12.1: Pressure in Fluids
Objectives
The student will:
- Understand density and be able to solve problems with density.
- Understand pressure and be able to solve problems with pressure.
Vocabulary
- density: The ratio of an object's mass divided by its volume.
- pressure: Force per unit area.
Introduction
In the language of physics, density is defined as the ratio of the mass of an object (or a material) \begin{align*}m\end{align*}
\begin{align*}\rho = \frac{m}{V}\end{align*}
This is different than in everyday English, where we tend to use “dense” only to describe heavy or thick objects. In physics, a massive object may have a small density, and a very light object may have a large density, depending on volume. For example, an empty aluminum soft drink can is very light compared to a full one. However, aluminum is 2.7 times denser than water. The typical mass of an empty soda can is about 15 grams, but the soft drink in it has a mass of about 336 grams. So, a small amount of a denser material can easily have a smaller mass than a large amount of a less dense material.
The typical units for density are kilograms per meter cubed \begin{align*}\text{kg/m}^3\end{align*}
In the language of physics, pressure is defined as force per unit area.
\begin{align*}P = \frac{F}{A}\end{align*}
This is also different than in everyday English, where we often use “pressure” to mean the same as “force” (i.e. “he was pressured to sign” is like “he was forced to sign”). The SI unit of pressure is called the pascal, abbreviated \begin{align*}\text{Pa}\end{align*}
Density and Specific Gravity
Measured in these grams per cubic centimeter, density is the same as specific gravity, defined as the ratio of the object's density to the density of water. The density of water (at \begin{align*}4.0^\circ C\end{align*}
\begin{align*}\rho_{~\text{water}} = \frac{1.00 ~ \text{g}}{\text{cm}^3} = \frac{1000.0 ~ \text{kg}}{\text{m}^3}\end{align*}
So, the specific gravity of water is 1.00, and the specific gravity of aluminum is 2.7. There are no units for specific gravity, because it is defined as the ratio of one density to another.
Illustrative Example 1
a. A United States 25-cent coin (a quarter) is composed of copper sandwiched between two coatings of nickel, Figure below.
United States quarter seen on edge.
The copper in a quarter occupies a volume of \begin{align*}5.84 ~ \text{cm}^3\end{align*}
Answer: \begin{align*}\rho_{~ \text{copper}} = \frac{m}{V} = \frac{5.198 ~ \text{g}}{0.584 ~ \text{cm}^3} = 8.90 \text{g/cm}^3\end{align*}
b. What is the specific gravity of copper?
Answer:
\begin{align*}sg = \frac{\text{density of material}}{\text{density of water}} = \frac{8.90 ~ \text{g/cm}^3}{1.00 ~ \text{g/cm}^3} = 8.90\end{align*}
Check Your Understanding
1a. The nickel of the quarter in Example 1 is \begin{align*}0.4723 ~ \text{g}\end{align*}
Answer:
\begin{align*}\rho_{~\text{copper}} = \frac{m}{V} = \frac{0.4723 ~ \text{g}}{0.0530 ~ \text{cm}^3} = 8.91 ~ \text{g/cm}^3\end{align*}
The densities of copper and nickel are almost the same (see Example 1). Does this surprise you? By comparison, the density of silver is \begin{align*}10.5 ~ \text{g/cm}^3\end{align*}
1b. The current value of copper is 0.8455 cents per gram, and the current value of nickel is 1.8931 cents per gram. What is the value of a United States 25-cent piece based on its metallic content?
Answer: \begin{align*}\text{Total value of copper} = \left(0.8455 \frac{cents}{g} \right)(5.198 \ g) = 4.394 \ cents\end{align*}
\begin{align*}\text{Total value of nickel} = \left(1.8931 \frac{cents}{g} \right)(0.4723 \ g) = 0.89411 \rightarrow 0.8941 \ cents\end{align*}
\begin{align*}\text{Total value} = 4.394 \ cents + 0.8941 \ cents = 5.288 \ cents\end{align*}.
This is not too surprising when you consider that it takes the same amount of paper to print a one-dollar bill as it does a 100 dollar bill!
