Planning of farm work system newly, the following procedure for each crop system will be useful.
At first, select the basic machinery like as tractor, combine and drying system.
The second, select each machine width or capacity. (See 5-2-3.)
The third, the Rate of work will be calculated by using the empirical data or the official data as followings.
1) Effective Field Capacity(EFC) is calculated by Operation width(W), Operation speed(V) (Table A-215b.) and Field efficiency(EF) (Table A-216.).
2) Coverage(CA) of each work is calculated by a) Working hour per day(Dt), b) Daily net working rate(NWR), c) Days of work period(DWP), d) Rate of available work days(ADR) (Table 33a.).
Economic selection finds that capacity which produces the lowest net cost. The increased ownership cost of high capacity machines are balanced against the increased operation costs and timeliness costs of low capacity machines.
(See
ASAE P496)Size selection of machinery is based on a combination of expected performance and expected costs. Both capacity and capital costs increase with size. At the same time performance improves, particularly with critical operations such as planting and harvesting. Delays in planting can reduce yields. Delays in harvest can reduce both quantity and quality of production.
See [Selection of tractor size]
in reference 5)
Simple capacity selection is made by estimating the number of days in the time span within which the operation should be accomplished, and by determining the probability of a working day in this time span. The required capacity of machinery or farm work for an area is
EFC = A / (AWD * Dn) Eq. 5-23
or C = A / (D * H * pwd)
Where:
symbol |
term |
unit |
example |
EFC |
Effective Field Capacity |
ha/h |
0.139 |
C |
Required machine capacity or farm work |
ha/h |
|
A |
Area |
ha |
10 |
AWD |
Available work days |
d |
12 |
D |
Number of days within the time span within which the operation should be accomplished |
d |
|
pwd |
Probability of a working day |
in decimal |
|
Dn |
Net work hours per day |
h/d |
6 |
H |
Expected time available for field work each day |
h/d |
The width of machinery will be shown by next equation.
Wt = 10 * EFC / (Vt * ef) Eq. 5-24
Where: example is rotary tillage.
symbol |
term |
unit |
example |
Wt |
Theoretical operation width |
m |
0.93 |
EFC |
Effective Field Capacity |
ha/h |
0.14 |
Vt |
Theoretical operation speed |
km/h |
2.0 |
ef |
Field efficiency |
in decimal |
0.75 |
Select minimum cost correspond to the planning farm scale, using the data of annual operation area vs. cost per ha.
See Table A-523. in Appendix
Implements are mounted on tractor and the capacities of them are not same, therefore it is not necessary to prepare the same number of these kind of implement. Required number of tractors and implement is obtained by next equation.
M >= A / CAS Eq. 5-25
Where:
symbol |
term |
unit |
example |
M |
Number of machine set |
- |
3 |
A |
Area |
ha |
27 |
CAS |
Coverage of one set |
ha |
9.1 |
Table 523. Required number of implement: Example
(Area: A = 27 ha and Available net working hour: ANWH=157.5 h)
Farm work with tractor |
Implement |
Work capacity: WC |
Number of operation: N |
Required number of implement |
h/ha |
times |
|||
Tillage |
Rotary |
4.3 |
1 |
2 |
Harrow |
Rotary |
3.7 |
2 |
|
Leveling |
Tooth harrow |
0.6 |
2 |
1 |
Seeding |
Grain drill |
2.9 |
1 |
1 |
Pressing |
Roller |
1.5 |
1 |
1 |
Work capacity of these five farm work is required as followings:
WCp =
ΣWCi * Ni =4.3 + 3.7*2 +0.6*2 +2.9 +1.5 = 17.3 h/haCoverage and number of tractor are shown as following from equation 5-12 and the above equation.
CAS = ANWH / WCp = 157.5 /17.3 = 9.1 ha
Required number of tractor: M = 3 >= A/CAS =27/9.1 = 2.97
Required number of each implement or work is obtained by next equation.
Mi >= A / CASi = A * WCi * Ni / ANWH Eq. 5-26
M1 >= 27*4.3/157.5 = 0.74
M2 >= 27*3.7*2/157.5 =1.27
M3 >= 27*0.6*2/157.5 =0.21
M4 >= 27*2.9/157.5 = 0.50
M5 >= 27*1.5/157.5 =0.26
Work-1 and 2 are operated by same implement, therefore number of rotary is M1+M2 >= 2.01. Required numbers of implement are shown in above table.
Exercise 5-10., 5-11., 5-12.
