Fluid Mechanics

Lessons

Senior 5 · Physics

Fluid Mechanics

4 lessons

Lesson 1 of 4

Pressure in Liquids and U-tube Manometers

25 min

Gas held down by water

Lake Kivu is up to 475 m deep, and below about 300 m its water holds huge amounts of dissolved methane and carbon dioxide. The gas stays dissolved because the weight of the water above presses on it: the pressure down there is dozens of times atmospheric pressure. On the lake, the KivuWatt plant pumps this deep water up to a barge; as the water rises, the pressure falls and the gas comes out of solution, like the bubbles in a bottle of soda when you open it. The methane is separated, cleaned and burned to generate electricity for Rwanda's grid. The physics of all this starts with one question: how does pressure change with depth?

Pressure at a depth

Pressure is force per unit area, $p = \frac{F}{A}$ , measured in pascals ( $1\,\mathrm{Pa} = 1\,\mathrm{N\,m^{-2}}$ ). In a fluid at rest, the pressure at a point acts equally in all directions and is always perpendicular to any surface it pushes on.

Deriving $p = p_0 + \rho gh$

Imagine a vertical column of the liquid, of cross-sectional area $A$ , reaching from the surface down to depth $h$ . It is at rest, so the vertical forces on it balance.

pA = p_0A + \rho Ahg \quad\Longrightarrow\quad p = p_0 + \rho gh

Here $p_{0}$ is the pressure at the surface (the atmospheric pressure for an open lake or tank) and $\rho gh$ is the extra pressure due to the liquid. Three consequences:

Pressure rises linearly with depth, by $\rho g$ per metre: about $9.8\,\mathrm{kPa}$ per metre of water. Every 10 m of water adds roughly one more atmosphere.
Pressure depends on depth, not on the shape or width of the container. All points at the same depth in one connected liquid at rest have the same pressure.
Dam walls are made thicker at the bottom, where the water pushes hardest.

The difference $p - p_{0}$ is called the gauge pressure; $p$ itself is the absolute pressure.

Definition

The hydrostatic pressure at depth $h$ in a liquid of density $\rho$ is $p = p_0 + \rho gh$ , where $p_{0}$ is the pressure at the surface. The gauge pressure $\rho gh$ is the pressure above atmospheric. Example: at a depth of 5.0 m in water the gauge pressure is $1000 \times 9.8 \times 5.0 = 49\,\mathrm{kPa}$ .

Worked example 1: pressure in Lake Kivu

Take the atmospheric pressure at the surface of Lake Kivu as 85 kPa and the density of the water as $1000\,\mathrm{kg\,m^{-3}}$ . Find the absolute pressure (a) at 50 m, (b) at 300 m, where the gas-rich layer begins. (c) How many times the surface pressure is (b)?

Solution

(a) $p = 85 \times 10^3 + 1000 \times 9.8 \times 50 = 85 \times 10^3 + 490 \times 10^3 = 575\,\mathrm{kPa}$ .

(b) $p = 85 \times 10^3 + 1000 \times 9.8 \times 300 = 85 \times 10^3 + 2940 \times 10^3 = 3.03 \times 10^6\,\mathrm{Pa}$ .

(c) $\frac{3.03 \times 10^6}{85 \times 10^3} = 36$ . (The salty deep water is slightly denser than 1000, so the true value is a little higher.)

The U-tube manometer

A manometer is a U-shaped tube containing a liquid, used to measure pressure differences. Connect one arm to a gas supply and leave the other open. The gas pushes the liquid down on its side until the liquid on the open side stands a height $h$ higher.

Points X and Y are at the same level in the same connected liquid, so they are at the same pressure. At X the pressure is the gas pressure $p$ ; at Y it is atmospheric pressure plus the column of height $h$ above it:

p = p_0 + \rho gh \qquad\text{so the gauge pressure is}\qquad p - p_0 = \rho gh

Water is used for small pressure differences (a 1 cm difference is about 98 Pa); a denser liquid gives a shorter column for larger pressures. A barometer is a tube closed at the top with a vacuum above the liquid, so $p_0 = \rho gh$ measures atmospheric pressure itself.

Comparing densities with two liquids

Pour water into a U-tube, then pour oil, which does not mix with water and floats on it, into one arm. Take the level of the oil–water interface. Below it, both arms contain only water, so the pressure at that level is the same in both arms. Above it, the left arm holds a height $h_{1}$ of oil and the right arm a height $h_{2}$ of water, both open to the air:

p_0 + \rho_1gh_1 = p_0 + \rho_2gh_2 \quad\Longrightarrow\quad \rho_1h_1 = \rho_2h_2

Worked example 2: two manometer readings

(a) A water manometer on a laboratory gas tap shows a difference in levels of 12.0 cm. Find the gauge pressure of the gas. (b) In a U-tube, a 10.0 cm column of oil balances an 8.5 cm column of water above the interface. Find the density of the oil.

Solution

(a) $p - p_0 = \rho gh = 1000 \times 9.8 \times 0.120 = 1.18 \times 10^3\,\mathrm{Pa}$ , about 1.2% of one atmosphere.

(b) $\rho_1 = \frac{\rho_2h_2}{h_1} = \frac{1000 \times 8.5}{10.0} = 850\,\mathrm{kg\,m^{-3}}$ .

Worked example 3: a barometer at altitude

A mercury barometer ( $\rho = 13\,600\,\mathrm{kg\,m^{-3}}$ ) in a town at high altitude reads 640 mm. Find the atmospheric pressure, and compare it with the sea-level reading of 760 mm.

