A tractor exerts a drawbar force of 1760 lb for 310 yards. What is the mechanica

Question

A tractor exerts a drawbar force of 1760 lb for 310 yards. What is the mechanical Work done by the tractor in MJ? If it takes the tractor 64 seconds to cover the 310 yards, what is machine's average drawbar power output in kW and hp? If the tractor is 18% efficient at converting the energy in diesel into mechanical Work (at the drawbar), what volume of diesel fuel (mL) is consumed during the process? Assuming current diesel prices apply, and that fuel costs are 30% of the overall cost of using the tractor, what is the approximate cost of the tractor's mechanical Work on a $/GJ basis?

Explanation / Answer

given

F = 1760 lb = 1760*4.448 = 7828.48 N

d = 310 yards = 310*0.9144 = 283.464 m

a) Workdone = F*d

= 7828.48*283.464

= 2.22*10^6 J

= 2.22 MJ

b) Power output = Workdone/time taken

= 2.22*10^6/64

= 34.7*10^3 W

= 34.7 kW

= 34.7*10^3/(746)

= 46.5 hp

c) we know, energy produced from 1 L of diesel = 10.7 KWH

= 10.7*3.6*10^6 J

= 3.85*10^7 J

so, volume of the diesel use, 0.18*V = W/Q

V = 2.22*10^6/(0.18*3.85*10^7)

= 0.320 L

= 320 mL

d)

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