{
"metadata": {
"name": "",
"signature": "sha256:855ede77f42d7d38570b25f9dfe540c158b87984c52d22c6171f4226086950d9"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 7: Supply Systems"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.1, Page Number: 145"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"V1 = 200 #initial volt of system(V)\n",
"V2 = 400 #final volt of the system(V)\n",
"\n",
"\n",
"#Calculation:\n",
"I1,I2,R1,R2 = symbols('I1 I2 R1 R2')\n",
"I2 = V1*I1/V2\n",
"W1 = 2*I1**2*2*R1 #Power loss in 200 V system\n",
"W2 = 2*I2**2*2*R2 #Power loss in 400 V system\n",
"# As power loss in the two cases is the same,\n",
"R22 = solve(W1/W2-1,R2)[0]\n",
"a = R22/R1 #ratio of cables' cross-section\n",
"v = 1/a #ratio of cables' volumes\n",
"s = (1-v)*100\n",
"\n",
"#Result:\n",
"print \"% Saving in feeder copper is \",s,\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"% Saving in feeder copper is 75 %\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.2, Page Number: 146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"#P = power upplied, W = power loss, \n",
"#I = current, V = voltage, R = resistance\n",
"\n",
"R,V = symbols('R V')\n",
"P1,I1,W1 = symbols('P1,I1,W1') #for 2-wire system\n",
"P2,I2,W2 = symbols('P2,I2,W2') #for 3-wire system\n",
"\n",
"#Calculation:\n",
"#Two-wire d.c. system:\n",
"P1 = V*I1\n",
"W1 = 2*I1**2*R\n",
"L1 = W1/P1*100 #% power loss\n",
"\n",
"#3-phase, 3-wire a.c. system:\n",
"P2 = math.sqrt(3)*V*I2\n",
"W2 = 3*I2**2*R\n",
"L2 = W2/P2*100 #% power loss\n",
"\n",
"I2 = solve(L1-L2,I2)[0]\n",
"#r2 = P2/P1\n",
"r2 = math.sqrt(3)*I2*V/(I1*V)\n",
"P1 = 2*P2\n",
"\n",
"#Result:\n",
"print '''Additional power which can be supplied at \n",
"unity p.f. by 3-phase, 3-wire a.c. system = 100%.'''"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Additional power which can be supplied at \n",
"unity p.f. by 3-phase, 3-wire a.c. system = 100%.\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.3, Page Number: 146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"#Let R be the resistance per conductor in each case\n",
"\n",
"R,V = symbols('R V')\n",
"P1,I1,W1 = symbols('P1,I1,W1') #for 3-wire dc system\n",
"P2,I2,W2 = symbols('P2,I2,W2') #for 4-wire ac system\n",
"\n",
"#Calculation:\n",
"\n",
"#3-wire d.c. system.:\n",
"#At balanced loads, implying no power loss in the neutral wire\n",
"P1 = 2*V*I1\n",
"W1 = 2*I1**2*R\n",
"L1 = W1/P1*100 #% power loss\n",
"\n",
"\n",
"#3-phase, 4-wire a.c. system.:\n",
"#At balanced loads, implying no power loss in the neutral wire\n",
"P2 = 3*V*I2\n",
"W2 = 3*I2**2*R\n",
"L2 = W2/P2*100 #% power loss\n",
"\n",
"I2 = solve(L1-L2,I2)[0]\n",
"r = P2/P1\n",
"#solving r, we get\n",
"P2 = 1.5*P1\n",
"\n",
"#Result:\n",
"print '''Extra power that can be supplied at unity power \n",
"factor by 3-phase, 4-wire a.c. system = 50%.'''"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Extra power that can be supplied at unity power \n",
"factor by 3-phase, 4-wire a.c. system = 50%.\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.4, Page Number: 148"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"R,V,phy = symbols('R V phy') #phy is power factor angle\n",
" #V is voltage between the conductors\n",
" #R is resistance per conductor \n",
"P1,I1,W1 = symbols('P1,I1,W1') #for Single phase 2-wire system\n",
