How to calculate the amount of reduced energy consumption in kilograms of standard oil per ton by reducing energy losses
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This post was last edited by jia2310 on 2016-1-21 09:36. Assuming a reduction in heat loss of 2610 MJ/h and a calorific value of the fuel gas of 35984 kJ/kg, 72.5 kg/h of fuel gas can be saved; this amounts to 1.74 tons of fuel gas saved per day, resulting in a reduction in the facility’s energy consumption by 0.12 kgeo/t. If the crude oil processing volume is 400,000 tons per month
I’ve calculated it. . . 1.74÷(450×〖10〗^4÷365) = 1.41×〖10〗^(-4) t/t = 0.141 kg/t. 0.141×35984÷41868 = 0.12 kgeo/t
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