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\(\bullet\) 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.

Short Answer

Expert verified
The performance coefficient is approximately 3.49.

Step by step solution

01

Understand the Problem

We're asked to calculate the performance coefficient of a cooling unit, given its power and efficiency specifications.
02

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.
03

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 \( \frac{1}{10} \times 746 = 74.6 \text{ watts} \).
04

Adjust for Efficiency

Since the efficiency is only 70\%, we find the actual electrical power consumption: \( W = \frac{74.6}{0.7} \approx 106.57 \text{ watts} \).
05

Convert Work to Btu/h

Convert watts to Btu/h by using the conversion factor: 1 watt \(\approx 3.412 \) Btu/h. So, \( W = 106.57 \times 3.412 \approx 363.44 \text{ Btu/h} \).
06

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, \( \text{COP} = \frac{1270}{363.44} \approx 3.49 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Entropy change due to driving. Premium gasoline pro- duces \(1.23 \times 10^{8} \mathrm{J}\) of heat per gallon when it is burned at a temperature of approximately \(400^{\circ} \mathrm{C}\) (although the amount can vary with the fuel mixture). If the car's engine is 25\(\%\) efficient, three- fourths of that heat is expelled into the air, typically at \(20^{\circ} \mathrm{C}\) . If your car gets 35 miles per gallon of gas, by how much does the car's engine change the entropy of the world when you drive 1.0 mile? Does it decrease or increase it?

.. You are designing a Carnot engine that has 2 mol of \(\mathrm{CO}_{2}\) as its working substance; the gas may be treated as ideal. The gas is to have a maximum temperature of \(527^{\circ} \mathrm{C}\) and a maximum pressure of 5.00 atm. With a heat input of 400 \(\mathrm{J}\) per cycle, you want 300 \(\mathrm{J}\) of useful work. (a) Find the temperature of the cold reservoir. (b) For how many cycles must this engine run to melt completely a 10.0 -kg block of ice originally at \(0.0^{\circ} \mathrm{C},\) using only the heat rejected by the engine?

\(\bullet\) (a) Calculate the theoretical efficiency for an Otto cycle engine with \(\gamma=1.40\) and \(r=9.50\) . (b) If this engine takes in \(10,000\) J of heat from burning its fuel, how much heat does it discard to the outside air?

A typical coal-fired power plant generates 1000 \(\mathrm{MW}\) of usable power at an overall thermal efficiency of 40\(\%\) . (a) What is the rate of heat input to the plant? (b) The plant burms anthracite coal, which has a heat of combustion of \(2.65 \times 10^{7} \mathrm{J} / \mathrm{kg} .\) How much coal does the plant use per day, if it operates continuously? (c) At what rate is heat ejected into the cool reservoir, which is the nearby river? (d) The river's temperature is \(18.0^{\circ} \mathrm{C}\) before it reaches the power plant and \(18.5^{\circ} \mathrm{C}\) after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. (e) By how much does the river's entropy increase each second?

A diesel engine performs 2200 \(\mathrm{J}\) of mechanical work and discards 4300 \(\mathrm{J}\) of heat each cycle. (a) How much heat must be supplied to the engine in each cycle? (b) What is the thermal efficiency of the engine?

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