5-3. Optimization of Farm System
Seek optimal investment by simulation
Select maximum d Profit / d capital
See Steepest gradient method:
SGM-001.doc
Machinery cost is decreased by replacing machine, which shows excessive capacity to lower capacity machine.
1. Select maximum CA of farm work operated by machinery:
Example = 21-Baler = 51.5 ha
2. Replace baler to smaller one in sheet step-02 from machinery table step-03:
Example 1.4m ->0.73m
3. Change EFC in sheet 1.field-capacity, and FRh in 2.Variable-cost
4. Calculate new CA, total cost per ha etc. in sheet fwtotal
See Table A-531-i. Improvement by replacing machinery in Appendix.
See Table A-531-ii. Improvement coverage of system in Appendix.
Exercise 5-8., 5-9.
Energy consumption will show the important index of energetic evaluation of farm work systems.
LCA (Life Cycle Assessment) is acceptable method of calculating energy consumption in industry or daily life.
Calculation of energy consumption of farm machinery or facilities by using input-output analysis of inter-industry (SANGYOU RENKANBUNSEKI) is also effective method for energetic evaluation.(ECU in the following table)
Total annual energy consumed for manufacturing will be obtained as following equations. (ECU in next table)
AEG = AFC * Yrate* ECU/1000 Eq. 5-27
where,
symbol |
term |
unit |
AEG |
Total annual energy consumed |
MJ |
AFC |
Annual fixed cost ($): Total |
$ |
Yrate |
Yen exchange rate |
Yen/$ |
ECU |
Energy conversion unit |
kJ / Yen |
Variable energy per ha(MJ/ha) will be calculated by using conversion factor of next table.
Table 532a. Conversion factor
Output energy of per ha = Yield * RCF =4500(kg/ha) * 14.9(MJ/kg) = 67 GJ/ha
symbol |
Conversion factor |
unit |
|
ECU |
Energy conversion unit for manufacturing the machinery by using input-output table of inter-industry |
48.1 |
kJ / Yen |
RCF |
Rice grain Conversion factor |
14.9 |
MJ/ kg |
GCF |
Gasoline Conversion factor |
35.2 |
MJ/ L |
KCF |
Kerosene Conversion factor |
37.3 |
MJ/ L |
DCF |
Diesel light oil Conversion factor |
38.5 |
MJ/ L |
ECF |
Electric power Conversion factor |
9.4 |
MJ / kWh |
by handbook of energy save: 1996 and food handbook: 1 kWh = 3.6 MJ
Example See
Rice-erg.xls: Step-C1, fwtotal-erg
Method of environmental evaluation is not yet completed now, but CO2 exhaust amount will show some index of environmental situation.
Table 532b. CO2 gas generation ratio
CO2 gas generation ratio |
unit |
|
Gasoline |
2.3587 |
kg / L |
Kerosene |
2.5284 |
kg / L |
Diesel light oil |
2.6444 |
kg / L |
Electric power |
0.42 |
kg / kWh |
5-4. Machinery Management
Profit will be calculated simply as next equation.
B = NI – I Eq. 5-28
where,
symbol |
term |
unit |
B |
Profit |
$ |
NI |
Net income |
$ |
I |
Investment |
$ |
Example
If we invest to the combine shown in the following table, then the profit of n year will be obtained as follows.(Assume interest rate = 0)
term |
Investment: Price |
Economic Life |
Operating Cost |
Custom Charge |
Effective Field Capacity |
Net work hour |
Daily Capacity |
Available work days |
Coverage |
symbol |
I |
L |
OC |
CC |
EFC |
Dn |
DC |
AWD |
CA |
unit |
$ |
year |
$/ha |
$/ha |
ha/h |
h/d |
ha/d |
d |
ha/year |
data |
130,000 |
8 |
300 |
1,800 |
0.2 |
5.0 |
1.0 |
30 |
30.0 |
Net income of the year: NIi = (CC-OC)*CA
n |
year |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
NIi |
Net income |
45,000 |
45,000 |
45,000 |
45,000 |
45,000 |
45,000 |
45,000 |
45,000 |
NI: ($) |
Total Net income |
45,000 |
90,000 |
135,000 |
180,000 |
225,000 |
270,000 |
315,000 |
360,000 |
B=NI-I: ($) |
Profit |
-85,000 |
-40,000 |
5,000 |
50,000 |
95,000 |
140,000 |
185,000 |
230,000 |
See fm-541.xls: Ex-9
This table shows that the profit will be obtained after 3 years and total profit of 8 years later is $230,000, and rate of profit (B / I) is 1.77.
