I have a question regarding energy calculation

Find the calorific value of diesel; multiplying this value by the mass of diesel gives the heat generated by that diesel in one day； Find the power of the electric heater, multiply it by the time, and that gives the amount of heat generated by electricity; when these two values are equal, you know how much electricity is required.

Using a diesel combustion heat value of 10,000 kcal/kg, 800 kg of diesel produces 8 million kcal of heat; 800*10,000/860 = 9,302 KW.

I agree with the views of the two people above, but I disagree with the approach suggested by the person in post 3. The original poster did not specify whether the diesel is of grade 0, hydrogenated diesel, or waste diesel used in industrial applications; therefore, it’s difficult to determine its calorific value

Since it is a diesel boiler, it can be converted in accordance with the \"GB∕T 2589-2008 General Rules for Calculating Comprehensive Energy Consumption\". There’s no need to worry about exactly what type of diesel it is; as long as it is diesel of a commercial grade, its calorific value is more or less the same. In accordance with the provisions of Appendix A in GB/T 2589-2008 \"General Rules for Calculating Comprehensive Energy Consumption\", the average lower heating value of diesel is 10,200 kcal/kg; the conversion factor to standard coal is 1.4571 kgce/kg. The equivalent energy value of electricity is 860 kcal/(kW·h), with a conversion factor to standard coal of 0.1229 kgce/(kW·h). Using these values, the number of kilowatt-hours equivalent to 1 kg of diesel is calculated as follows: 1.4571/0.1229 = 11.856 (kW·h)/kg of diesel. For 800 kg of diesel, the total amount is 800 x 11.856 = 9,484.8 kW·h. In other words, 800 kg of diesel is equivalent to 9,485 kilowatt-hours of electricity. The calculation done by that friend ahead is more or less the same. .

The thermal energy utilization efficiency of both should also be considered! Give priority to those with high efficiency