**Author:**

Johann Lainer, Senior Marketing Communications Specialist, Watlow

**Date**

09/30/2021

Troubleshooting any inefficiencies and reducing heat loss can improve the efficiency of heating systems and save costs. Here, it breaks down the factors related to heat loss and explains how to calculate predicted heat loss for your application.

Heat loss is the intentional or unintentional movement of heat from one material to another, which can happen through conduction, convection or radiation. Conduction often occurs when an insulated or uninsulated component is in direct contact with another component, convection occurs when a pipe, electric heater or other component has an air barrier around it, while radiation is when there is no contact and heat moves as waves.

**How heat loss happens**

With an understanding of how heat loss occurs, it’s important to then select power and temperature controllers that best suit your heating situation. One way to do this is to account for common heat loss factors. While every scenario will have different factors affecting heat loss and transfer, uninsulated surfaces, vertical or horizontal insulated surfaces and oil, paraffin or water surfaces are common areas of heat loss that design engineers must account for when selecting heating products.

Factors that influence heat loss can change when relocating a process or make changes to an assembly line. Targeting a facility’s heat loss factors and making necessary adjustments with controlling equipment or insulation strategies keeps your operations running at the same quality and consistency as it was before the change.

In addition, adjusting the temperature of a heater or protecting it from one or more of these heat loss factors can reduce the wattage usage of the system, saving a considerable amount in daily operating expenses.

**Emissivity and heat loss**

Emissivity is defined as the ability of a particular object or material to emit infrared energy. The emissivity of a heater, thermocouple and insulation can affect the heat loss by radiation.

This is another piece of the puzzle when determining how much heat is required for a specific application. It’s important to explore the emissivity of the material in the heating process and identify whether the material is a polished surface or has medium or heavy oxide. Even non-metals are emissive, so the specific heat of insulation materials must be checked to see how it can affect overall heat loss.

**Calculating heat loss**

When working to reduce heat loss, adopting a predictive mindset is advised. Watlow recommends calculating a process or piece of equipment’s expected heat loss based on the location, insulation type and material being heated.

To understand how to calculate heat loss, multiply the convection curve value by a factor of 1.29 for horizontal heating products, whereas vertical pipes should use the direct value of the curve. For bottom surfaces, multiply the curve by 0.63. However, this offers a rough guide and the calculation does not consider other heat loss factors that may affect your product. Original equipment manufacturers (OEMs) need a more accurate way to measure heat loss for their specific product, so alterations known as emissivity values need to be made.

Using the graph is helpful at high temperatures, but as the temperature reaches ambient, or over 21 degrees Celsius, it can be difficult to read. There are two general rules that can help you arrive at a more precise temperature reading.

Firstly, calculate losses from an uninsulated surface close to 1.0 emissivity by dividing the temperature rise above ambient by 200, and secondly, calculate losses from an insulated surface with the approximate thickness of one inch and a thermal conductivity K-value of 0.5 by dividing the rise above ambient temperature by 950.

**From heat loss, to heat gained**

Now we’ve covered how to calculate heat loss, it’s equally important to understand how to calculate is the amount of heat required for a given operation. Most electrical heating problems can be solved by determining the heat required to carry out the job. The heat requirement must be converted to electrical power and the most practical heater can then be selected for the job. Whether the problem is heating solids, liquids or gases, the approach to determining the power requirement remains the same.

Click image to enlarge

*Figure 2: Heat loss image*

Before calculating energy requirements, the heating problem must be clearly determined, paying careful attention to defining operating parameters. To achieve the most accurate result, OEMs must ensure key information is considered such as minimum start and finish temperatures expected, required time for start-up heating and process cycle times, weight and dimensions of both heated material and the containing vessel, as well as electrical requirements including voltage, and power controller type.

The thermal system being designed may not consider all the possible or unforeseen heating requirements, so it’s crucial to remember the safety factor, which increases heater capacity beyond calculated requirements.

The power required (kW) is the heat energy value (kWH) divided by the required start-up or working cycle time. The kW rating of the heater will be the greater of these values plus a safety factor. The total kWH required will be either the heat required for start-up or the heat required to maintain the desired temperature, depending on which calculated result is larger.

The calculation of start-up and operating requirements consists of several distinct parts that are best handled separately. However, a shorter method can be used for a quick estimate of the heat energy required.

Watlow offers a range of industrial heating equipment and technical guides to help design engineers understand and manage heat loss factors.

**Summary**

Heat loss is influenced by various reasons dependent on the material and application. However, the result remains the same — increased costs and more inefficiencies. Understanding what causes heat loss and knowing how to calculate the predicted loss of your application is key to maintaining productivity. OEMs can also take their design one step further by understanding how to calculate the required heat for an operation.