Solution

$p_0 = \rho gh = 13\,600 \times 9.8 \times 0.640 = 85.3\,\mathrm{kPa}$ . At sea level, $13\,600 \times 9.81 \times 0.760 = 101\,\mathrm{kPa}$ . The town's air pressure is about 84% of the sea-level value, typical of a height of about 1500 m.

Experiment 4.1: pressure and depth

Aim: to show that the pressure in a liquid increases in proportion to depth.

Apparatus: a tall clear container of water, a thistle funnel with a thin rubber sheet stretched over its mouth, rubber tubing, a U-tube water manometer, a metre rule.

Connect the funnel to one arm of the manometer with the tubing.
Lower the funnel into the water in steps of 5 cm. At each depth, record the difference $h$ in the manometer levels.
Turn the funnel to face sideways and then upwards at the same depth: the reading does not change, because pressure acts equally in all directions.
Plot $h$ against depth. You should get a straight line through the origin, of gradient slightly less than 1: the rubber sheet needs a little pressure to stretch.

Warning

Safety. School manometers and barometers should contain water or coloured oil, not mercury: mercury vapour is toxic and spills are hard to clean. Mop up spilled water at once, and keep water away from electrical sockets and apparatus. Never connect a manometer to a high-pressure supply.

Warning

Common mistakes. (1) Forgetting $p_{0}$ when the question asks for absolute pressure, or adding it when it asks for gauge pressure. (2) Measuring $h$ from the bottom of the tube instead of as the difference between the two liquid levels. (3) Thinking a wider tank gives more pressure at the same depth: only depth and density matter. (4) In a two-liquid U-tube, measuring both heights from the bottom instead of from the interface level.

Tip

In any U-tube problem, draw a horizontal line through the lowest interface between two different fluids. Below that line the same liquid fills both arms, so the pressures on the line are equal. Write "pressure on the left = pressure on the right" and the problem is solved.

Measuring Pressure With Barometers and Manometers (Professor Dave Explains). Watch how the height of a liquid column measures a pressure, and why a barometer uses mercury.

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Practice

Derive $p = p_0 + \rho gh$ for a liquid at rest.
Find the gauge pressure at the bottom of a water tank 2.5 m deep, and the force it exerts on a drain plug of area $20\,\mathrm{cm^2}$ .
A diver is 12 m below the surface of a lake where the atmospheric pressure is 85 kPa. Find the absolute pressure on the diver.
A water manometer connected to a gas supply shows a level difference of 25 cm, with the gas side lower. Find the gas pressure if $p_0 = 85\,\mathrm{kPa}$ .
In a U-tube, 12.0 cm of a liquid balances 9.6 cm of water above the interface. Find the liquid's density.
Explain why water barometers are not used, by finding the height of a water barometer at a pressure of 85 kPa.
A sealed bottle of soda opened at the shore of Lake Kivu fizzes more than the same bottle opened at sea level. Explain why, using ideas from this lesson.

Show answersHide answers

For a column of area $A$ and height $h$ at rest: $pA = p_0A + \rho Ahg$ , so $p = p_0 + \rho gh$ .
$\rho gh = 1000 \times 9.8 \times 2.5 = 24.5\,\mathrm{kPa}$ ; $F = 24.5 \times 10^3 \times 20 \times 10^{-4} = 49\,\mathrm{N}$ .
$85 \times 10^3 + 1000 \times 9.8 \times 12 = 85 + 117.6 = 203\,\mathrm{kPa}$ .
$p = p_0 + \rho gh = 85 \times 10^3 + 1000 \times 9.8 \times 0.25 = 87.5\,\mathrm{kPa}$ .
$\rho = \frac{1000 \times 9.6}{12.0} = 800\,\mathrm{kg\,m^{-3}}$ .
$h = \frac{p_0}{\rho g} = \frac{85 \times 10^3}{1000 \times 9.8} = 8.7\,\mathrm{m}$ : far too tall to be practical; mercury needs only 0.64 m.
The gas is held in solution by the pressure above the liquid. At the lake's high altitude the atmospheric pressure is lower than at sea level, so when the cap is removed the pressure on the drink falls further and more gas comes out of solution: the same reason gas escapes from deep lake water as it is pumped up.

Key points

$p = p_0 + \rho gh$ : pressure rises linearly with depth, the same at all points at one depth in a connected liquid.
Gauge pressure $= p - p_0 = \rho gh$ ; a manometer reads it as a level difference.
Two immiscible liquids in a U-tube: $\rho_1h_1 = \rho_2h_2$ , heights measured from the interface level.
A barometer measures $p_0 = \rho gh$ .

Sources and credits

14.2 Pressure and 14.3 Pascal's Principle and Hydraulics, from University Physics Volume 1 by William Moebs, Samuel J. Ling and Jeff Sanny, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Facts on Lake Kivu and KivuWatt: Lake Kivu, Wikipedia, licensed under CC BY-SA 4.0; facts only, summarised by Ganzaa.
Video: Measuring Pressure With Barometers and Manometers, Professor Dave Explains, embedded from YouTube.

The experiment, the three diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Gas held down by water

2Pressure at a depth

Deriving p=p0+ρgh p = p_0 + \rho gh p=p0​+ρgh

3The U-tube manometer

Comparing densities with two liquids

4Experiment 4.1: pressure and depth

Gas held down by water

Pressure at a depth

Deriving $p = p_0 + \rho gh$

The U-tube manometer

Experiment 4.1: pressure and depth