__Static fluids__

What is the hydrostatic force on the back of Grand Coulee Dam if the water in the reservoir is 150m deep and the width of the dam is 1200m.

Solution:

- Concepts:

The relationship between force and pressure

The pressure at a point below the surface of a liquid at rest in a constant gravitational field depends only on the depth d of that point and the pressure at the surface.

P = P_{0 }+ ρgd. - Reasoning:

The water behind the dam at rest. - Details of the calculation:

Let us denote the distance below the surface by y. The pressure on the side of the dam facing the water at depth y is P_{0 }+ ρgy. Here P_{0}is the atmospheric pressure (101 kPa) and ρ is the density of water. The force on a surface element of width w and height dy at depth y is

dF = (P_{0 }+ ρgy)dA = (P_{0 }+ ρgy)wdy.

The total force on this side of the dam points outward and has magnitude

∫_{0}^{150}dF = wP_{0}∫_{0}^{150}dy + wρg∫_{0}^{150}ydy = wP_{0}150 + wρg150^{2}/2.

(All quantities are measured in SI units.)

The total force on the side of the dam facing air is wP_{0}150, inward. The net force on the dam is F = wρg150^{2}/2, outward. With w = 1200m and ρ = 1000kg/m^{3}, F = 1.32*10^{11}N.

A pipe of length 180 m, open on one end and closed on the other, lies at the
bottom of a 200 m deep lake. A light movable piston is placed inside the
pipe. The space between the closed end of the pipe and the piston is
filled with air. The piston is in equilibrium 20 m away from the closed
end of the pipe.

The open end of the pipe is very slowly raised until the
pipe is brought into the vertical position, its closed end resting at the bottom
of the lake.

What is the height of the air column inside the vertical pipe?

Neglect the atmospheric pressure and assume the water temperature is the
same throughout the lake.

Solution:

- Concepts:

The ideal gas law, pressure in a liquid as a function of depth - Reasoning:

In equilibrium the air pressure in the pipe equals the water pressure, P_{water}= ρ_{water}gd, neglecting the atmospheric pressure. Here d is the depth below the surface of the water. - Details of the calculation:

Before: P_{water}= ρ_{water}g*200 m = nkT/(A 20 m).

Here A is the cross-sectional area of the pipe.

nkT/A = ρ_{water}g*4000 m^{2}= constant.

After: ρ_{water}g*(200 m - h) = nkT/(A h).

nkT/A = ρ_{water}g*(200 m - h)h.

4000 m^{2}= (200 m - h)h.

h = (100 ± √6000) m.

h = (100 - 20*√15) m = 22.5 m.

Only the solution h = 100 - 20*√15 represents a stable equilibrium.

d(P_{water}(h) - P_{air}(h))/dh must be positive for a stable equilibrium to exist.

d(P_{water}- P_{air})/dh = -ρ_{water}g + nkT/(Ah^{2}) = ρ_{water}g*(4000 m/h^{2}- 1).

For h = (100 - 20*√15) m, d(P_{water}- P_{air})/dh is positive but for h = (100 + 20*√15) m it is negative.

__Flowing fluids__

A cylindrical vessel of radius R = 0.1 m is filled with a liquid to a height of 0.5 m. It has a capillary tube 0.15 m long and radius r = 0.0002 m fixed horizontally at its bottom. Find the time in which the water level will fall to a height of 0.2 m. Assume an ideal fluid in a frictionless environment.

Solution:

- Concepts:

Bernoulli's equation and the continuity equation - Reasoning:

We have an ideal incompressible fluid - Details of the calculation:

We introduce points B and C as shown in the figure below.

The velocities at these two points are v

_{C}and v_{B}.

The continuity equation yields v_{C}A_{C}= v_{B}A_{B}.

A_{C}= πR^{2}= 0.0314 m^{2}, A_{B}= πr^{2}= 1.3*10^{-7}m^{2}.

v_{C}= 4*10^{-6}v_{B}.

The velocity of liquid at point C is much smaller than the velocity of the liquid at point B.

The pressure at points B and C is atmospheric pressure. Bernoulli's equation therefore yields

ρgh + ρv_{C}^{2}/2 = ρv_{B}^{2}/2.

Since v_{C}<< v_{B}then we have ρgh = ρv_{B}^{2}/2, v_{B}= (2gh)^{½}, v_{C}= 4*10^{-6}(2gh)^{½}.

We also have v_{C}= -dh/dt.

