Lampiran 4. Perhitungan nilai kerapatan massa, kerapatan partikel dan porositas tanah. BTKO = Berat tanah kering oven (massa tanah kering)

Lampiran 4. Perhitungan nilai kerapatan massa, kerapatan partikel dan porositas tanah Bagian saluran BTKO (gr) Volume Total (cm3) Volume Partikel (cm3) Bulk Density (gr/cm3) Particle Density (gr/cm3) Porositas (%) Dalam 217,11 192,325 120 1,13 2,71 58,30 Tepi kanan 193,11 192,325 125 1,004 2,57 60,93 Tepi kiri 207,11 192,325 120 1,07 2,58 58,52 Dimana:

BTKO = Berat tanah kering oven (massa tanah kering) Volume total = volume ring sample = π

= (3,14)(3,5 cm)2(5 cm) = 192,325 cm3

Dalam saluran

Kerapatan Massa (Bulk Density) Ms = 217,11 gr

Bd = Ms_Vt

= _192,325 217 ,11gr/cm3 _{= 1,13 gr/cm}3

Tepi kanan saluran

Kerapatan Massa (Bulk Density) Ms = 193,11 gr

Bd = Ms_Vt

Tepi kiri saluran Ms = 207,11 gr Bd = Ms_Vt

= _192,325 207 ,11 gr/cm3 _{= 1,07 gr/cm}3 Dalam saluran

Kerapatan Partikel (Particel density) Berat Tanah = 217,11 gr

Volume Tanah = 190 ml

Volume Air = 200 ml

Volume Air Tanah = 270 ml Pd = _{(volume tanah −volume pori )}berat tanah

Volume Ruang Pori = (volume air + volume tanah)- volume air tanah Volume Ruang Pori = (200ml+190ml) – 270 ml

= 120 ml Pd = _(200−130)217,11

= 2,71gr/cm3 Tepi kanan saluran

Kerapatan Partikel (Particel Density) Berat Tanah = 193,11 gr

Volume Tanah = 190 ml

Volume Air = 200 ml

Volume Air Tanah = 265 ml Pd = _{(volume tanah −volume pori )}berat tanah

Volume Ruang Pori = (volume air + volume tanah)- volume air tanah Volume Ruang Pori = (200ml+190ml) – 265 ml

= 125 ml Pd = _(200−125)193,11

= 2,57 gr/cm3 Tepi kiri saluran

Kerapatan Partikel (Particel Density) Berat Tanah = 207,11 gr

Volume Tanah = 200 ml

Volume Air = 200 ml

Volume Air Tanah = 280 ml Pd = _{(volume tanah −volume pori )}berat tanah

Volume Ruang Pori = (volume air + volume tanah)- volume air tanah Volume Ruang Pori = (200ml+200ml) – 280 ml

= 120 ml Pd = _(200−120)207,11 = 2,58 gr/cm3 Porositas Dalam saluran Porositas = (1- Bd Pd ) x 100% = (1- 1,13_2,71 ) x 100% = 58,30%

Tepi kanan saluran Porositas = (1- Bd

Pd ) x 100%

= (1- 1,004_2,57 ) x 100% = 60,93%

Tepi kiri saluran Porositas = (1- Bd

Pd ) x 100%

= (1- 1,07_2,58 ) x 100% = 58,52%

Lampiran 5. Perhitungan evaporasi

Hari ke - Penurunan tinggi air pada evapopan

(mm/hari) I 4 II 5 III 5 IV 5 V 5 VI 7 VII 6 Rata – rata 5,28 Penyelesaian : E = k x Ep K = 0,8 Ep = 4,71 mm/hari Jadi, E = k x Ep = 0,8 x 5,28 = 4, 224 mm/hari

Lampiran 6. Perhitungan debit dan koefisien rembesan Komponen kehilangan air Ulangan Volume (ml) Waktu (s) Debit (ml/s) Koefisien rembesan (mm/hari) Rataan koefisien (mm/hari) Perkolasi I 3220 60 53,67 8830,08 8925,12 II 3300 61 54,09 8899,2 III 3410 62 55 9046,08 Rembesan kanan I 1850 60 30,83 63123,84 64313,28 II 1930 61 31,63 64756,8 III 1970 62 31,77 65059,2 Rembesan kiri I 1700 60 28,33 58026,24 60062,4 II 1790 61 29,34 60082,56 III 1880 62 30,32 76705,92 Debit