"P2,I2,W2 = symbols('P2,I2,W2') #for 3-phase, 3-wire a.c. system\n",
"\n",
"\n",
"#Calculation:\n",
"#Single phase 2-wire system\n",
"P1 = 2*V*I1*math.cos(phy)\n",
"W1 = 2*I1**2*R\n",
"L1 = W1/P1*100 #% power loss\n",
"\n",
"#3-phase, 3-wire a.c. system. \n",
"P2 = math.sqrt(3)*V*I2*math.cos(phy)\n",
"W2 = 3*I2**2*R\n",
"L2 = W2/P2*100 #% power loss\n",
"\n",
"I22 = solve(L1-L2,I2)[0]\n",
"r = P1/(math.sqrt(3)*V*I22*math.cos(phy))\n",
"#As the power supplied by single phase, 2-wire is 200 kW,\n",
"P1 = 200\n",
"\n",
"#Result:\n",
"print \"Power supplied by 3-phase, 3-wire a.c. system is\",P1*r,\"kW\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Power supplied by 3-phase, 3-wire a.c. system is 400 kW\n"
]
}
],
"prompt_number": 38
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.5, Page Number: 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"MVA = 5 #power transmitted to load\n",
"n = 0.9 #efficiency\n",
"pf = 0.8 #power factor\n",
"V = 33 #receiving end voltage(V)\n",
"l = 50*10**3 #length of line(m)\n",
"s = 2.85*10**-8 #specific resistance of aluminium(ohm-m)\n",
"\n",
"#Calculation:\n",
"P = MVA*pf*10**6 #power transmitted(W)\n",
"W = 10*P/100 #line loss(W)\n",
"\n",
"\n",
"#(i) Single phase, 2-wire system:\n",
"I1 = MVA*10**3/V #Apparent power(A)\n",
"a1 = 2*I1**2*s*l/W #area of cross-section(m**2)\n",
"v1= 2*a1*l #volume of conductor required(m**3)\n",
"\n",
"#(ii) 3-phase, 3-wire system:\n",
"I2 = MVA*10**3/(math.sqrt(3)*V) #Apparent power(A)\n",
"a2 = 3*I2**2*s*l/W #area of cross-section(m**2)\n",
"v2 = 3*a2*l #volume of conductor required(m**3)\n",
"\n",
"#Result:\n",
"print \"(i) Single phase, 2-wire system, Vol. of conductor required is\",round(v1,2),\"m^3\"\n",
"print \"(ii)3-phase, 3-wire system, Vol. of conductor required\",round(v2,2),\"m^3\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) Single phase, 2-wire system, Vol. of conductor required is 16.36 m^3\n",
"(ii)3-phase, 3-wire system, Vol. of conductor required 12.27 m^3\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.6, Page Number: 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"V1 = 11*10**3 #Supply voltage(V)\n",
"pf = 0.8\n",
"R1 = 0.15 #Total resistance in ac system(ohm)\n",
"R2 = 0.05 #Total resistance in dc system(ohm)\n",
"r1 = 0.15 #voltage drop in ac system(%)\n",
"r2 = 0.25 #voltage drop in dc system(%)\n",
"\n",
"#Calculation:\n",
"#Single phase system:\n",
"I1 = r1*V1/R1 #line current(A)\n",
"P1 = V1*I1*pf/1000 #power received by consumer(kW)\n",
"\n",
"#D.C. two-wire system:\n",
"P2 = P1\n",
"V2 = (P2*10**3*R2/r2)**0.5 #supply voltage(V)\n",
"\n",
"#Result:\n",
"print \"The supply voltage is\",V2,\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The supply voltage is 4400.0 V\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.7, Page Number: 153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"I = 200 #current(A)\n",
"l = 1 #length of cable(km)\n",
"c1 = 5 #cost per unit(paise/kWh)\n",
"r1 = 10 #interrest & depreciation(%)\n",
"ro = 1.73*10**-6 #resistivity(ohm-m)\n",
"\n",
"#Calculation:\n",
"#If 'a' is the area os cross-section of the cnductor,\n",
"#cost of cable including installation = (20*a + 20) Rs./m.\n",
"\n",
"a = symbols('a')\n",
"R = ro*l*10**5/a\n",
"E = 2*I**2*R*8760/1000 #energy lost per annum\n",
"C = c1/100*E #cost of energy lost(Rs)\n",
"\n",
