{
"metadata": {
"name": "",
"signature": "sha256:af2e84ceb74c3bc3148f31759e2a692dba979671e7232787c666481af8571437"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 11 : SPILLWAYS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.1 pg : 538"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#Given\n",
"h = 1.2; \t\t\t\t#head of water\n",
"Cd = 2.2; \t\t\t\t#coefficient of discharge\n",
"rho = 1; \t\t\t\t#density of water\n",
"gamma_w = 9.81; \t\t\t\t#unit weigth of water\n",
"\n",
"q = Cd*h**1.5;\n",
"\n",
"#applying bernaulli's equation at u/s water surface at section A and B\n",
"#solving it by error and trial method we get\n",
"v1 = 13.7;v2 = 14.7;\n",
"d1 = 0.212;d2 = 0.197;\n",
"\n",
"F1 = gamma_w*d1**2*math.cos(math.radians(60))/2;\n",
"F2 = gamma_w*d2**2/2;\n",
"W = gamma_w*60*2*math.pi*3*((d1+d2)/2)/360;\n",
"Fx = rho*q*(v2-v1*math.cos(math.radians(60)))-F1/2+F2;\n",
"Fy = rho*q*(v1*math.sin(math.radians(60)))+F1*math.sin(math.radians(60))+W;\n",
"F = (Fx**2+Fy**2)**0.5;\n",
"F = round(F*100)/100;\n",
"\n",
"# Results\n",
"print \"Resultant force = %.2f kN/m.\"%(F);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Resultant force = 46.68 kN/m.\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.2 pg : 539"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\n",
"#Given\n",
"C = 2.4; \t\t\t\t#coefficient of discharge\n",
"H = 2; \t\t\t\t#head\n",
"L = 100; \t\t\t\t#length of spillway\n",
"wc = 8; \t\t\t\t#heigth of weir crest above bottom\n",
"g = 9.81; \t\t\t\t#acceleration due to gravity\n",
"\n",
"# Calculations\n",
"h = H+wc;\n",
"Q1 = C*L*H**(1.5); \t\t\t\t#neglecting approach velocity and end contractions\n",
"va = Q1/(h*L);\n",
"ha = va**2/(2*g);\n",
"Ha = ha+H;\n",
"Q = C*L*Ha**1.5;\n",
"Q = round(Q*10)/10;\n",
"\n",
"# Results\n",
"print \"discharge over oggy weir = %.2f cumecs.\"%(Q);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"discharge over oggy weir = 690.80 cumecs.\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.3 pg : 540"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#capacity of siphon\n",
"#head required in oggy spillway\n",
"#length of oggy weir required\n",
"\n",
"#Given\n",
"t = 6; \t\t\t\t#tail water elevation\n",
"h = 1; \t\t\t\t#heigth of siphon spillway\n",
"w = 4; \t\t\t\t#width of siphon spillway\n",
"hw = 1.5; \t\t\t\t#head water elevation\n",
"C = 0.6; \t\t\t\t#coefficient of discharge\n",
"Co = 2.25; \t\t\t\t#coefficient of discharge of oggy spillway\n",
"lo = 4; \t\t\t\t#length of oggy spillway\n",
"hc = 1.5; \t\t\t\t#head on weir crest\n",
"g = 9.81; \t\t\t\t#acceleration due to gravity\n",
"\n",
"# Calculations and Results\n",
"#part (a)\n",
"Q = C*h*w*(2*g*(t+hw))**0.5;\n",
"Q = round(Q*10)/10;\n",
"print \"capacity of siphon = %.2f cumecs.\"%(Q);\n",
"\n",
"#part (b)\n",
"h1 = (Q/(Co*lo))**(2./3);\n",
"h1 = round(h1*100)/100;\n",
"print \"head required in oggy spillway = %.2f m\"%(h1);\n",
"\n",
"#part (c)\n",
"L = Q/(Co*(hc)**1.5);\n",
"L = round(L*100)/100;\n",
"print \"length of oggy weir required = %.2f m.\"%(L);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"capacity of siphon = 29.10 cumecs.\n",
"head required in oggy spillway = 2.19 m\n",
"length of oggy weir required = 7.04 m.\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.4 pg : 540"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"\n",
"\t\t\t\t\n",
"#Given\n",
"rl = 435; \t\t\t\t#full reservior level\n",
"cl = 429.6; \t\t\t\t#level of centre of siphon\n",
"hfl = 435.85; \t\t\t\t#high flood level\n",
"hfd = 600; \t\t\t\t#high flood discharge\n",
"w = 4; \t\t\t\t#width of throat\n",
