Problem 93 - Physics for Scientists and Engineers with Modern Physics

Question

A fire hose for use in urban areas must be able to shoot a stream of water to a maximum height of \(33 \mathrm{~m}\). The water leaves the hose at ground level in a circular stream \(3.0 \mathrm{~cm}\) in diameter. What minimum power is required to create such a stream of water? Every cubic meter of water has a mass of \(1.00 \times 10^{3} \mathrm{~kg}\).

Step-by-Step Solution

Answer

The minimum power required is approximately 5837 watts.

Step 1: Understand the Problem

We need to determine the minimum power required to eject water from a hose to reach a height of 33 meters. Power is the rate of doing work, and here, the work done goes into elevating the water against gravity.

Step 2: Calculate the Volume Flow Rate

First, we'll calculate how much water needs to be pumped per second to reach the height of 33 meters. We start by determining the cross-sectional area of the water stream, which is given by the diameter: \[A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{3.0 \times 10^{-2}}{2}\right)^2 \approx 7.07 \times 10^{-4} \text{ m}^2\]Next, we use the kinematic equation to find the velocity:\[v^2 = u^2 + 2gh \rightarrow v^2 = 0 + 2 \times 9.81 \times 33 \rightarrow v = \sqrt{2 \times 9.81 \times 33} \approx 25.5 \text{ m/s}\]The flow rate \(Q\) is:\[Q = A \times v = 7.07 \times 10^{-4} \times 25.5 \approx 0.018 \text{ m}^3/\text{s}\]

Step 3: Calculate the Mass Flow Rate

The mass flow rate \( \dot{m} \) is given by:\[ \dot{m} = \rho \times Q = 1000 \times 0.018 = 18 \text{ kg/s}\]

Step 4: Determine the Power Required

Power required is the rate of work done, which is equal to the rate of change of potential energy, \( P = \dot{m} \cdot g \cdot h \):\[P = 18 \times 9.81 \times 33 \approx 5837 \text{ W}\]

Step 5: Conclusion

Therefore, the minimum power required to pump the water is approximately 5837 watts or 5.8 kilowatts.

Key Concepts

Power Calculation

In physics, power is all about how quickly work is done or energy is transferred. It's measured in watts (W). In the context of our exercise, we're interested in how much power is needed to shoot water out of a hose to a certain height. This involves calculating the rate at which work must be done against the force of gravity to lift the water to a height of 33 meters.

To find this, you need to know the mass flow rate of the water and the height it needs to reach. You then use these to find the power using the formula:

• Power, \( P = \dot{m} \times g \times h \)

here, \( \dot{m} \) is the mass flow rate, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \mathrm{m/s^2} \)), and \( h \) is the height. This dimension helps you understand how much energy is needed to maintain a constant flow of water to the height given, allowing you to assess what kind of pump would be required to handle it.

Fluid Mechanics

Fluid mechanics is the branch of physics concerned with the mechanics of fluids (liquids and gases) and the forces acting upon them. It helps us understand how fluids move and how forces affect them, which is crucial for problems involving the motion of liquids such as water ejected from a hose.

In our exercise, we use fluid mechanics principles to determine how water from a fire hose can reach a height of 33 meters. The velocity of the water as it exits the hose plays a crucial role in this process. By employing fluid motion equations, we can calculate the required exit velocity of water based on its kinetic energy and the gravitational potential energy it must overcome:

• Kinetic equation: \( v^2 = u^2 + 2gh \)

Here, \( v \) is the final velocity, \( u \) is the initial velocity (which is zero in this case), and \( h \) is the height the water needs to achieve.

Understanding these principles allows us to design and predict how systems like fire hoses perform under different conditions, ensuring the appropriate measures are in place for efficiency and effectiveness.

Mass Flow Rate

Mass flow rate is a pivotal concept in fluid dynamics that tells us how much mass passes a point in a system per unit time. It plays a crucial role in determining the amount of force and energy needed to keep fluid flowing at a specific rate and height.

In the exercise, we determined the mass flow rate by multiplying the density of the water by the volumetric flow rate, given by:

• Mass flow rate, \( \dot{m} = \rho \times Q \)

where \( \rho \) is the density of the fluid (water, in this case), and \( Q \) is the volume flow rate. It's expressed in kilograms per second (kg/s).

Having an accurate mass flow rate helps us understand the energy demand needed to maintain the water flow to reach the desired height. This understanding is crucial in applications such as firefighting, irrigation, and industrial processes involving fluid transfer, ensuring both efficiency and effectiveness of the pumping systems.