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{
"metadata": {
"name": "",
"signature": "sha256:06bc45481e0c8d1fd696ec9d96ac784607ece32d90669bf112f17cc78a27982d"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 15 : IRRIGATION CHANNEL 2 DESIGN PROCEDURE"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.1 pg : 739"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"#design the distributory umath.sing Laecey theory\n",
"\t\t\t\t\n",
"#Given\n",
"f = 0.85; \t\t\t\t#silt factor\n",
"AR = 3600.; \t\t\t\t#area for rabi\n",
"AK = 1400.; \t\t\t\t#area for kharif\n",
"delta_r = 0.135; \t\t\t\t#kor depth for rabi\n",
"delta_k = 0.19; \t\t\t\t#kor depth for kharif\n",
"tr = 4.; \t\t\t\t#kor period for rabi\n",
"tk = 2.5; \t\t\t\t#kor period for kharif\n",
"\n",
"# Calculations\n",
"Du_r = 8.64*tr*7/delta_r; \t\t\t\t#duty for rabi\n",
"Du_k = 8.64*tk*7/delta_k; \t\t\t\t#duty for kharif\n",
"q_r = AR/Du_r; \t\t\t\t#discharge for rabi\n",
"q_k = AK/Du_k; \t\t\t\t#discharge for kharif\n",
"Q = q_r; \t\t\t\t#math.since q_r>q_k\n",
"V = (Q*f**2/144)**(1./6);\n",
"A = Q/V;\n",
"P = 4.75*(Q)**0.5;\n",
"D = (P-(P**2-6.944*A)**0.5)/3.472;\n",
"S = f**(5./3)/(3340*Q**(1./6));\n",
"P = round(P*100)/100;\n",
"D = round(D*100)/100;\n",
"\n",
"# Results\n",
"print \"Bed slope = %.2e.\"%(S);\n",
"print \"Perimeter of channel section = %.2f m.\"%(P);\n",
"print \"Depth of channel section = %.2f m.\"%(D);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Bed slope = 2.03e-04.\n",
"Perimeter of channel section = 6.73 m.\n",
"Depth of channel section = 0.81 m.\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.2 pg : 740"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"#design an irrigation channel in alluvial soil by Laecy's theory\n",
"\t\t\t\t\n",
"#Given\n",
"Q = 15.; \t\t\t\t#Full supply discharge\n",
"f = 1.; \t\t\t\t#silt factor\n",
"s = 1./2; \t\t\t\t#side slope of channel\n",
"\n",
"\n",
"# Calculations\n",
"#from Laecey regime channel (Fig.15.4(b)) B and D is obtained as;\n",
"B = 15.1;\n",
"D = 1.38;\n",
"\t\t\t\t#also from Fig.15.5 we get slope as\n",
"S = 0.19/1000;\n",
"\n",
"# Results\n",
"print \"Width of channel section = %.2f m.\"%(B);\n",
"print \"Depth of channel section = %.2f m.\"%(D);\n",
"print \"Bed slope = %.2e.\"%(S);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Width of channel section = 15.10 m.\n",
"Depth of channel section = 1.38 m.\n",
"Bed slope = 1.90e-04.\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.3 pg : 740"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from numpy import array,zeros,linspace,float64\n",
"\n",
"#design and prepare the longitudnal section;schedule of area statistics and channel dimension of irrigation channel\n",
"\t\t\t\t\n",
"#Given\n",
"dl = 157.7; \t\t\t\t#datum level\n",
"fsl = 157.; \t\t\t\t#full supply level of parent channel\n",
"bl = 156.; \t\t\t\t#bed level of parent channel\n",
"kor_r = 4.; \t\t\t\t#kor period of rabi\n",
"kor_k = 2.5; \t\t\t\t#kor period of kharif\n",
"kord_r = 13.4; \t\t\t\t#kor depth of rabi\n",
"kord_k = 19.; \t\t\t\t#kor depth of kharif\n",
"s = 0.5; \t\t\t\t#side slope\n",
"m = 1.; \t\t\t\t#critical velocity ratio\n",
"N = 0.0225; \t\t\t\t#Kutter n\n",
"qo_r = 8.64*7*kor_r*100/kord_r; \t\t\t\t#outlet discharge for rabi(calculation is wrong in book)\n",
"qo_k = 8.64*7*kor_k*100/kord_k; \t\t\t\t#outlet discharge for kharif(calculation is wrong in book)\n",
"ca = 16000.; \t\t\t\t#culturable commanded area\n",
"Ir = 0.3; \t\t\t\t#intensity of irrigation in rabi\n",
"Ik = 0.125; \t\t\t\t#intensity of irrigation in rabi\n",
"\n",
"# Calculations and Results\n",
"Ar = Ir*ca; \t\t\t\t#area under rabi\n",
"Ak = ca*Ik; \t\t\t\t#area under kharif\n",
"q_r = Ar/qo_r;\n",
"q_k = Ak/qo_k;\n",
"q_r = round(q_r*100)/100;\n",
"q_k = round(q_k*100)/100;\n",
"print \"discharge neede for rabi crop = %.2f cumecs.\"%(q_r);\n",
"print \"discharge neede for kharif crop = %.2f cumecs.\"%(q_k);\n",
"print \"outlet discharge factor adopted = %i hectares per cumecs.\"%(qo_r);\n",
"\t\t\t\t#at km 5\n",
