14.7.2 Volume Stress and the Bulk Modulus

Young’s modulus characterizes the response of an object to being pulled in one direction. FIGURE 14.39 shows an object being squeezed in all directions. For example, objects under water are squeezed from all sides by the water pressure. The force per unit area $F / A$ applied to all surfaces of an object is called the volume stress. Because the force pushes equally on all sides, the volume stress (unlike the tensile stress) really is the same as pressure p.

The figure illustrates six force vector arrows that point toward the center of each of the faces of a cube with a cross-section area "A".

No material is perfectly rigid. A volume stress applied to an object compresses its volume slightly. The volume strain is defined as $Δ V / V$ . The volume strain is a negative number because the volume stress decreases the volume.

Volume stress, or pressure, is linearly proportional to the volume strain, much as the tensile stress is linearly proportional to the strain in a rod. That is,

\frac{F}{A} = p = - B \frac{Δ V}{V}

(14.42)

where $B$ is called the bulk modulus. The negative sign in Equation 14.42 ensures that the pressure is a positive number. Table 14.4 gives values of the bulk modulus for several materials. Smaller values of $B$ correspond to materials that are more easily compressed. Both solids and liquids can be compressed and thus have a bulk modulus, whereas Young’s modulus applies only to solids.

EXAMPLE 14.16 Compressing a sphere

A 1.00-m-diameter solid steel sphere is lowered to a 10,000 m depth in a deep ocean trench. By how much does its diameter shrink?

MODEL The water pressure applies a volume stress to the sphere.

SOLVE The water pressure at $d = 10, 000 m$ is

p = p_{0} + ρ g d = 1.01 \times 10^{8} Pa

where we used the density of seawater. The bulk modulus of steel, taken from Table 14.4, is $16 \times 10^{10} N / m^{2}$ . Thus the volume strain is

\frac{Δ V}{V} = - \frac{p}{B} = - \frac{1.01 \times 10^{8} Pa}{1.6 \times 10^{11} Pa} = - 6.3 \times 10^{- 4}

The volume of a sphere is $V = \frac{4}{3} π r^{3}$ . If the radius changes by the infinitesimal amount $d r$ , we can use calculus to find that the volume changes by

d V = \frac{4}{3} π d (r^{3}) = \frac{4}{3} π \times 3 r^{2} d r = 4 π r^{2} d r

Thus $Δ V = 4 π r^{2} Δ r$ is a quite good approximation for very small changes in the sphere’s radius and volume. Using this, the volume strain is

\frac{Δ V}{V} = \frac{4 π r^{2} Δ r}{\frac{4}{3} π r^{3}} = \frac{3 Δ r}{r} = - 6.3 \times 10^{- 4}

Solving for $Δ r$ gives $Δ r = - 1.05 \times 10^{- 4} m = - 0.105 mm$ . The diameter changes by twice this, decreasing 0.21 mm.

REVIEW The immense pressure of the deep ocean causes only a tiny change in the sphere’s diameter. You can see that treating solids and liquids as incompressible is an excellent approximation under nearly all circumstances.

CHAPTER 14 CHALLENGE EXAMPLE Draining a cone

A conical tank of radius $R$ and height H, pointed end down, is full of water. A small hole of radius $r$ is opened at the bottom of the tank, with $r ≪ R$ so that the tank drains slowly. Find an expression for the time $T$ it takes to drain the tank completely.

MODEL Model the water as an ideal fluid. We can use Bernoulli’s equation to relate the flow speed from the hole to the height of the water in the cone.

VISUALIZE FIGURE 14.40 is a pictorial representation. At this instant, the water height is $y$ and the top surface has radius $r_{1}$ . Points 1 and 2 are on a streamline from the surface to the exit hole. Because the tank drains slowly, we’ve assumed that the water velocity at the surface is always very close to zero: $v_{1} = 0$ . The pressure at the surface is $p_{1} = p_{atmos}$ . The water discharges into air, so we also have $p_{2} = p_{atmos}$ at the exit.

The figure illustrates a pictorial representation of water draining from a tank.