2. The specific gravity of steel is 7.8. What is the density of steel in grams per cubic centimeter?
Answer: \begin{align*}7.8 ~ \text{g/cm}^3\end{align*}. Do you see why this is so? Try the link below!
http://phet.colorado.edu/en/simulation/density
Illustrative Example 2
Let’s consider the crate in Figure below. The crate has a weight of 1500 N and it is initially placed on the ground. The area of the crate that rests upon the ground is \begin{align*}0.750 ~ \text{m}^2\end{align*}. The pressure on the crate (and the ground under the crate) is \begin{align*}P = \frac{F}{A} = \frac{1,500 \ N}{0.750 \ m^2} = 2,000 \frac{N}{m^2}\end{align*}.
If the crate is now turned on its narrow side, Figure below, the area which rests upon the ground is \begin{align*}0.40 ~ \text{m}^2\end{align*} and the pressure is \begin{align*}P = \frac{F}{A} = \frac{1,500 \ N}{0.400 \ m^2} = 3,750 \frac{N}{m^2}\end{align*}
Check Your Understanding
A woman of weight 540 N stands on the floor. The area of the bottom of each of her feet is \begin{align*}0.0135 \ m^2\end{align*}. What is the pressure on each of her feet?
Answer: Each foot supports half of her weight, so
\begin{align*}F = \frac{540 ~ \text{N}}{2} = 270 ~ \text{N} \rightarrow P = \frac{270 ~ \text{N}}{0.0135 ~ \text{m}^2} = 2.00 \times 10^4 \frac{\text{N}}{\text{m}^2}\end{align*}.
Does the answer seem rather large to you? Remember, the answer is given in units of \begin{align*}\frac{\text{N}}{\text{m}^2}\end{align*}. A square meter is a large area when compared to the area of a person’s foot. In Newtons per square centimeter, the answer is \begin{align*}2.00 \frac{\text{N}}{\text{cm}^2}\end{align*}.
Hydrostatic Pressure
Hydrostatic pressure is the pressure produced by a fluid under gravity when at equilibrium. Under hydrostatics pressure, a fluid remains motionless. For example, a vertical column of water confined behind a dam experiences hydrostatic pressure.
Figure below shows a picture of Hoover Dam and a typical cross section of a dam. Have you ever noticed that dams tend to be bottom heavy, that is, there is more reinforcement at the bottom of the dam than at the top of the dam? Why do you suppose this is?
Try this simple experiment! Poke a hole near the bottom of an empty soft drink can and another hole in the center of the can. Fill the can with water and water will flow out of each hole. But will the flow out of each hole have the same horizontal extent? No. The flow will resemble what we see in Figure below. Water will flow out of the bottom hole at a greater pressure (and velocity) than the top hole. The reason for this and the reason why dams are reinforced more at their base is hydrostatic pressure. Pressure increases with depth.
Figure below shows a column of water visualized as three “cubes” of water, one on top of the other. Each cube is identical, with length, width, and height equal to \begin{align*}d\end{align*} and of weight \begin{align*}mg\end{align*}. The entire column of cubes rests upon area \begin{align*}A\end{align*}. The normal force upon each cube increases with depth since each cube must support the weight of all the cubes above it. The surface area \begin{align*}A\end{align*} remains constant but the force acting upon this area increases with \begin{align*}mg, 2 mg, 3 mg\end{align*}, and so on. We will show that the pressures \begin{align*}P_1, P_2\end{align*}, and \begin{align*}P_3\end{align*} at the points shown in Figure below increase with the height, if the density of the fluid is constant. We define \begin{align*}P_0 = 0\end{align*}.
Consider the pressure at the bottom of each cube.
