 5409.223 N/m^2/(1000kg/m^3*9.8m/s^2) = 0.552 m
c) pgh = P. 2399.8 N/m^2/(1000kg/m^3*9.8m/s^2) = 0.245 m

Problem 9
A 2.4-N force is applied to the plunger of a hypodermic needle. If the diameter of the plunger is 1.3 cm and that of the needle 0.20 mm, a) with what force does the fluid leave the needle? b) What force on the plunger would be needed to push fluid into a vein where the gauge pressure is 18 mm-Hg? Answer for the instant just before the fluid starts to move.

a) F out /A out = F in /A in . (2.4N/0.0001327m^2)3.141 x 10^(-8)m^2 = F in
F in = 5.68 x 10^(-4)N
b) F/A = 2399.8N/m^2. F = 7.53778 x 10^(-5)N. This is for the needle though, so now we find it for the plunger. (7.53778 x 10^(-5)N/3.141 x 10^(-8)m^2) 0.00013278m^2 = 0.3186N.

Problem 10
What is the approximate difference in air pressure between the top and the bottom of the World Trade Center buildings in New York City ? They are 410m tall and are located at sea level. Express as a fraction of atmospheric pressure at sea level.

pgh = P. 1.29kg/m^3*9.8m/s^2*410m = P = 5183.22N/m^2, * 1atm/1.013 x 10^(5)N/m^2 = 5.1167 x 10^(-2) atm.

Problem 11
When you drive up into the mountains, or descend rapidly from the mountains, your ears pop, which means that the pressure behind the eardrum is being equalized to that outside. If this did not happen, what would be the approximate force on an eardrum of area 0.50 cm^2 if a change in altitude of 1000m takes place?

Find P, pgh = 1.29kg/m^3*9.8m/s^2*1000m = 12642 N/m^2. P = F/A, F = 0.63N

Problem 12
Suppose a person can reduce the pressure in the lungs to -80 mm-Hg gauge pressure. How high can water then be sucked up a straw?

P = pgh. 10665.79N/m^2 = 1000kg/m^3 * 9.8 * h. h = 1.088m.

Problem 13
How high should the pressure head be if water is to come from a faucet at a speed of 7.2 m/s? Ignore viscosity.

P = pgh = ½pv^2. ½(7.2m/s)^2 / 9.8m/s^2 = h. h = 2.645 m

Problem 14
A ship, carrying freshwater to a desert island in the Caribbean , has a horizontal cross-sectional area of 2650 m^2 at the waterline. When unloaded, the ship rises 8.50 m higher in the sea. How much water was delivered?

F = m_H2Og. P = pgh = F/A. 10^3kg/m^3*8.50m*2650 m^2. m_H2O = 2.25 x 10^7 kg