{
"metadata": {
"name": "",
"signature": "sha256:e3b742f593657a54a0584c9ec4333c4c34458354568e4b8a1e97b9d029596f5b"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Chapter 20: Transformers
"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 1, page no. 317
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"N1 = 500;# primary turns\n",
"N2 = 3000;# secondary turns\n",
"V1 = 240;# in Volts\n",
"\n",
"#calculation:\n",
" #For an ideal transformer, voltage ratio = turns ratio\n",
"V2 = V1*N2/N1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n secondary voltage \",round(V2,2),\"V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" secondary voltage 1440.0 V"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 2, page no. 317
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"tr = 2/7;# turns ratio\n",
"V1 = 240;# in Volts\n",
"\n",
"#calculation:\n",
" #A turns ratio of 2:7 means that the transformer has 2 turns on the primary \n",
" #for every 7 turns on the secondary\n",
"V2 = V1/tr\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n output voltage \",round(V2,2),\" V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" output voltage 840.0 V"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 3, page no. 317
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"tr = 8/1;# turns ratio\n",
"I1 = 3;# in Amperes\n",
"V1 = 240;# in Volts\n",
"\n",
"#calculation:\n",
" #A turns ratio of 8:1 means that the transformer has 28 turns on the \n",
" #primary for every 1turns on the secondary\n",
"V2 = V1/tr\n",
" #secondary current\n",
"I2 = I1*tr\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n secondary voltage is \",round(V2,2),\" V and secondary current is \", round(I2,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" secondary voltage is 30.0 V and secondary current is 24.0 A"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 4, page no. 318
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V1 = 240;# in Volts\n",
"V2 = 12;# in Volts\n",
"P = 150;# in Watts\n",
"\n",
"#calculation:\n",
"I2 = P/V2\n",
" #A turns ratio = Vp/Vs\n",
"tr = V1/V2# turn ratio\n",
" # V1/V2 = I2/I1\n",
" #current taken from the supply\n",
"I1 = I2*V2/V1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n turn ratio is \",round(tr,2),\" and current taken from the supply is \",round(I1,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" turn ratio is 20.0 and current taken from the supply is 0.63 A"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 5, page no. 318
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"S = 5000;# in VA\n",
"tr = 10;# turn ratio\n",
"V1 = 2500;# in Volts\n",
"\n",
"#calculation:\n",
" #A turns ratio of 8:1 means that the transformer has 28 turns on the primary for every 1turns on the secondary\n",
"V2 = V1/tr\n",
" #transformer rating in volt-amperes = Vs*Is\n",
"I2 = S/V2\n",
" #Minimum value of load resistance\n",
"RL = V2/I2\n",
" # tr = I2/I1\n",
"I1 = I2/tr\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)full-load secondary current is \",round(I2,2),\" A\"\n",
"print \"\\n (b)minimum load resistance is \",round(RL,2),\" ohm\"\n",
"print \"\\n (c) primary current is \",round(I1,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)full-load secondary current is 20.0 A\n",
"\n",
" (b)minimum load resistance is 12.5 ohm\n",
"\n",
" (c) primary current is 2.0 A"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 6, page no. 319
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V1 = 2400;# in Volts\n",
"V2 = 400;# in Volts\n",
"I0 = 0.5;# in Amperes\n",
"Pc = 400;# in Watts\n",
"\n",
"#calculation:\n",
" #Core loss (i.e. iron loss) P = V1*I0*cos(phi0)\n",
"pf = Pc/(V1*I0)\n",
"phi0 = math.acos(pf)\n",
" #Magnetizing component\n",
"Im = I0*math.sin(phi0)\n",
" #Core loss component\n",
"Ic = I0*math.cos(phi0)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)magnetizing component is \",round(Im,3),\" A and Core loss component is \",round(Ic,3),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)magnetizing component is 0.471 A and Core loss component is 0.167 A"
]
}
],
"prompt_number": 7
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 7, page no. 320
