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Variation of Pressure with Depth in a Fluid

If your ears have ever popped on a plane flight or ached during a deep dive in a swimming pool, you have experienced the effect of depth on pressure in a fluid. At the Earth’s surface, the air pressure exerted on you is a result of the weight of air above you. This pressure is reduced as you climb up in altitude and the weight of air above you decreases. Under water, the pressure exerted on you increases with increasing depth. In this case, the pressure being exerted upon you is a result of both the weight of water above you and that of the atmosphere above you. You may notice an air pressure change on an elevator ride that transports you many stories, but you need only dive a meter or so below the surface of a pool to feel a pressure increase. The difference is that water is much denser than air, about 775 times as dense.

Consider the container in Figure 11.9. Its bottom supports the weight of the fluid in it. Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure is the weight of the fluid $mg mg size 12{ ital "mg"} {}$ divided by the area $A A size 12{A} {}$ supporting it (the area of the bottom of the container):

P = \frac{mg}{A} . P = \frac{mg}{A} . size 12{P= {  { ital "mg"}  over  {A} } } {}

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We can find the mass of the fluid from its volume and density:

m = ρV . m = ρV . size 12{m=ρV} {}

11.13

The volume of the fluid $V V size 12{V} {}$ is related to the dimensions of the container. It is

V = Ah, V = Ah, size 12{V= ital "Ah"} {}

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where $A A size 12{A} {}$ is the cross-sectional area and $h h size 12{h} {}$ is the depth. Combining the last two equations gives

m = ρ Ah . m = ρ Ah . size 12{m=ρ ital "Ah"} {}

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If we enter this into the expression for pressure, we obtain

P = \frac{(ρ Ah) g}{A} . P = \frac{(ρ Ah) g}{A} . size 12{P= {  { left (ρ ital "Ah" right )g}  over  {A} } } {}

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The area cancels, and rearranging the variables yields

P = hρg . P = hρg . size 12{P=hρg} {}

11.17

This value is the pressure due to the weight of a fluid. The equation has general validity beyond the special conditions under which it is derived here. Even if the container were not there, the surrounding fluid would still exert this pressure, keeping the fluid static. Thus the equation $P = hρg P = hρg size 12{P=hρg} {}$ represents the pressure due to the weight of any fluid of average density $ρ ρ size 12{ρ} {}$ at any depth $h h size 12{h} {}$ below its surface. For liquids, which are nearly incompressible, this equation holds to great depths. For gases, which are quite compressible, one can apply this equation as long as the density changes are small over the depth considered. Example 11.4 illustrates this situation.

A container with fluid filled to a depth h. The fluid’s weight w equal to m times g is shown by an arrow pointing downward. A denotes the area of the fluid at the bottom of the container and as well as on the surface.

Figure 11.9 The bottom of this container supports the entire weight of the fluid in it. The vertical sides cannot exert an upward force on the fluid (since it cannot withstand a shearing force), and so the bottom must support it all.

Example 11.3

Calculating the Average Pressure and Force Exerted: What Force Must a Dam Withstand?

In Example 11.1, we calculated the mass of water in a large reservoir. We will now consider the pressure and force acting on the dam retaining water. (See Figure 11.10.) The dam is 500 m wide, and the water is 80.0 m deep at the dam. (a) What is the average pressure on the dam due to the water? (b) Calculate the force exerted against the dam and compare it with the weight of water in the dam (previously found to be $1.96 \times 10^{13} N 1.96 \times 10^{13}$ ).

Strategy for (a)

The average pressure $\bar{P} \bar{P}$ due to the weight of the water is the pressure at the average depth $\bar{h} \bar{h}$ of 40.0 m, since pressure increases linearly with depth.

