Case 1: two times a day for two hours

65 percentage has been chosen because it is the minimum water amount at which the crops can survive.

The water that comes out of the tank depends on how much it is necessary to irrigate the crops. In the best case, the crops are irrigated twice a day in the fresher hours, in the morning, and in the late evening. In the worst case, when the water income is only at 20%, the irrigation happens only three times a week. In the middle case, the water is given only once a day, in the morning. To simulate the water in and out of the tank has been considered a timing cycle of tank filling and emptying.

Considering irrigating almost one time a day, the cycle is daily, while in the worst case, the cycle is weekly. To design the water tank the transitory phase has not been considered. The transitory phase is the first time at which the tank is filled with water and enters the amount of water necessary to start the cycle. The volume that enters in the transitory moment is the initial volume.

So, the scenarios considered in the water tank design are:

- Case 1: the spring gives 13 m³/day, the required water amount for the maintenance of the gardens, and the water is distributed two times a day;

- Case 2: the spring gives 13 m³/day, the required water amount for the maintenance of the gardens, and the water is distributed one time a day;

- Case 3: the spring gives 20% of the required water amount equal to 2.6 m³/day and the water is distributed three times a week.

For each case has been calculated the minimum volume of compensation, the initial volume, and are being produced the graphs of the flow rate and the volume of water in and out of the tank

66

Figure 4.4: Flow rate in and out of the tank for Case 1

In the tank, the income water volume must be equal to the outcome water, indeed, the blue area of the income flow rate (13 m³/day) is equal to the orange area of the exiting water (3.25 m³/h*4). The flow rate can be expressed as the variation of the volume in time as in Formula 4.14:

𝑄(𝑡) = ^{𝑑𝑉(𝑡)}

𝑑𝑡 (4.14)

With

𝑄_𝑖𝑛(𝑡) = 𝑑𝑉_𝑖𝑛(𝑡)

𝑑𝑡 𝑎𝑛𝑑 𝑄_𝑜𝑢𝑡(𝑡) = 𝑑𝑉_𝑜𝑢𝑡(𝑡) 𝑑𝑡 Where:

- 𝑄𝑖𝑛 and 𝑉𝑖𝑛 are the flow rate and the volume of water which income in the tank;

- 𝑄𝑜𝑢𝑡 and 𝑉𝑜𝑢𝑡 are the flow rate and the volume of water which outcome from the tank;

The equation that represents the change of the volume inside the tank in time is Formula 4.15 where 𝑉₀is the initial volume, and the integral of the in and out flow rate is equal to the respective subtended area in Figure 4.4:

𝑉(𝑡) = 𝑉₀+ ∫ 𝑄(𝑡)𝑑𝑡 = 𝑉₀^𝑡 ₀+ ∫ 𝑄₀^𝑡 _𝑖𝑛(𝑡)𝑑𝑡 + ∫ 𝑄₀^𝑡 _𝑜𝑢𝑡(𝑡)𝑑𝑡 (4.15) If the considered period is the cycle duration, the t will be equal to 24: the difference between the volume in entrance and the exit will be 0 and the total volume will be equal to the initial volume 𝑉₀. To observe the volume variation in time, the extremes of the integral should be changed in the time

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50

0-1 1-2 2-3 3-4 4-5 5-6 6-7 7-8 8-9 9-10 10-11 11-12 12-13 13-14 14-15 15-16 16-17 17-18 18-19 19-20 20-21 21-22 22-23 23-24

Flow rate (m3/h)

Hour in a day Case 1 - Flow rate diagram

Flow rate in Flow rate out

67 interval inside the daily cycle that would be observed. At first, the initial volume was unknown and was not considered, so the Formula 4.15 became:

𝑉(𝑡) = ∫ 𝑄_𝑖𝑛(𝑡)𝑑𝑡

𝑡 0

+ ∫ 𝑄_𝑜𝑢𝑡(𝑡)𝑑𝑡

𝑡 0

The graphic solution of the equation is the volume diagram in Figure 4.5. The volume diagram has been obtained plotting the cumulative volume income and outcome in a daytime. When the water is required, the line of volume out is over the line of volume in. For that, to guarantee water availability, it is necessary an initial volume to start the water filling and leaving cycle. The change in volume is tabulated in Table 4.17.

