Problem 65 - University Physics with Modern Physics

Question

A large tank of water has a hose connected to it, as shown in Fig. P18.65. 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 \(4.20 \times\) \(10^{5}\) Pa. Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be \(1.00 \times 10^{5}\) 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

Answer

(a) \(v \approx 8.6 \, \text{m/s}\) for \(h = 3.50\, \text{m}\). (b) \(v \approx 7.9 \, \text{m/s}\) for \(h = 3.00\, \text{m}\), and \(v \approx 6.5 \, \text{m/s}\) for \(h = 2.00\, \text{m}\). (c) Flow stops at a lower \(h\), determined from pressure balance.

Step 1: Understand the Problem

We have a tank with water at height \(h\) and a compressed air chamber at the top. The air pressure changes as the water level decreases. We need to find the speed of water as it exits for different values of \(h\) and determine when the flow stops.

Step 2: Apply Bernoulli’s Equation

Bernoulli’s equation for the flow of water from a height \(h\) out of a tank is \[ p + \frac{1}{2}\rho v^2 + \rho gh = \, \text{constant} \], where \(p\) is the pressure, \(\rho\) is the density of the water, and \(v\) is the flow velocity.

Step 3: Determine Initial Conditions

When \(h = 3.50 \, \text{m}\), \(p = 4.20 \times 10^5 \, \text{Pa}\). Assume atmospheric pressure \(p_{a} = 1.00 \times 10^5 \, \text{Pa}\) and \(\rho = 1000 \, \text{kg/m}^3\).

Step 4: Calculate Velocity for \(h = 3.50 \, \text{m}\)

Using Bernoulli's principle, set the initial state energy equal to the state at the hose. For \(h = 3.50\, \text{m}\): \[ 4.20 \times 10^5 + 0 + 1000 \times 9.81 \times 3.50 = 1.00 \times 10^5 + \frac{1}{2} \times 1000 \times v^2 \]. Solve for \(v\), resulting in \(v = \sqrt{2(4.20 \times 10^5 - 1.00 \times 10^5 + 1000 \times 9.81 \times 3.50) / 1000}\). The approximate velocity when solved is \(v \approx 8.6 \, \text{m/s}\).

Step 5: General Formula for Speed of Flow

For any height \(h\), use the equation derived from Bernoulli's equation: \(v = \sqrt{2((P_1 - P_2)/\rho + gh)}\), where \(P_1\) is the initial pressure and \(P_2\) is the atmospheric pressure.

Step 6: Calculate Speed for Different Heights

For \(h = 3.00 \, \text{m}\): Substitute into the general formula to find \(v\approx 7.9 \, \text{m/s}\).For \(h = 2.00 \, \text{m}\): Substitute into the general formula to find \(v\approx 6.5 \, \text{m/s}\).

Step 7: Determine When the Flow Stops

Flow stops when the velocity \(v = 0\), which means the energy terms should balance without contributing potential for flow. \(P_1 = P_2 + \rho gh\), so set equal to solve: \[ (4.20 \times 10^5) / (1.00 \times 10^5) \times (3.50)^\kappa = 2.00 \times 10^5 + (9.81\times 1000\, \times \, h)\] and solve for \(h\).

Key Concepts

Fluid Dynamics

Fluid dynamics is a branch of physics concerned with the mechanics of fluid flow. It involves understanding how liquids and gases move and the forces that act upon them. In the context of the exercise, we are dealing with a tank of water where fluid dynamics principles help us predict how water will behave as it exits through a hose.

The principles of fluid dynamics include:

• Conservation of mass, which states that mass cannot be created or destroyed.
• Conservation of energy, which involves Bernoulli’s Equation in fluid dynamics.
• The study of flow conditions, such as laminar and turbulent flow.

In this scenario, we assume that the water flow from the tank remains steady, meaning it does not change over time. A steady flow assumption lets us apply Bernoulli's Equation for consistent calculations, as the conditions in the tank are assumed to not change abruptly.

Pressure Calculation

Pressure plays a crucial role in determining fluid flow. It is defined as a force exerted by a fluid per unit area. In our problem, we deal with different pressures: the absolute pressure of the compressed air on top of the water and the atmospheric pressure outside the tank.

To understand how these pressures affect the water flow, consider the following:

• Absolute Pressure: Defined as the total pressure at a specific point, including atmospheric pressure. In the tank, it's 4.20 x 10⁵ Pa when the water height is 3.50 m.
• Atmospheric Pressure: The pressure exerted by the weight of the atmosphere, approximated as 1.00 x 10⁵ Pa in this problem.
• Gauge Pressure: The difference between absolute pressure and atmospheric pressure, although it's not directly used here, understanding this concept helps in other similar problems.

By identifying these pressures in Bernoulli's equation, we can determine how they influence the velocity of fluid flow.

Velocity of Fluid Flow

Velocity in fluid dynamics refers to the speed at which fluid particles move from one point to another. When calculating the speed of water leaving the hose, it is essential to use Bernoulli's Equation effectively, which links pressure, potential energy due to height, and kinetic energy due to flow speed.

The velocity of fluid flow can be calculated using:

• The change in potential energy since the height (\( h \)) of the water in the tank affects the gravitational force acting on it.
• The difference in pressure between the compressed air and atmospheric pressure, which provides the driving force for the water to exit the hose.
• Kinetic energy, since increased pressure or height will transition into speed as the water exits.

For different water heights (\( h = 3.50 ext{m}, 3.00 ext{m}, 2.00 ext{m} \)), we use the formula derived from Bernoulli's principle:\[v = \sqrt{2\left(\frac{P_1 - P_2}{\rho} + gh\right)}\]Solving this equation for each height gives us the flow speed at those instances.

Conservation of Energy in Fluids

The principle of conservation of energy is also fundamental in fluid dynamics. It states that energy cannot be created or destroyed, only transformed from one form to another. Bernoulli’s Equation is an expression of this principle for flowing fluids.

In our exercise, conservation of energy allows us to understand:

• The balance between pressure energy, kinetic energy, and potential energy in a fluid.
• How changes in these energies affect fluid flow. For example, a decrease in potential energy (as water level drops) increases the kinetic energy (speed of water flow).
• Why as water exits the tank, the compressed air must do more work to maintain flow, thus changing its pressure.

At the point where the potential energy due to height reduces such that the air pressure and atmospheric pressure equalize, the flow will stop because there is no more potential energy to convert into kinetic energy. This scenario occurs when the pressures and the hydrostatic potential energy of the water (\( \rho gh \)) equalize, preventing further outflow.