diff options
Diffstat (limited to 'Electrical_Machines_by_M._V._Despande')
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_1.ipynb | 327 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_11.ipynb | 688 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_12.ipynb | 377 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_13.ipynb | 345 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_14.ipynb | 338 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_15.ipynb | 206 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_16.ipynb | 144 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_17.ipynb | 68 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_2.ipynb | 481 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_3.ipynb | 160 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_4.ipynb | 149 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_7.ipynb | 495 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_8.ipynb | 393 | ||||
| -rwxr-xr-x | Electrical_Machines_by_M._V._Despande/Chapter_9.ipynb | 129 |
14 files changed, 0 insertions, 4300 deletions
diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_1.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_1.ipynb deleted file mode 100755 index 4dfd3fc0..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_1.ipynb +++ /dev/null @@ -1,327 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:864737033e4858048b12f5380ee773a5abe813ec712aa20dce406f30836cc082"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 1 - SINGLE-PHASE TRANSFORMERS: PRINCIPLE AND CONSTRUCTION"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 5"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the (a) n ratio (b) load current referred to high voltage side (c) load impedance on low voltage side for full load (d) and impedance referred to high voltage side\n",
- "#Exa:1.1\n",
- "import math \n",
- "P_s=25000.#Supplied power (in VA)\n",
- "V_1=1910.#Voltage on primary side (in volt)\n",
- "V_2=240.#Voltage on secondary side (in volt)\n",
- "f=50.#frequency in hertz\n",
- "n=V_1/V_2\n",
- "print '%s %.2f' %('(a)n-ratio=',n)\n",
- "I_1=P_s/V_1\n",
- "print '%s %.1f' %('(b)load current referred to high voltage side (in A)=',I_1)\n",
- "I_2=P_s/V_2\n",
- "Z_2=V_2/I_2\n",
- "print '%s %.1f' %('(c)load impedance on low voltage side for full load (in ohm)=',Z_2)\n",
- "Z_1=Z_2*(n**2)\n",
- "print '%s %.f' %('(d)impedance referred to high voltage side(in ohm)=',Z_1)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)n-ratio= 7.96\n",
- "(b)load current referred to high voltage side (in A)= 13.1\n",
- "(c)load impedance on low voltage side for full load (in ohm)= 2.3\n",
- "(d)impedance referred to high voltage side(in ohm)= 146\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 6"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a) Power factor on no load (b) active current (c) magnetising current (d) copper loss in the primary winding (e) core loss\n",
- "#Exa:1.2\n",
- "import math,cmath\n",
- "from math import acos,sin\n",
- "V_1=3300.#Primary voltage (in volt)\n",
- "V_2=240.#Secondary voltage (in volt)\n",
- "I_0=2.#No load current (in A)\n",
- "P=60.#Power (in watt)\n",
- "R=0.8#Resistance of the low voltage winding (in ohm)\n",
- "Pf=P/(V_2*I_0)\n",
- "print '%s %.3f' %('(a)Power factor on no load=',Pf)\n",
- "I_c=I_0*Pf\n",
- "print '%s %.3f' %('(b)Active current(in A)=',I_c)\n",
- "theta=(acos(Pf)*57.3)\n",
- "I_m=I_0*sin(theta)\n",
- "print '%s %.3f' %('(c)magnetising current is(in A)=',I_m) #text answer is wrong\n",
- "Culoss=(I_0**2)*R\n",
- "print '%s %.1f' %('(d)copper loss in the primary winding is(in watt)=',Culoss)\n",
- "Coreloss=P-Culoss\n",
- "print '%s %.1f' %('(e)core loss(in watt)=',Coreloss)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Power factor on no load= 0.125\n",
- "(b)Active current(in A)= 0.250\n",
- "(c)magnetising current is(in A)= 1.821\n",
- "(d)copper loss in the primary winding is(in watt)= 3.2\n",
- "(e)core loss(in watt)= 56.8\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E3 - Pg 8"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find number of turns per limb on the (a) high voltage and (b) low voltage sides\n",
- "#Exa:1.3\n",
- "import math \n",
- "A=0.0386#cross sectional area of core(in m**2)\n",
- "B=1.#maximum flux density (in weber/m**2)\n",
- "f=50.#frequency (in hertz)\n",
- "V_1=3300.#voltage on primary side (in volt)\n",
- "V_2=240.#voltage on secondary side (in volt)\n",
- "C=B*A\n",
- "n_2=V_2/(4.44*C*f)\n",
- "T_2=n_2/2.\n",
- "print '%s %.f' %('(a)number of turns on low voltage side is=',T_2)\n",
- "T_1=T_2*V_1/V_2\n",
- "print '%s %.f' %('(b)number of turns on high voltage side is=',T_1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)number of turns on low voltage side is= 14\n",
- "(b)number of turns on high voltage side is= 193\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E4 - Pg 13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Calculate (a) equivalent resistance and reactance of low voltage side in terms of high voltage side (b) equivalent resistance and reactance of high voltage side in terms of low voltage side (c) total resistance and reactance of transformer in terms of high voltage side (d) total resistance and reactance of transformer in terms of low voltage side \n",
- "#Exa:1.4\n",
- "import math \n",
- "V_1=2200.#Primary side voltage(in volt)\n",
- "V_2=220.#secondary side voltage(in volt)\n",
- "f=50.#frequency(in hertz)\n",
- "r_1=1.25#Primary side resistance(in ohm)\n",
- "x_1=4.#Primary side reactance(in ohm)\n",
- "r_2=0.04#Secondary side resistance(in ohm)\n",
- "x_2=0.15#Secondary side reactance(in ohm)\n",
- "n=V_1/V_2\n",
- "R_2=(n**2.)*r_2\n",
- "print '%s %.f' %('(a)equivalent resistance of low voltage side in terms of high voltage side=',R_2)\n",
- "X_2=(n**2)*x_2\n",
- "print '%s %.f' %('\\nequivalent reactance of low voltage side in terms of high voltage side=',X_2)\n",
- "R_1=r_1/(n**2)\n",
- "print '%s %.4f' %('\\n(b)equivalent resistance of high voltage side in terms of low voltage side =',R_1)\n",
- "X_1=x_1/(n**2)\n",
- "print '%s %.2f' %('\\nequivalent reactance of high voltage side in terms of low voltage side =',X_1)\n",
- "R_t=r_1+R_2\n",
- "print '%s %.2f' %('\\n(c)total resistance of transformer in terms of high voltage side=',R_t)\n",
- "X_t=x_1+X_2\n",
- "print '%s %.f' %('\\ntotal reactance of transformer in terms of high voltage side=',X_t)\n",
- "R_e=r_2+R_1\n",
- "print '%s %.4f' %('\\n(d)total resistance of transformer in terms of low voltage side=',R_e)\n",
- "X_e=x_2+X_1\n",
- "print '%s %.2f' %('\\ntotal reactance of transformer in terms of low voltage side=',X_e)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)equivalent resistance of low voltage side in terms of high voltage side= 4\n",
- "\n",
- "equivalent reactance of low voltage side in terms of high voltage side= 15\n",
- "\n",
- "(b)equivalent resistance of high voltage side in terms of low voltage side = 0.0125\n",
- "\n",
- "equivalent reactance of high voltage side in terms of low voltage side = 0.04\n",
- "\n",
- "(c)total resistance of transformer in terms of high voltage side= 5.25\n",
- "\n",
- "total reactance of transformer in terms of high voltage side= 19\n",
- "\n",
- "(d)total resistance of transformer in terms of low voltage side= 0.0525\n",
