Problem 16 - Applied Physics

Question

Oil is pumped at \(25.0 \mathrm{~m}^{3} / \mathrm{min}\) into a tank \(10.0 \mathrm{~m}\) above the ground (1 \(\mathrm{L}\) of oil has a mass of \(0.68 \mathrm{~kg}\) ). (a) What power, in \(\mathrm{kW}\), must be delivered by the pump? (b) What is the increase in potential energy of the oil after \(10.0 \mathrm{~min} ?\) (c) Find the increase in potential energy of the oil after \(10.0 \mathrm{~min}\) if the tank is \(5.00 \mathrm{~m}\) above the ground.

Step-by-Step Solution

Answer

(a) 16.6 kW, (b) 16677000 J, (c) 8338500 J.

Step 1: Understand the Problem

We need to calculate the power required by a pump to move oil to a height, and also find the potential energy increase of the oil in a tank at different heights after a given time.

Step 2: Convert Units

Convert the volume flow rate from \(\mathrm{m^3/min}\) to \(\mathrm{m^3/s}\) by dividing by 60: \[25.0 \, \frac{\mathrm{m^3}}{\mathrm{min}} = \frac{25.0}{60} \, \frac{\mathrm{m^3}}{\mathrm{s}} = 0.4167 \, \mathrm{m^3/s}\]

Step 3: Calculate Mass Flow Rate

Since 1 \(\mathrm{m^3}\) of oil has a mass of 680 \(\mathrm{kg}\) (because 1 \(\mathrm{L} = 0.68 \, \mathrm{kg}\)), the mass flow rate is:\[0.4167 \, \mathrm{m^3/s} \times 680 \, \frac{\mathrm{kg}}{\mathrm{m^3}} = 283.33 \, \mathrm{kg/s}\]

Step 4: Calculate Power

Use the formula \(P = \rho g h \, \dot{V}\), where \(\rho = 680 \, \mathrm{kg/m^3}\), \(g = 9.81 \, \mathrm{m/s^2}\), \(h = 10 \, \mathrm{m}\), and \(\dot{V} = 0.4167 \, \mathrm{m^3/s}\):\[P = 0.4167 \, \mathrm{m^3/s} \times 680 \, \mathrm{kg/m^3} \times 9.81 \, \mathrm{m/s^2} \times 10 \, \mathrm{m} P = 16616.66 \, \mathrm{W} = 16.6 \, \mathrm{kW}\]

Step 5: Calculate Potential Energy for 10 min at 10 m Height

Use the formula for potential energy \(U = mgh\). The mass in 10 minutes is: \[m = 283.33 \, \mathrm{kg/s} \times 600 \, \mathrm{s} = 170000 \, \mathrm{kg}\] Then, \[U = 170000 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 10 \, \mathrm{m} = 16677000 \, \mathrm{J}\]

Step 6: Calculate Potential Energy for 10 min at 5 m Height

For a 5 m height, the potential energy is: \[U = 170000 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 5 \, \mathrm{m} = 8338500 \, \mathrm{J}\]

Key Concepts

Potential Energy

Potential energy is a fundamental concept in physics that describes the stored energy of an object due to its position or condition.
In this exercise, we're dealing with gravitational potential energy, which depends on an object's height and weight.
When oil is stored at a height, it possesses potential energy that can be calculated using the formula:
\[U = mgh\]
where \(U\) is the gravitational potential energy, \(m\) is the mass of the object (or fluid), \(g\) is the acceleration due to gravity (approximately \(9.81 \, \mathrm{m/s^2}\) on Earth), and \(h\) is the height above the reference point.

• This formula shows how an increase in mass, gravity, or height will result in a higher potential energy.
• In our given scenario, height is particularly crucial as it determines how much extra energy is stored due to gravity.

Understanding potential energy helps us determine how much work is needed to bring the oil to a specific height, informing both the energy requirements and efficiency of systems involving elevation.

Mass Flow Rate

The concept of mass flow rate is essential for analyzing fluid flow systems.
It is defined as the mass of a substance that passes through a given surface per unit time.
The mass flow rate helps us understand the amount of oil being transported by the pump each second, which is vital for calculating required power and potential energy.

• In this exercise, the volume flow rate is given as \(25.0 \, \mathrm{m^3/min}\), and with the density of oil known (\(680 \, \mathrm{kg/m^3}\)), we can find the mass flow rate:
• Convert the flow rate to \(\mathrm{m^3/s}\) to facilitate calculations aligned with the International System of Units (SI).
• The mass flow rate formula is \(\dot{m} = \rho \times \dot{V}\), where \(\rho\) is the density and \(\dot{V}\) is the volume flow rate.

This allows us to calculate how much mass moves through the system every second, serving as a stepping stone for further calculations in power and energy.

Energy Conversion

Energy conversion is a central theme in physics, especially in systems that transform energy from one form to another.
In this problem, we are looking at the conversion of mechanical energy used by the pump to potential energy of the elevated oil.

• The power provided to lift the oil is transformed into an increase in gravitational potential energy.
• The conversion is guided by the formula \(P = \dot{m}gh\), which links the power of the pump to the energy needed for lifting the oil.
• Efficiency in such a system is determined by how well the pump converts electrical or mechanical power into potential energy.

Recognizing that no real-world system is perfectly efficient, it's essential to account for losses due to friction, heat, and other factors that might not directly convert all energy to potential energy.

Physics Problem-Solving

Physics problem-solving involves applying foundational principles to work through exercises systematically.
This exercise illustrates the skills needed to tackle real-world problems using physics.

• Start by clearly understanding what the problem asks and identify the relevant formulas.
• Convert all units to a consistent system like SI (meters, kilograms, seconds) to ensure smooth calculations.
• Apply step-by-step calculations to find intermediate values such as mass flow rate and potential energy before proceeding to final answers.

Breaking the task into manageable parts is crucial for clarity and accuracy.
Also, checking each step helps prevent errors, leading to more reliable conclusions.
Keeping these techniques in mind can make solving even the most complex physics problems less daunting and more structured.