"#capital cost for 1 km length of the cable is = Rs. 20,000*a.\n",
"\n",
"V = r1/100*20000*a #Variable annual charge(Rs)\n",
"a1 = solve(C-V,a)[1]\n",
"\n",
"#Result:\n",
"print \"The most economical conductor size is\",round(a1,2),\"cm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The most economical conductor size is 1.74 cm**2\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.8, Page Number: 153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"P = 5 #line load(MW)]\n",
"V = 33 #kV\n",
"pf = 0.8 #power factor\n",
"c1 = 4 #paise/kWh\n",
"r1 = 10 #interest and depreciation(%)\n",
"ro = 10**-6 #ohm-cm\n",
"\n",
"\n",
"#Calculation:\n",
"a = symbols('a') #size of conductor(cm**2)\n",
"R = ro*l*10**5/a #ohm\n",
"I = P*10**3/(math.sqrt(3)*V*pf) #current (A)\n",
"E = 3*I**2*R*8760/1000 #kWh\n",
"C = c1/100*E #Annual cost of energy lost(R)\n",
"#The capital cost (variable) of the cable = 25000*a Rs/km\n",
"V = r1/100*25000*a #Rs\n",
"a1 = solve(C-V,a)[1]\n",
"\n",
"#Result:\n",
"print \"The most economical size of conductor is\",round(a1,2),\"cm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The most economical size of conductor is 0.71 cm**2\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.9, Page Number: 154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"I = 250 #current through the feeder(A)\n",
"c1 = 5 #Rs/kg\n",
"r1 = 10 #interest and depreciation(%)\n",
"c2 = 5 #paise/kWh\n",
"d = 8.93 #density of wire(gm/cm**3)\n",
"ro = 1.73*10**-8 #speciafic resistance(ohm-m)\n",
"\n",
"\n",
"#Calculation:\n",
"l = 1 #length of wire(1 m say)\n",
"a = symbols('a') #conductor size(m**2)\n",
"R = ro*l/a #ohm\n",
"E = 2*I**2*R*8760/1000 #kWh\n",
"C = c2/100*E #annual cost os energy lost(Rs)\n",
"m = 2*a*1*d*10**3 #Mass of 1 metre feeder\n",
"CC = c1*m #capital cost(Rs)\n",
"V = r1*CC/100 #Variable Annual charge(Rs)\n",
"a1 = solve(C-V)[1]\n",
"\n",
"#Result:\n",
"print \"The most economic size of conductor is\",round(a1*10000,2),\"cm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The most economic size of conductor is 3.26 cm**2\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 7.10, Page Number: 154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"l = 1 #line length(km)\n",
"V = 110 #supply voltage(kV)\n",
"\n",
"#Working hours(hrs) Load(MW) Power factor\n",
"t1 = 6; P1 = 20 ; pf1 = 0.8\n",
"t2 = 12; P2 = 5 ; pf2 = 0.8\n",
"t3 = 6; P3 = 6 ; pf3 = 0.8\n",
"\n",
"c1 = 6 #\n",
"n = 365 #no.of days used\n",
"r1 = 10 #interest and dep-reciation(%)\n",
"\n",
"#Calculation:\n",
"a = symbols(\"a\") #cross-section of conductor(cm**2)\n",
"R = 0.176/a #Resistance per km of each conductor(ohm)\n",
"#The load currents at various loads are :\n",
"#At 20 MW,\n",
"I1 = P1*1000/(math.sqrt(3)*V*pf1) #A\n",
"I2 = P2*1000/(math.sqrt(3)*V*pf2) #A\n",
"I3 = P3*1000/(math.sqrt(3)*V*pf3) #A\n",
"\n",
"E1= 3*R*1/1000*(I1**2*t1+I2**2*t2+I3**2*t3) #Energy loss per day in 3-phase line(kWh)\n",
"E = E1*n #energy lost per annum(kWh)\n",
"C = c1*E/100 #Annual cost of energy(Rs)\n",
"v = r1*6000*a/100 #variable annual charge(Rs/kWh)\n",
"a1 = solve(C-v,a)[1]\n",
"\n",
"\n",
"#Result:\n",
"print \"THe most economical size of the conductor is\",round(a1,2),\"cm**2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"THe most economical size of the conductor is 1.56 cm**2\n"
]
}
],
"prompt_number": 24
}
],
"metadata": {}
}
]
}