Question

A refrigerator is operated by a 0.25-hp (1 hp = \(746 \text { watts })\) motor. If the interior is to be maintained at \(4.50^{\circ} \mathrm{C}\) and the room temperature on a hot day is \(38^{\circ} \mathrm{C}\), what is the maximum heat leak (in watts) that can be tolerated? Assume that the coefficient of performance is \(50 \%\) of the maximum theoretical value. What happens if the leak is greater than your calculated maximum value?

Short answer

The maximum heat leak that can be tolerated by the refrigerator is approximately 370.31 watts. If the leak is greater than this value, the refrigerator will not maintain the interior temperature at 4.5°C, leading to an increase in interior temperature and potential spoilage of the food inside.

Step-by-step solution

Convert motor power from hp to watts

We are given that the motor operates at 0.25 hp. We need to convert this to watts using the conversion factor: 1 hp = 746 watts. Therefore, Motor power in watts = 0.25 hp × 746 watts/hp = \(187.5 \text{ watts}\)

Calculate the maximum theoretical coefficient of performance

The maximum theoretical coefficient of performance (COP) can be determined using the formula: COP = \(\frac{T_{low}}{T_{high} - T_{low}}\) where \(T_{low}\) is the interior temperature of the refrigerator (in Kelvin) and \(T_{high}\) is the room temperature (in Kelvin). First, let's convert the given temperatures to Kelvin: \(T_{low} = 4.5^{\circ}\mathrm{C} + 273.15 = 277.65 \mathrm{K}\) \(T_{high} = 38^{\circ}\mathrm{C} + 273.15 = 311.15 \mathrm{K}\) Now we can calculate the maximum theoretical COP: COP = \(\frac{277.65 \mathrm{K}}{311.15 \mathrm{K} - 277.65 \mathrm{K}}\) COP ≈ 3.95

Multiply maximum theoretical COP by the given coefficient

We are given that the actual coefficient of performance is 50% of the maximum theoretical value. So we need to multiply the calculated COP by 0.5: Actual COP = \(3.95 \times 0.5 = 1.975\)

Calculate the maximum heat leak

To calculate the maximum heat leak, we will use the formula: Heat Leak = Motor Power × Actual COP Heat Leak = \(187.5 \text{ watts} \times 1.975\) Heat Leak ≈ \(370.31 \text{ watts}\) The maximum heat leak that can be tolerated is approximately 370.31 watts. If the leak is greater than the calculated maximum value, the refrigerator will struggle to maintain the interior temperature at 4.5°C, and the interior temperature will rise, causing the food inside to spoil.

Key concepts

Refrigerator Efficiency

Efficiency in the context of refrigeration is essential. A refrigerator's job is to transfer heat from a cool space inside it to a warmer space outside. This process helps to keep our food cold, fresh, and away from spoiling. But refrigerators are not 100% efficient. Their efficiency is impacted by many factors including the temperature difference between the inside and outside environments.

A core part of understanding refrigerator efficiency is knowing how much energy it uses compared to how much it saves. This "energy in versus energy out" describes how effectively a device like a refrigerator can move heat. The efficiency of a refrigerator is often expressed in terms of its Coefficient of Performance (COP), which we explore further below. If a refrigerator were to exceed a certain efficiency, technical limits would be breached, and it wouldn't be sustainable.

Coefficient of Performance

The Coefficient of Performance (COP) is a measure of a refrigerator’s efficiency. COP tells us how well a refrigerator is doing its job of moving heat. It's given by the formula:

• COP = \(\frac{T_{low}}{T_{high} - T_{low}}\)

Here, \(T_{low}\) is the temperature inside the fridge in Kelvin, and \(T_{high}\) is the outside or room temperature in Kelvin.

The greater the COP, the more efficient the refrigerator is. However, actual COP is typically lower than the theoretical value due to physical limitations and inefficiencies, such as heat loss. In our exercise, the actual COP is half of the theoretical COP because of these inefficiencies. A lower actual COP means the refrigerator uses more energy to remove the same amount of heat than specified by the theoretical limit.

Heat Transfer

Heat transfer in a refrigerator is the process of moving heat from the cooler interior to the warmer exterior. The fridge does this efficiently with special substances called refrigerants, moving through coils and changing states.

• Refrigerants absorb heat from inside the fridge, cooling it down.
• The heat is then released outside as the refrigerant changes back.

A key factor is the motor power used to facilitate this process. An efficient transfer means less power is required to maintain the desired temperature, and the maximum heat leak is minimized.

If the heat leak into the fridge exceeds the calculated maximum, the refrigerator's motor struggles to transfer all the heat out. This can result in increased energy consumption and warmer interior temperatures, potentially spoiling stored food.

Motor Power Calculation

Understanding motor power calculation is vital for determining the refrigerator’s capacity to handle heat transfer tasks efficiently. In this exercise, the motor power was provided in horsepower (hp) and converted to watts, with the conversion factor being:

• 1 hp = 746 watts

This conversion allows us to calculate how much electrical power the motor uses—essential in predicting the fridge's performance and designing effective cooling systems.

Motor power impacts the refrigerator’s ability to effectively manage heat without utilizing extra energy. In the exercise, the motor’s power was 187.5 watts, dictating how much energy it can use to assist in transferring heat out of the interior of the fridge. If the motor needs more power due to higher heat leaks, it could mean higher electricity costs and lessening the appliance’s efficiency.