Recovery period will be obtained by next equation.
n = I / N1 Eq. 5-29
where,
symbol |
term |
unit |
Example |
n |
Recovery period |
year |
2.9 |
N1 |
Net income of one year |
$ |
45,000 |
I |
Investment |
$ |
130,000 |
This means we will get profit after 3 years use of the combine.
Machine employed in production may need to be replaced for one or more reasons.
1. A machine suffers accidental damage such that the cost of renewal is so great that a new machine is more economical.
2. The capacity of the existing machine is inadequate because of increased scale of production.
3. The machine is obsolete (see ASAE S495)
4. The machine is not expected to operate reliably. (Suffers considerable unanticipated downtime from random part failures.)
5. The cost of making an anticipated repair would increase the average unit accumulated cost above the expected minimum. Only capital costs and actual repair and maintenance costs need be accumulated.
For example, a $3000 machine is used 100 ha annually. It experiences the following end-of-year depreciation, interest (8% simple interest on average investment), and actual repair and maintenance costs in next table. Year 9 has the lowest unit cost and indicates the machine should be replaced with a similar machine at the end of year 9 if not before for other reasons. Inflation effects must be considered in making replacement decisions. Annual depreciation charges may be quite low or even negative in times of rapid inflation producing a premature minimum unit accumulated cost. In such instances replacement is better indicated by comparing the unit accumulated cost of the present machine with the projected costs for a potential successor machine. Optimum replacement time may be delayed beyond that time determined under more stable economic conditions. (See
ASAE-P496)Table 542. Average unit accumulated costs
year |
R&M costs |
Depr. |
Int. |
Tot. acc. Costs |
Acc. Use, ha |
Unit acc. Costs |
$ |
$ |
$ |
$ |
ha |
$/ha |
|
1 |
10 |
1000 |
200 |
1210 |
100 |
12.10 |
2 |
50 |
600 |
136 |
1996 |
200 |
9.98 |
3 |
70 |
400 |
96 |
2562 |
300 |
8.54 |
4 |
100 |
300 |
68 |
3030 |
400 |
7.58 |
5 |
200 |
200 |
48 |
3478 |
500 |
6.96 |
6 |
300 |
150 |
34 |
3962 |
600 |
6.60 |
7 |
350 |
125 |
23 |
4460 |
700 |
6.37 |
8 |
450 |
100 |
14 |
5024 |
800 |
6.28 |
9 |
550 |
25 |
9 |
5608 |
900 |
6.23 |
10 |
600 |
25 |
7 |
6240 |
1000 |
6.24 |
Total cost of several years for machinery is calculated as next equation , and machinery should be replaced at the year so that annual payment is minimum, which is called as economical life of devices.
AP(n) = { P +
Σ[(Rj + Qj) / (1+i)^j] - [Sn / (1+i)^n]} * {[i * (1+i)^n] / {(1+i)^n - 1} Eq. 5-30Where:
symbol |
term |
unit |
AP(n) |
Adjusted annual payments of worth after n year usage |
$ / year |
P |
Purchase price |
$ |
Rj |
Repairing cost in j year |
$ |
Qj |
Timely cost etc. in j year |
$ |
Sn |
Remaining value after n years |
$ |
i |
Annual interest |
in decimal |
Equation above 5-30 is induced from next equation.
AP(n) = P(n) + R(n) + Q(n) - S(n) Eq. 5-31
E = P * [(1+i)^n] Eq. 5-32
symbol |
term |
unit |
P(n) |
Annual present worth after n year |
$ / year |
R(n) |
Annual repairing cost after n year |
$ / year |
Q(n) |
Annual timeliness cost after n year |
$ / year |
S(n) |
Annual remaining value after n year |
$ / year |
E |
Final worth after n years |
$ |
P |
Present worth |
$ |
(1+i)^n |
Final worth factor or compound amount factor |
- |
1 / [(1+i)^n] |
Present worth factor |
- |
Table 542. Annual payment
: Example: Combine(P=5 M Yen, i =0.05)
Year |
P(n) |
Rn |
R(n) |
Remaining ratio |
Sn |
AP(n)* |
AP(n)** |
n |
k Yen |
k Yen |
k Yen |
% |
k Yen |
k Yen |
k Yen |
1 |
5,250 |
50 |
50 |
65 |
3,250 |
2,050 |
2,100 |
2 |
2,689 |
100 |
74 |
40 |
2,000 |
1,788 |
1,862 |
3 |
1,836 |
150 |
98 |
25 |
1,250 |
1,538 |
1,636 |
4 |
1,410 |
1,850 |
505 |
15 |
800 |
1,729 |
2,234 |
5 |
1,155 |
800 |
558 |
10 |
500 |
1,623 |
2,181 |
6 |
9,85 |
650 |
572 |
4 |
200 |
1,527 |
2,099 |
7 |
8,64 |
2,300 |
784 |
2 |
100 |
1,636 |
2,420 |
*: Assume Q(n) =0 **: Assume Q(n) = R(n)
In case of above table, it is recommendable to replace combine after 6 years or 3 years depending on evaluation of timeliness cost.
