/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A reservoir dam holds an \(8.00-... [FREE SOLUTION] | 91Ó°ÊÓ

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A reservoir dam holds an \(8.00-\mathrm{km}^{2}\) lake behind it. Just behind the dam, the lake is \(12.0 \mathrm{~m}\) deep. What is the water pressure \((a)\) at the base of the dam and \((b)\) at a point \(3.0 \mathrm{~m}\) down from the lake's surface? The area of the lake behind the dam has no effect on the pressure against the dam. At any point, \(P=\rho_{w} g h\). (a) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(12.0 \mathrm{~m})=118 \mathrm{kPa}\) (b) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(3.0 \mathrm{~m})=29 \mathrm{kPa}\)

Short Answer

Expert verified
Pressure at the base is 118 kPa; at 3m below surface, it's 29 kPa.

Step by step solution

01

Understanding the Formula for Pressure

The formula for computing the pressure at a certain depth in a fluid is given by \(P = \rho_w g h\). Here, \(\rho_w\) is the density of water (approximately \(1000 \text{ kg/m}^3\)), \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\)), and \(h\) is the depth of the fluid.
02

Calculating Pressure at the Base of the Dam

To find the water pressure at the base of the dam, substitute \(h = 12.0 \text{ m}\) into the formula. Thus, \(P = (1000 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(12.0 \text{ m})\). This gives us \(P = 117,720 \text{ Pa}\) or \(118 \text{ kPa}\) when expressed in kilopascals.
03

Calculating Pressure 3 Meters Below the Surface

For the pressure at a point 3 meters below the surface, substitute \(h = 3.0 \text{ m}\) into the same formula. Thus, \(P = (1000 \text{ kg/m}^3)(9.81 \text{ m/s}^2)(3.0 \text{ m})\). This results in \(P = 29,430 \text{ Pa}\) or \(29 \text{ kPa}\) in kilopascals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
Pressure is a measure of force applied over a surface area. In fluids, pressure is influenced by both the characteristics of the fluid and the depth at which you measure it. To calculate pressure at a specific point in a fluid, we use the formula:
  • \( P = \rho w \cdot g \cdot h \)
In this formula:
- \( P \) represents the pressure at a particular depth.
- \( \rho w \) is the density of the fluid, which, for water, is typically \( 1000 \text{ kg/m}^3 \).
- \( g \) is the gravitational acceleration, generally \( 9.81 \text{ m/s}^2 \).
- \( h \) is the depth of the fluid at the point where the pressure is measured.
The equation shows that pressure increases linearly with depth. So, the deeper you go, the greater the pressure is exerted by the water above.
Density of Water
The density of a fluid is a measure of how much mass it contains in a given volume. For water, the density is approximately \( 1000 \text{ kg/m}^3 \) at standard temperature and pressure. This value is a key factor in many calculations involving fluids, such as determining buoyancy and calculating pressure.
When solving problems in fluid mechanics, we usually use this standard density unless the problem specifies a different value. Since density is a constant factor in the pressure calculation formula, it ensures that pressure calculations for water can be quite straightforward.
Water's density being consistent allows us to make reliable predictions about how water will behave under different physical conditions, including varying depths.
Gravitational Acceleration
Gravitational acceleration, denoted as \( g \), is the rate at which objects accelerate towards the Earth due to gravity. It is approximately \( 9.81 \text{ m/s}^2 \). This constant plays a crucial role in fluid mechanics, especially when calculating pressure across different depths.
In the context of fluid pressure, gravitational acceleration helps determine how much force is applied by the fluid weight per unit area.
  • It affects the pressure component in the pressure calculation formula: \( P = \rho w \cdot g \cdot h \).
The greater the value of \( g \), the higher the pressure exerted by the fluid. This makes gravitational acceleration a vital concept when exploring how pressure varies with depth in lakes, oceans, or any body of fluid.
Depth-related Pressure
Depth-related pressure refers to the fact that pressure in a fluid increases with depth. You can visualize this by thinking of all the water molecules sitting atop each other, with each additional layer adding weight.
For instance, at the base of a dam, the pressure will be quite high because of the depth of the water column above.
  • As you decrease the depth, such as moving 3 meters down from the surface, the pressure measured is less because the column of water above that point is shorter.
This concept explains why divers feel more pressure as they descend deeper into the water and why structures like dams must be built to withstand high pressures at their base. Understanding depth-related pressure is essential in planning and constructing any structure that will interact with fluids.

