ClassX Video Lessons Youtube Channel The Engineering Mindset Cooling Load Calculation – Cold Room hvac
Welcome to an insightful exploration of how to calculate the cooling load for a cold room storage unit. This article will guide you through understanding what a cold room is, identifying the sources of heat that need to be removed, and performing a sample calculation with a safety factor. Finally, we’ll determine the appropriate size for the refrigeration unit to match the calculated cooling load.
A cold room is a specialized storage space designed to keep perishable goods like meat and vegetables fresh by slowing down their deterioration. This is achieved by removing heat through a refrigeration system, which maintains a controlled environment to preserve the goods effectively.
Several sources contribute to the heat that needs to be removed from a cold room:
Let’s walk through a simplified example of calculating the cooling load. For real-world applications, consider using design software like the Danfoss Coolselector tool for precision.
Consider a cold store measuring six meters long, five meters wide, and four meters high. The ambient air temperature is 30°C with 50% relative humidity, while the internal air is 1°C with 95% relative humidity. The walls, roof, and floor are insulated with 80mm polyurethane, having a U-value of 0.28 W/m²K. The ground temperature is 10°C.
Use the formula: Q = U × A × (T_out - T_in) × 24 / 1,000 to calculate the transmission load. Here, U is the U-value, A is the surface area, and T_out and T_in are the outside and inside temperatures, respectively. The total transmission cooling load is 23.8 kWh per day.
For product exchange, consider storing 4,000 kg of apples arriving daily at 5°C, while the store holds 20,000 kg. Use the formula: Q = M × CP × (T_enter - T_store) / 3,600 to find the cooling load, resulting in 16 kWh per day. Additionally, calculate product respiration heat using Q = M × resp / 3,600, yielding 10.5 kWh per day. The total product load is 26.5 kWh per day.
Estimate two people working for four hours daily, generating 270 watts of heat per hour. Use Q = people × time × heat / 1,000 to calculate 2.16 kWh per day. For lighting, with three 100-watt lamps running for four hours, the load is 1.2 kWh per day. The total internal load is 3.36 kWh per day.
For fan motors, with three fans rated at 200 watts each running for 14 hours, the load is 8.4 kWh per day. For defrosting, with a 1.2 kW heating element running for 30 minutes three times daily at 30% efficiency, the load is 0.54 kWh per day. The total equipment load is 8.94 kWh per day.
Estimate five air changes per day with a volume of 120 cubic meters. Use Q = changes × volume × energy × (T_out - T_in) / 3,600 to find the load, resulting in 9.67 kWh per day.
The total cooling load sums to 72.27 kWh per day. Applying a 20% safety factor, the total is 86.7 kWh per day. To determine the refrigeration capacity, divide the total daily load by the estimated runtime (14 hours), resulting in a required capacity of 6.2 kW.
By following these steps, you can effectively calculate the cooling load for a cold room and size the refrigeration unit accordingly. For further assistance, consider using tools like the Danfoss Coolselector for enhanced accuracy.
Engage in an interactive simulation where you identify and categorize different sources of heat within a cold room. This activity will help you visualize how each source contributes to the overall cooling load. Use this simulation to test your understanding of transmission, product, internal, equipment, and air infiltration loads.
Work in teams to design a cold room for a specific type of perishable goods. Calculate the cooling load using the formulas provided in the article, and present your design, including the chosen refrigeration unit size. This project will enhance your collaborative skills and deepen your understanding of practical applications.
Analyze a real-world case study of a cold room installation. Evaluate the cooling load calculations and the choice of refrigeration unit. Discuss any discrepancies or improvements that could be made. This activity will develop your critical thinking and analytical skills.
Participate in a hands-on workshop where you perform cooling load calculations using different scenarios and variables. This workshop will provide you with practical experience and confidence in applying theoretical knowledge to real-world situations.
Explore and use design software like the Danfoss Coolselector tool to perform cooling load calculations. Compare the software results with manual calculations to understand the benefits and limitations of using technology in engineering design.
Here’s a sanitized version of the YouTube transcript:
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Hello everyone, Paul here from TheEngineeringMindset.com. In this video, we’re going to learn how to calculate the cooling load for a cold room storage unit. We’ll cover what a cold room is, the sources of heat that need to be removed, and then we’ll work through an example calculation with a safety factor. Finally, we’ll size the refrigeration unit to match this cooling load.
