How large is the energy consumption of 40,000 kW?
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Hello everyone, I’ve been busy working on a project recently and have a question that needs help with. The purpose of this project is to design a reactor under supercritical water conditions, with the aim of minimizing energy consumption. After doing the calculations, I found that for a reactor with this processing capacity designed to remove phenol from wastewater at a rate of 100 m3/h, if the water is heated from 25°C and 1 atm to 425°C and 250 atm, the heating energy required, based on this flow rate, is 40,000 kilowatts. Is this energy consumption too high? I would like to ask the experienced seniors: does this project still have industrial feasibility? If it doesn’t work, how can it be improved? Thank you!
Looking at the absolute energy consumption alone, it is indeed very high: with a processing capacity of 100 m3/h, the energy consumption for electricity alone amounts to 40,000 kWh/h, resulting in an energy cost of 300 yuan per m3. Who can afford that? Did the original poster make a mistake and add an extra zero?
I miscalculated; it should be 47 KW.
Please help check if there are any mistakes, thank you. Heat the water to a supercritical state, from 25°C to 425°C. The flow rate is 100 m3/h; thus, the volume flow rate is 100 m3/h = 0.028 m3/s, which corresponds to a mass flow rate of 28 kg/s. The heat capacity of water, cp, is 4.2 kJ/Kg/K = 4200 J/Kg/K. The temperature difference is 425–25 = 400°C = 400 K. The power required for heating is calculated as follows: 28 kg/s * 4200 J/Kg/K * 400 K = 47,040,000 W = 47,040 kW, or approximately 47,000 kilowatts
Oh, I’m sorry, I miscalculated. . Your calculation is correct
It can’t be calculated this way; this formula is only applicable to single-phase systems. In your case, there is a phase change, and the specific heat of water in gaseous form is not as high as 4.2