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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter : Transformers"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 1 : pg 66"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"number of primary winding turns are 1720.0\n",
"number of secondary winding turns are 172.0\n"
]
}
],
"source": [
"# Example 4.1;NUMBER OF TURNS\n",
"#calculate the number of turns\n",
"# given :\n",
"e1=2200.;#voltage in volts\n",
"f=50.;#frequency in Hz\n",
"e2=220.;#voltage in volts\n",
"fd=1.6;#magnetic field in Tesla\n",
"a=3600.;#area in mm**2\n",
"#calculations\n",
"n1=(e1/(4.44*f*fd*a*10**-6));#number of turns\n",
"n2=(e2/(4.44*f*fd*a*10**-6));#number of turns\n",
"#results\n",
"print \"number of primary winding turns are\",round(n1)\n",
"print \"number of secondary winding turns are\",round(n2)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 2 : pg 68"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"part (a)\n",
"iron loss current is, (A)= 0.182\n",
"magnetising component is, (A)= 0.572\n",
" part (b)\n",
"iron loss current is, (A)= 0.1\n",
"magnetising component is, (A)= 0.49\n"
]
}
],
"source": [
"# Example 4.2;\n",
"#calculate the components of no load currents,magnetising and working components of exciting current \n",
"from math import sqrt\n",
"# given \n",
"print \"part (a)\"\n",
"nlw=2000.;#no load input watts\n",
"pv=11000.;#primary voltage\n",
"#calculations and results\n",
"Iw=nlw/pv;#current in amperes\n",
"Io=0.6;#current in amperes\n",
"Imu=sqrt(Io**2-Iw**2);#current in amperes\n",
"print \"iron loss current is, (A)=\",round(Iw,3)\n",
"print \"magnetising component is, (A)=\",round(Imu,3)\n",
"pf=0.2;#power factpr\n",
"Io=0.5;#current in amperes\n",
"Iw=Io*(pf);#current in amperes\n",
"Imu=Io*sqrt(1-pf**2);#magnetising component in amperes\n",
"print \" part (b)\"\n",
"print \"iron loss current is, (A)=\",Iw\n",
"print \"magnetising component is, (A)=\",round(Imu,3)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 3 : pg 68"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"current is, (A)= 33.43\n"
]
}
],
"source": [
"# Example 4.3;current\n",
"#calculate the current\n",
"from math import sqrt, acos, cos\n",
"# given \n",
"pf1=0.866;#power factor\n",
"pf2=0.1736;#power factor\n",
"#calculations\n",
"ph1=acos(pf1);#phase angle in radians\n",
"ph2=acos(pf2);#phase angle in radians\n",
"ir=120.;#current in amperes\n",
"n2=110;#number of turns\n",
"n1=440.;#number of turns\n",
"i2d=(n2/n1)*ir;#current in amperes\n",
"io=5.;#current in amperes\n",
"aioi2=ph2-ph1;#change in angle in degree\n",
"i1=sqrt(io**2+i2d**2+(2*io*i2d*cos(aioi2)));#current in amperes\n",
"#results\n",
"print \"current is, (A)=\",round(i1,2)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 4 : pg 69"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"total core losses is,(W) = 2900.0\n"
]
}
],
"source": [
"# Example 4.4;core losses\n",
"#calculate the core losses\n",
"# given \n",
"f=50.;#frquency in Hz\n",
"hl=650.;#hystresis loss\n",
"edl=400.;#eddy current loss\n",
"#calculations\n",
"A=hl/f;#parameter\n",
"B=edl/f**2;#parameter\n",
"Ph=A*2*f;#loss in watts\n",
"Pe=B*(2*f)**2;#loss in watts\n",
"pt=Ph+Pe;#total loss in watts\n",
"#results\n",
"print \"total core losses is,(W) = \",pt\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 5 : pg 71"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"part (a)\n",
"full load efficiency at 0.8 pf is,(%)= 96.165\n",
"part (b)\n",
"percentage of full load on which efficiency will be maximum is,(%)= 95.603\n"
]
}
],
"source": [
"# Example 4.5;\n",
"#calculate the efficiency and load for maximum efficiency \n",
"from math import sqrt\n",
"# given \n",
"cl=125.;#copper losses\n",
"fcl=2**2*cl;#full load copper losses\n",
