Problem 15 - Applied Physics

Question

Water is pumped at \(25 \overline{0} \mathrm{~m}^{3} / \mathrm{min}\) from a lake into a tank \(65.0 \mathrm{~m}\) above the lake. (a) What power, in \(\mathrm{kW}\), must be delivered by the pump? (b) What horsepower rating does this pump motor have? (c) What is the increase in potential energy of the water each minute?

Step-by-Step Solution

Answer

(a) 26,562.5 kW (b) 35,623.92 hp (c) 1,593,750,000 J

Step 1: Calculate the Work Done per Minute

First, we need to calculate the work done to lift the water. The work done is equal to the change in potential energy as the water is lifted to a certain height. The formula for work done (or change in potential energy) is \( W = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)), and \( h \) is the height. The volume flow rate is \(250 \mathrm{~m^3/min}\), and the density of water is \(1000 \mathrm{~kg/m^3}\), giving us a mass flow rate of \(250,000 \mathrm{~kg/min}\). Thus, work done per minute is: \[ W = 250,000 \times 9.81 \times 65 = 1,593,750,000 \mathrm{~J} \].

Step 2: Calculate the Power Delivered by the Pump

Power is the rate at which work is done. Therefore, power in watts is given by \( P = \frac{W}{t} \), where \( W \) is the work done and \( t \) is time in seconds. Since we have calculated work done per minute, convert it to seconds: \[ P = \frac{1,593,750,000}{60} = 26,562,500 \mathrm{~W} \] or \( 26,562.5 \mathrm{~kW} \).

Step 3: Convert the Power to Horsepower

The horsepower (hp) can be calculated using the conversion: \[ 1 \mathrm{~hp} = 745.7 \mathrm{~W} \]. Therefore, the pump's power converted to horsepower is \[ \frac{26,562,500}{745.7} = 35,623.92 \mathrm{~hp} \].

Step 4: Calculate the Increase in Potential Energy Each Minute

The increase in potential energy is simply the work done, which is already calculated in Step 1. Hence, the increase in potential energy per minute is \( 1,593,750,000 \mathrm{~J} \).

Key Concepts

Potential Energy

Potential energy is a critical concept in physics, especially when it comes to pumping water in systems like the one described in the exercise. It is the stored energy of an object due to its position relative to a zero potential energy position. In this context, the potential energy relates to the energy acquired by the water as it is elevated to a higher level.

The formula for calculating potential energy is given by:\[PE = mgh\]where:- \( m \) is the mass of the water,- \( g \) is the acceleration due to gravity (approximately \(9.81 \, \mathrm{m/s^2}\)),- \( h \) is the height in meters.

The scenario presented involves lifting a mass of water from the lake to a tank. By calculating the change in potential energy, one can determine the amount of work required to lift the water to a new height.

Energy Conversion

Energy conversion is the process of changing one form of energy into another. In the case of the pumping power calculation, the pump converts electrical energy into mechanical energy that does work to move water to a higher elevation.

The conversion from electrical energy to mechanical energy is not always 100% efficient due to energy losses as heat or due to friction within the pump system. Understanding energy conversion is crucial when determining the required power output to perform the given task efficiently.

By calculating the potential energy increase, you can estimate the theoretical minimum amount of energy that must be supplied to the pump for operation. The actual requirement, however, may require more energy due to practical inefficiencies.

Power in Watts

Power is the rate at which work is done or energy is transferred in a system. It is measured in watts (\( \mathrm{W} \)), which indicates how much energy is used or converted per unit time.

In the pumping example, the power necessary to lift water is calculated by dividing the work required by the time period over which it occurs. Using the relationship:\[P = \frac{W}{t}\]where \( W \) is the work done in joules and \( t \) is the time in seconds, we can find the power output required. To convert power from watts to kilowatts (\( \mathrm{kW} \)), divide by 1,000.

• Work done per minute: \(1,593,750,000 \mathrm{~J}\)
• Power in watts: \( \frac{1,593,750,000}{60} \approx 26,562,500\, \mathrm{W}\)
• Power in kilowatts: \( 26,562.5 \mathrm{~kW} \)

Understanding this helps in designing systems capable of fulfilling the power demand efficiently.

Horsepower Calculation

Horsepower is an additional unit of power used commonly, especially in machinery like motors, pumps, and cars. It is often used instead of watts because it provides a more intuitive sense of capacity for specific engines or motors.

The conversion factor between horsepower and watts is:\[1 \, \text{horsepower} = 745.7 \, \mathrm{W}\]Knowing the power in watts allows for an easy conversion:\[\text{Horsepower} = \frac{\text{Power in watts}}{745.7}\]For the pumping example, we have:\[\\text{Power in watts} = 26,562,500 \, \mathrm{W}\]which converts to:\[\text{Horsepower} \approx \frac{26,562,500}{745.7} = 35,623.92 \, \text{hp}\]Having this conversion allows engineers to select an adequate motor size for the pumping system, ensuring that it can handle the necessary load to lift the water effectively.