FIGURE 14.40 Full Alternative Text

SOLVE As the tank drains, the water height $y$ decreases from $H$ to 0. If we can find an expression for $d y / d t$ , the rate at which the water height changes, we’ll be able to find $T$ by integrating from “full tank” at $t = 0$ to “empty tank” at $t = T$ . Our starting point is the rate at which water flows out of the hole at the bottom—the volume flow rate $Q = v_{2} A_{2} = π r^{2} v_{2}$ , where $v_{2}$ is the exit speed.

We can relate $v_{2}$ to the water height $y$ by using Bernoulli’s equation to connect the conditions at the surface (point 1) to conditions at the exit (point 2):

p_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g y_{1} = p_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g y_{2}

With $p_{1} = p_{2}, v_{1} = 0, y_{1} = y, and y_{2} = 0$ at the bottom, Bernoulli’s equation simplifies to $ρ g y = \frac{1}{2} ρ v_{2}^{2}$ . Thus the exit speed of the water is

v_{2} = \sqrt{2 g y}

The exit speed decreases as the water height drops. With this result, the flow rate for water of height $y$ is

Q = π r^{2} \sqrt{2 g y}

The volume of water inside the tank is changing at the same rate as the volume flowing out, except with the opposite sign because the water volume is decreasing with time; that is,

\frac{d V_{water}}{d t} = - Q = - π r^{2} \sqrt{2 g y}

We need to relate $V_{water}$ to the height $y$ of the water surface. The volume of a cone is $V = \frac{1}{3} \times base \times height$ , so the cone of water has volume $V_{water} = \frac{1}{3} π r_{1}^{2} y$ . Based on the similar triangles in Figure 14.40, $r_{1} / R = y / H$ . Thus $r_{1} = (R / H) y$ and

V_{water} = \frac{π R^{2}}{3 H^{2}} y^{3}

Taking the time derivative, we find

\frac{d V_{water}}{d t} = \frac{d}{d t} [\frac{π R^{2}}{3 H^{2}} y^{3}] = \frac{π R^{2}}{H^{2}} y^{2} \frac{d y}{d t}

This relates the rate at which the volume changes to the rate at which the height changes.

With this information, our equation for the rate at which the volume is changing becomes

\frac{d V_{water}}{d t} = \frac{π R^{2}}{H^{2}} y^{2} \frac{d y}{d t} = - π r^{2} \sqrt{2 g y}

In preparation for integration, we need to get all the y’s on one side of the equation and $d t$ on the other. Rearranging gives

d t = - \frac{R^{2}}{r^{2} H^{2} \sqrt{2 g}} y^{3 / 2} d y

We need to integrate this from the beginning, with $y = H$ at $t = 0$ , to the moment the tank is empty, with $y = 0$ at $t = T$ :

\int_{0}^{T} d t = T = - \frac{R^{2}}{r^{2} H^{2} \sqrt{2 g}} \int_{H}^{0} y^{3 / 2} d y = \frac{R^{2}}{r^{2} H^{2} \sqrt{2 g}} \int_{0}^{H} y^{3 / 2} d y

The minus sign was eliminated by reversing the integration limits. Performing the integration gives us the desired result for the time to drain the tank:

\begin{array}{l} T & = \frac{R^{2}}{r^{2} H^{2} \sqrt{2 g}} \int_{0}^{H} y^{3 / 2} d y = \frac{R^{2}}{r^{2} H^{2} \sqrt{2 g}} {[\frac{2}{5} y^{5 / 2}]}_{0}^{H} \\ = \frac{2}{5} \frac{R^{2}}{r^{2}} \sqrt{\frac{H}{2 g}} \end{array}

REVIEW Making the tank larger by increasing $R$ or $H$ increases the time needed to drain. Making the hole at the bottom larger—a larger value of $r$ —decreases the time. These are as we would have expected, giving us confidence in our result.

14.7.2 Volume Stress and the Bulk Modulus

FIGURE 14.39 An object is compressed by pressure forces pushing equally on all sides.

EXAMPLE 14.16 Compressing a sphere

CHAPTER 14 CHALLENGE EXAMPLE Draining a cone

FIGURE 14.40 Pictorial representation of water draining from a tank.