"il=457.;#iron losses\n",
"pf=0.8;#power factor\n",
"kba=30.;#loss\n",
"#calculations and results\n",
"print \"part (a)\"\n",
"fle=((kba*pf)/((kba*pf)+(fcl+il)*10**-3))*100;#full load efficiency in %\n",
"print \"full load efficiency at 0.8 pf is,(%)=\",round(fle,3)\n",
"lme=kba*sqrt(il/fcl);#variable\n",
"pfl=(lme/kba)*100;#percentage of full load on which efficiency will be maximum \n",
"print \"part (b)\"\n",
"print \"percentage of full load on which efficiency will be maximum is,(%)=\",round(pfl,3)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 6 : pg 73"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"all day efficiency is,(%)= 95.31\n"
]
}
],
"source": [
"# Example 4.6;\n",
"#calculate the all day efficiency \n",
"#given\n",
"ef=0.98;#efficiency in %\n",
"kva=15;#kVA\n",
"pf=1;#power factor\n",
"#calculations\n",
"op=kva*pf;#output power in kW\n",
"ip=op/ef;#input power in kW\n",
"loss=ip-op;#loss in kW\n",
"cl=(loss*10**3)/2;#copper loss in W\n",
"il=cl;#iron loss in W\n",
"t1=12;#time in hours\n",
"p1=2;#power in kW\n",
"pf1=0.5;#power factor\n",
"y1=(p1)/pf1;#kVA\n",
"il1=il*t1;#loss in Wh\n",
"cl1=cl*((y1)/kva)**2*t1;#copper loss in Wh\n",
"top1=p1*t1;#kWht1=12;#time in hours\n",
"t2=6;#time in hours\n",
"p2=12;#power in kW\n",
"pf2=0.8;#power factor\n",
"y2=(p2)/pf2;#kVA\n",
"il2=il*t2;#iron loss in Wh\n",
"cl2=cl*((y2/kva)**2)*t2;#copper loss in Wh\n",
"top2=p2*t2;#kWh\n",
"t3=6;#time in hours\n",
"il3=il*t3;#iron loss Wh\n",
"tol=top1+top2;#iron loss kWh\n",
"til=(il1+il2+il3)*10**-3;#total iron loss in kWh\n",
"tcl=(cl1+cl2)*10**-3;#total copper loss in kWh\n",
"ade=((tol)/(tol+til+tcl))*100;#efficiency in %\n",
"#results\n",
"print \"all day efficiency is,(%)=\",round(ade,2)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 7 : pg 75"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"iron losses is,(kW)= 2.32\n"
]
}
],
"source": [
"# Example 4.7;iron losses\n",
"#calculate the iron losses\n",
"from math import sqrt\n",
"#given\n",
"kva=200;#kVA\n",
"pf=0.8;#power factor\n",
"rflo=kva*pf;#kW\n",
"ef=0.96;#efficiency\n",
"#calculations\n",
"ip=rflo/ef;#kW\n",
"tl=ip-rflo;#kW\n",
"e2=800;#volts\n",
"e1=6600;#volts\n",
"n21=((e2/sqrt(3))/e1);#turn ratiom\n",
"r1=4;#ohms\n",
"r2=0.05;#ohms\n",
"roe=(r1)*n21**2+r2;#ohms\n",
"fli=((kva*10**3)/(sqrt(3)*e2));#amperes\n",
"fcl=3*fli**2*roe;#kW\n",
"il=tl-(fcl)*10**-3;#kW\n",
"#results\n",
"print \"iron losses is,(kW)=\",round(il,2)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 8 : pg 78"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"part (a) \n",
"equivalent resistance referred to the primary is,(Ohm)= 7.05\n",
"equivalent reactance referred to the primary is,(Ohm)= 11.2\n",
"equivalent resistance referred to the secondary is,(Ohm)= 0.017625\n",
"equivalent reactance referred to the secondary is,(Ohm)= 0.028\n",
"equivalent impedance referred to the primary is,(Ohm)= 13.234\n",
"equivalent impedance referred to the secondary is,(Ohm)= 0.033\n",
"part (b) \n",
"total copper losses considering individual resistance is,(W)= 910.382\n",
"total copper losses consdering equivalent resistance (for primary) is,(W)= 910.382\n",
"total copper losses consdering equivalent resistance (for secondary) is,(W)= 910.382\n"
]
}
],
"source": [
"# Example 4.8;\n",
"#calculate the resistance,reactances and impedances and copper losses\n",
"from math import sqrt \n",
"#given\n",
"r1=3.45;#ohms\n",
"r2=0.009;#ohms\n",
"x1=5.2;#ohms\n",
"x2=0.015;#ohms\n",
"kva=100.;#kVA\n",
"e1=8800.;#volts\n",
"e2=440.;#volts\n",
"#calculations\n",
"i1=(kva*10**3)/e1;#in amperes\n",
"i2=(kva*10**3)/e2;#in amperes\n",
"k=e2/e1;#transformation ratio\n",
"ro1=r1+(r2/k**2);#ohms\n",
"xo1=x1+(x2/k**2);#ohms\n",
"ro2=r2+(k**2*r1);#ohms\n",
"xo2=k**2*xo1;#ohms\n",