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V = 240;# in Volts\n",
"I0 = 0.8;# in Amperes\n",
"P = 72;# in Watts\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
" #Power absorbed = total core loss, P = V*I0*cos(phi0)\n",
" #Ic = I0*cos(phi0)\n",
"Ic = P/V\n",
"pf = Ic/I0\n",
" #From the right-angled triangle in Figure 20.2(b) and using\n",
" #Pythagoras\u2019 theorem, \n",
"Im = (I0*I0 - Ic*Ic)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a) Core loss component is \",round( Ic,2),\" A\"\n",
"print \"\\n (b) power factor is \",round( pf,2),\"\"\n",
"print \"\\n (c)magnetizing component is \",round(Im,2),\" A\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a) Core loss component is 0.3 A\n",
"\n",
" (b) power factor is 0.37 \n",
"\n",
" (c)magnetizing component is 0.74 A"
]
}
],
"prompt_number": 8
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 8, page no. 321
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"S = 100000;# in VA\n",
"V1 = 4000;# in Volts\n",
"V2 = 200;# in Volts\n",
"N2 = 100;# sec turns\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
" #Transformer rating = V1*I1 = V2*I2\n",
" #primary current\n",
"I1 = S/V1\n",
" #secondary current\n",
"I2 = S/V2\n",
" #primary turns\n",
"N1 = N2*V1/V2\n",
" #maximum flux\n",
" #assuming E2 = V2\n",
"Phim = V2/(4.44*f*N2)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)primary current is \",round( I1,2),\" A and secondary current is \",round( I2,2),\" A\"\n",
"print \"\\n (b)number of primary turns is \",round( N1,2),\"\"\n",
"print \"\\n (c)maximum value of the flux is \",round(Phim*1000,2),\"mWb\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)primary current is 25.0 A and secondary current is 500.0 A\n",
"\n",
" (b)number of primary turns is 2000.0 \n",
"\n",
" (c)maximum value of the flux is 9.01 mWb"
]
}
],
"prompt_number": 2
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 9, page no. 322
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V1 = 250;# in Volts\n",
"A = 0.03;# in m2\n",
"N2 = 300;# sec turns\n",
"N1 = 25;# prim turns\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
" #e.m.f. E1 = 4.44*f*Phim*N1\n",
" #maximum flux density,\n",
"Phim = V1/(4.44*f*N1)\n",
" #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
" #maximum core flux density\n",
"Bm = Phim/A\n",
" #voltage induced in the secondary winding,\n",
"V2 = V1*N2/N1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)maximum core flux density \",round( Bm,2),\" T\"\n",
"print \"\\n (b)voltage induced in the secondary winding is \",round( V2,2),\" V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)maximum core flux density 1.5 T\n",
"\n",
" (b)voltage induced in the secondary winding is 3000.0 V"
]
}
],
"prompt_number": 10
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 10, page no. 323
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V1 = 500;# in Volts\n",
"V2 = 100;# in Volts\n",
"Bm = 1.5;# in Tesla\n",
"A = 0.005;# in m2\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
" #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
" #maximum core flux density\n",
"Phim = Bm*A\n",
" #e.m.f. E1 = 4.44*f*Phim*N1\n",
" #primary turns,\n",
"N1 = V1/(4.44*f*Phim)\n",
" #secondary turns,\n",
"N2 = V2*N1/V1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n no. of primary and secondary turns are \",round(N1,2),\" turns, and \",round(N2,2),\" turns respectively\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" no. of primary and secondary turns are 300.3 turns, and 60.06 turns respectively"
]
}
],
"prompt_number": 10
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 11, page no. 323
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"emfpt = 15;# in Volts\n",
"V1 = 4500;# in Volts\n",
"V2 = 225;# in Volts\n",
"Bm = 1.4;# in Tesla\n",
"f = 50;# in Hz\n",
"\n",
"#calculation:\n",
" #E.m.f. per turn, V1/N1 = V2/N2 = emfpt\n",
" #primary turns,\n",
"N1 = V1/emfpt\n",
" #secondary turns,\n",
"N2 = V2/emfpt\n",
" #e.m.f. E1 = 4.44*f*Phim*N1\n",
" #maximum flux density,\n",
"Phim = V1/(4.44*f*N1)\n",
" #Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core\n",
" #cross-sectional area\n",
