Question

Water is pumped steadily out of a flooded basement at a speed of \(5.30 \mathrm{~m} / \mathrm{s}\) through a uniform hose of radius \(9.70 \mathrm{~mm}\). The hose passes out through a window \(2.90 \mathrm{~m}\) above the water line. How much power is supplied by the pump?

Short answer

The total power supplied by the pump is calculated by adding the kinetic power and the potential power together.

Step-by-step solution

Calculation of mass flow rate

First, let's calculate the mass flow rate, denoted by \(\dot{m}\), using the formula \(\dot{m}=\rho AV\), where the density of water \(\rho = 1000 kg/m^3\), the cross-sectional area of the uniform hose \(A=\pi r^2\) where \(r = 9.7 × 10^{-3} m\) and the speed \(V = 5.30 m/s\). This yields \(\dot{m} = \rho \pi r^2 V\).

Calculation of kinetic energy power

Once you have calculated the mass flow rate, you can now find the power due to kinetic energy. Power for kinetic energy (\(P_k\)) is given by the formula \(P_k = ½ \dot{m} V^2\). So using the obtained mass flow rate and pump speed, plug in the values to compute this power.

Calculation of potential energy power

Next is to find the power due to potential energy or gravitational power. This is found using the formula \(P_p = \dot{m} g h\), where \(g = 9.81 m/s^2\) is the acceleration due to gravity and \(h = 2.90 m\) is the height above water line. Substitute the mass flow rate from step 1, gravity and height into the formula to get the gravitational power.

Sum of the two powers

Eventually, to find the total power provided by the pump, which is in joules per second or watt (W), add the kinetic power and the potential power together. This gives us the total power. i.e: \(P_{total} = P_k + P_p \).

Key concepts

Mass Flow Rate

Understanding the mass flow rate is critical when calculating pump power in fluid dynamics. The mass flow rate, \(\dot{m}\), represents the amount of mass passing through a given cross-section of a pipeline or area per unit time. It is a product of the fluid density \(\rho\), the flow area \(A\), and the velocity \(V\) of the fluid. The formula is given by \(\dot{m}=\rho AV\). In the context of our problem, the radius \(r\) of the hose and the speed of water establish a specific cross-sectional flow area and velocity, respectively. By calculating the mass flow rate, we determine the foundation for later assessing both the kinetic and potential energy power of the pumping system, which are pivotal for solving the exercise in question. Utilizing the mass flow rate provides insight into the volume of water the pump can move, affecting the design and power requirements of such systems.

For this specific exercise, the mass flow rate was found using the density of water, the given radius of the hose, and the flow velocity. The obtained value serves as a crucial input for the subsequent steps of calculating the energy powers associated with the flow.

Kinetic Energy Power

Kinetic energy power, \(P_k\), in a flowing fluid refers to the power due to the velocity of the mass flow. This is particularly important when fluids are moved through systems like pipes or pumps, as the velocity at which the fluid is being displaced has a direct impact on the energy requirements. The power associated with the kinetic energy of a flow can be expressed using the formula \(P_k = \frac{1}{2} \dot{m} V^2\). It is the product of one-half the mass flow rate and the square of the velocity, signifying how the energy required for maintaining a certain flow speed escalates with increasing flow rate or velocity.

In our example, after computing the mass flow rate, applying it to the kinetic energy power formula helps us estimate the part of the pump's power that is used to maintain the water's flow speed. The kinetic power is critical to consider because it contributes to the total power that the pump must supply and is a direct indicator of the energy conversion happening within the pump due to motion.

Potential Energy Power

When discussing potential energy power, \(P_p\), we're looking at the power required to move a mass against the force of gravity. The pump has to provide sufficient energy to lift water to a certain height, which in our textbook case, is the window \(2.90 m\) above the water line. The formula for calculating potential energy power in the context of pumping is given by \(P_p = \dot{m} gh\), where \(g\) is the acceleration due to gravity, and \(h\) is the height the fluid is pumped to. This gives a measure of how much power is needed to overcome gravitational pull.

The calculation of potential energy power based on the mass flow rate previously determined helps in ascertaining the ability of the pump to lift water against gravity. This energy is converted to potential energy stored in the elevated water, making it a vital part of the overall power requirement for the pump and playing a central role in the design of such fluid system.

Fluid Mechanics

The field of fluid mechanics is fundamental when examining the behavior of fluids within pumps and other systems. It is a branch of physics that deals with the motion of fluids (liquids and gases) and the forces on them. Fluid mechanics is divided into fluid statics—or hydrostatics, when dealing with liquids—and fluid dynamics. This topic is essential for the understanding of mass flow rate, kinetic energy power, and potential energy power, which all play a role in pumping systems.

The principles derived from fluid mechanics enable us to understand and predict how fluids will move under various conditions, and how this movement translates into the need for mechanical power from devices such as pumps. The application of fluid mechanics theories in the calculation of pump power, such as Bernoulli's equation and the conservation of mass and energy, provides a comprehensive detail in solving textbook problems like the one addressed here, encompassing aspects of flow velocity, pressure, and height in the power calculations.