Assignment 1 Questions Answers

Last update: 05.04.2016 TKP3501 Agricultural Mechanization Assignment 1 Question 1: A combine harvester having 400 cm c...

0 downloads 223 Views 394KB Size
Last update: 05.04.2016

TKP3501 Agricultural Mechanization Assignment 1 Question 1: A combine harvester having 400 cm cutter bar is operated at an average speed of 10 km/hr to harvest paddy crop. What is the theoretical field capacity of the combine? If we assumed the field efficiency of the combine is 80 percent, what is the effective field efficiency? {4 marks} Answers: Given; w = 4 m,

V = 10 km/hr,

Field Efficiency for combine harvester = 80%,

Time to finish the task, t = 5 hrs

Calculations TFC = (v x w)/10 = (4 x10)/10 = 4 ha/hr AFC or EFC = (v x w x FE)/10 = (4 x10 x 0.8)/10 = 3.2 ha/hr (based on FE = 80%) However, only 10 % time is lost in turning, etc, thus actual FE = 100 – 10% = 90%

If 10% time is lost in turning, loading and unloading, calculate the harvested area in 5 hours, and the farm machinery index (FMI). {1.5 marks} The FE at 90% is more practical since previous value at FE=80% was based on assumption. Therefore, the actual field capacity (FC) = TFC x FE (actual) = 4 ha/hr x 0.9 = 3.6 ha/hr. However, if we were based on the assumption value, the FC = 4 ha/hr x 0.8 = 3.2 ha/hr. The different was about 0.4 ha/hr which is considered is quite significant different. :. Total harvested area in 5 hours = 3.6 ha/hr x 5 hr = 18 ha (actual at FE = 90%) For assumption of FE at 80% = 3.2 ha/hr x 5 hr = 16 ha (at FE = 80%)

FMI = total time – turning or reverse OR STOP TIME / total time = 5 hr or 300 hr – 10/100*300 / 300 = 90%

List down three factors contributed to the low efficiency during such operation. {0.5 marks} Low efficiency during operation Turning time, loading and unloading, speed,

Last update: 05.04.2016

What can be expected when the speed is reduced by half from the initial speed? Briefly explain, how you can improve the actual field efficiency? {2 marks}

Answers V reduce from 10 to 5 km/hr TFC = half, = 2 ha/hr, EFC = 1.6 ha/hr. Reduction by 50 % of the TFC and EFC.

Possible factors 1. 2. 3. 4. 5. 6. 7. 8.

Machine: breakdown, stop for loading and unloading, Operator: efficiency, effectiveness, human factors Terrain: undulating terrain Landscape/soil: soft soil Crop: type of crop, method of harvesting, the yield of the crop, Field pattern Speed of machine Soil texture

How to improve actual FE (Kecekapan ladang)? 1. 2. 3. 4.

Reduce the turning during such operation. Possible make a non-stop and follow the perimeter of the field Upkeep the machine before the operation to avoid any breakdown in the field. Standard of operation-The combine harvester required a regular cleaning after each operation. Pre-plan the harvesting operation to optimize the time and area coverage Unloading the yield on-the-go with the help of a tractor and trailer.

Last update: 05.04.2016

If we assume the field is perfectly square, calculate how many path of the combine made to harvest the crop? Please give a brief explanation. (Tips: consider the area and the width of the combine). {2 marks} (Bonus question) If we assume the field is perfectly square, calculate how many path of the combine made to harvest the crop? Please give a brief explanation. (Tips: consider the area and the width of the combine) Given:1 Area covered under 90% of FE = 18 ha, Width of combine = 4 m We know that 1 ha = 10,000 m2 thus equal to 180,000 m2, thus for a square field = √180,000 = 424.26 m 424.26 m 424.26 m

18 ha

Therefore, 424.26 m / 4 m of combine cutter width = 106.06 paths or ~106 paths For 16 ha = 100 paths The actual field may not exactly square thus, the number of path is depend on the shape of the field, the heading of the harvesting operation. The effective width also not always maintained at 4 m, and sometime during operation, it’s reduced less than 4 m wide.

