Problem: Determine the power being produced by a tractor if it developed 6.67 kN of force while traveling at a speed of 9.0 km/hr.

Solution:

Note: Appendix II includes conversions from customary to SI units. The answer to this problem could be converted to horsepower by using the appropriate conversion. Horsepower equals:

16.68 kW 0.7456999

Problem: Determine the useable power for a 120 kW naturally aspirated natural gas irrigation engine that will be operated with a fan and radiator. It is located at an elevation of 1,000 m and will operate in an air temperature of 38°C and on continuous duty.

Solution: The process is the same as customary units. The only difference is in the values used. For this example direct conversions are used, Table 7.1.

Table 7.1. |
Direct conversions for derating factors. | |

Accessories |
10% other than fan | |

5% fan and radiator | ||

Temperature |
1% or each 5.6°C above 29°C for gasoline | |

1% for each 2.7°C above 29°C for diesel | ||

Altitude |
3% for each 305 m above 152 m | |

Type of service |
10% for intermittent loads | |

20% continuous loads |

The useable power is determined by determining the total amount of power loss. Power loss (%) = Lacc + Lalt + Ltemp + Lduty

305 m

Ltemp = 0.01 x i 472 I = 0.01 x 1.906... = 0.019 L duty = 0.20

Total loss = 0.05 + 0.083 + 0.019 + 0.20 = 0.662 Useable power = 120 kW - (120 kW x 0.662) = 120 kW - 79.44 = 40.55 kW

All of the power losses for this engine are 79.44 kW, leaving 40.55 kW of useable power.

Was this article helpful?

## Post a comment