1. Dasar saluran (perkolasi) Ulangan I Diketahui : V = 3220 ml t = 60 detik Penyelesaian : Q = V/t = 3220 𝑚𝑚𝑚𝑚 60 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 53,67 ml / detik Ulangan II Diketahui : V = 3300 ml t = 61 detik Penyelesaian : Q = V/ t = 3300 𝑚𝑚𝑚𝑚 61 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 54,09 ml / detik

Ulangan III Diketahui : V = 3410 ml t = 62 detik Penyelesaian : Q = V/ t = 3410 𝑚𝑚𝑚𝑚 62 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 55 ml / detik

2. Dinding kanan saluran Ulangan I Diketahui : V = 1850 ml t = 60 detik Penyelesaian : Q = V/ t = 1850 𝑚𝑚𝑚𝑚 60 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 30,83 ml / detik Ulangan II Diketahui : V = 1930 ml t = 61 detik Penyelesaian : Q = V/ t = 1930 𝑚𝑚𝑚𝑚 61𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 31,63 ml / detik

Ulangan III Diketahui : V = 1970 ml t = 62 detik Penyelesaian : Q = V/ t = 1970 𝑚𝑚𝑚𝑚 62 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 31,77 ml / detik

3. Dinding kiri saluran Ulangan I Diketahui : V = 1700 ml t = 60 detik Penyelesaian : Q = V/ t = 1700 𝑚𝑚𝑚𝑚 60 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 28,33 ml / detik Ulangan II Diketahui : V = 1790 ml t = 61 detik Penyelesaian : Q = V/ t = 1790 𝑚𝑚𝑚𝑚 61𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 29,34 ml / detik

Ulangan III Diketahui : V = 1880 ml t = 62 detik Penyelesaian : Q = V/ t = 1880 𝑚𝑚𝑚𝑚 62𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 30,32 ml / detik Koefisien rembesan

1. Dasar saluran (perkolasi) Ulangan I Diketahui : Q = 53, 67 ml / detik L = 20 cm h = H1 + L = 15 + 20 = 35 cm A = 20 cm x 150 cm Penyelesaian : k = 𝑞𝑞𝐿𝐿 ℎ𝐴𝐴 = 53,67 𝑑𝑑 10−6 𝑚𝑚3/ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 20 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 58,30 % 35 𝑑𝑑 10−2 _{𝑚𝑚 𝑑𝑑 20 𝑑𝑑 10}−2 _{𝑚𝑚 𝑑𝑑 150 𝑑𝑑 10}−2_𝑚𝑚 = 5,959 x 10-5 m / detik = 5,959 x 10-5 x 1000 x 86400 = 5148,576 mm/hari

Ulangan II Diketahui : Q = 54,09 ml / detik L = 20 cm h = H1 + L = 15 + 20 = 35 cm A = 20 cm x 150 cm Penyelesaian : k = 𝑞𝑞𝐿𝐿 ℎ𝐴𝐴 = 54,09 𝑑𝑑 10−6 𝑚𝑚3/ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 20 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 58,30 % 35 𝑑𝑑 10−2 _{𝑚𝑚 𝑑𝑑 20 𝑑𝑑 10}−2 _{𝑚𝑚 𝑑𝑑 150 𝑑𝑑 10}−2_𝑚𝑚 = 6,006 x 10-5 m / detik = 6,006 x 10-5 x 1000 x 86400 = 5189,184 mm/hari Ulangan III Diketahui : Q = 55 ml / detik L = 20 cm h = H1 + L = 15 + 20 = 35 cm A = 20 cm x 150 cm Penyelesaian : k = 𝑞𝑞𝐿𝐿 ℎ𝐴𝐴 = 55 𝑑𝑑 10−6 𝑚𝑚3/ 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 20 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 58,30 % 35 𝑑𝑑 10−2 _{𝑚𝑚 𝑑𝑑 20 𝑑𝑑 10}−2 _{𝑚𝑚 𝑑𝑑 150 𝑑𝑑 10}−2_𝑚𝑚

= 6,107 x 10-5 m / detik = 6,107 x 10-5 x 1000 x 86400 = 5276,448 mm/hari

2. Dinding/tebing kanan saluran Ulangan I

Diketahui : Q = 30,83 ml / detik

q = debit per satuan panjang dinding/tebing saluran = 30,83 𝑑𝑑 10−6𝑚𝑚3/𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 150 𝑑𝑑 10−2 _𝑚𝑚 = 2,055 x 10-5 m3/m.detik d = 40 cm H1 = 15 cm Penyelesaian : k = 𝑞𝑞2𝑑𝑑 𝐻𝐻12 = 2,055 𝑑𝑑 10−5 𝑚𝑚3/𝑚𝑚.𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 2 𝑑𝑑 40 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 60,93 % (15 𝑑𝑑 10−2₎2 _𝑚𝑚2 = 4,451 x 10-4 m / detik = 4,451 x 10-4 x 1000 x 86400 = 38456,64 mm/hari Ulangan II Diketahui : Q = 31,63 ml / detik

q = debit per satuan panjang dinding/tebing saluran = 31,63 𝑑𝑑 10−6𝑚𝑚3/𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘

150 𝑑𝑑 10−2 _𝑚𝑚

H1 = 15 cm Penyelesaian : k = 𝑞𝑞2𝑑𝑑 𝐻𝐻12 = 2,108 𝑑𝑑 10−5 𝑚𝑚3/𝑚𝑚.𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 2 𝑑𝑑 40 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 60,93 % (15 𝑑𝑑 10−2₎2 _𝑚𝑚2 = 4,566 x 10-4 m / detik = 4,566 x 10-4 x 1000 x 86400 = 39450,24 mm/hari Ulangan III Diketahui : Q = 31,77 ml / detik

q = debit per satuan panjang dinding/tebing saluran = 31,77 𝑑𝑑 10−6𝑚𝑚3/𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 150 𝑑𝑑 10−2 _𝑚𝑚 = 2,118 x 10-5 m3/m.detik d = 40 cm H1 = 15 cm Penyelesaian : k = 𝑞𝑞2𝑑𝑑 𝐻𝐻₁2 = 2,118 𝑑𝑑 10−5 𝑚𝑚3/ 𝑚𝑚.𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 2 𝑑𝑑 40 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 60,93 % (15 𝑑𝑑 10−2₎2 _𝑚𝑚2 = 4,588 x 10-4 m / detik = 4,588 x 10-4 x 1000 x 86400 = 39640,32 mm/hari

3. Dinding/tebing kiri saluran Ulangan I

q = debit per satuan panjang dinding/tebing saluran = 28,33 𝑑𝑑 10−6𝑚𝑚3/𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 150 𝑑𝑑 10−2 _𝑚𝑚 = 1,889 x 10-5 m3/m.detik d = 40 cm H1 = 15 cm Penyelesaian : k = 𝑞𝑞2𝑑𝑑 𝐻𝐻₁2 = 1,889 𝑑𝑑 10−5 𝑚𝑚3/𝑚𝑚.𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 2 𝑑𝑑 40 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 58,52 % (15 𝑑𝑑 10−2₎2 _𝑚𝑚2 = 3,930 x 10-4 m / detik = 3,930 x 10-4 x 1000 x 86400 = 33955,2 mm/hari Ulangan II Diketahui : Q = 29,34 ml / detik

q = debit per satuan panjang dinding/tebing saluran = 29,34 𝑑𝑑 10−6𝑚𝑚3/𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 150 𝑑𝑑 10−2 _𝑚𝑚 = 1,956 x 10-5 m3/m.detik d = 40 cm H1 = 15 cm Penyelesaian : k = 𝑞𝑞2𝑑𝑑 𝐻𝐻₁2 = 1,956 𝑑𝑑 10−5 𝑚𝑚3/ 𝑚𝑚.𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 2 𝑑𝑑 40 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 58,52 % (15 𝑑𝑑 10−2₎2 _𝑚𝑚2

= 4,069 x 10-4 x 1000 x 86400 = 35156,16 mm/hari

Ulangan III

Diketahui : Q = 37, 46 ml / detik

q = debit per satuan panjang dinding/tebing saluran = 30,32 𝑑𝑑 10−6𝑚𝑚3/𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 150 𝑑𝑑 10−2 _𝑚𝑚 = 2, 021 x 10-5 m3/m.detik d = 40 cm H1 = 15 cm Penyelesaian : k = 𝑞𝑞2𝑑𝑑 𝐻𝐻₁2 = 2,497𝑑𝑑 10−5 𝑚𝑚3/𝑚𝑚.𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 𝑑𝑑 2 𝑑𝑑 40 𝑑𝑑 10−2 𝑚𝑚 𝑑𝑑 58,52 % (15 𝑑𝑑 10−2₎2 _𝑚𝑚2 = 4,205 x 10-4 m / detik = 4,205 x 10-4 x 1000 x 86400 = 36331,2 mm/hari

Lampiran 7. Gambar

1. Pengisian tanah pada saluran

3. Tinggi air dalam saluran konstan 15 cm