- "\n",
- "total reactance of transformer in terms of low voltage side= 0.19\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E5 - Pg 14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a) terminal voltage on load (b) voltage on load at high voltage terminals (c) efficiency of transformer\n",
- "#Exa:1.5 \n",
- "import math,cmath \n",
- "n=10.#ratio of high voltage to low voltage\n",
- "V_1=200.#Voltage on low voltage side(in volt)\n",
- "x_m=231.#Magnetising resistance(in ohms)\n",
- "r_c=400.#Core loss resistance(in ohms)\n",
- "r_e=0.1#Equivalent resistance referred to low voltage side(in ohms)\n",
- "x_e=0.5#Equivalent reactance referred to low voltage side(in ohms)\n",
- "r_l=7.9#Load resistance(in ohms)\n",
- "x_l=5.5#Load reactance(in ohms)\n",
- "I_m=V_1/x_m\n",
- "I_c=V_1/r_c\n",
- "I_0=I_c+(1j*I_m)\n",
- "R_l=r_l+r_e\n",
- "X_l=x_l+x_e\n",
- "I=V_1/(R_l+(1j*X_l))\n",
- "I_1=I+I_0\n",
- "V_2=V_1-I*(r_e+(1j*x_e))\n",
- "v=193.;#math.sqrt(V_2*conj(V_2))\n",
- "print '%s %.1f' %('(a)terminal voltage on load(in volt)=',v)\n",
- "V=v*n\n",
- "print '%s %.2f' %('(b)voltage on load at hgih voltage terminals(in volt)=',V)\n",
- "P_o=v*I.real\n",
- "P_i=V_1*I_1.real\n",
- "eff=(P_o/P_i)*100\n",
- "print '%s %.1f' %('(c)efficiency of transformer is(in %)=',eff)\n",
- "#answers are rounded off due to calculation error "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)terminal voltage on load(in volt)= 193.0\n",
- "(b)voltage on load at hgih voltage terminals(in volt)= 1930.00\n",
- "(c)efficiency of transformer is(in %)= 93.6\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E6 - Pg 15"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a) impedance (b) % resistance and reactance\n",
- "#Exa:1.6\n",
- "import math\n",
- "P_s=500000.#Power supplied(in VA)\n",
- "V_1=2200.#Voltage on primary side(in volt)\n",
- "V_2=500.#Voltage on secondary side(in volt)\n",
- "f=50.#frequency(in hertz)\n",
- "r=0.01#Resistance of transformer(in ohms)\n",
- "z=0.1#impedance of transformer(in %)\n",
- "I=P_s/V_2\n",
- "Z=z*V_2/I\n",
- "print '%s %.2f' %('(a)Impedance(in ohms)=',Z)\n",
- "R=(I*r/V_2)*100\n",
- "print '%s %.f' %('(b) Resistance(in %)=',R)\n",
- "x=math.sqrt(Z**2-r**2)\n",
- "X=(x*I/V_2)*100\n",
- "print '%s %.1f' %('Reactance(in %)=',X)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Impedance(in ohms)= 0.05\n",
- "(b) Resistance(in %)= 2\n",
- "Reactance(in %)= 9.8\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_11.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_11.ipynb deleted file mode 100755 index 3dc20568..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_11.ipynb +++ /dev/null @@ -1,688 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:5e134845e769c2130f3443709b1fd511a00a2fa7ed994543b6a235ec502411af"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 11 - THREE PHASE INDUCTION MOTORS:PRINCIPLES AND CHARACTERISTICS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find (a)Number of poles and (b)% slip\n",
- "#Exa:11.1\n",
- "import math \n",
- "f=50.#Frequency(in hertz)\n",
- "n=960.#Speed of induction motor on full load(in r.p.m)\n",
- "n_s=1000.#Synchronous speed(in r.p.m)\n",
- "p=(f*120.)/(n_s)\n",
- "print '%s %.f' %('(a)Number of poles is=',p)\n",
- "s=n_s-n\n",
- "S=(s/n_s)*100.\n",
- "print '%s %.f' %('(b)Slip is(in%)=',S)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Number of poles is= 6\n",
- "(b)Slip is(in%)= 4\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find (a)Speed of motor (b)%Slip\n",
- "#Exa:11.2\n",
- "p=6.#Number of poles\n",
- "f_s=50.#Stator frequency(in c/s)\n",
- "f_r=2.#Rotor frequency(in c/s)\n",
- "n_s=(120.*f_s)/p\n",
- "n=(f_r*120.)/p\n",
- "s=n_s-n\n",
- "print '%s %.f' %('Speed of motor(in r.p.m)=',s)\n",
- "S=(n/n_s)*100\n",
- "print '%s %.f' %('Slip(in %)=',S)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Speed of motor(in r.p.m)= 960\n",
- "Slip(in %)= 4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E3 - Pg 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate (a)Number of poles (b)Slip (c)Slip for full load torque if total resistance in rotor circuit is doubled\n",
- "#Exa:11.3\n",
- "n=970.#Speed of induction motor(in r.p.m)\n",
- "f=50.#Frequency(in hertz)\n",
- "n_s=1000.#Synchronous speed(in r.p.m)\n",
- "p=(f*120.)/n_s\n",
- "print '%s %.f' %('(a)Number of poles=',p)\n",
- "s=((n_s-n)/n_s)*100\n",
- "print '%s %.f' %('(b)Slip(in%)=',s)\n",
- "S=((s/100)*2)*100\n",
- "print '%s %.f' %('(c)Required slip(in%)=',S)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Number of poles= 6\n",
- "(b)Slip(in%)= 3\n",
- "(c)Required slip(in%)= 6\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E4 - Pg 234"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate (a)Mechanical power output (b)Torque (c)Maximum Torque (d)Speed at maximum torque (e)Power output when torque is maximum\n",
- "#Exa:11.4\n",
- "import math \n",
- "p=4.#Number of poles\n",
- "V=440.#Voltage of motor(in volts)\n",
- "f=50.#Frequency(in hertz)\n",
- "n_s=1500.#Synchronous speed(r.p.m)\n",
- "sp=1440.#Speed of motor at load(in r.p.m)\n",
- "s=4.#Slip at full load(in %)\n",
- "t=1.8#Stator to rotor turns ratio\n",
- "R_r=0.1#Resistance of rotor per phase(in ohms)\n",
- "X_r=0.8#Reactance of rotor per phase at standstill(in ohms)\n",
- "r_r=R_r*(t**2.)#Rotor resistance referred to stator(in ohms)\n",
- "x_r=X_r*(t**2.)#Reactance of rotor at stanstill referred to stator(in ohms)\n",
- "E=V/(math.sqrt(3.))\n",
- "P=((s/100.)*(E**2.)*r_r)/((r_r**2.)+((s/100.)**2.)*(x_r**2.))\n",
- "T=(3.*P)/(2.*(math.pi)*(n_s/60.))\n",
- "P_M=(3.*P*sp)/n_s\n",
- "print '%s %.f' %('(a)Mechanical power output(in watt)=',P_M)\n",
- "print '%s %.f' %('(b)Torque(in N-m)=',T)\n",
- "s_m=R_r/X_r\n",
- "N=n_s*(1-s_m)\n",
- "P_1=((s_m)*(E**2)*(r_r))/((r_r**2.)+((s_m**2)*(x_r**2)))\n",
- "T_m=(3.*P_1)/(2.*(math.pi)*(n_s/60.))\n",
- "print '%s %.1f' %('(c)Speed at maximum torque(in r.p.m)=',N)\n",
- "print '%s %.f' %('(d)Maximum torque(in N-m)=',T_m)\n",
- "P_o=(3*P_1*N)/n_s\n",
- "print '%s %.f' %('(e)Output power at maximum torque(in watt)=',P_o)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Mechanical power output(in watt)= 20814\n",
- "(b)Torque(in N-m)= 138\n",
- "(c)Speed at maximum torque(in r.p.m)= 1312.5\n",
- "(d)Maximum torque(in N-m)= 238\n",
- "(e)Output power at maximum torque(in watt)= 32677\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E5 - Pg 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Speed of the motor (b)Speed at which torque will be maximum (c)Ratio of maximum to full load torque\n",
- "#Exa:11.5\n",
- "import math\n",
- "V=3300.#Voltage supplied to induction motor(in volts)\n",
- "p=10.#Number of poles\n",
- "f=50.#frequency(in hertz)\n",
- "R_r=0.015#Rotor resistance per phase(in ohms)\n",
- "X_r=0.25#Standstill reactance per phase(in ohms)\n",
- "s=2.5#Slip(in %)\n",
- "n_s=(f*120.)/p\n",
- "n=n_s*(1.-(s/100.))\n",
- "print '%s %.f' %('(a)Speed of the motor(in r.p.m)=',n)\n",
- "S=R_r/X_r\n",
- "N=n_s*(1-S)\n",
- "print '%s %.f' %('(b)Speed at which torque will be maximum(in r.p.m)=',N)\n",
- "T_f=(s/100.)*R_r/((R_r**2.)+(((s/100.)**2.)*(X_r**2.)))\n",
- "T_m=S*R_r/((R_r**2.)+((S**2.)*(X_r**2.)))\n",
- "#R=T_m/T_f\n",
- "R=2.316\n",
- "print '%s %.3f' %('(c)Ratio of maximum to full load torque=',R)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Speed of the motor(in r.p.m)= 585\n",
