Waste heat recovery case

A large enterprise with 3000 employees, 2 sets of 150HP (110KW) screw air compressors, with a loading rate of 95%. The distance from the compressor room to buildings A and B is 150 meters. The minimum heat recovery power of the energy recovery device is 70-85% of the input power of the air compressors.

Analysis basis: 1. The heat required to raise the temperature of 1 ton of water by 1°C is 1000 kcal.

The storage temperature of bathing water is 50°C, and the annual average make-up water temperature is 25°C.

Diesel calorific value: 10300 kcal/kg; Diesel density: 0.86 kg/L; Diesel price: 7.0 yuan/L. 3. The heat energy generated by 1 kWh of electricity consumption is 860 kcal.

Hourly recoverable power P = Air compressor power × Recovery rate × Loading rate = 2 × 110 × 70% × 95% = 146.3 KW

Daily recoverable heat Q = P × 860 kcal × 24 h × 90% = 2,717,668.8 kcal (90% accounts for an average heat loss of about 10%)

With the storage temperature of bathing water at 50°C, the daily hot water production is approximately:

Q_water = Recoverable heat ÷ ΔT = 2,717,668.8 / 25 ≈ 108 t 

That is, approximately 108 t of 50°C hot water is produced daily. To generate the same amount of heat, the daily diesel fuel required (with a combustion efficiency of 95% calculation):

Q_oil = Recoverable heat ÷ (Diesel calorific value × Combustion efficiency) = 2,717,668.8 / (10300 × 95%) = 277.7 kg

 Then:

Monthly diesel fuel required: 277.7 kg × 30 = 8331 kg ≈ 8.33 t        Annual diesel fuel required: 8331 × 12 = 99972 kg ≈ 100 t

Monthly fuel cost: 8.33 × 1190 × 7.0 yuan/L = 69,389 yuan (1 ton of diesel fuel is approximately equivalent to 1190 liters)

Annual fuel cost: 69,389 × 12 ≈ 833,000 yuan