Question

A water-powered backup sump pump uses tap water at a pressure of \(3.00 \mathrm{~atm}\left(p_{1}=3 p_{\mathrm{atm}}=\right.\) \(3.03 \cdot 10^{5} \mathrm{~Pa}\) ) to pump water out of a well, as shown in the figure \(\left(p_{\text {well }}=p_{\text {ttm }}\right)\). This system allows water to be pumped out of a basement sump well when the electric pump stops working during an electrical power outage. Using water to pump water may sound strange at first, but these pumps are quite efficient, typically pumping out \(2.00 \mathrm{~L}\) of well water for every \(1.00 \mathrm{~L}\) of pressurized tap water. The supply water moves to the right in a large pipe with cross-sectional area \(A_{1}\) at a speed \(v_{1}=2.05 \mathrm{~m} / \mathrm{s}\). The water then flows into a pipe of smaller diameter with a cross-sectional area that is ten times smaller \(\left(A_{2}=A_{1} / 10\right)\). a) What is the speed \(v_{2}\) of the water in the smaller pipe, with area \(A_{2} ?\) b) What is the pressure \(p_{2}\) of the water in the smaller pipe, with area \(A_{2} ?\) c) The pump is designed so that the vertical pipe, with cross-sectional area \(A_{3}\), that leads to the well water also has a pressure of \(p_{2}\) at its top. What is the maximum height, \(h,\) of the column of water that the pump can support (and therefore act on ) in the vertical pipe?

Short answer

Answer: The speed of water in the smaller pipe is 20.5 m/s, the pressure in the smaller pipe is 9.4 x 10^4 Pa, and the maximum height the pump can support is 9.58 m.

Step-by-step solution

Find the speed of water in the smaller pipe (v2)

The equation of continuity states that the product of the cross-sectional area and the speed of the fluid remains constant in a pipe. Given that the cross-sectional area of the smaller pipe is 10 times smaller than that of the larger pipe (\(A_2 = A_1/10\)), we have: $$A_1v_1 = A_2v_2$$ Solve for \(v_2\): $$v_2 = \frac{A_1v_1}{A_2}$$ We also know \(A_2 = A_1/10\). Therefore, we can write: $$v_2 = \frac{A_1v_1}{A_1/10} = 10v_1$$ Plug in the given value for \(v_1 = 2.05~m/s\): $$v_2 = 10(2.05~m/s)$$ $$v_2 = 20.5~m/s$$

Determine the pressure in the smaller pipe (p2)

We can use Bernoulli's theorem to find the pressure in the smaller pipe. For the larger pipe, we have: $$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$$ Now, we need to solve for \(p_2\): $$p_2 = p_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2$$ We are given that \(p_1 = 3\;\text{atm} = 3.03\times10^5~\mathrm{Pa}\), we have calculated \(v_1 = 2.05\;\text{m/s}\) and \(v_2 = 20.5\;\text{m/s}\). The density of water can be considered as \(\rho = 1000~\mathrm{kg/m^3}\). Plug in the values: $$p_2 = 3.03\times10^5~\mathrm{Pa} + \frac{1}{2}(1000~\mathrm{kg/m^3})(2.05~\mathrm{m/s})^2 - \frac{1}{2}(1000~\mathrm{kg/m^3})(20.5~\mathrm{m/s})^2$$ $$p_2 = 3.03\times10^5~\mathrm{Pa} - 2.09\times10^5~\mathrm{Pa}$$ $$p_2 = 9.4\times10^4~\mathrm{Pa}$$