Before we begin, I want to thank our partners at Danfoss for sponsoring this video. They offer a wide range of solutions for cold rooms that can help you comply with future refrigerant and energy regulations without sacrificing performance. Their free Coolselector tool will assist you in applying the principles discussed today. You can download it for free at Coolselector.Danfoss.com.
So, what is a cold room and how does it work? A cold room is designed to store perishable goods, such as meat and vegetables, to slow down their deterioration and keep them fresh for as long as possible. Heat accelerates this deterioration, so we need to cool the food by removing heat, which is achieved through a refrigeration system. This system allows us to accurately and automatically control the temperature to preserve the goods.
Now, where does the heat come from that we need to remove? Typically, 5 to 15% comes from transmission loads, which is the thermal energy transferred through the roof, walls, and floor into the cold room. Heat always flows from hot to cold, so the interior of the cold room, being much colder than its surroundings, will always have heat trying to enter. If the cold room is exposed to direct sunlight, additional corrections will be needed.
Next, we have product loads, which account for approximately 55 to 75% of the cooling load. This includes the heat introduced into the cold room when new products enter, as well as the energy required to cool, freeze, and further chill after freezing. If you’re just cooling products, you only need to consider the sensible heat load. However, if you’re freezing, you must account for the latent heat as a phase change occurs.
Additionally, packaging will also generate heat, and if you’re cooling fruits and vegetables, these products are still alive and will generate some heat that needs to be removed.
The internal loads account for around 10 to 20% of the cooling load. This includes heat generated by people working in the cold room, as well as lighting and equipment like forklifts. You need to consider the equipment used by staff members, how much heat they generate, and how long they and the equipment will be in the store each day.
The refrigeration equipment in the room contributes about 1 to 10% of the total cooling load. For this, you’ll want to know the rating of the fan motors and estimate their daily runtime. Additionally, any heat transferred into the space due to defrosting the evaporator must be accounted for.
Lastly, air infiltration adds another 1 to 10% to the cooling load. This occurs when the door is opened, allowing heat to transfer into the space through the air. Ventilation may also be necessary for fruits and vegetables, which produce carbon dioxide, and this air needs to be cooled.
Now, let’s run a simplified example of a cooling load calculation. If you’re doing this for a real-world application, I recommend using design software like the Danfoss Coolselector app for speed and accuracy.
First, let’s start with the transmission load. The dimensions of our cold store are six meters long, five meters wide, and four meters high. The ambient air temperature is 30 degrees Celsius at 50% relative humidity, while the internal air is at one degree Celsius with 95% relative humidity. The walls, roof, and floor are insulated with 80 millimeters of polyurethane, with a U-value of 0.28 watts per meter squared per Kelvin, and the ground temperature is 10 degrees Celsius.
To calculate the transmission load, we’ll use the formula Q = U × A × (T_out – T_in) × 24 / 1,000. The U-value is known, A is the surface area of the walls, roof, and floor, and we already know the temperatures. We divide by 1,000 to convert the value from watts to kilowatts.
Calculating the area A is straightforward. You need to find the size of each internal wall, the roof, and the floor. The floor must be calculated separately from the walls and roof due to different temperature differences affecting heat transfer.
If the floor isn’t insulated, a different formula based on empirical data will be needed. Combining the walls and roof gives us a daily heat gain of 22 kilowatt-hours per day, and the floor adds another 1.8 kilowatt-hours per day. Therefore, the total transmission cooling load is 23.8 kilowatt-hours per day. Remember, if your cold room is in direct sunlight, you must account for the sun’s energy as well.
Next, we will calculate the cooling load from product exchange, which is the heat pulled into the cold room from new products entering at a higher temperature. For this example, we’ll be storing apples. There are 4,000 kilograms of new apples arriving each day at a temperature of five degrees Celsius, while the store maintains a hold of 20,000 kilograms of apples.
Using the formula Q = M × CP × (T_enter – T_store) / 3,600, we can calculate the cooling load. Dropping in the numbers gives us 16 kilowatt-hours per day.