"A = Phim/Bm\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)no. of primary and secondary turns are \", N1,\" turns, and \", N2,\" turns respectively\"\n",
"print \"\\n (b)cross-sectional area is \", round(A,4),\"m2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)no. of primary and secondary turns are 300.0 turns, and 15.0 turns respectively\n",
"\n",
" (b)cross-sectional area is 0.0483 m2"
]
}
],
"prompt_number": 11
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 12, page no. 324
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"N1 = 2000;# prim turns\n",
"N2 = 800;# sec turns\n",
"I0 = 5;# in Amperes\n",
"pf0 = 0.20;# power factor\n",
"I2 = 100;# in Amperes\n",
"pf2 = 0.85;# power factor\n",
"\n",
"#calculation:\n",
" #Let I01 be the component of the primary current which provides the restoring mmf. Then I01*N1 = I2*N2\n",
"I01 = I2*N2/N1\n",
" #If the power factor of the secondary is 0.85\n",
"phi2 = math.acos(pf2)\n",
" #If the power factor on no-load is 0.20,\n",
"phi0 = math.acos(pf0)\n",
"I1h = I0*math.cos(phi0) + I01*math.cos(phi2)\n",
"I1v = I0*math.sin(phi0) + I01*math.sin(phi2)\n",
" #Hence the magnitude of I1\n",
"I1 = (I1h*I1h + I1v*I1v)**0.5\n",
"pf1 = math.cos(math.atan(I1v/I1h))\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n Primary current is \", round(I1,2),\" A, and Power factor is \",round(pf1,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" Primary current is 43.58 A, and Power factor is 0.8"
]
}
],
"prompt_number": 12
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 13, page no. 328
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"N1 = 600;# prim turns\n",
"N2 = 150;# sec turns\n",
"R1 = 0.25;# in ohms\n",
"R2 = 0.01;# in ohms\n",
"X1 = 1.0;# in ohms\n",
"X2 = 0.04;# in ohms\n",
"\n",
"#calculation:\n",
"tr = N1/N2# turn ratio\n",
"vr = tr# voltage ratio = turn raio, vr = V1/V2\n",
" #equivalent resistance Re\n",
"Re = R1 + R2*(vr**2)\n",
" #equivalent reactance, Xe\n",
"Xe = X1 + X2*(vr**2)\n",
" #equivalent impedance, Ze\n",
"Ze = (Re*Re + Xe*Xe)**0.5\n",
" #cos(phie) = Re/Ze\n",
"pfe = Re/Ze\n",
"phie = math.acos(pfe)\n",
"phied = phie*180/math.pi# in \u00b0(deg)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the equivalent resistance referred to the primary winding is \",round( Re,2),\" ohm\"\n",
"print \"\\n (b)the equivalent reactance referred to the primary winding is \",round( Xe,2),\" ohm\"\n",
"print \"\\n (c)the equivalent impedance referred to the primary winding is \",round( Ze,2),\" ohm\"\n",
"print \"\\n (d)phase angle is \",round( phied,2),\"deg\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the equivalent resistance referred to the primary winding is 0.41 ohm\n",
"\n",
" (b)the equivalent reactance referred to the primary winding is 1.64 ohm\n",
"\n",
" (c)the equivalent impedance referred to the primary winding is 1.69 ohm\n",
"\n",
" (d)phase angle is 75.96 deg"
]
}
],
"prompt_number": 13
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 14, page no. 329
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V1 = 200;# in Volts\n",
"V2 = 400;# in Volts\n",
"V2L = 387.6;# in Volts\n",
"S = 5000;# in VA\n",
"\n",
"#calculation:\n",
" #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
"reg = (V2 - V2L)*100/V2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the regulation of the transformer is \",round(reg,2),\" percent \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the regulation of the transformer is 3.1 percent "
]
}
],
"prompt_number": 14
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 15, page no. 329
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"VnL = 240;# in Volts\n",
"reg = 2.5;# in percent\n",
"\n",
"#calculation:\n",
" #regulation =(No-load secondary voltage -\u0003 terminal voltage on load)*100/no-load secondary voltage in %\n",
"VL = VnL - reg*VnL/100\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the load voltage at which the mechanism operates is \",round(VL,2),\" V \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the load voltage at which the mechanism operates is 234.0 V "
]
}
],
"prompt_number": 15
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 16, page no. 331
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"S = 200000;# in VA\n",