-dh/dt = 4*10^{-6}(2gh)^{½}, -dh/h^{½}= = 4*10^{-6}(2g)^{½}dt,

∫_{0.2}^{0.5}= 1.8*10^{-5}t, 2(0.5^{½}– 0.2^{½}) = 1.8*10^{-5}t, t = 2.9*10^{4}s = 8 h.

The volume flow rate of water through a horizontal pipe is 2 m^{3}/min.
Determine the speed of flow at a point where the diameter of the pipe is

(a)
10 cm,

(b) 5 cm.

Solution:

- Concepts:

The equation of continuity - Reasoning:

Area 1 * v_{1}= Area 2 * v_{2}, since water is incompressible. - Details of the calculation:

Al = V. Adl/dt = dV/dt = volume flow rate. (A is the cross-sectional area and l is the length of a section of pipe.)

dl/dt = v = flow speed.

(πd^{2}/4)v = 2 m^{3}/60 s. v = (0.042/d^{2}) m/s with d measured in m.

d = 10 cm: v = 4.24 m/s,

d = 5 cm: v = 16.98 m/s

A Venturi tube may be used as a fluid flow meter. If the difference in
pressure P_{1 }- P_{2 }= 21 kPa, find the fluid flow rate in m^{3}/s
given that the radius of the outlet tube is 1 cm, the radius of the inlet tube
is 2 cm, and the fluid is gasoline (ρ = 700 kg/m^{3}).

Solution:

- Concepts:

Bernoulli's equation: P + ρgh + ½ρv^{2 }= constant. - Reasoning:

If we neglect friction, we have a conservative system. Bernoulli's equation is derived assuming the mechanical energy of the system is conserved. If a fluid or a gas, which is not being compressed, is flowing in a steady state, then the pressure depends on the speed of the fluid or the gas. The faster the fluid is flowing, the lower is the pressure at the same height. - Details of the calculation:

P_{1 }+ ρgh_{1 }+ ½ρv_{1}^{2 }= P_{2 }+ ρgh_{2 }+ ½ρv_{2}^{2}.

h is constant, so P_{1 }+ ½ρv_{1}^{2 }= P_{2 }+ ½ρv_{2}^{2}.

P_{1 }- P_{2 }= ½ρv_{2}^{2 }- ½ρv_{1}^{2}.

21 kPa = (350 kg/m^{3})(v_{2}^{2 }- v_{1}^{2}).

From the equation of continuity we have Area 1 * v_{1}= Area 2 * v_{2}.

v_{1 }= (A_{2}/A_{1})v_{2}.

Inserting this into the above equation we have(1 - (A_{2}/A_{1})^{2})v_{2}^{2}= (21000/350)(m/s)^{2}.

(A_{2}/A_{1})^{2 }= ¼^{2 }= 1/16. v_{2}^{2 }= (21000/350)(16/15)(m/s)^{2 }= 64 (m/s)^{2}.

v_{2 }= 8 m/s.

The fluid flow rate therefore is v_{2}A_{2}= (8 m/s)π(0.01 m)^{2}= 0.0025 m^{3}/s.

Water circulates throughout a house in a hot-water heating system. If the water is flowing at a speed of 0.2 m/s through a 5.0 cm-diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 1 cm-diameter pipe on the second floor 5.0 m above the basement? Assume the pipes do not divide into branches. (1 atm = 101 kPa)

Solution:

- Concepts:

Bernoulli's equation: P + ρgh + ½ρv^{2 }= constant

The equation of continuity: Area 1 * v_{1}= Area 2 * v_{2}. - Reasoning:

If we neglect friction, we have a conservative system. Bernoulli's equation is derived assuming the mechanical energy of the system is conserved. If a fluid or a gas, which is not being compressed, is flowing in a steady state, then the pressure depends on the speed of the fluid or the gas. The faster the fluid is flowing, the lower is the pressure at the same height. - Details of the calculation:

P_{1 }+ ρgh_{1 }+ ½ρv_{1}^{2 }= P_{2 }+ ρgh_{2 }+ ½ρv_{2}^{2}.

P_{2 }= P_{1 }+ ρg(h_{1}– h_{2})_{ }+ ½ρ(v_{1}^{2 }– v_{2}^{2}).

v_{2}= v_{1}*25 = 5 m/s.

v_{1}^{2 }– v_{2}^{2}= -24.96 (m/s)^{2}.^{}P_{2 }- P_{1}= -1000 kg/m^{3}*9.8 m/s^{2}*5 m - 500 kg/m^{3}*24.96 (m/s)^{2}= -61480 Pa.

1 atmosphere = 101 kPa.

P_{2 }= (3 - 0.61) atm = 2.39 atm = 241 kPa.