Understanding Cooling Capacity in TR: Calculating for Water from 23°C to 10°C

In this article, we will delve into the concept of cooling capacity in tons of refrigeration (TR) and how to calculate it using a specific example. We will go through step-by-step calculations to help you understand the process thoroughly.

Introduction to Cooling Capacity in TR

Cooling capacity in tons of refrigeration (TR) is a measure of the refrigeration capacity of a cooling system. One ton of refrigeration is the amount of energy required to freeze 1 ton (2000 pounds) of water in 24 hours at 0°C (32°F).

Formula for Calculating Cooling Capacity

The formula to calculate the cooling capacity Q in joules is given by the following equation:

Q mcΔT

Q is the heat removed in joules.
m is the mass of water in kilograms (kg).
c is the specific heat capacity of water, approximately 4.186 kJ/kg°C (kilojoules per kilogram per degree Celsius).
ΔT is the change in temperature in degrees Celsius (°C).

Step-by-Step Calculation

Convert Volume of Water to Mass

The density of water is approximately 1 kg/L (kilogram per liter). Therefore, for 5 liters of water:

m 5 , text{kg}

Calculate the Change in Temperature

The initial temperature is 23°C and the final temperature is 10°C. The change in temperature ΔT is calculated as follows:

ΔT 23 - 10 13 , text{°C}

Calculate the Heat Removed

The heat removed Q can be calculated using the formula:

Q mcΔT 5 , text{kg} times 4.186 , text{kJ/kg°C} times 13 , text{°C}

Q 5 times 4.186 times 13 272.09 , text{kJ}

Convert Q to kWh

Since 1 kWh (kilowatt-hour) 3.6 MJ (megajoules), the conversion is:

Q frac{272.09 , text{kJ}}{3600 , text{kJ/kWh}} approx 0.07556 , text{kWh}

Convert kWh to TR

Since 1 ton of refrigeration (TR) is approximately equal to 3.517 kW:

text{Cooling Capacity TR} frac{0.07556 , text{kWh}}{1 , text{hour}} times frac{1 , text{TR}}{3.517 , text{kW}} approx 0.0215 , text{TR}

Conclusion

The cooling capacity required to cool 5 liters of water from 23°C to 10°C is approximately 0.0215 TR.

Considering Temperature Changes and Specific Heat Capacities

It is important to note that the specific heat of water changes slightly with temperature. For a more accurate calculation, specific heat values at different temperatures should be used. However, for simplicity, we used an average value of 4.18 kJ/kg°C, as the change is minimal between 23°C and 10°C.

If we consider the specific heat of water at 25°C (4.180 kJ/kg°C) and 10°C (4.192 kJ/kg°C), the calculation would be:

Q 5 , text{kg} times 4.180 times (23 - 10) 271.7 , text{kJ}

Converting 271.7 kJ/hr to TR:

text{Cooling Capacity TR} approx 0.02146 , text{TR}

Useful Links

Learn more about Water - Thermal Properties Use the Kilojoules/Hour to Tons Refrigeration Conversion Calculator

Understanding the concept of cooling capacity in TR is crucial for designing and selecting appropriate refrigeration systems. By following the steps outlined in this article, you can accurately determine the cooling requirements for various applications.