How to Calculate Heat

Do you know how to determine the amount of energy required to heat a substance to a specific temperature? If not, we’ve got a simple method to help you determine the energy required for almost any heating application! Let’s start with the “total heat equation”. You can see, this equation has three different components. One by one, we will cover each component so that you can determine exactly how to calculate the total energy required, in kilowatts, for your application. First, though, I want to talk about the difference between start up and operational heat requirements. The startup or pre-heat is defined as the amount of heat required to bring a system from its initial temperature to the final operating conditions in a specified length of time. Operational heat is defined as the amount of heat required per unit of time (usually an hour) once the system is at its specified operating temperature and doing the work you need it to do. The majority of the energy is on the operational side. So, our focus will be to determine the KWs required on this portion of the heating process. The first component is QA which is the heat energy absorbed by the materials. This equation allows us to determine the total KWs required for a specific application. It uses 4 different values:
•Media weight in lbs.
•Specific heat in BTU/lbs- °F
•Temperature change in degrees F
•Heat up time in hrs.
In order to solve for QA, there are three simple steps you should follow:
1. Gather information 2. Determine needed values and convert to appropriate units 3. Solve the equation

Let’s say that you have an oven that is used to heat 500 lbs. worth of granite blocks in a batch process. The oven is 7.5 ft wide, 10 ft long, and 7.5 ft in height. It has steel walls with 2-inch insulation. In this scenario, we need to determine the amount of heat (in kilowatts) absorbed by the granite as they are being heated from 50°F to 230°F in 2 hours. Now, we can determine needed values and convert to appropriate units. Looking online, we are able to determine that granite has a specific heat value of 0.189 BTU/lb*F. Solving for temperature rise, or the change in temperature, is fairly simple. This is just the final temperature minus the initial value. So 230°F – 50°F equals 180° delta.

If the first two steps are done correctly all you’ve got to do is plug in the values (see formula in video).
So, you can see we are putting in the mass, the Cp, the delta T, and the time required, do some simple math… and you now know the KWs absorbed per hour for your heating application! In this scenario, we end up with a total energy required of 2.5 KWs.

Heat loss can have significant impact when determining the KW required. To find the heat loss, you need:
•Area of the heated container
•Material/construction of the container
•Difference to the ambient

Heat loss curves allow you to determine the loss in wattage per square foot per hour. After you know that value, all you have to do is plug your values into the following equation (see formula in video). This gives you the total kilowattage lost per hour due to heat loss.

You can look up these curves online…or you can plug these parameters into a computer program. The application engineering team at Valin runs these types of calculations all the time. If you would like help with these, you can contact us using the contact information provided below.

Here is the heat loss curve that best fits our application (see graph in video). We can utilize the curve to determine the value of the wattage lost per square foot per hour. In our scenario, due to the 2-inch thick insulation and 160° difference to the ambient (assuming 70°F ambient and 230°F oven temp), the curve gives us a value of 5 watts per square foot per hour. Next, we will need to calculate the area of the heated container. Because we already have the dimensions, this is easy to do. It’s just (see formula in video). Using this formula, we can determine the area of the oven is 562.5 square feet.

The safety factor is the final component. This basically ensures that you’ll have more than enough energy available for your application. We recommend adding a safety factor of at least 10%... sometimes up to 20% if the conditions are sketchy or there are many unknown variables.

In order to determine what a 20% safety factor would be, all you have to do is add your heat absorbed (QA) and heat lost (QL) values together and multiply it by 20% (see formula in video). Doing this, we get a safety factor value of 0.7 KW.

Once you’ve calculated the safety factor, you’re pretty much good to go! All that’s left is to add the three numbers together. In our scenario, we get 4.2 Kw/ hour for the total heat required.

If you'd like to speak to one of Valin's process heat experts call (855) 737-4718 or visit www.valin.com to learn more.

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