Problem 18 - College Physics

Question

\(\bullet\) A cooling unit for chilling the water of an aquarium gives specifications of 1\(/ 10\) hp and 1270 \(\mathrm{Btu} / \mathrm{h}\) . Assuming the unit produces its 1\(/ 10\) hp at 70.0\(\%\) efficiency, calculate its perform- ance coefficient.

Step-by-Step Solution

Answer

The performance coefficient is approximately 3.49.

Step 1: Understand the Problem

We're asked to calculate the performance coefficient of a cooling unit, given its power and efficiency specifications.

Step 2: Define Relevant Formulas

The performance coefficient (COP) for a cooling device is defined as \( \text{COP} = \frac{Q_c}{W} \) where \(Q_c\) is the cooling energy provided, and \(W\) is the work input.

Step 3: Convert Horsepower to Watts

First, convert the power from horsepower to watts. We know 1 horsepower equals approximately 746 watts. Then, for \( \frac{1}{10} \) hp, we compute \( \frac{1}{10} \times 746 = 74.6 \text{ watts} \).

Step 4: Adjust for Efficiency

Since the efficiency is only 70\%, we find the actual electrical power consumption: \( W = \frac{74.6}{0.7} \approx 106.57 \text{ watts} \).

Step 5: Convert Work to Btu/h

Convert watts to Btu/h by using the conversion factor: 1 watt \(\approx 3.412 \) Btu/h. So, \( W = 106.57 \times 3.412 \approx 363.44 \text{ Btu/h} \).

Step 6: Calculate Performance Coefficient (COP)

Now, use \( \text{COP} = \frac{Q_c}{W} \) where \( Q_c = 1270 \text{ Btu/h} \) and \( W = 363.44 \text{ Btu/h} \). So, \( \text{COP} = \frac{1270}{363.44} \approx 3.49 \).

Key Concepts

Cooling Efficiency

Cooling efficiency is a vital aspect when examining devices designed to reduce temperatures, such as refrigerators or air conditioners. It refers to how effectively the device uses energy to provide the desired cooling effect. High cooling efficiency means that the device uses less energy to achieve the same cooling power compared to less efficient devices.
Understanding this concept helps consumers choose appliances that save energy and costs in the long run. When choosing a cooling unit, looking at its specifications allows one to estimate how much energy the unit will require to operate efficiently. This is particularly critical in industries or homes where energy costs can accumulate significantly.

COP Calculation

The Coefficient of Performance (COP) is a measure that expresses the efficiency of a cooling or heating system. It shows the ratio of useful heating or cooling provided to the electrical energy consumed.
You can determine the COP using the formula:

• \[ \text{COP} = \frac{Q_c}{W} \]

Where \( Q_c \) is the amount of heat removed from the cooling space and \( W \) is the work or energy consumed by the device.
A higher COP value signifies a more efficient system. By comparing COP values, you can assess different systems and choose the one that offers the best energy use for the required cooling or heating task.

Conversion of Units

Converting units is a common task in physics, especially when dealing with power, energy, and work. In this concept, you often change between different units of measurement to ensure consistency and accuracy.
For example, converting horsepower to watts is essential because many calculations use watts as a standard unit for power:

• 1 horsepower (hp) is approximately 746 watts.

Ensure all units are consistent when working with formulas.
In our case, converting watts to British Thermal Units per hour (Btu/h) becomes important where you use:

• 1 watt is approximately 3.412 Btu/h.

Such conversions help blend different sections of a calculation that uses various unit standards.

Thermodynamics

Thermodynamics lays the foundation for understanding how energy is transferred and transformed between systems. In the context of cooling units, it explains how refrigeration cycles remove heat from one area and expel it elsewhere.
Key principles include:

• Energy conservation, indicating energy cannot be created or destroyed, only transformed.
• Energy efficiency, concerning how much useful work is obtained from energy input.

Understanding thermodynamic principles helps us evaluate device performance, enabling us to improve system designs for better efficiency and functionality.

Physics Problem Solving

Solving physics problems involves identifying, understanding, and applying the correct formulas and principles to derive solutions. This requires breaking down complex problems into manageable steps and consistently applying fundamental physics laws.
Effective problem solving includes:

• Defining all known variables correctly.
• Recognizing applicable physics principles, such as energy conservation or efficiency calculations.
• Converting between units for consistency.

By approaching problems methodically, we can ensure accurate solutions that adhere to the rules of physics, providing insights and understanding of how systems work in practical scenarios.