- "(b)Speed at which torque will be maximum(in r.p.m)= 564\n",
- "(c)Ratio of maximum to full load torque= 2.316\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E6 - Pg 236"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate (a)Speed at which mechanical power from rotor will be maximum (b)Maximum power\n",
- "#Exa:11.6\n",
- "import math\n",
- "p=4.#Number of poles\n",
- "f=50.#Frequency(in hertz)\n",
- "V=440.#Supplied voltage to induction motor(in volts)\n",
- "R_r=0.1#Rotor resistance per phase(in ohm)\n",
- "X_r=0.8#Rotor reactance at standstill per phase(in ohm)\n",
- "t=1.3#Ratio of stator turns per phase to rotor turns per phase\n",
- "a=R_r/X_r\n",
- "s=(-(a**2.))+math.sqrt(1.+(a**2.))\n",
- "n_s=120.*f/p \n",
- "#N=n_s*(1.-s)\n",
- "N=1334.5\n",
- "print '%s %.1f' %('(a)Required speed(in r.p.m)=',N)\n",
- "r=R_r*t\n",
- "x=X_r*t\n",
- "E=V/math.sqrt(3.)\n",
- "#P_m=(3.*s*(E**2.)*r*(1.-s))/((r**2.)+((s**2.)+(x**2.)))\n",
- "P_m=62.72\n",
- "print '%s %.2f' %('(b)Maximum power(in kwatts)=',P_m)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Required speed(in r.p.m)= 1334.5\n",
- "(b)Maximum power(in kwatts)= 62.72\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E7 - Pg 236"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find Current per phase in the rotor (a)when rotor is at standstill and star connected impedance of 4.1+%i2 per phase is connected in series with rotor (b)when rotor runs at 3% slip with short circuit at the slip rings\n",
- "#Exa:11.7\n",
- "import math,cmath \n",
- "V=69.28#Induced e.m.f(in volts)\n",
- "r=0.9#Resistance of rotor per phase(in ohm)\n",
- "x=6.#Standstill rectance of rotor per phase(in ohm)\n",
- "z=4.1+(1j*2)\n",
- "s=3#Slip(in%)\n",
- "V_r=V/math.sqrt(3.)\n",
- "R_r=r+z.real\n",
- "X_r=(1j*2.)+(1j*x)\n",
- "Z=R_r+X_r\n",
- "I_r=V_r/Z\n",
- "print ('(a)Current when rotor is at standstill=',I_r)\n",
- "E=(s/100.)*V_r\n",
- "Imp=r+(1j*(s/100.)*x)\n",
- "i_r=E/Imp\n",
- "print ('(b)Current when rotor runs at 3% slip=',i_r)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "('(a)Current when rotor is at standstill=', (2.2471250926661392-3.595400148265823j))\n",
- "('(b)Current when rotor runs at 3% slip=', (1.2820136746620923-0.25640273493241844j))\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E8 - Pg 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)% reduction in stator voltage and (b)the power factor of the rotor circuit\n",
- "#Exa:11.8\n",
- "import math\n",
- "R_r=0.02#Rotor resistance per phase(in ohm)\n",
- "X_r=0.1#Rotor reactance per phase(in ohm)\n",
- "s=4#Slip(in%)\n",
- "S=100.-s\n",
- "T_f=((s/100.)*R_r)/((R_r**2.)+(((s/100.)**2.)*(X_r**2.)))\n",
- "S_r=1-(.5*(S/100.))\n",
- "T=(S_r*R_r)/((R_r**2.)+((S_r**2.)*(X_r**2.)))\n",
- "Re=(1-math.sqrt(T_f/T))*100.\n",
- "print '%s %.2f' %('(a)% reduction in stator voltage(in %)=',Re)\n",
- "pf=R_r/(math.sqrt((R_r**2.)+((S_r**2.)*(X_r**2.))))\n",
- "print '%s %.2f' %('(b)Power factor=',pf)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)% reduction in stator voltage(in %)= 24.24\n",
- "(b)Power factor= 0.36\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E9 - Pg 242"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)the rotor copper loss per phase if motor is running at slip of 4% (b)Mechanical power developed\n",
- "#Exa:11.9\n",
- "P_i=100000.#Input power(in watt)\n",
- "P_sc=2000.#Stator copper loss(in watt)\n",
- "s=4.#slip(in %)\n",
- "P_r=P_i-P_sc\n",
- "P_rc=((s/100.)*P_r)/3.\n",
- "print '%s %.f' %('(a)Rotor copper lossper phase(in watt)=',P_rc)\n",
- "P_m=P_r-(P_rc*3)\n",
- "print '%s %.f' %('(b)Mechanical power developed(in watt)=',P_m)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Rotor copper lossper phase(in watt)= 1307\n",
- "(b)Mechanical power developed(in watt)= 94080\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E10 - Pg 242"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate (a)% slip (b)Rotor copper loss (c)Output from the rotor (d)Efficiency \n",
- "#Exa:11.10\n",
- "import math \n",
- "V=440.#Supplied voltage(in volts)\n",
- "f=50.#frequency(in hertz)\n",
- "p=6.#Number of poles\n",
- "n=960.#Speed of motor(in r.p.m)\n",
- "P_i=50000.#Input power(in watt)\n",
- "P_wf=1800.#Winding and friction losses(in watt)\n",
- "P_s=1200.#Stator losses(in watt)\n",
- "n_s=(120.*f)/p\n",
- "S=((n_s-n)/n_s)*100.\n",
- "print '%s %.f' %('(a)% slip=',S)\n",
- "P_r=P_i-P_s\n",
- "P_rc=(S/100.)*P_r\n",
- "print '%s %.f' %('(b)Rotor copper loss(in watt)=',P_rc)\n",
- "P_o=P_r-P_rc-P_wf\n",
- "print '%s %.f' %('(c)Output of rotor(in watt)=',P_o)\n",
- "eff=(P_o/P_i)*100.\n",
- "print '%s %.f' %('(d)Efficiency(in%)=',eff)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)% slip= 4\n",
- "(b)Rotor copper loss(in watt)= 1952\n",
- "(c)Output of rotor(in watt)= 45048\n",
- "(d)Efficiency(in%)= 90\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E11 - Pg 243"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Equivalent rotor current per phase (b)Stator current per phase (c)Power factor (d)Rotor input (e)Rotor copper losses (f)Torque (g)Mechanical power output from rotor (h)Stator input (i)Efficiency \n",
- "#Exa:11.11\n",
- "import math,cmath\n",
- "from math import cos,atan,sqrt\n",
- "V=440.#Voltage supplied(in volts)\n",
- "p=8.#Number of poles\n",
- "f=50.#Frequency(in hertz)\n",
- "r1=0.2#Stator resistance(in ohm)\n",
- "x1=1.2#Stator reactance(in ohm)\n",
- "r2=0.3#Equivalent resistance of rotor referred to stator(in ohm)\n",
- "x2=1.2#Equivalent reactance of rotor referred to stator(in ohm)\n",
- "r_m=150.#Magnetising resistance(in ohms)\n",
- "x_m=18.#Magnetising reactance(in ohms)\n",
- "P_wf=750.#Winding and friction losses(in watt)\n",
- "s=0.04#Slip\n",
- "n_s=(f*120.)/(p*60.)\n",
- "y1=1./r_m\n",
- "y2=1./(1j*x_m)\n",
- "y3=1./((r2/s)+(1j*x2))\n",
- "Y=y1+y2+y3\n",
- "Z=1./Y\n",
- "Z_t=Z+(r1+(1j*x1))\n",
- "E=V*Z/(Z_t)\n",
- "z3=1./y3\n",
- "i2=E/z3\n",
- "print '(a)Rotor current per phase(in A)=',i2\n",
- "i1=V/Z_t\n",
- "print '(b)Stator current per phase(in A)=',i1\n",
- "#pf=cos(atan(-(Z_t.imag)/(Z_t.real))*57.3)*57.3\n",
- "pf=0.793\n",
- "print '(c)Power factor=',pf\n",
- "#P_r=(i2*(i2.conjugate()))*(r2/s)\n",
- "P_r=20.569\n",
- "print '(d)Rotor input(in kwatt)=',P_r\n",
- "#P_rc=(i2*(i2.conjugate()))*r2\n",
- "P_rc=0.822\n",
- "print '(e)Rotor copper loss(in kwatt)=',P_rc\n",
- "#T=3*P_r/(2*math.pi*n_s)\n",
- "T=785.67\n",
- "print '(f)Torque(in N-m)=',T\n",
- "#P_me=P_r-P_rc-(P_wf/3)\n",
- "P_me=19497\n",
- "print '(g)Mechanical output from rotor(in watts per phase)=',P_me\n",
- "#P_st=V*((i1*(i1.conjugate()))**.5)*pf\n",
- "P_st=21713\n",
- "print '(h)Stator input(watts per phase)=',P_st\n",
- "#eff=(P_me/P_st)*100\n",
- "eff=89.77\n",
- "print '(i)Efficiency(in %)=',eff#Caption:Find (a)Equivalent rotor current per phase (b)Stator current per phase (c)Power factor (d)Rotor input (e)Rotor copper losses (f)Torque (g)Mechanical power output from rotor (h)Stator input (i)Efficiency \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Rotor current per phase(in A)= (49.1649692861-14.7144544702j)\n",
- "(b)Stator current per phase(in A)= (48.8875753093-36.52322494j)\n",
- "(c)Power factor= 0.793\n",
- "(d)Rotor input(in kwatt)= 20.569\n",
- "(e)Rotor copper loss(in kwatt)= 0.822\n",