"zo1=sqrt(ro1**2+xo1**2);#ohms\n",
"zo2=sqrt(ro2**2+xo2**2);#ohms\n",
"tcl=i1**2*r1+i2**2*r2;#in watts\n",
"tcl1=i1**2*ro1;#in watts\n",
"tcl2=i2**2*ro2;#in watts\n",
"#results\n",
"print \"part (a) \"\n",
"print \"equivalent resistance referred to the primary is,(Ohm)=\",ro1\n",
"print \"equivalent reactance referred to the primary is,(Ohm)=\",xo1\n",
"print \"equivalent resistance referred to the secondary is,(Ohm)=\",ro2\n",
"print \"equivalent reactance referred to the secondary is,(Ohm)=\",xo2\n",
"print \"equivalent impedance referred to the primary is,(Ohm)=\",round(zo1,3)\n",
"print \"equivalent impedance referred to the secondary is,(Ohm)=\",round(zo2,3)\n",
"print \"part (b) \"\n",
"print \"total copper losses considering individual resistance is,(W)=\",round(tcl,3)\n",
"print \"total copper losses consdering equivalent resistance (for primary) is,(W)=\",round(tcl1,3)\n",
"print \"total copper losses consdering equivalent resistance (for secondary) is,(W)=\",round(tcl2,3)\n",
"#copper losses are calculated wrong in the textbook\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example 9 : pg 82"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" part (a)\n",
"magnetising component of no load current (Im) is,(A)= 0.87\n",
"working component of no load current (Iw) is,(A)= 0.5\n",
"resistance for primary side (Rm) is,(Ohm)= 400.0\n",
"reactance for primary ohms (Xm) is,(Ohm)= 230.94\n",
"impedence for primary side (X01) is,(Ohm)= 0.31\n",
"part (b)\n",
"percentage regulation on lagging load is,(%)= 3.55\n",
"percentage regulation on leading load is,(%)= -0.15\n",
"part (c)\n",
"efficiency at full load is,(%)= 94.53\n",
"efficiency at half load is,(%)= 92.96\n"
]
}
],
"source": [
"# Example 4.9;\n",
"#calculate the parameter of primary side ,regulation and efficiency\n",
"from math import sqrt\n",
"#given\n",
"po=100.;#watts\n",
"v1=200.;#volts\n",
"io=1;#amperes\n",
"#calculations and results\n",
"ocpf=po/(v1*io);#open circuit power factor\n",
"sinpf=sqrt(1-ocpf**2);#\n",
"im=io*sinpf;#in amperes\n",
"iw=io*ocpf;#current in amperes\n",
"rm=v1/iw;#ohms\n",
"xm=v1/im;#in ohms\n",
"vs=15.;#volts\n",
"ia=10.;#amperes\n",
"zo2=vs/ia;#in ohms\n",
"wa=85.;#watts\n",
"ro2=wa/(ia)**2;#ohms\n",
"e2=400.;#volts\n",
"e1=200.;#volts\n",
"k=e2/e1;#transformation ratio\n",
"zo1=zo2/k**2;#ohms\n",
"ro1=ro2/k**2;#ohms\n",
"xo1=sqrt(zo1**2-ro1**2);#ohms\n",
"print \" part (a)\"\n",
"print \"magnetising component of no load current (Im) is,(A)=\",round(im,2)\n",
"print \"working component of no load current (Iw) is,(A)=\",iw\n",
"print \"resistance for primary side (Rm) is,(Ohm)=\",round(rm,2)\n",
"print \"reactance for primary ohms (Xm) is,(Ohm)=\",round(xm,2)\n",
"print \"impedence for primary side (X01) is,(Ohm)=\",round(xo1,2)\n",
"print \"part (b)\"\n",
"kva=4000;#kVA\n",
"i2=kva/e2;#in amperes\n",
"xo2=sqrt(zo2**2-ro2**2);#ohms\n",
"pf=0.8;# power factor\n",
"vlag=i2*(ro2*pf+xo2*sqrt(1-pf**2));#in volts\n",
"prld=(vlag*po)/e2;#\n",
"vlag1=i2*(ro2*pf-xo2*sqrt(1-pf**2));#in volts\n",
"prld1=(vlag1*po)/e2;#\n",
"print \"percentage regulation on lagging load is,(%)=\",round(prld,2)\n",
"print \"percentage regulation on leading load is,(%)=\",round(prld1,2)\n",
"print \"part (c)\"\n",
"cl=85;#copper losses\n",
"nloss=100;#no load losses\n",
"fll=cl+nloss;#full load losses\n",
"pf=0.8;#power factor\n",
"flo=kva*pf;#efficiency \n",
"effl=flo/(flo+fll);#efficiency \n",
"hll=(1./2)**2*cl+nloss;#loss in watts\n",
"op=(1./2)*kva*pf;#ouput power in watts\n",
"efhl=op/(hll+op);#efficiency at half load\n",
"print \"efficiency at full load is,(%)=\",round(effl*100,2)\n",
"print \"efficiency at half load is,(%)=\", round(efhl*100,2)\n"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
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|