Farm Power And Machinery (1+1)

Terms used for evaluation of an implements

Theoretical field capacity
It is the rate of field coverage of the implement, based on 100 per cent of time at the rated speed and covering 100 per cent of its rated width.

W x S
Theoretical field capacity = ---------- ………… ha/hr
10

Where,
W – Width of the implement in meters (m)
S - Speed of operation in Kilometers per hour (Km/hr)

Effective field capacity
It is the actual area covered by the implement based on its total time consumed and its actual width of operation.
Field efficiency

It is the ratio of effective field capacity and theoretical field capacity expressed in percent.
Effective field capacity
Field efficiency = --------------------------------
Theoretical field capacity
Effective field capacity is calculated as follows
S x W x E
C = -----------------

1000

Where,

C - Effective field capacity, hectare per hr (ha/hr).
S - Speed of travel in km per hour.
W - Theoretical width of cut of the machine in metre, and
E - Field efficiency in per cent.


Problem 1:
A 5 x 20 cm double action disc harrow is operated by a tractor having a speed of 5 km/h. Calculate the actual field capacity, assuming the field efficiency of 80 percent.

Solution:
Size of the harrow (width) = 5 x 20 = 100 cm = 1 meter

S x W x E

Actual field capacity, C = -----------------

1000

= (1 x 5 x 80) / 1000

Actual field capacity = 0.4 ha/hr.


Problem 2:
A 3 x 30 cm plough is moving at a speed of 4 km/h. calculate how much time it take to plough 500 x 500 m field when the field efficiency is 70 %.

Solution:
Width of the plough = 3 x 30 = 90 cm = 0.9 m

S x W x E

Effective field capacity, C = -----------------

1000

= (0.9 x 4 x 70)/ 1000

Effective field capacity, C = 0.25 ha/hr = 2500 m2/hr, ( 1 ha = 10,000 m2 )
Time required = 500 x 500/ 2500 = 100 hr.


Problem 3:
A 4 bottom 40 cm mould board plough is operating at 5.5 km/h speed with 75 % field efficiency . Calculate the rate of doing work in hectares per hour.

Solution :
Width of the plough = 4 x 40 = 160 cm = 1.6 m
Rate of doing work = 1.6 x 5.5 x 75/1000
Rate of doing work = 0.66 ha/h


Problem 4:
An indigenous plough has a 20 cm wide furrow at the top and 10 cm depth. Calculate the volume of soil handled per day 8 hours if the speed of working is 2.5 km/h.

Solution:
Furrow cross section = 10 x20/2 = 100 cm2
Distance traveled in 8 hours = 8 x 2.5 x 1000 = 20,000 m
Volume of soil handled = 20000 x 100/ 10000
Volume of soil handled = 200 m3