Last update: 05.04.2016 Question #2: Refer to the excel data - 10 marks A data from yield monitoring was used to log the information about the machine performance during the harvesting operation. The time for each of the operation was automatically recorded using the GPS data set. Therefore, from the data provided, the operator will be able to know the machine efficiency and the field machine index for the purpose of improving the farm machinery management. Using the data provided in the excel sheet; I.

Calculate the simple statistic of the parameters in the table.

II.

What is the average between times in actual vs theoretical operation time and the average of the FMI for this machine?

Days

T1

T2

T3

T4

T5

T1-T4-T5

T1-T5

FMI

Δtime (T1T2)

1

3.97

2.24

1.74

0.11

0.88

2.98

3.09

0.96

1.73

2

4.53

2.84

1.69

0.12

0.96

3.45

3.57

0.97

1.69

3

2.09

1.08

1.01

0.04

0.48

1.57

1.61

0.98

1.01

4

4.3

2.34

1.97

0.09

1.05

3.16

3.25

0.97

1.96

5

3.57

1.81

1.76

0.07

0.87

2.63

2.7

0.97

1.76

6

3.89

1.92

1.97

0.07

0.85

2.97

3.04

0.98

1.97

7

4.04

2.2

1.84

0.07

1.06

2.91

2.98

0.98

1.84

8

6.19

3.25

2.94

0.1

1.84

4.25

4.35

0.98

2.94

9

8.92

5.17

3.75

0.21

2.54

6.17

6.38

0.97

3.75

10

6.06

3.16

2.9

0.36

1.66

4.04

4.4

0.92

2.9

11

9.33

5.23

4.1

0.37

2.98

5.98

6.35

0.94

4.1

12

1.22

0.42

0.8

0.01

0.6

0.61

0.62

0.98

0.8

13

10.03

7.28

2.75

0.33

1.86

7.84

8.17

0.96

2.75

14

5.84

4.44

1.4

0.16

1.09

4.59

4.75

0.97

1.4

15

5.62

4.08

1.54

0.44

1.83

3.35

3.79

0.88

1.54

16

7.61

5.3

2.31

0.18

2.19

5.24

5.42

0.97

2.31

17

6.54

4.21

2.32

0.16

1.89

4.49

4.65

0.97

2.33

18

7.17

4.7

2.48

0.16

2

5.01

5.17

0.97

2.47

19

6.55

3.93

2.62

0.13

1.95

4.47

4.6

0.97

2.62

20

9.86

3.82

6.04

0.16

4.8

4.9

5.06

0.97

6.04

Total

117

69

48

3

33

81

84

19

48

Min

1.22

0.42

0.80

0.01

0.48

0.61

0.62

0.88

0.80

Max

10.03

7.28

6.04

0.44

4.80

7.84

8.17

0.98

6.04

Avg.

5.87

3.47

2.40

0.17

1.67

4.03

4.20

0.96

2.40

Last update: 05.04.2016 III.

Plot the graph between the total turning time vs FMI for this machine. In the graph, state the formula of the best fit of regression line and

IV.

From the equation, what is the FMI index at turning time of 0.05, 0.2 and 0.5 second? Discuss your finding. Note: Print out the result table in case of hard copy submission.

Turning time, sec

FMI

0.05 0.1 0.15

0.975 0.974 0.971

Last update: 05.04.2016 0.2 0.25 0.3 0.35 0.4 0.45

0.965 0.957 0.945 0.931 0.915 0.895

0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0.873 0.848 0.821 0.790 0.757 0.722 0.683 0.642 0.598 0.552 0.503

How to submit? Hardcopy during the next class OR by email, before the class hour on 29th Nov 2016. The assignment must be typed and no handwriting.