Question

A typical \(200-\mathrm{m}^{2}\) house can be cooled adequately by a 3.5 -ton air conditioner whose COP is \(4.0 .\) Determine the rate of heat gain of the house when the air conditioner is running continuously to maintain a constant temperature in the house.

Short answer

Answer: The rate of heat gain of the house is approximately 12,306 W.

Step-by-step solution

Convert the cooling capacity from tons to power units

Cooling capacity is given in tons, so we need to convert it to watts. We use the conversion factor: 1 ton = 12000 BTU/hr By definition of a ton, we know that 1 ton of cooling capacity is equivalent to the power required to melt 1 ton (2000 lb) of ice in a 24-hour period. Since 1 BTU is the amount of heat energy needed to raise the temperature of 1 lb of water by 1°F, we also know that 1 ton corresponds to: $$ 2000 \, \mathrm{lb} \cdot \frac{144 \, \mathrm{Btu}}{\mathrm{lb} \cdot ^{\circ}\mathrm{F}} \cdot 1 ^{\circ}\mathrm{F} = 288,000 \, \mathrm{Btu/day} \,. $$ Dividing this by 24 hours per day will give us the power required to cool the room in BTU/hr: $$ \frac{288,000 \, \mathrm{Btu}}{24 \, \mathrm{h}} = 12,000 \, \mathrm{Btu/h} \,. $$ Now, we can convert the 3.5-ton cooling capacity into BTU/hr: $$ 3.5 \, \mathrm{ton} \times 12,000 \, \frac{\mathrm{Btu}}{\mathrm{h \, ton}} = 42,000 \, \mathrm{Btu/h} \,. $$ The conversion factor for BTU/hr to watts is: $$ 1 \, \mathrm{Btu/h} = 0.293 \, \mathrm{W} \,. $$ So, the cooling capacity in watts is: $$ 42,000 \, \mathrm{Btu/h} \times 0.293 \, \frac{\mathrm{W}}{\mathrm{Btu/h}} = 12,306 \, \mathrm{W} \,. $$

Finding the rate of heat removal from the house

We are given the COP (Coefficient of Performance) of the air conditioner and its cooling capacity in watts. We can use the following formula to find the rate of heat removal: $$ \mathrm{COP} = \frac{\mathrm{Rate \, of \, heat \, removal}}{\mathrm{Power \, input}} \,. $$ We know the cooling capacity (rate of heat removal) is 12,306 W and the COP is 4.0. Therefore, we can find the power input: $$ \mathrm{Power \, input} = \frac{\mathrm{Rate \, of \, heat \, removal}}{\mathrm{COP}} = \frac{12,306 \, \mathrm{W}}{4.0} = 3,076.5 \, \mathrm{W} \,. $$

Determine the rate of heat gain of the house

Since the air conditioner is maintaining a constant temperature, the rate of heat gain must be equal to the rate of heat removal. Therefore, the rate of heat gain of the house is: $$ \\\mathrm{Rate \, of \, heat \, gain} = \mathrm{Rate \, of \, heat \, removal} = 12,306 \, \mathrm{W} $$ In conclusion, the rate of heat gain of the house when the air conditioner is running continuously to maintain a constant temperature in the house is approximately 12,306 W.

Key concepts

Ton to BTU/hr Conversion

When dealing with air conditioning units, it's common to encounter the term 'ton' as a measure of cooling capacity. However, it's essential to convert this to a more universally recognized measurement such as BTU/hr (British Thermal Units per hour) to calculate the heat exchange rates effectively. The key conversion factor is that one ton of cooling capacity is equivalent to 12,000 BTU/hr.

This metric originates from the amount of energy required to melt one ton of ice over a 24-hour period. Understanding this conversion is critical because it allows for the determination of an air conditioner's power in terms that can be used for further calculations, such as determining the rate of heat gain in a building, which is a crucial step in sizing HVAC systems appropriately for the spaces they are meant to condition.

By knowing the conversion rate, we can then translate a 3.5-ton air conditioner's capacity to 42,000 BTU/hr, providing a solid starting point for evaluating the system's performance. Using the formula \( 3.5 \, \mathrm{tons} \times 12,000 \, \frac{\mathrm{BTU}}{\mathrm{h \, ton}} = 42,000 \, \mathrm{BTU/h} \), we arrive at the capacity in terms of energy per unit of time, which is essential for HVAC calculations.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a dimensionless measure that helps to compare the efficiency of heating or cooling systems. It represents the ratio of useful heat movement per work input, reflecting energy efficiency in heat pumps and air conditioners. A higher COP indicates a more efficient device, meaning it provides more heating or cooling power without consuming as much energy.

In a practical sense, if an air conditioner has a COP of 4.0, it means that for every unit of electrical power input, it provides four units of cooling power. This is calculated using the formula: \(\mathrm{COP} = \frac{\mathrm{Rate \, of \, heat \, removal}}{\mathrm{Power \, input}}\). It's vital to understand COP because it directly influences operational costs and environmental impact. For the aforementioned 3.5-ton air conditioner with a COP of 4.0, we found the power input to be 3,076.5 W. This means the unit is relatively efficient, producing a significant amount of cooling for the consumed power, an important consideration for anyone concerned about energy-saving and cutting costs on electricity bills.

Heat Removal in Thermodynamics

Heat removal is a fundamental concept in thermodynamics, particularly within the context of air conditioning and refrigeration. In these systems, heat removal is associated with transferring heat from a cooler space to a warmer space, which is contrary to the natural flow of heat.

The process relies on refrigerants and the refrigeration cycle to remove heat from an indoor environment, thus lowering the temperature. Understanding the rate of heat gain is essential as it determines the thermal load that the air conditioning system must handle to keep the indoor temperature constant.

For our 200-square-meter house, calculating the rate of heat gain is a matter of equating it to the rate of heat removal of the air conditioner, since the system is running continuously to maintain a steady temperature. This relationship is highlighted by the equation \(\mathrm{Rate \, of \, heat \, gain} = \mathrm{Rate \, of \, heat \, removal} = 12,306 \, \mathrm{W}\). Thus, the capacity of the air conditioner must match the thermal load from the house to achieve a stable interior climate, a crucial consideration when designing or upgrading HVAC systems.