"h = 2; \t\t\t\t#heigth of throat\n",
"C = 0.65; \t\t\t\t#coefficient of discharge\n",
"g = 9.81; \t\t\t\t#acceleration due to gravity\n",
"\n",
"# Calculations\n",
"H = hfl-cl;\n",
"Q = C*w*h*(2*g*H)**0.5;\n",
"n = hfd/Q;\n",
"n = round(n*100)/100;\n",
"\n",
"# Results\n",
"print \" number of siphons units required = %.2f.hence provide 11 siphons units.\"%(n);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" number of siphons units required = 10.42.hence provide 11 siphons units.\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.5 pg : 541"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from numpy import arange,zeros\n",
"\n",
"#design oggy spillway for concrete gravity dam\n",
"\n",
"#Given\n",
"rbl = 250; \t\t\t\t#avarage river bed level\n",
"rlc = 350; \t\t\t\t#R.L of spillway crest\n",
"s = 0.75; \t\t\t\t#slope on downstream side\n",
"Q = 6500; \t\t\t\t#discharge\n",
"L = 5*9; \t\t\t\t#length of spillway\n",
"Cd = 2.2; \t\t\t\t#coefficient of discharge\n",
"t = 2; \t\t\t\t#thickness of each pier\n",
"\n",
"#step 1. computation of design head\n",
"H = (Q/(Cd*L))**(2./3);\n",
"P = rlc-rbl;\n",
"\n",
"#P/H = 6.15,which is<1.33;it is a high overflow spillway\n",
"\n",
"#H+P/H = 7.15>1.7; hence discharge coefficient is not affected by downstream apron interface\n",
"\n",
"Kp = 0.01;\n",
"Ka = 0.1;\n",
"N = 4;\n",
"He = 17.5; \t\t\t\t#assumed\n",
"Le = L-2*(N*Kp+Ka)*He;\n",
"He1 = (Q/(Cd*Le))**(2./3);\n",
"He1 = round(He1*100)/100;\n",
"#He1 is almost equal to He\n",
"print \"crest profile will be designed for Hd = %.2f m.\"%(He1);\n",
"\n",
"#step 2. determination of d/s profile\n",
"\n",
"#equating the slope of d/s side and derivative of profile equation suggested by WES\n",
"x = 27.03;\n",
"y = 0.04372*x**1.85;\n",
"print \"downstream profile:\";\n",
"x = arange(1,27)\n",
"y = zeros(26)\n",
"for i in range(26):\n",
" y[i] = 0.04372*x[i]**1.85;\n",
" y[i] = round(y[i]*1000)/1000;\n",
"\n",
"print \"x y\";\n",
"for i in range(26):\n",
" print \"%i %.2f\"%(x[i],y[i]);\n",
"\n",
"print \"27.03 19.48\";\n",
"#step 3. determination of u/s profile\n",
"# math.cosidering equation for vertical u/s face and Hd = 17.58\n",
"\n",
"print \"upstream profile:\";\n",
"x = [-0.5, -0.1, -1.5, -2.0, -3.0, -4.0, -4.75];\n",
"y = zeros(7)\n",
"for i in range(7):\n",
" if i==6:\n",
" y[i] = 0.0633*(x[i]+4.75)**1.85+2.2151-1.2643*(x[i]+4.75)**0.625;\n",
" y[i] = round(y[i]*1000)/1000;\n",
" continue\n",
" \n",
" y[i] = 0.0633*(x[i]+4.7466)**1.85+2.2151-1.2643*(x[i]+4.7466)**0.625;\n",
" y[i] = round(y[i]*1000)/1000;\n",
"\n",
"print \"x y\";\n",
"for i in range(7):\n",
" print \"%.2f %.2f\"%(x[i],y[i]);\n",
"\n",
"\n",
"#step 4.design of d/s bucket\n",
"\n",
"R = P/4;\n",
"print \"radius of bucket = %i m.\"%(R);\n",
"print \"bucket will subtend angle of 60 degree at the centre.\";\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"crest profile will be designed for Hd = 17.58 m.\n",
"downstream profile:\n",
"x y\n",
"1 0.04\n",
"2 0.16\n",
"3 0.33\n",
"4 0.57\n",
"5 0.86\n",
"6 1.20\n",
"7 1.60\n",
"8 2.05\n",
"9 2.55\n",
"10 3.10\n",
"11 3.69\n",
"12 4.34\n",
"13 5.03\n",
"14 5.77\n",
"15 6.55\n",
"16 7.38\n",
"17 8.26\n",
"18 9.18\n",
"19 10.15\n",
"20 11.16\n",
"21 12.21\n",
"22 13.31\n",
"23 14.45\n",
"24 15.63\n",
"25 16.86\n",
"26 18.13\n",
"27.03 19.48\n",
"upstream profile:\n",
"x y\n",
"-0.50 0.01\n",
"-0.10 -0.00\n",
"-1.50 0.14\n",
"-2.00 0.25\n",
"-3.00 0.60\n",
"-4.00 1.20\n",
"-4.75 2.21\n",
"radius of bucket = 25 m.\n",