- "(f)Torque(in N-m)= 785.67\n",
- "(g)Mechanical output from rotor(in watts per phase)= 19497\n",
- "(h)Stator input(watts per phase)= 21713\n",
- "(i)Efficiency(in %)= 89.77\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E12 - Pg 245"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Equivalent rotor current per phase (b)Stator current per phase (c)Power factor (d)Rotor input (e)Rotor copper losses (f)Torque (g)Mechanical power output from rotor (h)Stator input (i)Efficiency.Solve it by APPROXIMATE equivalent circuit method \n",
- "#Exa:11.12\n",
- "import math,cmath\n",
- "from math import sqrt,cos,atan\n",
- "V=440.#Voltage supplied(in volts)\n",
- "p=8.#Number of poles\n",
- "f=50.#Frequency(in hertz)\n",
- "r1=0.2#Stator resistance(in ohm)\n",
- "x1=1.2#Stator reactance(in ohm)\n",
- "r2=0.3#Equivalent resistance of rotor referred to stator(in ohm)\n",
- "x2=1.2#Equivalent reactance of rotor referred to stator(in ohm)\n",
- "r_m=150.#Magnetising resistance(in ohms)\n",
- "x_m=18.#Magnetising reactance(in ohms)\n",
- "P_wf=750.#Winding and friction losses(in watt)\n",
- "s=0.04#Slip\n",
- "I2=V/((r1+(r2/s))+(1j*x1)+(1j*x2))\n",
- "print '(a)Equivalent rotor current per phase(in A)=',I2\n",
- "y1=1./r_m\n",
- "y2=1./(1j*x_m)\n",
- "I_o=V*(y1+y2)\n",
- "I_1=I2+I_o\n",
- "print '(b)Stator current per phase(in A)=',I_1\n",
- "#pf=cos(atan(I_1.imag/I_1.real)*57.3)*57.3\n",
- "pf=0.804\n",
- "print '(c)Power factor=',pf\n",
- "#P_r=(I2*I2.conjugate())*(r2/s)\n",
- "P_r=22.317\n",
- "print '(d)Rotor input(in kwatt)=',P_r\n",
- "#P_rc=(I2*I2.conjugate())*r2\n",
- "P_rc=892\n",
- "print '(e)Rotor copper losses(in watts)=',P_rc\n",
- "#T=P_r/(2.*math.pi*((f*120.)/(p*60.)))\n",
- "T=852.42\n",
- "print '(f)Torque(in N-m)=',T\n",
- "#P_me=P_r-P_rc-(P_wf/3.)\n",
- "P_me=21175\n",
- "print '(g)Mechanical power output from rotor(in watts per phase)=',P_me\n",
- "#P_si=V*pf*((I_1*I_1.conjugate()))*5\n",
- "P_si=24100\n",
- "print '(h)Stator input(in watts per phase)=',P_si\n",
- "#e=(P_me/P_si)*100.\n",
- "e=87.86\n",
- "print '(i)Efficiency (in %)=',e"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Equivalent rotor current per phase(in A)= (52.0830130669-16.2336664105j)\n",
- "(b)Stator current per phase(in A)= (55.0163464002-40.6781108549j)\n",
- "(c)Power factor= 0.804\n",
- "(d)Rotor input(in kwatt)= 22.317\n",
- "(e)Rotor copper losses(in watts)= 892\n",
- "(f)Torque(in N-m)= 852.42\n",
- "(g)Mechanical power output from rotor(in watts per phase)= 21175\n",
- "(h)Stator input(in watts per phase)= 24100\n",
- "(i)Efficiency (in %)= 87.86\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 6,
- "metadata": {},
- "source": [
- "Example E13 - Pg 246"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Equivalent rotor current (b)Stator current (c)Power factor (d)Stator input (e)Rotor input (f)Efficiency\n",
- "#Exa:11.13\n",
- "import math,cmath\n",
- "from math import sqrt,cos,atan\n",
- "V=440#Voltage supplied(in volts)\n",
- "f=50#frequency(in hertz)\n",
- "Z_s=1.5+(1j*3)#Stator impedance per phase(in ohms)\n",
- "Z_r=1.6+(1j*1)#Rotor impedance per phase(in ohms)\n",
- "Z_m=3+(1j*40)#Magnetising impedance per phase(in ohms)\n",
- "P_wf=300#Friction and winding loss(in watt)\n",
- "s=0.04#Slip\n",
- "Z=40+(1j*1)\n",
- "z=Z*Z_m/(Z+Z_m)\n",
- "Zt=z+Z_s\n",
- "I1=(V/sqrt(3))/Zt\n",
- "E=(V/sqrt(3))-(I1*Z_s)\n",
- "I2=E/Z\n",
- "print '(a)Equivalent Rotor current(in A)=',I2\n",
- "print '(b)Stator current(in A)=',I1\n",
- "#pf=cos(atan(Zt.imag/Zt.real))\n",
- "pf=0.7\n",
- "print '(c)Power factor=',pf\n",
- "#P_s=sqrt(3)*V*(I1*I1.conjugate())*pf\n",
- "P_s=4486.5\n",
- "print '(d)Stator input(in watt)=',P_s\n",
- "#P_r=3*(I2*I2.conjugate())*(Z_r.real/s)\n",
- "P_r=3709.6\n",
- "print '(e)Rotor input(in watt)=',P_r\n",
- "P_ro=P_r*(1-s)\n",
- "P_me=P_ro-P_wf\n",
- "#e=(P_me/P_s)*100\n",
- "e=72.68\n",
- "print '(f)Efficiency(in%)=',e"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Equivalent Rotor current(in A)= (5.66994104723-0.356954905389j)\n",
- "(b)Stator current(in A)= (5.87947093141-6.02010508394j)\n",
- "(c)Power factor= 0.7\n",
- "(d)Stator input(in watt)= 4486.5\n",
- "(e)Rotor input(in watt)= 3709.6\n",
- "(f)Efficiency(in%)= 72.68\n"
- ]
- }
- ],
- "prompt_number": 13
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_12.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_12.ipynb deleted file mode 100755 index b6e144f8..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_12.ipynb +++ /dev/null @@ -1,377 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:79705a61e88262ad8a41adb7ddf642b0c3f67b202f89d8ad31de55fbf850a47d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 12 - THREE-PHASE INDUCTION MOTORS: OPERATION AND TESTING"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 255"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Calculate (a)No load power factor (b)Core and friction loss (c)r_m (d)power factor on short circuit (e)Equivalent impedance in series circuit (f)Rotor resistance referred to stator (g)Stator leakage reactance (h)Rotor leakage reactance referred to stator\n",
- "#Exa:12.1\n",
- "import math \n",
- "from math import sqrt\n",
- "P=3000.#Power of motor(in watt)\n",
- "V=415.#Voltage supplied(in volts)\n",
- "f=50.#Frequency(in hertz)\n",
- "p=6.#Number of poles\n",
- "pf=0.8#Power factor\n",
- "I_n=3.5#No load current(in A)\n",
- "P_n=250.#Power input on no load test(in watt)\n",
- "r_s=1.5#Stator resistance per phase(in ohm)\n",
- "V_r=115.#Reduced voltage applied at short circuit test(in volts)\n",
- "I_s=13.#Current supplied on short circuit test(in A)\n",
- "P_s=1660.#Voltage applied at short circuit test(in watt)\n",
- "#pfn=P_n/(sqrt(3.)*V*I_n)\n",
- "pfn=84.3\n",
- "print ('(a)Noload power factor(degrees)=',pfn)\n",
- "P_wf=P_n-(3.*(I_n**2.)*r_s)\n",
- "print ('(b)Core and friction loss(in watt)=',P_wf)\n",
- "#r_m=(V/sqrt(3.))/(I_n*pfn)\n",
- "r_m=684.5\n",
- "print ('(c)Resistance(in ohms)=',r_m)\n",
- "#pfs=P_s/(sqrt(3.)*V_r*I_s)\n",
- "pfs=0.64\n",
- "print ('(d)Power factor on short circuit=',pfs)\n",
- "#Ze=(V/sqrt(3.))/((I_s*V)/V_r)\n",
- "Ze=5.11\n",
- "print ('(e)Equivalent impedance in series circuit(in ohms)=',Ze)\n",
- "#R=(Ze*pfs)-r_s\n",
- "R=1.77\n",
- "print ('(f)Rotor resistance referred to stator(in ohms)=',R)\n",
- "#X=(sqrt((Ze**2.)-((Ze*pfs)**2.)))\n",
- "X=3.92\n",
- "print ('(g)Stator leakage reactance(in ohms)=',X)\n",
- "#x=X/2.\n",
- "x=1.96\n",
- "print ('(h)Rotor leakage reactance referred to stator(in ohms)=',x)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "('(a)Noload power factor(degrees)=', 84.3)\n",
- "('(b)Core and friction loss(in watt)=', 194.875)\n",
- "('(c)Resistance(in ohms)=', 684.5)\n",
- "('(d)Power factor on short circuit=', 0.64)\n",
- "('(e)Equivalent impedance in series circuit(in ohms)=', 5.11)\n",
- "('(f)Rotor resistance referred to stator(in ohms)=', 1.77)\n",
- "('(g)Stator leakage reactance(in ohms)=', 3.92)\n",
- "('(h)Rotor leakage reactance referred to stator(in ohms)=', 1.96)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 256"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Starting current (b)Starting torque\n",
- "#Exa:12.2\n",
- "import math,cmath\n",
- "from math import cos,sin\n",
- "P=3000.#Power of motor(in watt)\n",
- "V=415.#Voltage supplied(in volts)\n",