Using the definition of density, we can rewrite the mass \begin{align*}m\end{align*} in terms of the volume \begin{align*}V\end{align*} and density \begin{align*}\rho\end{align*} as \begin{align*}m = \rho V\end{align*}. The volume of each cube is \begin{align*}d^3\end{align*} and the area of the any side of the cube is \begin{align*}d^2\end{align*}. Substituting \begin{align*}m = \rho V = \rho d^3\end{align*} and \begin{align*}d^2\end{align*} for \begin{align*}A\end{align*} in each pressure equation we have \begin{align*}P_1 = \frac{F}{A} = \frac{\rho d^3 \ g}{d^2}; P_2 = \frac{F}{A} = \frac{2\rho d^3 \ g}{d^2}; P_3 = \frac{F}{A} = \frac{3 \rho d^3 \ g}{d^2} \rightarrow P_1 = \rho gd; P_2 = 2 \rho gd; P_3 = 3 \rho gd\end{align*}.
In each case, \begin{align*}d, 2d, 3d\end{align*} represent the depth of the water as measured from the top of the column where \begin{align*}P_0 = 0\end{align*}. We can replace \begin{align*}d, 2d, 3d\end{align*} with \begin{align*}\Delta h\end{align*}, where \begin{align*}\Delta h\end{align*} represents the height difference in any two positions along the column. We can then express the corresponding difference in pressure as \begin{align*}P_2-P_1 = \Delta P = \rho g \Delta h\end{align*}. If we arbitrarily set \begin{align*}P_1 = 0\end{align*}, we can write \begin{align*}P = \rho gh\end{align*}. The pressure depends on the density and height (or depth) of the column of fluid. We will always assume that \begin{align*}g\end{align*} is constant. If the density is constant, the pressure varies only with the height \begin{align*}h\end{align*}. Follow the link below for a demonstration.
http://demonstrations.wolfram.com/FluidPressure/
Check Your Understanding
1. Lake Pontchartrain in Louisiana, Figure below, has an area of approximately \begin{align*}1600 \ km^2\end{align*} and an average depth of 4.0 m. The average density of the water in the lake is very close to the density of fresh water.
a. What is the average pressure at the bottom of the lake?
Answer:
\begin{align*}P = \rho gh = \left(1000 \frac{kg}{m^2} \right) \left(9.81 \frac{m}{s^2} \right)(4.0 \ m) = 39,240 \rightarrow 3.9 \times 10^4 \frac{N}{m^2}\end{align*}
b. A tall thin tank of water has dimensions of radius 30 cm and height 4.5 m. Compare the pressure at the bottom of the tank to the average pressure at the bottom of Lake Pontchartrain.
Answer: The pressure at the bottom of the tank is greater. No calculation is necessary since the depth of the water tank is greater than the depth of Lake Pontchartrain. Remember that pressure gives force per unit area (how much force a given area experiences). The fact that Lake Pontchartrain has 6.4 billion cubic meters of water compared to the 1.3 cubic meters of water in the tank is irrelevant. The question to consider is: How much force does each square meter at the bottom of Lake Pontchartrain experience?
Illustrative Example 3
A campsite has an open water storage tank whose bottom is elevated 3.00 m above a water faucet.The depth of water in the tank is 1.25 meters.
a. Determine the water pressure at the bottom of the tank. See Figure below.
Answer:
Since the water depth in the tank is 1.25 m
\begin{align*}P_2 = \rho gh = 1000 \frac{kg}{m^3} \left(9.81 \frac{m}{s^2} \right)(1.25 \ m) = 12,262.5 \rightarrow 1.23 \times 10^4 \frac{N}{m^2}\end{align*},
b. Determine the water pressure at the faucet.
Answer;
The hose that runs between the water tank and the faucet has a vertical displacement of 3.00 m. The pressure at the faucet is therefore no different than the pressure 4.25 m directly beneath the water level of the tank. (Imagine the hose extending to the water level of the tank as in Figure below.)
The total elevation from the water surface to the faucet is \begin{align*}1.25 \ m + 3.00 \ m = 4.25 \ m\end{align*}.
The pressure at the faucet is difference in pressure \begin{align*}P_2-P_1\end{align*}. We set \begin{align*}P_1 = 0\end{align*} since \begin{align*}h = 0.00 \ m\end{align*} and \begin{align*}P_2 = \rho gh = 1000 \frac{kg}{m^3} \left(9.81 \frac{m}{s^2} \right)(4.25 \ m) = 41692.5 \rightarrow 4.17 \times 10^4 \frac{N}{m^2}\end{align*}
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