b) 36%
c) 41.18%
d) 52.94%
Answer: d
Explanation:
According to figure
0.64x = 17; x=26.56
D=12.5 KN, R+4.5 KN, s=20%
P=12.5 + 4.5=17 KN; But useful pull is only 12.5 KN
Percent Loss = \(\frac{26.56-12.5}{26.56}\) *100 = 52.94%.
A two-wheel drive tractor pulls an implement which requires a draft of 12.5 KN. The motion resistance of the tractor is 4.5 KN and the slip of the driver wheel is 20%. The transmission efficiency is 0.8. the percentage of power lost in converting engine power into drawbar power is ______
a) 26.47%
b) 36%
c) 41.18%
d) 52.94%
Answer: d
Explanation:
According to figure
0.64x = 17; x=26.56
D=12.5 KN, R+4.5 KN, s=20%
P=12.5 + 4.5=17 KN; But useful pull is only 12.5 KN
Percent Loss = \(\frac{26.56-12.5}{26.56}\) *100 = 52.94%.
b) 36%
c) 41.18%
d) 52.94%
Answer: d
Explanation:
According to figure
0.64x = 17; x=26.56
D=12.5 KN, R+4.5 KN, s=20%
P=12.5 + 4.5=17 KN; But useful pull is only 12.5 KN
Percent Loss = \(\frac{26.56-12.5}{26.56}\) *100 = 52.94%.
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