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Chapter 18: Problem 59

A large tank of water has a hose connected to it (Fig. P18.59). 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 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 m? (b) As water flows out of the tank, \(h\) decreases. Calculate the speed of flow for \(h\) = 3.00 m and for \(h\) = 2.00 m. (c) At what value of h does the flow stop?

Short Answer

Expert verified
(a) Speed for 3.50 m: Calculate using Bernoulli. (b) Use same method for 3.00 m, 2.00 m. (c) Flow stops when pressure difference is zero (use formula).

Step by step solution

01

Understanding Bernoulli's Equation

Bernoulli's equation relates the speed of fluid flow to the pressure difference and height difference in a fluid system. It is given as:\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \]where \(P\) is the pressure, \(\rho\) is the density of the fluid (for water, \(1000 \text{ kg/m}^3\)), \(v\) is the velocity, and \(h\) is the height. We'll assume the velocity of air above the water is negligible, so we focus on pressure.
02

Speed of Flow for h = 3.50 m

When \( h = 3.50 \text{ m} \), the pressure of the air is \( P = 4.20 \times 10^5 \text{ Pa} \). The speed can be determined by setting Bernoulli's equation between the water surface and the hose exit. Assuming atmospheric pressure at the exit:\[ P + \rho gh = P_a + \frac{1}{2} \rho v^2 \]Solving for \( v \) gives:\[ v = \sqrt{\frac{2(P - P_a + \rho gh)}{\rho}} \]Substitute \( P = 4.20 \times 10^5 \text{ Pa} \), \( P_a = 1.00 \times 10^5 \text{ Pa} \), \( \rho = 1000 \text{ kg/m}^3 \), \( h = 3.50 \text{ m} \) to find \( v \).
03

Speed Calculation for Different h

Repeat the process for \( h = 3.00 \text{ m} \) and \( h = 2.00 \text{ m} \). Use the fact that the air expands isothermally, using \( P_1V_1 = P_2V_2 \) to find the new air pressures, and substitute them into the Bernoulli's equation. Calculate each speed just like in Step 2.
04

Determining Stopping Condition

Find the condition where no flow occurs by using Bernoulli's equation. The flow stops when \( P = P_a + \rho gh \). Rearrange the equation to solve for the height \( h \) when the pressures equalize.

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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 a branch of physics that deals with the behavior of liquids and gases in motion. This encompasses studying how forces, such as pressure differences, affect the motion of fluids in various situations. Understanding fluid dynamics is essential for solving problems where fluids like water or air are in motion, like in a sealed tank with water and compressed air.
  • Key Concepts: The main factors influencing fluid dynamics include fluid density, velocity, pressure, and flow rate.
  • Application: By applying principles of fluid dynamics, such as Bernoulli's equation, we can calculate how fast a fluid will move in different conditions.
For instance, when water flows through a hose from a tank, fluid dynamics helps us understand how the pressure and height affect the water's speed.
Pressure Difference
Pressure difference is a crucial factor in determining fluid flow. It refers to the change in pressure from one point in a system to another. In the context of a tank with water and air pressure, the pressure difference between the top of the water and the atmospheric pressure outside affects how fast the water exits through the hose.
  • Understanding Pressure: Pressure is commonly measured in Pascals (Pa). It is defined as force per unit area.
  • Effects on Flow: A greater pressure inside the tank compared to external atmospheric pressure typically results in faster fluid flow.
Bernoulli's equation uses this pressure difference to help calculate fluid speed, by relating it to the kinetic energy of the fluid as it flows from one point to another.
Isothermal Process
An isothermal process is a thermodynamic process that occurs at a constant temperature. For example, when air inside a tank above water expands isothermally, its temperature does not change despite changes in pressure or volume.
  • Key Characteristics: During an isothermal process, any added heat results in changes in the pressure and volume rather than temperature.
  • Application: By treating the air's expansion as isothermal, we can apply the equation \( P_1V_1 = P_2V_2 \) to find new pressures as the fluid dynamics change in the tank.
Thus, understanding this process allows us to calculate the changes in pressure that occur as water leaves the tank and air expands.
Velocity Calculation
Calculating the velocity of fluid flow is essential in applications involving fluid dynamics, like determining how fast water exits a tank. Bernoulli's principle is helpful for these calculations, as it considers the energy conservation in a flowing fluid.
  • Bernoulli's Equation: The equation \( P + \rho gh = P_a + \frac{1}{2} \rho v^2 \) relates pressure and gravitational potential energy to fluid velocity.
  • Using Known Values: By knowing the pressure, density \( \rho \), and height \( h \), the equation can be rearranged to solve for velocity \( v \).
  • Real-World Example: When \( h = 3.50 \text{ m} \), the pressure difference allows us to compute the speed at which water exits the tank by substituting values into the equation.
Understanding how to calculate velocity ensures accurate predictions of fluid behavior in complex systems.

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Most popular questions from this chapter

The atmosphere of Mars is mostly CO\(_2\) (molar mass 44.0 g/mol) under a pressure of 650 Pa, which we shall assume remains constant. In many places the temperature varies from 0.0\(^\circ\)C in summer to -100\(^\circ\)C in winter. Over the course of a Martian year, what are the ranges of (a) the rms speeds of the CO\(_2\) molecules and (b) the density (in mol/m\(^3\)) of the atmosphere?

A light, plastic sphere with mass \(m\) = 9.00 g and density \(\rho\) = 4.00 kg/m\(^3\) is suspended in air by thread of negligible mass. (a) What is the tension \(T\) in the thread if the air is at 5.00\(^\circ\)C and \(p\) = 1.00 atm? The molar mass of air is 28.8 g/mol. (b) How much does the tension in the thread change if the temperature of the gas is increased to 35.0\(^\circ\)C? Ignore the change in volume of the plastic sphere when the temperature is changed.

Calculate the volume of 1.00 mol of liquid water at 20\(^\circ\)C (at which its density is 998 kg/m\(^3\)), and compare that with the volume occupied by 1.00 mol of water at the critical point, which is 56 \(\times\) 10\({^-}{^6}\) m\(^3\). Water has a molar mass of 18.0 g/mol.

An automobile tire has a volume of 0.0150 m\(^3\) on a cold day when the temperature of the air in the tire is 5.0\(^\circ\)C and atmospheric pressure is 1.02 atm. Under these conditions the gauge pressure is measured to be 1.70 atm (about 25 lb/in.\(^2\)). After the car is driven on the highway for 30 min, the temperature of the air in the tires has risen to 45.0\(^\circ\)C and the volume has risen to 0.0159 m\(^3\). What then is the gauge pressure?

We have two equal-size boxes, \(A\) and \(B\). Each box contains gas that behaves as an ideal gas. We insert a thermometer into each box and find that the gas in box \(A\) is at 50\(^\circ\)C while the gas in box \(B\) is at 10\(^\circ\)C. This is all we know about the gas in the boxes. Which of the following statements must be true? Which could be true? Explain your reasoning. (a) The pressure in \(A\) is higher than in \(B\). (b) There are more molecules in \(A\) than in \(B\). (c) A and B do not contain the same type of gas. (d) The molecules in \(A\) have more average kinetic energy per molecule than those in \(B\). (e) The molecules in \(A\) are moving faster than those in \(B\).
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