Figure 4.5: Volume diagram, the change of water entering and leaving the tank Table 4.17: Change in volume tabulated values in Case 1.

Change in volume inside the tank: Case 1

Time Ve Cum

Ve Vu Cum

Vu

Cum Ve-Cum Vu

h m³ m³ m³ m³ m³

1 0.54 0.54 0 0 0.5417

2 0.54 1.08 0 0 1.0833

3 0.54 1.63 0 0 1.6250

4 0.54 2.17 0 0 2.1667

5 0.54 2.71 0 0 2.7083

6 0.54 3.25 0 0 3.2500

7 0.54 3.79 3.25 3.25 0.5417

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Flow rate (m3/h)

Hours in a day

Case 1 - Volume diagram

Volume in Volume out

68

8 0.54 4.33 3.25 6.5 -2.1667

9 0.54 4.88 0 6.5 -1.6250

10 0.54 5.42 0 6.5 -1.0833

11 0.54 5.96 0 6.5 -0.5417

12 0.54 6.50 0 6.5 0.0000

13 0.54 7.04 0 6.5 0.5417

14 0.54 7.58 0 6.5 1.0833

15 0.54 8.13 0 6.5 1.6250

16 0.54 8.67 0 6.5 2.1667

17 0.54 9.21 0 6.5 2.7083

18 0.54 9.75 0 6.5 3.2500

19 0.54 10.29 3.25 9.75 0.5417

20 0.54 10.83 3.25 13 -2.1667

21 0.54 11.38 0 13 -1.6250

22 0.54 11.92 0 13 -1.0833

23 0.54 12.46 0 13 -0.5417

24 0.54 13.00 0 13 0.0000

The minimum compensation volume has been calculated as the sum of the maximum deviation in negative and positive of the cumulative income and outcome volume difference. Graphically, the maximum difference between the line of the volume out and the line of volume in, is represented in Figure 4.6, by the red lines. It occurs at 6. a.m. and 6 p.m., as the maximum difference in positive between the in and out cumulative volume, and at 8 a.m. and 8 p.m., as the minimum difference in negative. These values are highlighted in Table 4.17.

Figure 4.6: The maximum deviations C¹ and C² between the volume aout and the volume in.

0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Flow rate (m3/h)

Hours in a day Case 1 - Volume diagram

Volume in Volume out

C¹

C²

C¹ C²

69 The formula for calculating the minimum volume is in Formula 4.16.

𝐶_𝑚𝑖𝑛 = 𝐶₁+ |𝐶₂ | = 3.25 𝑚³ + |−2.17 | 𝑚³= 5.42 𝑚³ (4.16) The initial required volume inside the tank is equal to the maximum deviation in negative, corresponding to |𝐶2 | = 2.17 𝑚³ .

Considering the initial volume, the line of volume out will stay under the line of volume in.

Graphically, it is obtained by moving up the income volume line of the initial volume 𝑉0, and it is represented in Figure 4.7. The dashed area is the variation of water volume inside the tank during the daytime. When the green line (𝑉₀ plus the income volume) and the orange line (the outcome volume) are in touch, it means that inside the tank the water level is zero.

Figure 4.7: Change in volume in the daytime in Case 1. The dashed area in green is the water volume variation inside the tank.

The results are resume in Table 4.18.

Table 4.18: Case 1: water distribution, initial volume V0 and minimum volume Cmin.

Times per day

Hours per time

Volume out per hour (m³)

V⁰ (m³)

C^min (m³)

2 2 3.25 2.17 5.42