How much heat does it take to make coke from a ton of wet coal with a moisture content of 11%?
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As title: Please give an expert answer! How much heat does it take to make coke from a ton of wet coal with a moisture content of 11%?
I hope someone can upload the detailed calculation process: lol, I will give you lots of points
I can calculate the heat consumption, but for the question, I think the conditions are not complete, because the heat consumption has a great relationship with the coking time. The longer the coking time, the greater the heat consumption.
Reply to the 3rd floor: You first assume that it is 24 hours, so that we can calculate based on 24 hours.
I thought about it for two days, and I can deduce the heat consumption of wet coal to coke based on the calculation formula of gas flow. In this way, we can first assume that the turnover time is τ, which can be omitted in the calculation process. There are two calculation methods for the consumption of returned gas per unit time (that is, the heating gas flow rate). One is based on the percentage of returned gas in the gas generation. The formula is:: V=NB(1-Mt)wη/τ………………(1) where: V—gas flow rate, m3/h N—number of carbonization chambers B—coal loading in each carbonization chamber, t Mt—moisture content of coal entering the furnace, % w—gas production per ton of dry coal, 320 to 330 m3/t dry coal, average 325 m3/t dry coal η—percentage of recycled gas in gas production, % (empirical value is 0.46) τ—turnover time, h It can also be calculated based on the heat consumption of coking, the formula is:: V=q (N/τ)B×1000/Q………………(2) where: V—gas flow, m3/hq—coke oven heat consumption, kJ/kg N—number of carbonization chambers τ—turnover time, h B—coal loading in each carbonization chamber, t Q—low calorific value of coke oven gas, kJ/m3 Since (1) = (2), that is, NB (1-Mt) × w × η/τ = q (N/τ) B × 1000/Q, so coking heat consumption: q=（1-Mt）wη×Q/1000 =（1-0.11）×325×0.46×17750/1000 =2361 kJ/kg
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