See Table A-542c. AHP: Example in replacement of tractor in Appendix.
5-5. Exercise
Exercise 5-1.
Obtain Maximum number of workers, Total Man-hours per ha, Minimum coverage, Annual cost per ha at farm scale of 10 ha, and Farm work cost index at farm scale 10, 30ha in next table.Assume, Sales per ha = 12,015 $/ha
No. |
Work |
M |
Nw |
MH |
AFC |
VCa |
CA |
AC-10ha |
AC-30ha |
AC-CA |
CI-10ha |
CI-30ha |
CI-CA |
- |
- |
h/ha |
$ |
$/ha |
ha |
$/ha |
$/ha |
$/ha |
- |
||||
1 |
Tillage |
1 |
1 |
3.5 |
0 |
79 |
93 |
79 |
79 |
79 |
0.7 |
0.7 |
0.7 |
2 |
Puddling |
1 |
1 |
3.3 |
1,350 |
74 |
42 |
119 |
282 |
2.3 |
|||
Nursery |
0 |
1,230 |
- |
||||||||||
3 |
Transplanting |
1 |
2 |
9.5 |
4,050 |
100 |
29 |
||||||
4 |
Caring crop |
1 |
1 |
12.5 |
324 |
153 |
13 |
185 |
164 |
203 |
1.5 |
1.4 |
1.7 |
5 |
Chemical application |
1 |
3 |
5.7 |
549 |
118 |
87 |
173 |
137 |
203 |
1.4 |
1.1 |
1.7 |
6 |
Harvest |
1 |
2 |
33.3 |
4,500 |
330 |
6.5 |
780 |
480 |
1026 |
6.5 |
4.0 |
8.5 |
7 |
Drying |
0 |
0 |
0.0 |
0 |
865 |
- |
865 |
865 |
865 |
7.2 |
7.2 |
7.2 |
8 |
Husking |
0 |
0 |
0.0 |
0 |
288 |
- |
288 |
288 |
288 |
2.4 |
2.4 |
2.4 |
9 |
Water manage |
0 |
1 |
0.0 |
0 |
269 |
- |
269 |
269 |
269 |
2.2 |
2.2 |
2.2 |
max |
sum |
sum |
sum |
min |
sum |
sum |
sum |
sum |
sum |
sum |
|||
Work system |
1 |
10,773 |
3,506 |
3,865 |
5,172 |
43 |
Where,
symbol |
term |
unit |
TOW |
Type of work: M= Machine, C= Contract, L= Manual |
- |
M, Nw |
No. of machine, workers |
- |
WC |
Work capacity |
h/ha |
MH |
Man-hours per ha |
h/ha |
AFC |
Annual fixed cost |
$ |
VCa |
Variable cost per ha |
$/ha |
CA |
Covered area |
ha |
AC-* |
Annual cost per ha at farm scale of * |
$/ha |
CI |
Cost index (x100): = Cost per ha/ Sales per ha |
- |
Exercise 5-2
. Obtain the maximum total profit and the maximum profit per ha of a farm work system give in next table.(Assume the original machinery set only available)
symbol |
term |
unit |
Example |
PRa-max |
Maximum Profit per ha: |
$/ha |
|
PSa |
Sales per ha |
$/ha |
12,015 |
ATCa-ca |
Cost per ha at area = coverage |
$/ha |
|
AFC |
Annual fixed cost |
$ |
10,773 |
CA |
Coverage |
ha |
10.1 |
VCa |
Effective Field Capacity |
$/ha |
3,438 |
PR-max |
Maximum Total Profit |
$ |
|
PS-ca |
Total Sales at area = coverage |
$ |
|
ATC-ca |
Total Cost at area = coverage |
$ |
PRa-max = PSa - ATCa-ca = PSa - [AFC/CA + VCa]
PR-max = PS-ca - ATC-ca = PS-ca - [AFC + VCa * CA]
Exercise 5-3. When annual operation area is larger than the coverage, we need to supply the additional machinery or worker, and machinery cost is calculated accordingly.
