Problem 85 - College Physics

Question

Figure P9.85 shows a water tank with a valve. If the valve is opened, what is the maximum height attained by the stream of water coming out of the right side of the tank? Assume \(h=10.0 \mathrm{~m}, L=2.00 \mathrm{~m}\), and \(\theta=30.0^{\circ}\), and that the cross-sectional area at \(A\) is very large compared with that at \(B\).

Step-by-Step Solution

Answer

The maximum height the stream of water will reach is 9.0 m.

Step 1: Determination of velocity using Bernoulli’s Equation

We begin with Bernoulli's equation, which when considering points A (top of the water in the tank) and B (outlet of the hose), can be written as: \( P_A + \frac{1}{2}\rho v_A^2 + \rho g h_A = P_B + \frac{1}{2}\rho v_B^2 + \rho g h_B \). Since the cross-sectional area at \(A\) is very large compared to \(B\), we can assume that the velocity of water at \(A\) is negligible compared to \(B\). Hence \(v_A\) approximates to 0. Thus, the equation simplifies to \( P_A + \rho g h_A = P_B + \frac{1}{2}\rho v_B^2 + \rho g h_B \). Here, \(P_A\) is the atmospheric pressure at points \(A\) and \(B\), so \(P_A = P_B\), and hence the equation would be simplified to: \( \rho g h_A = \frac{1}{2}\rho v_B^2 + \rho g h_B \). Solving the equation for \(v_B\) (velocity of the water at the outlet) we get \(v_B = \sqrt{2g(h_A - h_B)}\).

Step 2: Calculate \(v_B\)

The velocity of the water at the outlet \(v_B\) can be calculated using the values given in the problem: \(h_A = 10m\) and \(h_B = L sin \theta = 2m sin 30° = 1m\). So \(v_B = \sqrt{2g(h_A - h_B)} = \sqrt{2* 9.8 m/s^2 (10m - 1m)} = \sqrt{176.4 m^2/s^2} = 13.3 m/s.

Step 3: Maximum Height Calculation

To determine the maximum height attained by the water stream, we can equate the kinetic energy at the outlet (since this will convert to potential energy at the maximum height) to the gravitational potential energy at the maximum height. So \( \frac{1}{2} m v_B^2 = m g h_{max} \). Solving for \(h_{max} = \frac{v_B^2}{2g}\).

Step 4: Calculate \(h_{max}\)

Now we can substitute the value of \(v_B\) into the equation to get \(h_{max} = \frac{(13.3 m/s)^2}{2* 9.8 m/s^2} = 9.0m\). So the maximum height attained by water is 9.0m.

Key Concepts

Understanding Fluid Mechanics

Fluid mechanics is a branch of physics that studies the behavior of fluids (liquids and gases) and the forces on them. It can be quite fascinating to see fluid mechanics in action, particularly in the case of water flowing from a tank. When a valve opens and water gushes out, it's an example of fluid motion that can be understood using the principles of fluid mechanics.

These principles allow us to describe the water's flow using mathematical equations. In this case, Bernoulli's principle is particularly useful. It states that in a flowing fluid, the total energy along a streamline (a path the fluid elements follow) remains constant if the flow is steady and the fluid is incompressible and no energy is added or lost. The energy we look at here consists of three parts: the fluid's potential energy (related to its height above a reference level), kinetic energy (related to its velocity), and pressure energy.

So, in the exercise where the water travels from a high point in the tank to the outlet at a lower point, it's Bernoulli's principle that predicts the behavior of the water as it changes from having high potential energy and low kinetic energy at the top to having higher kinetic energy and reduced potential energy as it flows out.

Kinetic and Potential Energy in Fluid Flow

Let's delve into the energy transformations involved when a fluid flows. Kinetic energy is what comes to mind when we think of something in motion. For fluids, it's the energy due to the fluid's velocity. Potential energy, on the other hand, is tied to an object's position or height in a gravitational field and in the context of fluids, this often relates to the height above a certain reference point.

When we look at the exercise with water flowing from the tank, the gravitational potential energy that the water has initially (due to its height) is converted into kinetic energy as it falls. Bernoulli's equation elegantly demonstrates this energy conversion by equating the sum of kinetic energy, potential energy, and pressure energy at one point in the fluid with the sum of those energies at another point. Indeed, these energy exchanges are responsible for the water stream reaching a specific height once the valve is open.

The Equation of Continuity

The equation of continuity is a fundamental principle in fluid mechanics and one that's vital for understanding the flow of fluids in situations like the one described in the exercise. It asserts that for an incompressible, steady flow, the mass flowing into a pipe must equal the mass flowing out. This can be expressed mathematically as \( A_1v_1 = A_2v_2 \) where \( A \) represents the cross-sectional area and \( v \) is the velocity of the fluid at points 1 and 2.

In simpler terms, if you have a hose with varying widths, the water must speed up as it moves through narrower sections to compensate for the reduced space. This makes sense if you think about it: Since the water can't compress, more of it must move through the narrow part quickly to keep the flow constant. This principle also suggests why, in the exercise, we assume that the tank's top, with a large cross-sectional area, has negligible water speed compared to the stream of water shooting out of the narrow valve.