Find the maximum height of the water column (h) supported by the pump

We are given that the vertical pipe has a pressure of \(p_2\) at its top. To find the maximum height it can support, we can use the concept of hydrostatic pressure: $$\Delta p = p_2 - p_\text{well} = \rho gh$$ Where \(\Delta p\) is the pressure difference, and \(g\) is the acceleration due to gravity (\(9.81~\mathrm{m/s^2}\)). In this case, the pressure difference is equal to the pressure at the top of the vertical pipe (\(p_2\)). Solve for \(h\): $$h = \frac{\Delta p}{\rho g}$$ $$h = \frac{9.4\times10^4\;\text{Pa}}{1000\;\text{kg/m}^3 \times 9.81\;\text{m/s}^2}$$ $$h = 9.58~\mathrm{m}$$ The speed of the water in the smaller pipe is \(20.5\;\mathrm{m/s}\), the pressure in the smaller pipe is \(9.4\times10^4\;\mathrm{Pa}\), and the maximum height the pump can support is \(9.58~\mathrm{m}\).

Key concepts

Continuity Equation

When dealing with fluids that flow in pipes, it's important to consider the Continuity Equation. This equation comes from the principle of conservation of mass. It states that the mass flow rate must remain constant from one cross section of a pipe to another, assuming no leaks and a steady state flow.

For incompressible fluids, like water, this mass flow rate simplifies to the product of cross-sectional area and flow speed remaining constant. So, if water flows through a pipe that gets narrower, it must speed up to keep the mass flow consistent.

The mathematical expression of this is:

• Mass flow rate: \(A_1v_1 = A_2v_2\)

In the given exercise, since the smaller pipe's area is one-tenth of the larger pipe's area, the speed of water must be ten times faster in the smaller pipe. Thus, the Continuity Equation allows us to calculate the speed in a narrow pipe using known values from a wider pipe.

Bernoulli's Principle

Bernoulli's Principle provides an insightful way to understand how the speed of a fluid affects its pressure. According to Bernoulli, in a streamline flow, the sum of the pressure energy, kinetic energy per unit volume, and gravitational potential energy per unit volume is constant.

Here's the basic form of the equation:

• \(p_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = p_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2\)

In horizontal flow scenarios where height remains constant or isn't significantly different, the terms involving gravity cancel out.
In our exercise, Bernoulli's equation helps derive the pressure in the smaller pipe (\(p_2\)) by factoring in the speeds \(v_1\) and \(v_2\) as well as the known pressure \(p_1\).

The faster the fluid goes, the lower its pressure. That's why, in this problem, \(p_2\) turns out to be less than \(p_1\) due to the increased speed \(v_2\). By balancing energy forms, Bernoulli's Principle is a powerful tool in fluid dynamics.

Hydrostatic Pressure

Hydrostatic pressure plays an essential role when a fluid is at rest, or moving vertically under gravitational effects, such as in the vertical pipe of the given exercise. This concept describes the pressure at any given depth within a fluid and is particularly relevant when considering pressure differences.

The pressure in a fluid at depth is given by:

• \(p = p_0 + \rho gh\)

where \(p_0\) is the atmospheric pressure on the surface, \(\rho\) the fluid density, \(g\) the acceleration due to gravity, and \(h\) the height of the fluid column.

In our problem, the goal is to calculate how high the pump can lift water using its pressure \(p_2\). By using the formula \(h = \frac{(p_2 - p_{\text{well}})}{\rho g}\), where \(p_{\text{well}}\) is the baseline atmospheric pressure, we find the maximum height a pump can achieve. Hydrostatic principles reveal that, for a given pressure, the taller the column, the greater the effort needed to support it.

Fluid Dynamics

Fluid Dynamics is the broader field of physics that studies the behavior of fluid flow, including liquids and gases. It encompasses all the concepts discussed above and deals with how fluids move, interact with boundaries, and react to forces.

Various factors influence fluid dynamics:

• Viscosity: the measure of a fluid's resistance to deformation. Affects flow rate and energy loss due to friction.
• Velocity: how fast the fluid particles travel within the medium.
• Pressure: forces exerted by fluid particles on the container, influencing flow rate.

In our sump pump scenario, fluid dynamics principles help design efficient systems for moving water using another water source. By integrating continuity and Bernoulli's principles, we can craft pipelines that optimize speed and minimize pressure drop. This enables crucial devices like backup pumps to function with great efficiency.