FP&MM: Lesson 28 to 32

Lesson 28 to 32

Between 1960 and 2010, the growth rate in power in Indian farms was 3.5 per cent and reached to 1.60 kW/ha (Fig. 1). In 1960-61 major contribution (92.31%) in farm power was from animate power (human + draught animal), where as in 2008-09 the major share was that of mechanical and electrical power (85.30%), Fig. 2. Over the years the contribution of animate source of power, especially that of draught animals, has been going down drastically. This shows that the additional need of farm power is being met through mechanical and electrical sources of power. This trend is going to continue in future also. For adoption of higher level of technology to perform complex operations within time constraints and with comfort and dignity to the operators, mechanical power becomes essential. Thus, the extent of use of mechanical power serves as an indicator of acceptance of higher level of technology on farms. Food grains productivity in India has increased from 0.710 t/ha in 1960-61 to 1.856 t/ha in 2008.09, while farm power availability has increased from 0.296 kW/ha to 1.600 kW/ha during the same period. Thus, food grains productivity is positively associated with unit power availability in Indian agriculture (Fig. 3). The relationship between food grains productivity and unit farm power availability for the period 1960-61 to 2008-09 were estimated by log linear function, with highly significant value of coefficient of determination (R²) as following:

Y_fgs= 674.18 Ln(X_ps) + 1480.8

R² = 0.989

Where,

Y_fgs = food grains productivity of India, kg/ha, and

X_ps = power availability in India, kW/ha

This indicates that productivity and unit power availability is associated linearly. It is also evident that farm power input has to be increased further to achieve higher food grains production, the composition of farm power from different sources to be properly balanced to meet its timely requirement for various farm operations.

L 28 to 32 fig.1

L 28 to 32 fig.2

L 28 to 32 fig.3

Some Calculations

Q. No. 1: Find out optimum tractor power required for draw bar operation and transportation for paddy wheat crop rotation in light soil for field of area of 10 ha.

FC% = 25.84 , U = 0.7, H = 8 h/day, Pt = 9400 Rs/kW

Crop	Implement	A (ha)	E (kWh/ha)	r	n	K	Y	V	Sc	Nt
Paddy	Disc Harrow	10	6.187	0.5	2	0.00625	51.98	500	4	2	4.713
	Cultivator	10	3.324	0.5	2	0.00625	51.98	500	4	2	2.536
	Planker	10	3.037	0.5	2	0.00625	51.98	500	4	2	2.317
Wheat	Disc Harrow	10	6.187	0.5	2	0.00456	43.42	580	4	2	3.631
	Cultivator	10	3.324	0.5	2	0.00456	43.42	580	4	2	1.953
	Planker	10	3.037	0.5	2	0.00456	43.42	580	4	2	1.785
	Seed Drill	10	1.248	0.5	1	0.00456	43.42	580	2	1	1.158
											Total 18.093

Power required for transportation of material for Paddy – What rotation

Crop

W (t/ha)

A(ha)

Distance (kM)

Labour (Rs/h)

(100x 0.27 x Dis x W x A x Labour)/(FC% x Pt x r)

Paddy

7.44

0.4

2.071

Wheat

5.765

0.4

1.579

Optimum tractor power requirement, P = [18.093+2.071+1.579]^1/2= 4.663 kW

Q. No. 2: Find out the least width of disc harrow, cultivator, planter & seed drill in light soil for Paddy wheat rotation. Given Fc% of disc harrow, cultivator & planker is 24.2% and FC% for seed drill = 23.09%, P = Price of implement per unit width

Implement				Width (m)
Implement		Paddy	Wheat	Width (m)
Disc Harrow	0.00416	185.89	132.77	1.151
Cultivator	0.0187	154.91	110.64	2.228
Planker	0.0206	185.89	132.77	2.566
Seed Drill	0.00614	0	312.40	1.385

Q. No. 3: Find out optimum trator power required for drawbar operations for cotton wheat crop rotation in light soil. FC% = 25.8%, U=0.7, H=8 hr/day, Pt = 9400Rs./kW

Crop	Implement	A (ha)	E (kWh/ha)	r	n	K	Y	V	Sc	Nt
Wheat	Disc Harrow	10	6.187	0.5	2	0.00456	43.42	580	4	2	3.631
	Cultivator	10	3.324	0.5	2	0.00456	43.42	580	4	2	1.953
	Planker	10	3.037	0.5	2	0.00456	43.42	580	4	2	1.785
	Seed Drill	10	1.248	0.5	1	0.00456	43.42	580	2	1	1.158
Cotton	Disc Harrow	10	6.187	0.5	2	0.0351	2.05	2200	4	2	2.434
	Cultivator	10	3.324	0.5	2	0.0351	2.05	2200	4	2	1.309
	Planker	10	3.037	0.5	2	0.0351	2.05	2200	4	2	1.197
	Planter	10	1.645	0.5	1	0.0351	2.05	2200	2	1	0.889
											Total 14.346

Optimum tractor power requirement, P = [14.346+0.225+1.579]^1/2= 4.020kW

Power required for transportation of material for cotton – What rotation

Crop

W (t/ha)

A(ha)

Distance (kM)

Labour (Rs/h)

(100x 0.27 x Dis x W x A x Labour)/(FC% x Pt x r)

Cotton

0.809

0.4

0.225

Wheat

5.765

0.4

1.579

ptimum tractor power requirement, P = [14.346+0.225+1.579]^1/2= 4.020kW

Q. No. 4: Find out the least cost width of farm implements in light soil for cotton wheat rotation. FC% of disc harrow, cultivator planker = 24.2% FC% for seed drill = 23.19% p= price of implement per unit width

Implement				Width (m)
Implement		Wheat	Cotton	Width (m)
Disc Harrow	0.00416	132.701	73.996	0.927
Cultivator	0.0187	110.584	61.664	1.795
Planker	0.0206	132.701	73.996	2.066
Seed Drill	0.00614	312.237	-	1.384
Planter	0.00586	-	198.982	1.079

Q. No. 5: Find out the time for drawbar operation and transportation in heavy soil for cotton wheat rotation for field of 10 ha. Consider distance for transfer rotation=10 km and r= 0.8

Crop	Implement	A(ha)	N	E (kWh/ha)	P (kW)	LCF	VW (t/ha)
Wheat	Disc Harrow	10	4	6.178	4.389	0.5	5.67	112.59	43.63
	Cultivator	10	3	3.324	4.389	0.5	-	45.43	-
	Planker	10	4	3.037	4.389	0.5	-	55.36	-
	Seed Drill	10	1	1.248	4.389	0.5	-	5.68	-
Cotton	Disc Harrow	10	3	6.178	4.389	0.5	0.81	84.45	6.22
	Cultivator	10	8	3.324	4.389	0.5	-	60.58	-
	Planker	10	3	3.037	4.389	0.5	-	41.52	-
	Planter	10	1	1.645	4.389	0.5	-	7.49	-

Total Tractor running time in h/year, T = 462.99 h

The cost of Power per hour can be calculated as

L 28 to 32 eq.7

Last modified: Tuesday, 25 March 2014, 4:57 AM

Farm Power & Machinery Management 3(2+1)

Lesson 28 to 32

Navigation