- "f=50.#Frequency(in hertz)\n",
- "p=6.#Number of poles\n",
- "pf=0.8#Power factor\n",
- "pfs=0.64#Power factor on short circuit\n",
- "pfn=0.1#No load power factor\n",
- "I_n=3.5#No load current(in A)\n",
- "P_n=250.#Power input on no load test(in watt)\n",
- "r_s=1.5#Stator resistance per phase(in ohm)\n",
- "V_r=115.#Reduced voltage applied at short circuit test(in volts)\n",
- "I_s=13.#Current supplied on short circuit test(in A)\n",
- "P_s=1660.#Voltage applied at short circuit test(in watt)\n",
- "n_s=1000.#Synchronous speed(in r.p.m)\n",
- "R2=1.77#Rotor resistance referred to stator(in ohms)\n",
- "I_st=I_s*V/(V_r)\n",
- "print '%s %.1f'%('(a)Starting current(in A)=',I_st)\n",
- "I_i=I_st*(cos(pfs)+(1j*(sin(pfs)))*57.3)*57.3\n",
- "I_o=I_n*(cos(pfn)+(1j*(sin(pfn)))*57.3)*57.3\n",
- "I_2=I_i-I_o\n",
- "P_ri=3.*(I_2*I_2.conjugate())*R2\n",
- "#T=P_ri/(2.*math.pi*(n_s/60.))\n",
- "T=98.17\n",
- "print '(b)Starting torque(in N-m)=',T"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Starting current(in A)= 46.9\n",
- "(b)Starting torque(in N-m)= 98.17\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E5 - Pg 265"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find starting current in terms of full load current\n",
- "#Exa:12.5\n",
- "import math \n",
- "s=0.04#Slip\n",
- "a=1.#Starting torque T_st/Full load torque(T_fl) are equal\n",
- "I_s=math.sqrt(a/s)\n",
- "print '%s %.f %s' %('Starting current is=',I_s,'times the full load current')"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Starting current is= 5 times the full load current\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E7 - Pg 267"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find starting torque in terms of full load torque\n",
- "#Exa:12.7\n",
- "import math \n",
- "s=0.03#slip\n",
- "a=2.5#Ratio of supply current to full load current\n",
- "b=5.#Ratio of short circuit current to full load current\n",
- "x=math.sqrt(a/b)\n",
- "T=(x**2.)*(b**2.)*s\n",
- "print '%s %.3f' %('Starting torque is (below) times the full load torque=',T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Starting torque is (below) times the full load torque= 0.375\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E8 - Pg 267"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the % tappings required\n",
- "#Exa:12.8\n",
- "import math \n",
- "s=0.04#Slip\n",
- "a=4.#Ratio of short circuit current to full load current\n",
- "b=40.#Starting torque to full load torque(in%)\n",
- "x=math.sqrt((b/100.)/(s*(a**2.)))*100.\n",
- "print '%s %.f' %('% tappings required is(in%)=',x)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "% tappings required is(in%)= 79\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E9 - Pg 268"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the line current at start\n",
- "#Exa:12.9\n",
- "import math \n",
- "P=5000.#Power supplied to induction motor(in watts)\n",
- "V=415.#Voltage supplied to motor(in volts)\n",
- "f=50.#frequency(in hertz)\n",
- "e=0.85#Efficiency\n",
- "pf=0.8#Power factor lagging\n",
- "b=5.#Ratio of short circuit current to full load current\n",
- "P_i=P/e\n",
- "I_fl=P_i/(math.sqrt(3.)*V*pf)\n",
- "I_l=(1./3.)*b*I_fl\n",
- "print '%s %.2f' %('Line current(in A)=',I_l)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Line current(in A)= 17.05\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E11 - Pg 271"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Calculate external resistance per phase \n",
- "#Exa:12.11\n",
- "p=6.#Number of poles\n",
- "f=50.#Frequency(in hertz)\n",
- "r=0.25#Resistance per phase(in ohms)\n",
- "n_1=965.#Speed on full load(in r.p.m)\n",
- "n_2=800.#Reduced speed(in r.p.m)\n",
- "n_s=(120.*f)/p\n",
- "s_1=(n_s-n_1)/n_s\n",
- "s_2=(n_s-n_2)/n_s\n",
- "R=((s_2*r)/s_1)-r\n",
- "print '%s %.2f' %('Required external resistance per phase(in ohms)=',R)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required external resistance per phase(in ohms)= 1.18\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E12 - Pg 282"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the dimensions of D and L\n",
- "#Exa:12.12\n",
- "import math\n",
- "P=7.5#Power of induction motor(in KW)\n",
- "p=4.#Number of poles\n",
- "f=50.#frequency(in hertz)\n",
- "V=415.#Voltage applied of motor(in volts)\n",
- "e=0.88#Efficiency\n",
- "pf=0.87#Power factor\n",
- "b=2.5#Ratio of pull out torque to full load torque\n",
- "c=1.75#Ratio of starting to full load torque\n",
- "n=1440.#Speed of motor(in r.p.m)\n",
- "ac=23000.#Ampere conductors per meter\n",
- "k=0.955\n",
- "B=0.45#flux per pole(in wb/m**2)\n",
- "n_s=(120.*f)/(60.*p)\n",
- "S=P/(e*pf)\n",
- "D=165.#Choosing(in mm)\n",
- "#L=(S*(10.**3.))/(1.11*k*(math.pi**2.)*B*ac*n_s*(10.**(-3.))*(D**2.)*(10.**(-6.)))\n",
- "L=140\n",
- "print '%s %.f' % ('D (mm)=',D)\n",
- "print '%s %.f' % ('L (mm)=',L)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "D (mm)= 165\n",
- "L (mm)= 140\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_13.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_13.ipynb deleted file mode 100755 index dddb1b64..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_13.ipynb +++ /dev/null @@ -1,345 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:2b1993056b74a99105284d68a1d4fc60ea8b947e16782baac998356eb3473399"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 13 - SYNCHRONOUS MACHINES"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the frequency of voltage generated\n",
- "#Exa:13.1\n",
- "p=16.#Number of poles\n",
- "n=375.#Speed of alternator(in r.p.m)\n",
- "f=(p*n)/120.\n",
- "print '%s %.f' %('Frequency of voltage generated(in c/s)=',f)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Frequency of voltage generated(in c/s)= 50\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find (a)speed (b)number of poles\n",
- "#Exa:13.2\n",
- "f1=25.#Frequency of motor(in hertz)\n",
- "f2=60.#Frequency of generator(in hertz)\n",
- "p=10.#Number of poles\n",
- "N=(120.*f1)/p\n",
- "print '%s %.f' %('(a)Speed(in r.p.m)=',N)\n",
- "P=(f2*120.)/(N)\n",
- "print '%s %.f' %('(b)Number of poles=',P)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Speed(in r.p.m)= 300\n",
- "(b)Number of poles= 24\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E3 - Pg 293"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find distribution factor of winding\n",
- "#Exa:13.3\n",
- "import math\n",
- "ns=18.#Number of slots\n",
- "ph=3.#Number of phases\n",
- "p=2.#Number of poles\n",
- "m=ns/(ph*p)\n",
- "P_p=ns/p\n",
- "theta=180./P_p\n",
- "#k_b=math.sin(m*(theta/2.)*57.3)/(m*math.sin(theta/2.)*57.3)\n",
- "k_b=0.9597\n",
- "print '%s %.4f' %('Distribution factor=',k_b)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Distribution factor= 0.9597\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E4 - Pg 294"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find coil span factor\n",
- "#Exa:13.4\n",
- "import math\n",
- "s=9.#Number of slots\n",
- "theta=180./s\n",
- "#k_p=math.cos(theta/2.)*57.3\n",
- "k_p=0.9848\n",
- "print '%s %.4f' %('Coil span factor=',k_p)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Coil span factor= 0.9848\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E5 - Pg 295"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)frequency (b)Phase e.m.f (c)Line e.m.f\n",
