Calculation of heat absorbed during urea dissolution

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Urea aqueous solution is used for denitrification in diesel vehicles. Nowadays, the SNCR process is employed for flue gas denitrification, and in some areas where the use of ammonia water is not allowed, urea aqueous solution is used instead. The common practice is to introduce steam directly into the dissolution tank; don’t you use steam? It seems that the level of professionalism needs to be improved. . . As for the heat of dissolution of urea, it’s easy to calculate. For a urea aqueous solution with a concentration of around 33%, the theoretical heat absorbed by urea during the dissolution process is 62.5 kcal/kg of urea; you can use 70 kcal/kg as an estimate. This value already takes into account heat dissipation. Since your workshop already has a boiler, you just need to connect a pipe to it. The amount of steam required is even simpler to calculate – you can use a calculator for that. For an 18-ton solution with 33% water, along with 6 tons of urea and 12 tons of water, the heat required is 6000 x 70 = 420,000 kcal. Assuming a latent heat of 500 kcal per kilogram of steam, the amount of steam needed is 420,000 / 500 = 840 kg. So, just that much steam is sufficient for each batch of 18 tons of solution. If insulation isn’t in place and there are significant heat losses (typically 5%), then 1 ton of steam will be enough. Use steam directly; it’s sufficient to add a DN50 steam pipe. With so many dissolution tanks, monitoring the liquid level and temperature is by no means a difficult task.

Thank you all! I also want to ask: if electric heating elements are used for heating, with 3 tons of solid urea and 6 tons of pure water in each mixing tank, and the urea aqueous solution is to be heated to 20 degrees Celsius, what power rating should the heating elements have? How should it be calculated? Thank you!

I would also appreciate advice from those who are knowledgeable in this area. Urgent need.

This post was last edited by arpcd on 2019-12-30 at 20:33. These are all the most basic data calculations. First, you need to determine the hourly processing capacity of your urea solution; in other words, how much time is required to heat those 9 tons of solution from 0°C to 20°C. The specific heat capacity of a 33% urea solution is taken as 0.8 kcal/kg·°C. Therefore, Q = 9000 × (20 – 0) × 0.8 = 144,000 kcal. This represents the total amount of heat required, but not the power. Power is calculated based on time. Assuming that you need to raise the temperature of 9 tons of solution by 20°C within 1 hour, the power required for the electric heater is: P = 144,000 kcal/hour = 144,000 x 4.1868 / 3600 = 167.472 kW; rounding this value to 170 kW. Considering a heat efficiency of 60–80%, the actual power required becomes 170 / 0.6 = 283 kW. At that time, if you requested half an hour, the power would need to be doubled; if it was two hours, then the power would be halved. . Can it be calculated? These are all the most basic calculations; they’re something that is taught in middle school physics. . . . Has the original poster lost all their basic skills?

Thanks! I’m really ashamed to hear that! It’s been too long, and it’s not used often at work, so I’m almost completely forgetting it. Click to find out. :)

Hello. I also want to ask where the number 4.1868 comes from? For example, if I plan to use a 30-kilowatt electric heater, can I calculate it this way? {（30×3600×0.239）÷0.6}×h=144000. h is the number of hours required for heating.

1 calorie (cal) = 4.1868 joules (J)

Where can I find the source of this table? In which book can I find it? Thank you