Msys = INT(Aa / CAS + 1)
FCa = AFCs * Msys / Aa
Where,
symbol |
term |
unit |
Example |
Msys |
Number of machinery set of system |
- |
|
Aa |
Annual farm work area |
ha |
|
INT |
Function of getting integer |
- |
- |
CAS |
Coverage of one set |
ha |
10.1 |
FCa |
Fixed cost per ha |
$/ha |
|
AFCs |
Annual fixed cost of one set |
$ |
4,500 |
Fill the blank columns of next table.
Annual farm work area |
No. of set |
Annual total fixed cost |
Fixed cost per ha |
Aa |
Msys |
AFC |
FCa |
(ha) |
- |
US$ |
($/ha) |
1 |
1 |
4,500 |
4,500 |
5 |
|||
10 |
|||
15 |
|||
20 |
|||
30 |
Exercise 5-4
. Obtain Annual Total fixed cost(AFC) of the farm work system from next table of each farm work. Examine cost per ha(ATCa) of them.Variable cost per ha(VCa)=3,439$/ha
Annual fixed cost of each farm work(AFC) |
No. of set |
Farm work system |
||||||||||||
Aa |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
AFC |
FCa |
VCa |
ATCa |
|
ha |
$ |
$ |
$/ha |
$/ha |
$/ha |
|||||||||
1 |
0 |
1,350 |
4,050 |
324 |
549 |
4,500 |
0 |
0 |
0 |
1 |
10,773 |
10,773 |
3,439 |
14,212 |
5 |
0 |
1,350 |
4,050 |
324 |
549 |
4,500 |
0 |
0 |
0 |
1 |
10,773 |
2,155 |
3,439 |
5,593 |
10 |
0 |
1,350 |
4,050 |
324 |
549 |
4,500 |
0 |
0 |
0 |
1 |
3,439 |
|||
15 |
0 |
1,350 |
4,050 |
648 |
549 |
9,000 |
0 |
0 |
0 |
2 |
3,439 |
|||
20 |
0 |
1,350 |
4,050 |
648 |
549 |
9,000 |
0 |
0 |
0 |
2 |
3,439 |
|||
25 |
0 |
1,350 |
4,050 |
648 |
549 |
13,500 |
0 |
0 |
0 |
3 |
3,439 |
|||
30 |
0 |
1,350 |
8,100 |
972 |
549 |
13,500 |
0 |
0 |
0 |
3 |
24,471 |
816 |
3,439 |
4,254 |
Exercise 5-5
. Obtain the break-even point of area, using next table.
symbol |
term |
unit |
Example |
PSa |
Sales per ha |
$/ha |
12,015 |
Abp |
Break-even point of area |
ha |
|
AFC |
Annual total fixed cost |
$ |
10,773 |
VCa |
Total variable cost per ha |
$/ha |
3,439 |
Abp = AFC / (PSa - VCa)
Exercise 5-6
. When the farm work period schedule is given as following table,obtain the coverage of these farm work.
ß ---------------------- S ---------------------------------------------à |
|||||||
Work-1 |
ß ------------------S1 ------------------à |
||||||
Work-2 |
ß -------------- S2 --------------à |
||||||
Work-3 |
ß ----------- S3 ---------à |
||||||
ß --------------- S12 --------------------------à |
|||||||
ß ---------------------------- S23 ---------------------à |
|||||||
ß --------- s1 -------à |
ß ----------- s2 ------------à |
ß --- s3 -à |
Where,
Symbol |
term |
Work Capacity: WC |
|
h |
h/ha |
||
S |
Total available working hour |
200 |
|
S1 |
Available working hour for work-1 |
140 |
WC1 = 8.0 |
S2 |
Available working hour for work-2 |
70 |
WC2 = 1.0 |
S3 |
Available working hour for work-3 |
56 |
WC3 = 1.0 |
S12 |
Available working hour for work-1 and 2 |
175 |
|
S23 |
Available working hour for work-2 and 3 |
91 |
WCp = WC1 + WC2 + WC3
si = S * WCi / WCp
CA = min [S / WCp, S1 / WC1, S2 / WC2, S3 / WC3, S12 / (WC1 + WC2), S23 / (WC2 + WC3)]
Exercise 5-7.