"Pc = 1500;# in Watt\n",
"Pi = 1000;# in Watt\n",
"pf = 0.85;# power factor\n",
"\n",
"#calculation:\n",
" #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
" #Efficiency = 1 - losses/input power\n",
" #Full-load output power = V*I*pf\n",
"Po = S*pf\n",
" #Total losses\n",
"Pl = Pc + Pi\n",
" #Input power = output power + losses\n",
"PI = Po + Pl\n",
" #efficiency\n",
"eff = 1-(Pl/PI)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the transformer efficiency at full load is \",round(eff,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the transformer efficiency at full load is 0.99"
]
}
],
"prompt_number": 16
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 17, page no. 331
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"S = 200000;# in VA\n",
"Pc = 1500;# in Watt\n",
"Pi = 1000;# in Watt\n",
"pf = 0.85;# power factor\n",
"\n",
"#calculation:\n",
" #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
" #Efficiency = 1 - losses/input power\n",
" #Half full-load power output = V*I*pf/2\n",
"Po = S*pf/2\n",
" #Copper loss (or I*I*R loss) is proportional to current squared\n",
" #Hence the copper loss at half full-load is\n",
"Pch = Pc/(2*2)\n",
" #Iron loss = 1000 W (constant)\n",
" #Total losses\n",
"Pl = Pch + Pi\n",
" #Input power at half full-load = output power at half full-load + losses\n",
"PI = Po + Pl\n",
" #efficiency\n",
"eff = (1-(Pl/PI))*100\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the transformer efficiency at half full load is \",round(eff,2),\" percent\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the transformer efficiency at half full load is 98.41 percent"
]
}
],
"prompt_number": 17
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 18, page no. 332
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"S = 400000;# in VA\n",
"R1 = 0.5;# in Ohm\n",
"R2 = 0.001;# in Ohm\n",
"V1 = 5000;# in Volts\n",
"V2 = 320;# in Volts\n",
"Pi = 2500;# in Watt\n",
"pf = 0.85;# power factor\n",
"\n",
"#calculation:\n",
" #Rating = 400 kVA = V1*I1 = V2*I2\n",
" #Hence primary current\n",
"I1 = S/V1\n",
" #secondary current\n",
"I2 = S/V2\n",
" #Total copper loss = I1*I1*R1 + I2*I2*R2,\n",
"Pcf = I1*I1*R1 + I2*I2*R2\n",
" #On full load, total loss = copper loss + iron loss\n",
"Plf = Pcf + Pi\n",
" # full-load power output = V2*I2*pf\n",
"Pof = S*pf\n",
" #Input power at full-load = output power at full-load + losses\n",
"PIf = Pof + Plf\n",
" #Efficiency = output power/input power = (input power\u2014losses)/input power\n",
" #Efficiency = 1 - losses/input power\n",
"efff = (1-(Plf/PIf))*100\n",
"\n",
" #Half full-load power output = V*I*pf/2\n",
"Poh = S*pf/2\n",
" #Copper loss (or I*I*R loss) is proportional to current squared\n",
" #Hence the copper loss at half full-load is\n",
"Pch = Pcf/(2*2)\n",
" #Iron loss = 2500 W (constant)\n",
" #Total losses\n",
"Plh = Pch + Pi\n",
" #Input power at half full-load = output power at half full-load + losses\n",
"PIh = Poh + Plh\n",
" #efficiency\n",
"effh = (1-(Plh/PIh))*100\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the transformer efficiency at full load is \", round(efff,2),\" percent\"\n",
"print \"\\n (b)the transformer efficiency at half full load is \", round(effh,2),\" percent\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the transformer efficiency at full load is 97.91 percent\n",
"\n",
" (b)the transformer efficiency at half full load is 97.88 percent"
]
}
],
"prompt_number": 18
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 19, page no. 333
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"S = 500000;# in VA\n",
"Pcf = 4000;# in Watt\n",
"Pi = 2500;# in Watt\n",
"pf = 0.75;# power factor\n",
"\n",
"#calculation:\n",
" #Let x be the fraction of full load kVA at which the efficiency is a maximum.\n",
" #The corresponding total copper loss = (4 kW)*(x**2)\n",
" #At maximum efficiency, copper loss = iron loss Hence\n",
"x = (Pi/Pcf)**0.5\n",
" #Hence the output kVA at maximum efficiency\n",
"So = x*S\n",
" #Total loss at maximum efficiency\n",
"Pl = 2*Pi\n",
" #Output power\n",
"Po = So*pf\n",
" #Input power = output power + losses\n",
"PI = Po + Pl\n",
" #Efficiency = output power/input power = (input power\u00e2\u20ac\u201dlosses)/input power\n",