- "#Exa:13.5\n",
- "import math \n",
- "from math import sin,cos\n",
- "ph=3.#Number of phases \n",
- "p=16.#Number of poles\n",
- "sl=144.#Number of slots\n",
- "cs=10.#Number of conductors per slot\n",
- "B=0.04#Flux per pole(in wb)\n",
- "n=375.#Speed of alternator(in r.p.m)\n",
- "C_s=160.#Coil Span(in degrees)\n",
- "f=(p*n)/120.\n",
- "print '%s %.f' %('(a)Frequency(in hertz)=',f)\n",
- "ct=(sl*cs)/2.\n",
- "nt=ct/ph\n",
- "m=sl/(p*ph)\n",
- "P_p=sl/p\n",
- "theta=180./P_p\n",
- "k_b=sin(m*(theta/2.)*57.3)/(m*sin(theta/2.)*57.3)\n",
- "k_p=cos(theta/2.)*57.3\n",
- "#E_ph=4.44*B*f*nt*k_b*k_p\n",
- "E_ph=2014.22\n",
- "print '%s %.2f' %('(b)Phase e.m.f(in volts)=',E_ph)\n",
- "#E_l=math.sqrt(3.)*E_ph\n",
- "E_l=3488.73\n",
- "print '%s %.2f' %('(c)Line e.m.f(in volts)=',E_l)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Frequency(in hertz)= 50\n",
- "(b)Phase e.m.f(in volts)= 2014.22\n",
- "(c)Line e.m.f(in volts)= 3488.73\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E6 - Pg 295"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find number of armature conductors in series per phase\n",
- "#Exa:13.6\n",
- "import math\n",
- "p=10.#Number of poles\n",
- "ph=3.#Number of phases\n",
- "n=600.#Speed of alternator(in r.p.m)\n",
- "sl=90.#Number of slots\n",
- "Vl=6600.#Line voltage(in volts)\n",
- "B=0.1#Flux per pole(in wb)\n",
- "cs=160.#Coil span(in degrees)\n",
- "kb=0.9597#Distribution factor\n",
- "kp=0.9848#Pitch factor\n",
- "v_ph=Vl/math.sqrt(3.)\n",
- "f=(p*n)/120.\n",
- "m=sl/(p*ph)\n",
- "T=2.*v_ph/(4.44*kb*kp*B*f)\n",
- "print '%s %.f' %('Number of armature conductors in series per phase=',T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Number of armature conductors in series per phase= 363\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E12 - Pg 310"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Calculate synchronous reactance and synchronous impedance per phase\n",
- "#Exa:13.12\n",
- "import math\n",
- "V=3300.#Voltage of alternator(in volts)\n",
- "f=50.#Frequency(in hertz)\n",
- "r=0.4#Effective resistance per phase(in ohm)\n",
- "I_f=20.#Field current(in ohms)\n",
- "I_fl=300.#Full load current(in A)\n",
- "e=1905.#Voltage induced on open circuit(in volts)\n",
- "Zs=e/I_fl\n",
- "Xs=math.sqrt((Zs**2.)-(r**2.))\n",
- "print '%s %.2f' %('Impedance(in ohms)=',Zs)\n",
- "print '%s %.3f' %('Synchronous reactance=',Xs)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Impedance(in ohms)= 6.35\n",
- "Synchronous reactance= 6.337\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E13 - Pg 310"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Estimate terminal voltage for (a)same excitation (b)Load current at 0.8 power factor lagging\n",
- "#Exa:13.13\n",
- "import math\n",
- "from math import sin,acos,sqrt\n",
- "P=1000.#Power of alternator(in KVA)\n",
- "V=3300.#Voltage of alternator(in volts)\n",
- "ph=3.#Phase of alternator\n",
- "pf=0.8#Power factor lagging\n",
- "r=0.5#Resistance per phase(in ohms)\n",
- "x=6.5#Reactance per phase(in ohms)\n",
- "V_ph=V/math.sqrt(3.)\n",
- "I=(P*1000.)/(math.sqrt(3.)*V)\n",
- "#Eo=(((V_ph+(I*r*pf)+(I*x*sin(acos(pf)))*57.3)*57.3**2.)+(((I*x*pf)-(I*r*sin(acos(pf)))*57.3)*57.3**2))*5\n",
- "Eo=2792.4\n",
- "print '(a)Required terminal voltage(in volts)=',Eo \n",
- "#v=((Eo**2)-(((I*r*sin(acos(pf))*57.3)*57.3+(I*x*pf))**2.))+((I*x*sin(acos(pf))*57.3)*57.3-(I*r*pf))*5\n",
- "v=5621\n",
- "print '(b)Required voltage at given load current(in volts)=',v"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Required terminal voltage(in volts)= 2792.4\n",
- "(b)Required voltage at given load current(in volts)= 5621\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_14.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_14.ipynb deleted file mode 100755 index eec38008..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_14.ipynb +++ /dev/null @@ -1,338 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c8dd7a51eecf312e58a69358826d2f1f3a8a019a2bd49069e671f1c4a91ceb24"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 14 - SYNCHRONOUS MACHINES: GENERATORS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 318"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the excitation voltage in per unit\n",
- "#Exa:14.1\n",
- "import math,cmath\n",
- "from math import sin,cos\n",
- "pf=0.9#Power factor\n",
- "Xd=1.#Direct axis synchronous reactance(in per unit)\n",
- "Xq=0.6#Quadrature axis synchronous reactance(in per unit)\n",
- "V=1.#Terminal voltage(in volts)\n",
- "ang=49.#Phase angle between Ia and excitation voltage(in degrees)\n",
- "Ia=0.9-(1j*0.436)#Armature current(in A)\n",
- "v=(1j)*Ia*Xq\n",
- "E=V+v\n",
- "Id=(Ia*Ia.conjugate())*sin(ang)*57.3*5\n",
- "Iq=(Ia*Ia.conjugate())*cos(ang)*57.3*5\n",
- "#Ef=(E*E.conjugate())+(Id*(Xd-Xq))*5\n",
- "Ef=1.672\n",
- "print 'Excitation voltage (in per unit)=',Ef"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Excitation voltage (in per unit)= 1.672\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 319"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find regulation by (a)Two reaction method, and (b)Synchronous impedance method\n",
- "#Exa:14.2\n",
- "import math,cmath\n",
- "pf=0.9#Power factor\n",
- "Xd=1.#Direct axis synchronous reactance(in per unit)\n",
- "Xq=0.6#Quadrature axis synchronous reactance(in per unit)\n",
- "V=1.#Terminal voltage(in volts)\n",
- "ang=49.#Phase angle between Ia and excitation voltage(in degrees)\n",
- "Ia=0.9-(1j*0.436)#Armature current(in A)\n",
- "E=1.6742083#Excitation voltage(in per unit)\n",
- "#re=(E-V)*100./V\n",
- "re=67.4\n",
- "print '(a)Regulation by two reaction method(in%)=',re \n",
- "Ef=V+(1j*Ia*Xd)\n",
- "#RE=(((Ef*Ef.conjugate()))-V)*100./V*5.\n",
- "RE=69.4\n",
- "print '(b)Regulation by Synchronous impedance method(in%)=',RE"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Regulation by two reaction method(in%)= 67.4\n",
- "(b)Regulation by Synchronous impedance method(in%)= 69.4\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E3 - Pg 323"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find Regulation and resultant excitation\n",
- "#Exa:14.3\n",
- "import math\n",
- "from math import sin,acos\n",
- "pf=0.8#Power factor lagging\n",
- "P=1000.#Power of Synchronous generator(in KVA)\n",
- "Eo=1.25#No load voltage(in per unit)\n",
- "V=6600.#Voltage of Synchronous generator(in volts)\n",
- "f=50.#Frequency(in hertz)\n",
- "Fe=1.#Field excitation to produce terminal voltage(in per unit)\n",
- "Fa=1.#Field excitation to produce full load current(in per unit)\n",
- "#Ft=math.sqrt(((Fe+(Fa*sin(acos(pf)))*57.3)*57.3**2.)+((Fa*pf)**2.))\n",
- "Re=(Eo-Fa)*100./Fa\n",
- "Ft=1.788\n",
- "print 'Resultant excitation(in per unit)=',Ft\n",
- "print 'regulation(in %)=',Re"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resultant excitation(in per unit)= 1.788\n",
- "regulation(in %)= 25.0\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E5 - Pg 326"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the regulation of the machine\n",
- "#Exa:14.5\n",
- "Vf=400.#Full load voltage(in volts)\n",
- "Vr=480.#No load voltage(in volts)\n",
- "Re=(Vr-Vf)*100./Vf\n",
- "print '%s %.f' %('Regulation of the machine(in %)=',Re)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Regulation of the machine(in %)= 20\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E6 - Pg 332"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Synchronising power on full load (b)Synchronising torque\n",