Obtain Annual cost at farm scale 1, 10, 20 ha and at coverage, Sales amount of product, Cost index at the farm scale, no. 2,3,4 in next table.
No. |
System |
Type |
Country |
AFC |
VCa |
CA |
AC-01ha |
AC-10ha |
AC-30ha |
AC-CA |
PSa |
$ |
$/ha |
ha |
$/ha |
$/ha |
$/ha |
$/ha |
$/ha |
||||
1 |
FS01-J |
TE |
J |
10,773 |
3,506 |
6.5 |
14,279 |
4,583 |
3,865 |
5,163 |
12,015 |
2 |
FS01-X |
TE |
J |
10,000 |
3,000 |
10 |
12,000 |
||||
3 |
FS01-Y |
TE |
J |
12,000 |
2,500 |
20 |
12,000 |
||||
4 |
FS01-Z |
TE |
J |
15,000 |
2,000 |
30 |
12,000 |
Where,
symbol |
term |
unit |
1 |
2 |
3 |
4 |
FS |
Farm scale of system |
ha |
6.5 |
10 |
20 |
30 |
Nw-max |
Number of workers available |
- |
3 |
3 |
3 |
3 |
TMH |
Total Man-hours per ha |
h/ha |
68 |
50 |
40 |
30 |
AFC |
Annual fixed cost |
$ |
||||
VCa |
Variable cost per ha |
$/ha |
||||
CA |
Covered area |
ha |
||||
AC-* |
Annual cost per ha at farm scale of * |
$/ha |
- |
- |
- |
- |
PSa |
Sales per ha |
$/ha |
||||
PRa |
Profit per ha |
$/ha |
6,852 |
|||
Abp |
Break-even point |
ha |
1.27 |
1.11 |
1.26 |
1.50 |
CI |
Cost index (x100): = Cost / Sales |
- |
43 |
Exercise 5-8
. We have farm work system data of our theme experiment as following table. What and how farm work should be improved for more coverage.
Exercise 5-9. Discuss the idea and plan for more economical farm work system in case of 10ha farm scale.
Farm work |
Daily Capacity |
No. of machine set |
Working days |
Rate of available day |
Available work days |
Coverage of one set |
|
DC |
M |
DWP |
ADR |
AWD |
CAS |
||
No. |
Name |
ha/d |
- |
d |
% |
d |
ha |
1 |
Tillage |
1.83 |
1 |
70 |
73 |
51.1 |
93.5 |
2 |
Puddling |
1.92 |
1 |
30 |
73 |
21.9 |
42.0 |
3 |
Transplanting |
1.34 |
1 |
30 |
73 |
21.9 |
29.4 |
4 |
Caring crop |
0.44 |
1 |
47 |
65 |
30.6 |
13.5 |
5 |
Chemical application |
2.92 |
1 |
47 |
65 |
30.6 |
89.5 |
6 |
Harvest |
0.33 |
1 |
47 |
65 |
30.6 |
10.1 |
7 |
Drying |
1 |
47 |
65 |
30.6 |
1000.0 |
Exercise 5-10
. Obtain the required capacity of machinery and the width of machine(Wt), when data are given in next table.
symbol |
term |
unit |
rotary tillage |
EFC |
Effective Field Capacity |
ha/h |
|
A |
Area |
ha |
15 |
AWD |
Available work days |
d |
12 |
Dn |
Net work hours per day |
h/d |
6 |
Wt |
Theoretical operation width |
m |
|
Vt |
Theoretical operation speed |
km/h |
2.0 |
ef |
Field efficiency |
in decimal |
0.75 |
EFC = A / (AWD * Dn)
Wt = 10 * EFC / (Vt * ef)
Exercise 5-11.
How many machinery sets are necessary to next farm.
symbol |
term |
unit |
example |
M |
Number of machine set |
- |
|
A |
Area |
ha |
50 |
CAS |
Coverage of one set |
ha |
13 |
Exercise 5-12.
: How many tractors and implements are required for the following farm work by 35PS tractor (Area: A = 50 ha and Available net working hour: ANWH=250 h)
Farm work |
Implement |
Work capacity: WC |
Number of operation: N |
Required number of implement |
h/ha |
times |
|||
Tillage |
Rotary |
5 |
1 |
|
Harrow |
Rotary |
4 |
2 |
|
Leveling |
Tooth harrow |
1 |
2 |
|
Seeding |
Grain drill |
3 |
1 |
|
Pressing |
Roller |
2 |
1 |
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