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A large tank of water has a hose connected to it (Fig. P18.61). The tank is sealed at the top and has compressed air between the water surface and the top. When the water height \(h\) has the value \(3.50 \mathrm{~m}\), the absolute pressure \(p\) of the compressed air is \(4.20 \times 10^{5} \mathrm{~Pa}\). Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be \(1.00 \times 10^{5} \mathrm{~Pa}\). (a) What is the speed with which water flows out of the hose when \(h=3.50 \mathrm{~m} ?\) (b) As water flows out of the tank, \(h\) decreases. Calculate the speed of flow for \(h=3.00 \mathrm{~m}\) and for \(h=2.00 \mathrm{~m} .\) (c) At what value of \(h\) does the flow stop?

Step by step solution

01

Bernoulli's Principle

First, using Bernoulli's equation which is \(P + \frac{1}{2}蟻v^2 + 蟻gh = const\), where \(P\) is the pressure, \(v\) is the velocity, \(蟻\) is the density of the fluid, and \(h\) is the height. In this case, apply this to both inside the tank and at the exit of the hose. Inside the tank water is at rest, so \(v = 0\). This gives us : \(P_{air} + 蟻_{water}gh = P_{atm} + \frac{1}{2}蟻_{water}v^2\).

02

Solve for water velocity when h=3.5m

Plug in given values and isolate \(v\). \(P_{air} = 4.20 \times 10^{5} Pa\), \(P_{atm} = 1.00 \times 10^{5} Pa\), \(蟻_{water} = 1000 kg/m^3\), and \(g = 9.8 m/s^2\). Solving for \(v\) when \(h = 3.50 m\) gives \(v = \sqrt{\frac{2(P_{air} + 蟻_{water}gh - P_{atm})}{蟻_{water}}}\). Calculate the values.

03

Solve for water velocity when h=3.0m and h=2.0m

Apply the same formula as step 2 with values of \(h = 3.00 m\) and \(h = 2.00 m\).

04

Determine when the water flow stops

The water will stop flowing when the pressure of the air equals the atmospheric pressure because then the forces will balance out. Therefore, \(h\) when \(P_{air} = P_{atm}\). Use the ideal gas law \(PV = nRT\) assuming \(nRT = const\) (temperature and amount of air mass is constant), so \(P_1V_1 = P_2V_2\). Substituting \(V = Ah\) (A is the cross-sectional area of the tank and h is height), we get \(P_{1}h_{1} = P_{2}h_{2}\) also since \(P_2 = P_{atm}\), solve for \(h_2\).

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Key Concepts

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

Fluid Dynamics

Fluid dynamics is the study of how fluids (liquids and gases) move. In our exercise, water flows from a tank through a hose, which is a good example of fluid dynamics at work. The key to understanding this concept is realizing that fluid flow can change under different conditions. There are several factors that affect fluid movement, such as:

• Viscosity: This measures how thick or sticky a fluid is. Water has low viscosity, which allows it to flow easily.
• Flow rate: This is the volume of fluid that moves through a section of the system per unit of time.
• Flow type: Fluids can have laminar (smooth and orderly) or turbulent (rough and chaotic) flow. In this case, since the air in the tank is compressed and the water flows out, it鈥檚 more likely to be turbulent if the flow speed increases.
• Bernoulli鈥檚 principle: An important part of fluid dynamics, it tells us how fluid pressure, fluid velocity, and height are related.

By understanding these factors, it becomes easier to analyze and predict how the water behaves as it exits the hose.

Pressure

Pressure is a measure of how much force is applied over a certain area. In our problem, pressure plays a crucial role in determining how fast the water exits the hose. The compressed air in the tank creates pressure on the water surface. This pressure is higher than atmospheric pressure, thus forcing water out through the hose.
The critical points about pressure in fluid dynamics include:

• Absolute pressure: This is the total pressure experienced at any point. In our tank, the absolute pressure includes both the atmospheric pressure and the pressure from the compressed air.
• Gauge pressure: This measures the pressure relative to atmospheric pressure. In many problems, knowing this is enough because atmospheric pressure is constant at sea level.
• Pressure gradient: Differences in pressure between two points can set fluids in motion. In the tank, pressure differences push the water through the hose.

Understanding the relationships among these pressure concepts, and how they interact with fluid dynamics, is crucial for solving problems like this one.

Velocity of Fluid Flow

The velocity of fluid flow refers to the speed at which the fluid moves through a section of the system. This velocity is governed by several factors:

• Bernoulli鈥檚 equation: This equation relates the velocity of a fluid to pressure and height, showing that a decrease in pressure or an increase in height accompanies an increase in the fluid's velocity.
• Continuity equation: This states that the flow rate must remain constant in a closed system. Thus, if a fluid enters a narrower section of a pipe, it has to speed up to maintain flow rate.

In the problem鈥檚 context, when Bernoulli鈥檚 equation is applied at two points (inside the tank and at the hose outlet), it allows us to calculate the exit velocity of water by understanding how the pressures and heights at these points differ.
For example, as water leaves the tank, the height decreases, affecting the velocity according to the equation. As the height further decreases to points like 3.0 and 2.0 meters, the velocity will change accordingly. This interplay is critical in determining how fast water flows out through the hose.

Ideal Gas Law

The ideal gas law is an important relation in physics that connects pressure, volume, and temperature of a gas. Its formula is expressed as:
\( PV = nRT \)
Where:

• \( P \) is the pressure.
• \( V \) is the volume.
• \( n \) is the number of moles of gas.
• \( R \) is the gas constant.
• \( T \) is the temperature.

This law assumes that the gas behaves ideally, meaning there are no intermolecular forces and the gas molecules occupy negligible space. In this problem:

• The air above the water is treated as an ideal gas with a constant temperature as the water flows out.
• The compressed air鈥檚 pressure affects how it expands as the water leaves, showing that pressure and volume have an inverse relationship when temperature is constant (Boyle鈥檚 Law).

In the exercise, we use this law to determine when the water flow will stop. This happens when the compressed air pressure drops to equal the atmospheric pressure, allowing us to find the specific water height where this occurs.