Next, we calculate the product respiration heat generated by living products. For this example, I’ve used 1.9 kilojoules per kilogram per day as an average. The formula for this is Q = M × resp / 3,600. Using the numbers, we find it comes out to 10.5 kilowatt-hours per day. Summing the exchange and respiration loads gives us a total of 26.5 kilowatt-hours per day for the product section.
Now, let’s calculate the internal loads from people working in the cold room. We’ll estimate two people working for four hours a day, generating around 270 watts of heat per hour. The formula is Q = people × time × heat / 1,000. Plugging in the numbers gives us 2.16 kilowatt-hours per day.
Next, we calculate the heat generated by lighting. If we have three lamps at 100 watts each running for four hours a day, the total heat generated is 1.2 kilowatt-hours per day. Summing the people and lighting loads gives us a total internal load of 3.36 kilowatt-hours per day.
Now, we calculate the heat generation from the fan motors in the evaporator. With three fans rated at 200 watts each running for 14 hours per day, we find the total comes out to 8.4 kilowatt-hours per day.
Next, we calculate the heat load caused by defrosting the evaporator. If our electric heating element is rated at 1.2 kilowatts and runs for 30 minutes three times a day with a transfer efficiency of 30%, we find that this contributes 0.54 kilowatt-hours per day.
The total equipment load is the fan heat load plus the defrost heat load, which totals 8.94 kilowatt-hours per day.
Now, we need to calculate the heat load from air infiltration. We’ll estimate five volume air changes per day due to the door being opened. The volume is 120 cubic meters, and each cubic meter of air provides around two kilojoules of heat per degree Celsius of temperature difference. Using the formula Q = changes × volume × energy × (T_out – T_in) / 3,600, we find this comes out to 9.67 kilowatt-hours per day.
Finally, we calculate the total cooling load by summing all the calculations we’ve performed, which totals 72.27 kilowatt-hours per day. We should also apply a safety factor to account for errors and variations from design. A typical safety factor is 10 to 30%, and I’ve chosen 20%. Multiplying the cooling load by a safety factor of 1.2 gives us a total cooling load of 86.7 kilowatt-hours per day.
The last step is to calculate the refrigeration capacity needed to handle this load. A common approach is to average the total daily cooling load by the runtime of the refrigeration unit. Estimating the unit to run for 14 hours per day, we find that the refrigeration unit needs a capacity of 6.2 kilowatts to meet our cooling load.
That’s it for this video! Before we wrap up, I want to thank Danfoss once again and remind you to try out everything you’ve learned today by downloading their free Coolselector tool at Coolselector.Danfoss.com. Thank you for watching! If you found this helpful, please like, subscribe, and share, and leave your comments and questions below. Don’t forget to follow us on social media and check out our website, TheEngineeringMindset.com.
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This version maintains the essential information while removing any unnecessary or informal language.
Cooling – The process of removing heat from a system or substance to lower its temperature. – The cooling system in the reactor is designed to maintain optimal temperatures during operation.
Load – The amount of power or energy required by a system or component to perform its function. – Engineers must calculate the load requirements to ensure the power supply can handle peak demands.
Refrigeration – The process of removing heat from a space or substance to lower its temperature, typically using a refrigerant cycle. – The refrigeration unit is crucial for preserving the integrity of temperature-sensitive materials in the lab.
Heat – A form of energy transfer between systems or objects with different temperatures, often resulting in temperature change. – The heat generated by the machinery must be efficiently dissipated to prevent overheating.
Calculation – The process of determining a numerical result using mathematical methods and formulas. – Accurate calculation of stress factors is essential for designing safe and reliable structures.
Internal – Located or occurring within a system or structure. – The internal pressure of the vessel must be monitored to prevent potential hazards.
Product – A substance or material resulting from a chemical or physical process. – The final product of the reaction was analyzed to ensure it met the required specifications.
Transmission – The process of conveying energy, signals, or data from one place to another. – The transmission of electrical power over long distances requires efficient conductors to minimize losses.
Infiltration – The unintentional or accidental entry of air, water, or other substances into a system or structure. – Reducing air infiltration in buildings can significantly improve energy efficiency.
Equipment – The necessary tools, machinery, or devices required for a specific task or operation. – The laboratory equipment must be calibrated regularly to ensure accurate experimental results.