Solution for (a)

The average pressure due to the weight of a fluid is

\bar{P} = \bar{h} ρg . \bar{P} = \bar{h} ρg . size 12{P=hρg} {}

11.18

Entering the density of water from Table 11.1 and taking $\bar{h} \bar{h} size 12{h} {}$ to be the average depth of 40.0 m, we obtain

\begin{array}{l} \bar{P} & = & (40.0 m) (10^{3} \frac{kg}{m^{3}}) (9.80 \frac{m}{s^{2}}) \\ = & 3.92 \times 10^{5} \frac{N}{m^{2}} = 392 kPa. \end{array} \begin{array}{l} \bar{P} & = & (40.0 m) (10^{3}) (9.80) \\ = 3.92 \times 10^{5} \end{array}

11.19

Strategy for (b)

The force exerted on the dam by the water is the average pressure times the area of contact:

F = \bar{P} A . F = \bar{P} A . size 12{F= {overline  {P}} A} {}

11.20

Solution for (b)

We have already found the value for $\bar{P} \bar{P} size 12{ { bar {P}}} {}$ . The area of the dam is $A = 80.0 m \times 500 m = 4.00 \times 10^{4} m^{2} A = 80.0 m \times 500 m = 4.00 \times 10^{4} size 12{A="80" "." 0`m times "500"`m=4 "." "00" times "10" rSup { size 8{4} } `m rSup { size 8{2} } } {}$ , so that

\begin{array}{l} F & = & (3.92 \times 10^{5} {N/m}^{2}) (4.00 \times 10^{4} m^{2}) \\ = & 1.57 \times 10^{10} N. \end{array} \begin{array}{l} F & = & (3.92 \times 10^{5} \\ = 1.57 \times 10^{10} \end{array} alignl { stack {
 size 12{F= \( 3 "." "92" times "10" rSup { size 8{5} } `"N/m" rSup { size 8{2} }  \)  \( 4 "." "00" times "10" rSup { size 8{4} } `m rSup { size 8{2} }  \) }  {} #
"    "=1 "." "57" times "10" rSup { size 8{"10"} } `N "."  {}
} } {}

11.21

Discussion

Although this force seems large, it is small compared with the $1.96 \times 10^{13} N 1.96 \times 10^{13} size 12{1 "." "96" times "10" rSup { size 8{"13"} } `N} {}$ weight of the water in the reservoir—in fact, it is only $0.0800% 0.0800%$ of the weight. Note that the pressure found in part (a) is completely independent of the width and length of the lake—it depends only on its average depth at the dam. Thus the force depends only on the water’s average depth and the dimensions of the dam, not on the horizontal extent of the reservoir. In the diagram, the thickness of the dam increases with depth to balance the increasing force due to the increasing pressure.epth to balance the increasing force due to the increasing pressure.

A two-dimensional view of a dam with dimensions L and h is shown. Force F at h is shown by a horizontal arrow. The force F exerted by water on the dam is F equals average pressure p bar into area A and pressure in turn is average height h bar into density rho into acceleration due to gravity g.

Figure 11.10 The dam must withstand the force exerted against it by the water it retains. This force is small compared with the weight of the water behind the dam.

Atmospheric pressure is another example of pressure due to the weight of a fluid, in this case due to the weight of air above a given height. The atmospheric pressure at the Earth’s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth’s rotation (this creates weather “highs” and “lows”). However, the average pressure at sea level is given by the standard atmospheric pressure $P_{atm} P_{atm} size 12{P rSub { size 8{"atm"} } } {}$ , measured to be

1 atmosphere (atm) = P_{atm} = 1.01 \times 10^{5} {N/m}^{2} = 101 kPa . 1 atmosphere (atm) = P_{atm} = 1.01 \times 10^{5} . size 12{1`"atmosphere"` \( "atm" \) =P rSub { size 8{"atm"} } =1 "." "01" times "10" rSup { size 8{5} } `"N/m" rSup { size 8{2} } ="101"`"kPa"} {}

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This relationship means that, on average, at sea level, a column of air above $1.00 m^{2} 1.00$ of the Earth’s surface has a weight of $1.01 \times 10^{5} N 1.01 \times 10^{5} size 12{1 "." "01" times "10" rSup { size 8{5} } `N} {}$ , equivalent to $1 atm 1 atm$ . (See Figure 11.11.)

Figure shows a column of air exerting a weight of one point zero one times ten to the power five newtons on a rectangular patch of ground of one square meter cross section.

Figure 11.11 Atmospheric pressure at sea level averages

1 . 01 \times 10^{5} Pa 1 . 01 \times 10^{5} size 12{1 "." "01" times "10" rSup { size 8{5} } `"Pa"} {}

(equivalent to 1 atm), since the column of air over this

1 m^{2} 1 size 12{1`m rSup { size 8{2} } } {}

, extending to the top of the atmosphere, weighs

1 . 01 \times 10^{5} N 1 . 01 \times 10^{5} size 12{1 "." "01" times "10" rSup { size 8{5} } " N"} {}

Example 11.4

Calculating Average Density: How Dense Is the Air?

