Question

a crawler tractor is operated by a 180-hp diesel engine. calculate the probable gallons of fuel consumed per hr for each of the given conditions: a) when operating at an average of 60 percent of its capacity for 50 min per hr. b) when operating at 100 percent of its capacity for 15 min per hr, 60 percent capacity for 30 min per hr, and 20 percent of its capacity for 15 min per hr.

Answer 1

3.59

4.32

Step-by-step explanation:

We find the time factors and engine factor to solve for this.

Engine factor = (100%*15/60) + (60%*30/60) + (20%*15/60)

= 1x0.25 + 0.6x0.5 + 0.2x0.25

= 0.25 + 0.30 + 0.05

= 0.6

A. We find time factor

= 50/60 = 0.83 minutes

We then get consumption

= Time factor x engine x hp x .04

= 0.83 x 0.6 x 180 x 0.04

= 3.585

3.59 gallons in 1 hr

B. Time factor = 60/60 = 1

Consumption =

1x0.6x180x0.04

= 4.32 gallons in 1 hour