1. FLUIDS Pressure (P = F/A) The relationship is → P = Po + gh
The deeper you travel down in a fluid, the larger the pressure.
(you feel this on your ears as you go deep into a swimming pool.)
The relationship is → P = Po + gh
P = absolute pressure (Pa)
Po = atmospheric pressure (Pa)
 = density of the fluid (kg/m3)
g = 9.8 m/s2
h = depth (m)

2. How deep must you be under H2O to experience a pressure due to the H2O of 1 atm?
P = 2.0 atm = 2.02 x 105 Pa
Po = 1 atm = 1.01 x 105 Pa
 = 1000 kg/m3
g = 9.8 m/s2
h = ? m
P = Po + gh
2.02 x 105 = 1.01 x (1000) (9.8)h
h = 10.3 m
Hg = 13,600 kg/m3
Repeat for Hg.
P = Po + gh
2.02 x 105 = 1.01 x (13,600) (9.8)h
h = 0.76 m

3. Pascal's principle Pressure applied to an enclosed fluid is
transmitted undiminished to every part of
the fluid, as well as to the walls of the container.
A common application of this is a hydraulic lift used to raise a car off the ground so it can be repaired at a garage. A small force applied to a small-area piston is transformed to a large force at a large-area piston. If a car sits on top of the large piston, it can be lifted by applying a relatively small force to the smaller piston, the ratio of the forces being equal to the ratio of the areas of the pistons.

4. Rate of flow v1A1 = v2A2 (275) π(1.8)2 = (v2) π(1.1)2
The rate of flow of a fluid through any vessel
(volume per unit time) will remain constant. As a result, the velocity of the fluid is inversely proportional to the cross-sectional area of the
vessel. (flow rate = vA)
v1A1 = v2A2
A liquid with a density of flows through two horizontal sections of tubing joined end to end. In the first section, the pipe radius is 1.8 cm and the flow speed is 275 cm/s. In the second section, the cross-sectional area is pipe radius is 1.1 cm. Calculate the flow speed in the smaller section.
(275) π(1.8)2 = (v2) π(1.1)2
since A =πr2
v2 = 736 cm/s

5. Bernoulli’s Equation P1 + gy1 + ½v12 = P2 + gy2 + ½v22
Conservation of energy still applies for fluids in pipes.
P1 + gy1 + ½v12 = P2 + gy2 + ½v22
P = absolute pressure (Pa)
gy = GPE of the fluid (Pa)
y = height (m)
½v2 = KE of the fluid (Pa)
Of course energy is not measured in Pa, but the units will
work out, because the volume of fluid in the vessel is constant.

6. → P = (1.4 x 106)/(.36) P = F/A P = 3.9 x 106 Pa P = Po + gh
A deep sea vessel has a window of area 0.36 m2. The window can withstand a force of 1.4 x 106 N before breaking. Calculate the maximum depth below the surface the vessel can go. (The air pressure inside remains constant at kPa) →
P = (1.4 x 106)/(.36)
P = F/A
P = 3.9 x 106 Pa
This is the absolute pressure the window can withstand
P = Po + gh
3.9 x 106 = 101, (1000)(9.8)h
h = 387 m

7. v1A1 = v2A2 (10) π(.3)2 = (v2) π(.15)2 since A =πr2 v2 = 40 cm/s
Blood normally flows with an average speed of about 10 cm/s in the large arteries whose radii are about 0.3 cm. Suppose that a small section of the artery is reduced in radius by one-half due to the thickening of its walls (atherosclerosis). Calculate the speed past the restriction.
v1A1 = v2A2
since A =πr2
(10) π(.3)2 = (v2) π(.15)2
v2 = 40 cm/s
Water is pumped at a rate of 24 cc/s through a 0.5 cm radius pipe on the main floor of a house to a 0.35 cm radius pipe in a solar hot water collector 4.0 m above on the roof. If the pressure in the pipe on the roof is 1.2 x 105 N/m2, what is the pressure in the larger pipe on the main floor?

8. Pfloor = 1.59 x 105 Pa Proof = 1.2 x 105 N/m2 H2O = 1000 kg/m3
yroof = 4.0 m
rroof = 0.35 cm
rfloor = 0.5 cm
flow rate = 24 cm3/s
First determine the speed of the water
in the pipe on the roof and on the
main floor
On the roof
24 cm3/s = v(π)(.35)2
vroof = 62.4 cm/s = m/s
flow rate = vA
On the main floor
24 cm3/s = v(π)(.5)2
vfloor = 30.6 cm/s = m/s
Pfloor + gyfloor + ½vfloor2 = Proof + gyroof + ½vroof2
Pfloor + (1000)(9.8)(0) + ½(1000)(.306)2 =
1.2 x (1000)(9.8)(4) + ½(1000)(.624)2
Pfloor = 1.59 x 105 Pa