Caterpillar Performance Handbook, January 2017, SEBD0351-47

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Edition 47 10-21
Mining & Off-HighwayTrucks Mechanical PowerTrain Efficiencies
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
English
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
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
English
GMW (lb) × TR × Speed (mph)
retarding = ______________________________
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 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.