- "#Exa:14.6\n",
- "import math\n",
- "from math import sqrt,atan,sin,acos,cos\n",
- "P=5000.#Power ofan alternator(in KVA)\n",
- "f=50.#Frequency(in hertz)\n",
- "p=6.#Number of poles\n",
- "V=11000.#Voltageof alternator(in volts)\n",
- "pf=0.8#Power factor\n",
- "c=3.#Mechanical degree of print '%s %.2f' %lacement(in degrees)\n",
- "Xs=5.#Synchronous reactance per phase(in ohms)\n",
- "Vph=V/sqrt(3.)\n",
- "ns=(120.*f)/p\n",
- "If=(P*1000.)/(sqrt(3.)*V)\n",
- "E=sqrt(((Vph*pf)**2.)+(((Vph*sin(acos(pf))*57.3)*57.3+(If*Xs))**2.))\n",
- "a=atan(((Vph*sin(acos(pf))*57.3)*57.3+(If*Xs))/(Vph*pf))*57.3\n",
- "b=a-acos(pf)*57.3\n",
- "#Ps=(E*Vph*cos(b)*57.3*sin(c)*57.3)/Xs\n",
- "Ps=437.89\n",
- "print '%s %.2f' %('(a)Synchronising Poweron full load(in kwatt per phase)=',Ps)\n",
- "#Ts=(Ps*3.)/(2.*math.pi*(ns/60.))\n",
- "Ts=13569.55\n",
- "print '%s %.2f' %('(b)Synchronising Torque(in Nm)=',Ts)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Synchronising Poweron full load(in kwatt per phase)= 437.89\n",
- "(b)Synchronising Torque(in Nm)= 13569.55\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E9 - Pg 338"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Armature current of second machine (b)Power factor of ecach machine\n",
- "#Exa:14.9\n",
- "import math\n",
- "from math import tan,acos,atan,cos\n",
- "L=1000.#Total load(in KW)\n",
- "V=6600.#Total voltage(in volts)\n",
- "pf=0.8#Power factor\n",
- "Ia=50.#Armature current(in A)\n",
- "L1=L/2.\n",
- "Ia1=(L1*1000.)/(math.sqrt(3.)*V)\n",
- "#pf1=Ia1/Ia\n",
- "pf1=0.875\n",
- "a1=acos(pf1)*57.3\n",
- "b=tan(a1)*57.3\n",
- "P1=L1*b\n",
- "Pl=L*tan(acos(pf)*57.3)*57.3\n",
- "P2=P1-Pl\n",
- "#pf2=cos(atan(P2/L1)*57.3)*57.3\n",
- "pf2=0.726\n",
- "#Ia2=Ia1/pf2\n",
- "Ia2=60.25\n",
- "print '%s %.2f' %('(a)Armature current of second machine(in A)=',Ia2)\n",
- "print '%s %.3f %.3f ' %('(b)Power factor of both machines=',pf1,pf2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Armature current of second machine(in A)= 60.25\n",
- "(b)Power factor of both machines= 0.875 0.726 \n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E10 - Pg 339"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Load supplied by second machine and its power factor (b)Power factor of total load\n",
- "#Exa:14.10\n",
- "import math\n",
- "from math import tan,acos,cos,atan\n",
- "P1=300.#Lighting load(in KW)\n",
- "P2=500.#Industrial load(in KW)\n",
- "P3=200.#Industrial load(in KW)\n",
- "P4=100.#Load(in KW)\n",
- "Pa=500.#Power supplied by first machine(in KW)\n",
- "pf1=0.8\n",
- "pf2=0.707\n",
- "pf3=0.9\n",
- "pfa=0.8\n",
- "La=P1+P2+P3+P4\n",
- "Lr=(P2*tan(acos(pf1))*57.3)*57.3+(P3*tan(acos(pf2))*57.3)*57.3+(P4*tan(acos(pf3))*57.3)*57.3\n",
- "#Pb=La-Pa\n",
- "Pb=600\n",
- "Prl=Pa*(tan(acos(pfa))*57.3)*57.3\n",
- "Pc=Lr-Prl\n",
- "#pfb=cos(atan(Pc/Pb)*57.3)*57.3\n",
- "pfb=0.924\n",
- "#pfl=cos(atan(Lr/La)*57.3)*57.3\n",
- "pfl=0.87\n",
- "print '%s %.f %s %.3f' %('(a)Load supplied by second machine(in KW)=',Pb,'\\n its power factor=',pfb)\n",
- "print '%s %.2f' %('(b)Power factor of load=',pfl)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Load supplied by second machine(in KW)= 600 \n",
- " its power factor= 0.924\n",
- "(b)Power factor of load= 0.87\n"
- ]
- }
- ],
- "prompt_number": 7
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_15.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_15.ipynb deleted file mode 100755 index e0f4e5ec..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_15.ipynb +++ /dev/null @@ -1,206 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:6d44b7d9a62e13bf09e74de115b200d0be29281afd3a21e32eba9bc9017e30f6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 15 - SYNCHRONOUS MOTORS "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E6 - Pg 357"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Input power(in KVA) (b)Power factor\n",
- "#Exa:15.6\n",
- "import math\n",
- "from math import sin,acos\n",
- "V=440.#Voltage of circuit(in volts)\n",
- "f=50.#Frequency(in hertz)\n",
- "I=30.#Current taken by circuit(in A)\n",
- "pf=0.8#Power factor\n",
- "Pl=10.#Load supplied(in KW)\n",
- "e=0.85#Efficiency\n",
- "Pi=Pl/e\n",
- "Ii=Pi*1000./(math.sqrt(3.)*V)\n",
- "Ia=I*pf\n",
- "Ir=I*sin(acos(pf)*57.3)*57.3\n",
- "i=Ii+Ia\n",
- "It=math.sqrt((Ii**2.)+(Ir**2.))\n",
- "#pfm=Ii/It\n",
- "pfm=11.76\n",
- "#Wi=math.sqrt(3.)*V*It/(1000.)\n",
- "Wi=18.06\n",
- "print '%s %.2f' %('(a)Input power (KVA)=',Wi)\n",
- "print '%s %.2f' %('(b)Power factor (kW)=',pfm)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Input power (KVA)= 18.06\n",
- "(b)Power factor (kW)= 11.76\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E7 - Pg 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:(a)How much KVA should be supplied by synchronous motor (b)Power factor of synchronous motor\n",
- "#Exa:15.7\n",
- "import math \n",
- "from math import tan,acos,cos,atan\n",
- "Pm=40#Power absorb by motor(in Kw)\n",
- "Pl=300#Load connected in parallel with motor(in KW)\n",
- "pfm=0.85#Power factor of motor\n",
- "pfl=0.9#Power factor on load\n",
- "Pt=Pl+Pm\n",
- "Pr=Pt*tan(acos(pfl)*57.3)*57.3\n",
- "Pri=Pl*tan(acos(pfm)*57.3)*57.3\n",
- "#Ps=Pri-Pr\n",
- "Ps=185.92\n",
- "#pf=cos(atan(Ps/Pm)*57.3)*57.3\n",
- "pf=0.883\n",
- "print '%s %.2f' %('(a)Power supplied by synchronous motor(in KVA)=',Ps)\n",
- "print '%s %.3f' %('(b)Power factor of synchronous machine=',pf)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Power supplied by synchronous motor(in KVA)= 185.92\n",
- "(b)Power factor of synchronous machine= 0.883\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E8 - Pg 358"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: (a)Power alternator can supply (b)Power factor of synchronous motor (c)Load taken by motor\n",
- "#Exa:15.8\n",
- "import math \n",
- "from math import tan,acos,cos,atan\n",
- "P=500#Load supplied by alternator(inKW)\n",
- "pf=0.8#Power factor \n",
- "e=0.9\n",
- "L=P/pf\n",
- "Ps=L-P\n",
- "print '%s %.f' %('(a)Power alternator can supply(in KW)=',Ps)\n",
- "Pr=P*tan(acos(pf)*57.3)*57.3\n",
- "#pfm=cos(atan(Pr/Ps)*57.3)*57.3\n",
- "pfm=0.316\n",
- "print '%s %.3f' %('(b)Power factor of synchronous motor=',pfm)\n",
- "l=Ps*e\n",
- "print '%s %.1f' %('(c)Load taken by motor(in Kw)=',l)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Power alternator can supply(in KW)= 125\n",
- "(b)Power factor of synchronous motor= 0.316\n",
- "(c)Load taken by motor(in Kw)= 112.5\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E9 - Pg 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find efficiency of machine\n",
- "#Exa:15.9\n",
- "import math\n",
- "P=50000.#Power of alternator(in KVA)\n",
- "V=11.#Voltage of alternator(in Kv)\n",
- "pf=0.8#Power factor\n",
- "r=0.01#Resistance of stator winding per phase(in ohms)\n",
- "Wc=150.#Copper loss(in KW)\n",
- "Wf=100.#Friction loss(in KW)\n",
- "Ww=250.#Winding loss(in KW)\n",
- "Wco=200.#Core loss(in KW)\n",
- "We=15.#Excitor loss(in KW)\n",
- "Is=(P*1000.)/(math.sqrt(3.)*V*1000.)\n",
- "Ps=(Is**2.)*3.*(r/1000.)\n",
- "Ws=(0.5*Ps)\n",
- "Lt=Ps+Ws+Wc+Wf+Ww+Wco+We\n",
- "Po=P*pf\n",
- "Pi=Po+Lt\n",
- "e=Po*100./Pi\n",
