How is the energy associated with low-temperature water heat exchange converted into the energy consumption of the device?
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The 750,000 tons per year gasoline hydrogenation unit is equipped with low-temperature water heat exchangers to recover the low-temperature heat from the unit; how can this energy be converted into part of the unit’s energy consumption? I seek advice from experts! !
The specific heat of water is 1 kJ/kg·°C. The flow rate of hot water at low temperatures is expressed in kilograms; there is also a temperature difference between the inlet and outlet. The heat absorbed is equal to specific heat * flow rate * (temperature of outlet water – temperature of inlet water), with the result being in kJ. 1 j = 4.18 cal, 1 calorie = 1000 cal = 1 kJ, and 1 kilogram of standard oil equals 10,000 calories. By dividing this value by the instantaneous feed rate, we obtain the amount of standard oil in kilograms per ton.