Question

An open-topped tank completely filled with water has a release valve near its bottom. The valve is \(1.0 \mathrm{~m}\) below the water surface. Water is released from the valve to power a turbine, which generates electricity. The area of the ton of the tank \(A_{1}\) is 10 times the cross sectional area, \(A_{y}\) of the valve opening. Calculate the speed of the water as it exits the valve. Neglect friction and viscosity. In addition, calculate the speed of a drop of water released from rest at \(h=1.0 \mathrm{~m}\) when it reaches the elevation of the valve Compare the two speeds.

Short answer

Answer: The speed of the water exiting the valve is approximately \(4.43 \mathrm{~m/s}\). This speed is the same as the speed of a water droplet falling from rest at the same height, which is also approximately \(4.43 \mathrm{~m/s}\).

Step-by-step solution

Identify the variables

The variables given in the exercise are - Height from the water surface (\(h\)): 1 m - Area of the top of the tank (\(A_1\)): 10 times the cross-sectional area of the valve opening (\(A_v\)) - Cross-sectional area of valve opening (\(A_v\)): unknown - Speed of the water as it exits the valve (\(v\)): unknown - Speed of a droplet from rest at \(h = 1.0 \mathrm{~m}\) when reaching the elevation of the valve (\(v_2\)): unknown

Apply Bernoulli's equation to the open top and the valve

Bernoulli's equation for an incompressible fluid is given by \(P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2\) where \(P\) is the pressure, \(\rho\) is the density of water, \(v\) is the flow speed and \(h\) is the height. At the open top (\(1\)): \(P_1 = P_{atm}\), \(v_1 = 0\), and \(h_1 = 1 \mathrm{~m}\). At the valve (\(2\)): \(P_2 = P_{atm}\), \(v_2\) is to be determined, and \(h_2 = 0 \mathrm{~m}\). Substitute the values into Bernoulli's equation: \(P_{atm} + \frac{1}{2} \rho (0)^2 + \rho g (1 \mathrm{~m}) = P_{atm} + \frac{1}{2} \rho v_2^2 + \rho g (0 \mathrm{~m})\)

Calculate the speed of the water as it exits the valve (\(v\))

Solve the Bernoulli's equation for \(v_2\): \(\rho g (1 \mathrm{~m}) = \frac{1}{2} \rho v_2^2\) \(v_2^2 = 2g(1 \mathrm{~m})\) \(v_2 = \sqrt{2g(1 \mathrm{~m})}\) Using \(g = 9.81 \mathrm{~m/s^2}\), we get \(v_2 = \sqrt{2(9.81 \mathrm{~m/s^2})(1 \mathrm{~m})} \approx 4.43 \mathrm{~m/s}\)

Calculate the speed of a water droplet falling from rest at \(h\) when it reaches the elevation of the valve

Let's use the conservation of mechanical energy: \(E_{initial} = E_{final}\) Since the droplet is falling from rest, the initial kinetic energy is 0. Thus, \(Mg(1 \mathrm{~m}) + 0 = 0 + \frac{1}{2}Mv_{droplet}^2\) Solve for \(v_{droplet}\): \(v_{droplet} = \sqrt{2g(1 \mathrm{~m})}\) Again, using \(g = 9.81 \mathrm{~m/s^2}\), we get \(v_{droplet} = \sqrt{2(9.81 \mathrm{~m/s^2})(1 \mathrm{~m})} \approx 4.43 \mathrm{~m/s}\)

Compare the two speeds

The speed of the water as it exits the valve is \(4.43 \mathrm{~m/s}\), and the speed of a drop of water falling from rest at the same height is also \(4.43 \mathrm{~m/s}\). The two speeds are the same.

Key concepts

Fluid Dynamics

When we think about how liquids move and behave, we talk about fluid dynamics, which is an important part of physics. This field specifically focuses on how fluids, which can be liquids or gases, flow and how forces affect them. To imagine this, consider water flowing through a pipe or blood pumping through your veins.

Fluid dynamics deals with various aspects of the fluid flow such as velocity, pressure, density, and temperature as functions of space and time. One fundamental principle within fluid dynamics is Bernoulli's principle, which states that an increase in the speed of a fluid occurs simultaneously with a decrease in its pressure or a decrease in the fluid's potential energy. This principle is extremely helpful in understanding how fluids behave in different scenarios, such as in the exercise where we examine water being released from a tank to turn a turbine.

Pressure and Flow Rate

In the context of fluid dynamics, the terms pressure and flow rate are central concepts. Pressure is simply the force applied perpendicular to the surface of an object per unit area over which that force is distributed, and flow rate refers to the volume of fluid that passes through a given surface every second.

For fluids moving through areas of different sizes, like in our exercise where water passes from the large tank into the smaller valve, the flow rate is conserved. This is described by the equation of continuity, which for our case can be written as A₁v₁ = A_vv_v, where A is the cross-sectional area and v is the flow speed. It's important to note that this is an application of the conservation of mass principle. As area decreases, like with the valve opening in the exercise, the speed of the fluid must increase for the mass flow rate to remain constant.

Conservation of Energy

Conservation of energy is a fundamental concept not just in fluid dynamics, but in all of physics. It states that the total energy in an isolated system remains constant over time. This is known as the principle of the conservation of energy. Energy can change in form, such as from potential to kinetic energy, but the total amount of energy remains constant.

In the exercise, we used this principle to determine the speed of a water droplet falling from the top of the tank. Initially, all the energy is gravitational potential energy (since the droplet starts from rest), which then is completely converted to kinetic energy as the droplet falls, because there is no air resistance. This conservation principle allows us to predict the speed of the water droplet as it reaches the bottom, just as we can predict the speed of the water exiting the valve using Bernoulli's principle. Interestingly, as the exercise shows, the two speeds are equal, demonstrating the direct relationship between the physics of falling objects and fluid flow.