"ca = 8000; \t\t\t\t#culturable area\n",
"Ar = Ir*ca; \t\t\t\t#area under rabi\n",
"q_r = Ar/qo_r;\n",
"l = 0.5 \t\t\t\t#total loss after 5 km\n",
"q = q_r+l; \t\t\t\t#total discharge\n",
"dq = 1.1*q; \t\t\t\t#desigm discharge\n",
"S = 1./4000; \t\t\t\t#slope\n",
"B = array([5.5, 4.9, 4.55]); \t\t\t\t#Bed width\n",
"D = array([0.73, 0.79, 0.84]); \t\t\t\t#water depth\n",
"Vo = array([0.448, 0.472, 0.488]); \t\t\t\t#critical velocity\n",
"A = zeros(3)\n",
"V = zeros(3)\n",
"m = zeros(3)\n",
"print \"Bed width water depth area velocity critical velocity C.V.R\";\n",
"for i in range(3):\n",
" A[i] = B[i]*D[i]+D[i]**2/2;\n",
" V[i] = dq/A[i];\n",
" m[i] = V[i]/Vo[i];\n",
" A[i] = round(A[i]*100)/100;\n",
" V[i] = round(V[i]*1000)/1000;\n",
" m[i] = round(m[i]*100)/100;\n",
" print \"%.2f %.2f %.2f %.2f %.2f %.2f\"%(B[i],D[i],A[i],V[i],Vo[i],m[i]);\n",
"\n",
"B = 4.55;\n",
"D = 0.84;\n",
"print \"hence take B = %.2f .; D = %.2f m.\"%(B,D);\n",
"\t\t\t\t#at km 4\n",
"q = round(q*100)/100;\n",
"print \"discharge at 5 km = %.2f cumecs.\"%(q);\n",
"ca = 10000; \t\t\t\t#culturable area\n",
"Ar = Ir*ca; \t\t\t\t#area under rabi\n",
"q_r = Ar/qo_r;\n",
"l = 0.5 \t\t\t\t#total loss below 5 km\n",
"P = B+D*5**0.5; \t\t\t\t#wetted perimeter\n",
"l1 = P*1000*2/1000000; \t\t\t\t#loss between 5 km and 4km\n",
"l2 = l1+l;\n",
"q = q_r+l2;\n",
"dq = 1.1*q;\n",
"q = round(q*1000)/1000;\n",
"print \"discharge at 4 km = %.2f cumecs\"%(q);\n",
"print \"other discharge are calculated and are tabulated as:\";\n",
"x = linspace(1,5,6)\n",
"A1 = array([4800, 4200, 3600, 3300, 3000, 2400],dtype=float64);\n",
"A2 = array([2000, 1750, 1500, 1375, 1250, 1000],dtype=float64);\n",
"S = array([22.5, 22.5, 22.5, 24, 24, 25]);\n",
"B = array([5.5, 5.2, 4.85, 4.7, 4.55, 4.55]);\n",
"D = array([1.04, 1.007, 0.975, 0.945, 0.915, 0.840]);\n",
"dq = array([3.56, 3.17, 2.8, 2.6, 2.4, 2.02]);\n",
"V = array([0.570, 0.555, 0.538, 0.530, 0.521, 0.484]);\n",
"m = array([1.015, 1, 1, 1, 1, 0.992]);\n",
"print \"Below km area to irrigate rabi area to irrigate kharif bed slope bed width water depth design discharge velocity C.V.R\";\n",
"for i in range(6):\n",
" print \"%8i %i %i %.2f %.2f %.2f\\\n",
" %.2f %.2f %.2f\"%(x[i],A1[i],A2[i],S[i],B[i],D[i],dq[i],V[i],m[i]);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"discharge neede for rabi crop = 2.66 cumecs.\n",
"discharge neede for kharif crop = 2.51 cumecs.\n",
"outlet discharge factor adopted = 1805 hectares per cumecs.\n",
"Bed width water depth area velocity critical velocity C.V.R\n",
"5.50 0.73 4.28 0.47 0.45 1.05\n",
"4.90 0.79 4.18 0.48 0.47 1.02\n",
"4.55 0.84 4.17 0.48 0.49 0.99\n",
"hence take B = 4.55 .; D = 0.84 m.\n",
"discharge at 5 km = 1.83 cumecs.\n",
"discharge at 4 km = 2.17 cumecs\n",
"other discharge are calculated and are tabulated as:\n",
"Below km area to irrigate rabi area to irrigate kharif bed slope bed width water depth design discharge velocity C.V.R\n",
" 1 4800 2000 22.50 5.50 1.04 3.56 0.57 1.01\n",
" 1 4200 1750 22.50 5.20 1.01 3.17 0.56 1.00\n",
" 2 3600 1500 22.50 4.85 0.97 2.80 0.54 1.00\n",
" 3 3300 1375 24.00 4.70 0.94 2.60 0.53 1.00\n",
" 4 3000 1250 24.00 4.55 0.92 2.40 0.52 1.00\n",
" 5 2400 1000 25.00 4.55 0.84 2.02 0.48 0.99\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.4 pg : 744"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from numpy import roots\n",
"\n",
"#Given\n",
"B = 5.; \t\t\t\t#bed width\n",
"t = 2.; \t\t\t\t#top width of banks\n",
"h = 2.92; \t\t\t\t#heigth of banks from bed\n",
"n = 1.5;\n",
"\n",
"#sectional area of digging = sectional area of two banks\n",
"#By+zy**2 = 2(h-y)+2n(h-y)**2\n",
"#substituting the values and on simplificatio we get\n",
"s = [1,-13.26,18.59]\n",
"y = roots(s)[1];\n",
"#from this we get y = 11.666556 and 1.5934436.\n",
"#taking y = 1.5934436;\n",
"y = round(y*10)/10;\n",
"\n",
"# Results\n",
"print \"economical depth of cutting = %.2f m.\"%(y);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"economical depth of cutting = 1.60 m.\n"
]
}
],
"prompt_number": 4
}
],
"metadata": {}
}
]
}
|
