Caterpillar Performance Handbook, January 2017, SEBD0351-47

Mechanical PowerTrain Efficiencies

Mining & Off-HighwayTrucks

MECHANICAL POWERTRAIN EFFICIENCIES In selling against electric drive trucks, power train efficiency is an important consideration. To better illus- trate the advantages of mechanical drive performance, grade horsepower, power train efficiency, and retarding horsepower should be compared to electric drive trucks. Grade horsepower can be calculated by the follow- ing formula: Metric GMW (kg) × TR × Speed (km/h) grade HP = ______________________________ 273.75 GMW (lb) × TR × Speed (mph) grade HP = ______________________________ 375 where TR (total resistance) = Rolling resistance + Grade resistance (expressed as a decimal) English example 700,000 lb GMW, 2% rolling resistance, +8% actual grade at 8.2 mph would require 1530 HP 700,000 × (.02 + .08) × 8.2 ___________________________ = 1530 HP 375 Metric example 317 520 kg GMW, 2% rolling resistance, +8% actual grade at 13.2 km/h would require 1530 HP 317 520 × (.02 + .08) × 13.2 ___________________________ = 1530 HP 273.75 We then calculate power train efficiency by dividing grade horsepower by the gross horsepower produced by the engine. Most electric drive trucks run at constant maximum horsepower while under load. Mechanical drive trucks, however, lug the engine and may produce somewhat less than maximum horsepower. Engine power curves must be utilized to determine exact horse- power produced. Example 1530 grade horsepower ____________________ × 100 = 85% power train 1800 gross engine HP efficiency English

This exercise illustrates the effect of an efficient mechanical drive power train and should yield results in the 80-85% efficiency range. The same calculation for electric drive trucks would be lower (70-78% range) with a maximum efficiency of about 78% for the most com- mon systems. Likewise, retarding horsepower being consumed by the retarding system can be calculated by the follow- ing formula: Metric GMW (kg) × TR × Speed (km/h) retarding HP = ______________________________ 273.75 where TR (total resistance) = Rolling resistance + Grade resistance (expressed as a decimal) English example 700,000 lb GMW, 2% rolling resistance, –8% actual grade at 14.7 mph would equate to –1646 HP 700,000 × (.02 – .08) × 14.7 ___________________________ = 1646 HP 375 Metric example 317 520 kg GMW, 2% rolling resistance, –8% actual grade at 23.6 km/h would equate to –1646 HP 317 520 × (.02 – .08) × 23.6 ____________________________ = 1646 HP 273.75 This formula is intended for use in determining horse- power being consumed in the field based on field mea- surements. It is not intended to indicate how fast trucks should be operated on grade. Only job conditions, proper operating procedure, and good judgement should deter- mine safe operating speeds during retarder use. English GMW (lb) × TR × Speed (mph) = ______________________________ 375 retarding

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Edition 47 10-21