" #Efficiency = 1 - losses/input power\n",
" #Maximum efficiency\n",
"effm = (1 - Pl/PI)*100\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the output kVA at maximum efficiency is \",round(So/1000,2),\"kVA\"\n",
"print \"\\n max. efficiency is \",round(effm,2),\" pecent\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the output kVA at maximum efficiency is 395.28 kVA\n",
"\n",
" max. efficiency is 98.34 pecent"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 20, page no. 335
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"tr = 4;# turn ratio\n",
"RL = 100;# in Ohms\n",
"\n",
"#calculation:\n",
" #the equivalent input resistance,\n",
"Ri = RL*(tr**2)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the equivalent input resistance is \",round(Ri,2),\" ohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the equivalent input resistance is 1600.0 ohm"
]
}
],
"prompt_number": 20
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 21, page no. 335
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"R1 = 112;# in Ohms\n",
"RL = 7;# in Ohms\n",
"\n",
"#calculation:\n",
" #The equivalent input resistance, R1 of the transformer needs to be 112 ohm for maximum power transfer.\n",
" #R1 = RL*(tr**2)\n",
" # tr = N1/N2 turn ratio\n",
"tr = (R1/RL)**0.5\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the optimum turns ratio is \",tr,\": 1.0\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the optimum turns ratio is 4.0 : 1.0"
]
}
],
"prompt_number": 3
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 22, page no. 335
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"tr = 5;# turn ratio\n",
"R1 = 150;# in Ohms\n",
"\n",
"#calculation:\n",
" #The equivalent input resistance, R1 of the transformer needs to be 150 ohm for maximum power transfer.\n",
" #R1 = RL*(tr**2)\n",
"RL = R1/(tr**2)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the optimum value of load resistance is \",round(RL,2),\" ohm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the optimum value of load resistance is 6.0 ohm"
]
}
],
"prompt_number": 22
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 23, page no. 335
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V1 = 220;# in Volts\n",
"V2 = 1760;# in Volts\n",
"V = 220;# in Volts\n",
"RL = 1280;# in Ohms\n",
"R = 2;# in Ohms\n",
"\n",
"#calculation:\n",
" #Turns ratio, tr = N1/N2 = V1/V2\n",
"tr = V1/V2\n",
" #Equivalent input resistance of the transformer,\n",
" #R1 = RL*(tr**2)\n",
"R1 = RL*(tr**2)\n",
" #Total input resistance\n",
"Rin = R + R1\n",
" # Primary current\n",
"I1 = V1/Rin\n",
" #For an ideal transformer V1/V2 = I2/I1,\n",
"I2 = I1*tr\n",
" #Power dissipated in load resistor RL\n",
"P = I2*I2*RL\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a) primary current flowing is \",round(I1,2),\" A\"\n",
"print \"\\n (b) power dissipated in the load resistor is \",round(P,2),\" W\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a) primary current flowing is 10.0 A\n",
"\n",
" (b) power dissipated in the load resistor is 2000.0 W"
]
}
],
"prompt_number": 23
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 24, page no. 336
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"tr = 25;# teurn ratio\n",
"V = 24;# in Volts\n",
"R1 = 15000;# in Ohms\n",
"Rin = 15000;# in ohms\n",
"\n",
"#calculation:\n",
" #Turns ratio, tr = N1/N2 = V1/V2\n",
" #For maximum power transfer R1 needs to be equal to 15 kohm\n",
"RL = R1/(tr**2)\n",
" #The total input resistance when the source is connected to the matching transformer is\n",
"Rt = Rin + R1\n",
" #Primary current,\n",
"I1 = V/Rt\n",
" #N1/N2 = I2/I1\n",
"I2 = I1*tr\n",
" #Power dissipated in load resistor RL\n",
"P = I2*I2*RL\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a) the load resistance is \",round(RL,2),\"ohm\"\n",
"print \"\\n (b) power dissipated in the load resistor is \",round(P*1000,2),\"mW\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a) the load resistance is 24.0 ohm\n",
"\n",
" (b) power dissipated in the load resistor is 9.6 mW"
]
}
],
"prompt_number": 4
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 25, page no. 337
