Question

How many acres can be covered by a harrow of 1.5 m width in a day of 8 hours with bullock power. If each spike of the harrow is giving 1 kg resistance when there are 50 spikes. What power would be necessary for the bullocks to pull the harrow?

(a) 0.544 KW

(b) 0.987 KW

(c) 0.123 KW

(d) 0.333 KW

The question was asked in exam.

My question is based upon Harrows in chapter Harrows, Cultivators and Intercultural Implements of Farm Machinery

Answer 1

Right choice is (a) 0.544 KW

For explanation: Area covered/hr = 1.5*4*1000 m^3

Area per 8 hours = 6000*8 m^2 = 4.8 ha

Total draft = 50 kg

Power, KW = \(\frac{Draft*speed}{1000}=\frac{50*9.8*4*1000}{60*60*1000}\) = 0.544.