Metric Problems

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.

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