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Most popular questions from this chapter

A mass (or load) acting downward on a piston confines a fluid of density \(\rho\) in a closed container, as shown in Fig. 13-2. The combined weight of the piston and load on the right is \(200 \mathrm{~N}\), and the cross-sectional area of the piston is \(A=8.0 \mathrm{~cm}^{2}\). Find the total pressure at point- \(B\) if the fluid is mercury and \(h=25 \mathrm{~cm}\left(\rho_{\mathrm{Hg}}=13600 \mathrm{~kg} / \mathrm{m}^{3}\right) .\) What would an ordinary pressure gauge read at \(B ?\) Recall what Pascal's principle tells us about the pressure applied to the fluid by the piston and atmosphere: This added pressure is applied at all points within the fluid. Therefore, the total pressure at \(B\) is composed of three parts: Pressure of the atmosphere \(=1.0 \times 10^{5} \mathrm{~Pa}\) Pressure due to the piston and load \(=\frac{F_{W}}{A}=\frac{200 \mathrm{~N}}{8.0 \times 10^{-4} \mathrm{~m}^{2}}=2.5 \times 10^{5} \mathrm{~Pa}\) Pressure due to the height \(h\) of fluid \(=h p g=0.33 \times 10^{5} \mathrm{~Pa}\) In this case, the pressure of the fluid itself is relatively small. We have Total pressure at \(B=3.8 \times 10^{5} \mathrm{~Pa}\) The gauge pressure does not include atmospheric pressure. Therefore, Gauge pressure at \(B=2.8 \times 10^{5} \mathrm{~Pa}\)

A glass of water has a \(10-\mathrm{cm}^{3}\) ice cube floating in it. The glass is filled to the brim with cold water. By the time the ice cube has completely melted, how much water will have flowed out of the glass? The sp gr of ice is \(0.92\).

In a hydraulic press such as the one shown in Fig. \(13-3\), the large piston has cross-sectional area \(A_{1}=200 \mathrm{~cm}^{2}\) and the small piston has cross-sectional area \(A_{2}=5.0 \mathrm{~cm}^{2} .\) If a force of \(250 \mathrm{~N}\) is applied to the small piston, find the force \(F_{1}\) on the large piston. By Pascal's principle, Pressure under large piston \(=\) Pressure under small piston \(\quad\) or \(\quad \frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}\) so that $$F_{1}=\frac{A_{1}}{A_{2}} F_{2}=\frac{200}{5.0} 250 \mathrm{~N}=10 \mathrm{kN}$$ Note that atmospheric pressure acting on both pistons cancels out of the calculation.

A barrel will rupture when the gauge pressure within it reaches \(350 \mathrm{kPa}\). It is attached to the lower end of a vertical pipe, with the pipe and barrel filled with oil \(\left(\rho=890 \mathrm{~kg} / \mathrm{m}^{3}\right)\). How long can the pipe be if the barrel is not to rupture? From \(P=\rho g h\) we have $$ h=\frac{P}{\rho g}=\frac{350 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}}{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(890 \mathrm{~kg} / \mathrm{m}^{3}\right)}=40.1 \mathrm{~m} $$

A tank containing oil of sp \(\mathrm{gr}=0.80\) rests on a scale and weighs \(78.6 \mathrm{~N}\). By means of a very fine wire, a \(6.0 \mathrm{~cm}\) cube of aluminum, sp \(\mathrm{gr}=2.70\), is submerged in the oil. Find \((a)\) the tension in the wire and \((b)\) the scale reading if none of the oil overflows.

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