"bucket will subtend angle of 60 degree at the centre.\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.6 pg : 562"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"\t\t\t\t#design length and depth of stilling bamath.sin\n",
"\t\t\t\t\n",
"#Given\n",
"q = 1; \t\t\t\t#discharge of spillway\n",
"Cd = 0.7; \t\t\t\t#coefficient of discharge\n",
"h1 = 10; \t\t\t\t#heigth of crest above downstream silting bamath.sin\n",
"g = 9.81; \t\t\t\t#acceleration due to gravity\n",
"Cv = 0.9; \t\t\t\t#coefficient of velocity\n",
"\n",
"# Calculations\n",
"h = (3*q/(2*Cd*(2*g)**0.5))**(2./3);\n",
"H = h1+h/2;\n",
"vt = (2*g*H)**0.5;\n",
"v1 = Cv*vt;\n",
"y1 = q/v1;\n",
"F1 = v1/(g*y1)**0.5;\n",
"\t\t\t\t#F>1, flow is super-critical\n",
"y2 = 1;\n",
"v2 = q/y2;\n",
"F2 = v2/(g*y2)**0.5; \t\t\t\t#<1\n",
"y2 = (y1/2)*((1+8*F1**2)**0.5-1);\n",
"de = y2-1;\n",
"le = 5*(y2-y1);\n",
"de = round(de*1000)/1000;\n",
"le = round(le*10)/10;\n",
"\n",
"# Results\n",
"print \"stilling bamath.sin should be depressed by %.2f m.\"%(de);\n",
"print \"length of stilling bamath.sin = %.2f m.\"%(le);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"stilling bamath.sin should be depressed by 0.58 m.\n",
"length of stilling bamath.sin = 7.50 m.\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.7 pg : 563"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"\n",
"\t\t\t\t\n",
"#Given\n",
"q = 7.83; \t\t\t\t#discharge through spillway\n",
"w = 12.5; \t\t\t\t#width of fall\n",
"d = 2.; \t\t\t\t#depth of water in downstream\n",
"g = 9.8;\n",
"\n",
"y1 = 0.5;\n",
"v1 = q/y1;\n",
"F1 = v1/(g*y1)**0.5;\n",
"\n",
"#F>1,flow is super-critical\n",
"\n",
"# Calculations\n",
"v2 = q/d;\n",
"F2 = v2/(g*d)**0.5;\n",
"y2 = (y1/2)*((1+8*F1**2)**0.5-1);\n",
"de = y2-d;\n",
"le = 5*(y2-y1);\n",
"de = round(de*100)/100;\n",
"le = round(le*10)/10;\n",
"\n",
"# Results\n",
"print \"stilling bamath.sin should be depressed by %.2f m.\"%(de);\n",
"print \"length of stilling bamath.sin = %.2f m.\"%(le); \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"stilling bamath.sin should be depressed by 2.76 m.\n",
"length of stilling bamath.sin = 21.30 m.\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.8 pg : 564"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"\n",
"\t\t\t\t\n",
"#Given\n",
"Ag = 5*2.5; \t\t\t\t#area of gate\n",
"miu = 0.25; \t\t\t\t#coefficient of friction\n",
"w = 0.5; \t\t\t\t#weigth of gate\n",
"h = 2; \t\t\t\t#head of water over crest\n",
"g = 9.81; \t\t\t\t#acceleration due to gravity\n",
"gamma_w = 1000; \t\t\t\t#unit weigth of water\n",
"\n",
"\n",
"# Calculations\n",
"m = w*g*1000;\n",
"F = gamma_w*Ag*h*h*g/10;\n",
"ff = miu*F;\n",
"tf = (m+ff)/1000;\n",
"\n",
"# Results\n",
"print \"force to be exerted to lift the gate = %.2f kN.\"%(tf);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"force to be exerted to lift the gate = 17.17 kN.\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.9 pg : 564"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"\n",
"\t\t\t\t\n",
"#Given\n",
"q = 19; \t\t\t\t#dischrge through spillway\n",
"E = 1; \t\t\t\t#energy loss\n",
"\n",
"\t\t\t\t#from energy loss equation;E = (y2-y1)**3/4y2y1; and solving it we get\n",
"\t\t\t\t#x = 0.5*(-1+(1+294.39*(x-1)**9/64*x**3))\n",
"\t\t\t\t#by trial and error method x = 2.806\n",
"x = 2.806;\n",
"y1 = 4*x/(x-1)**3;\n",
"y2 = x*y1;\n",
"y1 = round(y1*1000)/1000;\n",
"y2 = round(y2*1000)/1000;\n",
"print \"depth of flow at both end of jumps = %.2f m and %.2f m. respectively.\"%(y1,y2);\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"depth of flow at both end of jumps = 1.91 m and 5.35 m. respectively.\n"
]
}
],
"prompt_number": 11
}
],
"metadata": {}
}
]
}