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V1 = 320;# in Volts\n",
"V2 = 250;# in Volts\n",
"S = 20000;# in VA\n",
"\n",
"#calculation:\n",
" #Rating = 20 kVA = V1*I1 = V2*I2\n",
" #Hence primary current, I1\n",
"I1 = S/V1\n",
" #secondary current, I2\n",
"I2 = S/V2\n",
" #Hence current in common part of the winding\n",
"I = I2 - I1\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n current in common part of the winding is \", round(I,2),\" A\"\n",
"print \"\\n primary current and secondary current are \",round(I1,2),\" A and \",round(I2,2),\" A respectively\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" current in common part of the winding is 17.5 A\n",
"\n",
" primary current and secondary current are 62.5 A and 80.0 A respectively"
]
}
],
"prompt_number": 25
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 26, page no. 339
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"V1a = 200;# in Volts\n",
"V2a = 150;# in Volts\n",
"V1b = 500;# in Volts\n",
"V2b = 100;# in Volts\n",
"\n",
"#calculation:\n",
" #For a 200 V:150 V transformer, xa\n",
"xa = V2a/V1a\n",
" #volume of copper in auto transformer\n",
"vca = (1 - xa)*100# of copper in a double-wound transformer\n",
" #the saving is\n",
"vsa = 100 - vca\n",
" #For a 500 V:100 V transformer, xb\n",
"xb = V2b/V1b\n",
" #volume of copper in auto transformer\n",
"vcb = (1 - xb)*100# of copper in a double-wound transformer\n",
" #the saving is\n",
"vsb = 100 - vcb\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)For a 200 V:150 V transformer, the saving is \", round(vsa,2),\" percent\"\n",
"print \"\\n (b)For a 500 V:100 V transformer, the saving is \", round(vsb,2),\" percent\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)For a 200 V:150 V transformer, the saving is 75.0 percent\n",
"\n",
" (b)For a 500 V:100 V transformer, the saving is 20.0 percent"
]
}
],
"prompt_number": 26
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 27, page no. 340
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"N1 = 500;# prim turns\n",
"N2 = 50;# sec turns\n",
"VL = 2400;# in Volts\n",
"\n",
"#calculation:\n",
" #For a star-connection, VL = Vp*(3**0.5)\n",
"VL1s = VL\n",
" #Primary phase voltage\n",
"Vp1s = VL1s/(3**0.5)\n",
" #For a delta-connection, VL = Vp\n",
" #N1/N2 = V1/V2, from which,\n",
" #secondary phase voltage, Vp2s\n",
"Vp2s = Vp1s*N2/N1\n",
"VL2d = Vp2s\n",
"\n",
" #For a delta-connection, VL = Vp\n",
"VL1d = VL\n",
" #primary phase voltage Vp1d\n",
"Vp1d = VL1d\n",
" #Secondary phase voltage, Vp2d\n",
"Vp2d = Vp1d*N2/N1\n",
" #For a star-connection, VL = Vp*(3**0.5)\n",
"VL2s = Vp2d*(3**0.5)\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n the secondary line voltage for star and delta connection are \",round(Vp2s,1),\" V \"\n",
"print \" and \",round(VL2s,0),\" V respectively\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" the secondary line voltage for star and delta connection are 138.6 V \n",
" and 416.0 V respectively\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Example 28, page no. 343
"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"import math\n",
"#initializing the variables:\n",
"N1 = 1;# prim turns\n",
"N2 = 60;# sec turns\n",
"I1 = 300;# in amperes\n",
"Ra = 0.15;# in ohms\n",
"R2 = 0.25;# in ohms\n",
"\n",
"#calculation:\n",
" #Reading on the ammeter,\n",
"I2 = I1*(N1/N2)\n",
" #P.d. across the ammeter = I2*RA, where RA is the ammeter resistance\n",
"pd = I2*Ra\n",
" #Total resistance of secondary circuit\n",
"Rt = Ra + R2\n",
" #Induced e.m.f. in secondary\n",
"V2 = I2*Rt\n",
" #Total load on secondary\n",
"S = V2*I2\n",
"\n",
"\n",
"#Results\n",
"print \"\\n\\n Result \\n\\n\"\n",
"print \"\\n (a)the reading on the ammeter is \",round(I2,2),\" A \"\n",
"print \"\\n (b)potential difference across the ammeter is \",round(pd,2),\" V \"\n",
"print \"\\n (c)total load (in VA) on the secondary is \",round(S,2),\" VA \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"\n",
" Result \n",
"\n",
"\n",
"\n",
" (a)the reading on the ammeter is 5.0 A \n",
"\n",
" (b)potential difference across the ammeter is 0.75 V \n",
"\n",
" (c)total load (in VA) on the secondary is 10.0 VA "
]
}
],
"prompt_number": 28
}
],
"metadata": {}
}
]
}