Calculate the average density of the atmosphere, given that it extends to an altitude of 120 km. Compare this density with that of air listed in Table 11.1.

Strategy

If we solve $P = hρg P = hρg size 12{P=hρg} {}$ for density, we see that

\bar{ρ} = \frac{P}{hg} . \bar{ρ} = \frac{P}{hg} . size 12{ { bar  {ρ}}= {  {P}  over  { ital "hg"} } } {}

11.23

We then take $P P size 12{P} {}$ to be atmospheric pressure, $h h size 12{h} {}$ is given, and $g g size 12{g} {}$ is known, and so we can use this to calculate $\bar{ρ} \bar{ρ} size 12{ { bar {ρ}}} {}$ .

Solution

Entering known values into the expression for $\bar{ρ} \bar{ρ} size 12{ { bar {ρ}}} {}$ yields

\bar{ρ} = \frac{1 . 01 \times 10^{5} {N/m}^{2}}{(120 \times 10^{3} m) (9.80 {m/s}^{2})} = 8 . 59 \times 10^{- 2} {kg/m}^{3} . \bar{ρ} = \frac{1 . 01 \times 10^{5}}{(120 \times 10^{3}} = 8 . 59 \times 10^{- 2} . size 12{ { bar  {ρ}}= {  {1 "." "01" times "10" rSup { size 8{5} } `"N/m" rSup { size 8{2} } }  over  { \( "120" times "10" rSup { size 8{3} } `m \)  \( 9 "." "80"`"m/s" rSup { size 8{2} }  \) } } =8 "." "59" times "10" rSup { size 8{ - 2} } `"kg/m" rSup { size 8{3} } } {}

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Discussion

This result is the average density of air between the Earth’s surface and the top of the Earth’s atmosphere, which essentially ends at 120 km. The density of air at sea level is given in Table 11.1 as $1.29 {kg/m}^{3} 1.29 size 12{1 "." "29"`"kg/m" rSup { size 8{3} } } {}$ —about 15 times its average value. Because air is so compressible, its density has its highest value near the Earth’s surface and declines rapidly with altitude.

Example 11.5

Calculating Depth Below the Surface of Water: What Depth of Water Creates the Same Pressure as the Entire Atmosphere?

Calculate the depth below the surface of water at which the pressure due to the weight of the water equals 1.00 atm.

Strategy

We begin by solving the equation $P = hρg P = hρg size 12{P=hρg} {}$ for depth $h h size 12{h} {}$ :

h = \frac{P}{ρg} . h = \frac{P}{ρg} . size 12{h= {  {P}  over  {ρg} } } {}

11.25

Then we take $P P size 12{P} {}$ to be 1.00 atm and $ρ ρ size 12{ρ} {}$ to be the density of the water that creates the pressure.

Solution

Entering the known values into the expression for $h h size 12{h} {}$ gives

h = \frac{1 . 01 \times 10^{5} {N/m}^{2}}{(1 . 00 \times 10^{3} {kg/m}^{3}) (9.80 {m/s}^{2})} = 10 . 3 m . h = \frac{1 . 01 \times 10^{5}}{(1 . 00 \times 10^{3}} = 10 . 3 . size 12{h= {  {1 "." "01" times "10" rSup { size 8{5} } `"N/m" rSup { size 8{2} } }  over  { \( 1 "." "00" times "10" rSup { size 8{3} } `"kg/m" rSup { size 8{3} }  \)  \( 9 "." "80"`"m/s" rSup { size 8{2} }  \) } } ="10" "." 3`m} {}

11.26

Discussion

Just 10.3 m of water creates the same pressure as 120 km of air. Since water is nearly incompressible, we can neglect any change in its density over this depth.

What do you suppose is the total pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water’s surface affect the pressure below? The answer is yes. This seems only logical, since both the water’s weight and the atmosphere’s weight must be supported. So the total pressure at a depth of 10.3 m is 2 atm—half from the water above and half from the air above. We shall see in Pascal’s Principle that fluid pressures always add in this way.