- "print '%s %.1f' %('Efficieny of machine(in %)=',e)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Efficieny of machine(in %)= 97.5\n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_16.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_16.ipynb deleted file mode 100755 index e95791d2..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_16.ipynb +++ /dev/null @@ -1,144 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ba24a460b17bf7079ac4093071a48c79f514703fb164d80eae9ed12c922c8957"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 16 - SINGLE-PHASE INDUCTION MOTORS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Input Current (b)Power factor (c)Input power (d)Torque due to forward revolving field (e)Torque due to backward revovlving field (f)Net torque (g)Output And (h)Efficiency\n",
- "#Exa:16.1\n",
- "import math,cmath\n",
- "from math import cos,atan\n",
- "Pi=750.#Power of Single phase induction motor(in Watts)\n",
- "p=4.#Number of poles\n",
- "f=50.#Frequency(in hertz)\n",
- "V=230.#Voltage supplied to motor(in volts)\n",
- "R1=2.#Resistance of stator(in ohm)\n",
- "X1=2.6#Reactance of stator(in ohm)\n",
- "Wf=25.#Friction and winding loss(in Watts)\n",
- "R2=3.8#Resistance of rotor(in ohm)\n",
- "X2=2.6#Reactance of rotor(in ohm)\n",
- "Xm=56.#Magnetising Reactance(in ohms)\n",
- "r2=1.9#Imaginary resistance of rotor(in ohm)\n",
- "x2=1.3#Imaginary reactance of rotor(in ohm)\n",
- "xm=28.#Imaginary magnetising reactance(in ohm)\n",
- "s=0.05#Slip\n",
- "Z1=R1+(1j*X1)\n",
- "Z2=((1j*xm)*((r2/s)+(1j*x2)))/((r2/s)+(1j*(x2+xm)))\n",
- "Z3=((1j*xm)*((r2/(2-s))+(1j*x2)))/((r2/(2-s))+(1j*(x2+xm)))\n",
- "Z=Z1+Z2+Z3\n",
- "#I=V/Z\n",
- "I=9.28\n",
- "print '(a)Input Current(in A)=',I\n",
- "#pf=cos(atan(Z.imag/Z.real)*57.3)*57.3\n",
- "pf=9.581\n",
- "print '(b)Power factor=',pf\n",
- "#Wp=V*pf*(I*I.conjugate())*5\n",
- "Wp=1127.8\n",
- "print '(c)Input power(in watts)=',Wp\n",
- "z2=math.sqrt(((r2/s)**2)+((x2)**2))\n",
- "v2=(I*I.conjugate())*(Z2*Z2.conjugate())\n",
- "i2=v2/z2\n",
- "z3=math.sqrt(((r2/(2-s))**2)+((x2)**2))\n",
- "v3=(I*I.conjugate())*(Z3*Z3.conjugate())\n",
- "i3=v3/z3\n",
- "#Tf=((i2)**2.)*(r2/s)\n",
- "Tf=919.8\n",
- "print '(d)Torque due to forward field(in Nm)=',Tf\n",
- "#Tb=(i3**2.)*(r2)/(2.-s)\n",
- "Tb=83.9\n",
- "print '(e)Torque due to backward field(in Nm)=',Tb\n",
- "#T=Tf-Tb\n",
- "T=835.9\n",
- "print '(f)Torque(in Nm)=',T\n",
- "#Wo=(T*(1.-s))-Wf\n",
- "Wo=794.1\n",
- "print '(g)Output(in Watts)=',Wo\n",
- "#e=(Wo/Wp)*100.\n",
- "e=62.8\n",
- "print '(h)Efficiency(in %)=',e"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Input Current(in A)= 9.28\n",
- "(b)Power factor= 9.581\n",
- "(c)Input power(in watts)= 1127.8\n",
- "(d)Torque due to forward field(in Nm)= 919.8\n",
- "(e)Torque due to backward field(in Nm)= 83.9\n",
- "(f)Torque(in Nm)= 835.9\n",
- "(g)Output(in Watts)= 794.1\n",
- "(h)Efficiency(in %)= 62.8\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - 374"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find equivalent circuit resistance\n",
- "#Exa:16.2\n",
- "Wc=60.#Core loss(in watts)\n",
- "a=90.#Voltage across first rotor is 90% of applied voltage(in %)\n",
- "V=230.#Voltage applied to motor(in volts)\n",
- "v=V*(a/100.)\n",
- "Ic=Wc/v\n",
- "#rc=v/Ic\n",
- "rc=713.8\n",
- "print '%s %.1f' %('Equivalent circuit resistance(in ohms)=',rc)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Equivalent circuit resistance(in ohms)= 713.8\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_17.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_17.ipynb deleted file mode 100755 index 26f8601a..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_17.ipynb +++ /dev/null @@ -1,68 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:48e2a30220ca8f0e5ace71d8aaa9abff99fd46fdaaae4dbc00c0c67cefe15ce3"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 17 - AC COMMUTATOR MOTORS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 393"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Speed (b)Power factor of motor\n",
- "#Exa:17.1\n",
- "import math\n",
- "r=25.#Resistance of motor(in ohms)\n",
- "P=2.#Number of poles\n",
- "l=0.4#Inductance of motor(in henry)\n",
- "n=1800.#Speed of motor(in r.p.m)\n",
- "V=230.#Voltage supplied(in volts)\n",
- "Il=1.#Load current(in A)\n",
- "f=50.#Frequency(in hertz)\n",
- "fr=(P*n)/120.\n",
- "Eb=V-(r*Il)\n",
- "Er=Eb/fr\n",
- "fac=(math.sqrt((V**2.)-((Il*2.*(math.pi)*f*l)**2.))-(Il*r))/Er\n",
- "n=fac*(120./P)\n",
- "print '%s %.f' %('(a)Speed(in rpm)=',n)\n",
- "pf=math.sqrt((V**2.)-((Il*2.*(math.pi)*f*l)**2.))/V\n",
- "print '%s %.2f' %('(b)Power factor=',pf)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Speed(in rpm)= 1472\n",
- "(b)Power factor= 0.84\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_2.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_2.ipynb deleted file mode 100755 index e32de0a7..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_2.ipynb +++ /dev/null @@ -1,481 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:3db427b2374be2ac418b99afcc7665dd62ff13b794f960dfd18538ca6cd385f7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 2 - SINGLE-PHASE TRANSFORMERS: OPERATION AND TESTING"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 20"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)all day efficiency (b)commercial efficiency on full load (c)efficiency on half load\n",
- "#Exa:2_1\n",
- "import math\n",
- "P_s=50.#Power supplied(in kVA)\n",
- "V_1=440.#Primary side voltage(in volt)\n",
- "V_2=220.#Secondary side voltage(in volt)\n",
- "t_1=6.#Full load(in hours)\n",
- "t_2=2.#50% load(in hours)\n",
- "Cu_1=2.#Copper loss on full load(in KW)\n",
- "Fe=1.#Iron losses(in KW)\n",
- "E_1=P_s*t_1#Energy used on full load(in watt-hours)\n",
- "E_2=0.5*P_s*t_2#Energy used on half load(in watt-hours)\n",
- "Cu_2=Cu_1*0.25#Copper losses on half load(in watts)\n",
- "E=(Cu_1*t_1)+(Cu_2*t_2)+(Fe*24.)#Energy lost on losses(in watt-hours)\n",
- "eff_1=(E_1+E_2)/(E_1+E_2+E)*100.\n",
- "print '%s %.f' %('(a)All day efficiency(in%)=',eff_1)\n",
- "eff_2=(E_2)/(E_2+Cu_1+Fe)*100\n",
- "print '%s %.1f' %('(b)commercial efficiency on full load(in%)=',eff_2)\n",
- "eff_3=(0.5*E_2)/(0.5*E_2+Cu_2+Fe)*100\n",
- "print '%s %.1f' %('(c)efficiency on half load(in%)=',eff_3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)All day efficiency(in%)= 90\n",
- "(b)commercial efficiency on full load(in%)= 94.3\n",
- "(c)efficiency on half load(in%)= 94.3\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 21"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Efficiency of transformer at half load at 0.8 power factor lagging (b)At what load will the efficiency be maximum and maximum efficiency?\n",
- "#Exa:2.2\n",
- "import math\n",
- "P_s=25000.#Power supplied(in VA)\n",
- "V_1=3300.#Voltage on primary side(in volts)\n",
- "V_2=230.#Voltage on secondary side(in volts)\n",
- "f=50.#frequency(in hertz)\n",
- "P_i=300.#Iron loss(in watt)\n",
- "P_c=400.#Copper loss(in watt)\n",
- "pf=0.8#Power factor\n",
- "Cu=P_c*(0.5**2.)#Copper loss on half load\n",
- "P_o=P_s*0.5*pf#Output of transformer on half load \n",
- "eff=(P_o)/(P_o+Cu+P_i)*100.\n",
- "print '%s %.1f' %('(a)Efficiency of transformer at half load(in %)=',eff)\n",
- "x=math.sqrt(P_i/P_c)*20000.\n",
- "print '%s %.1f' %('(b)Load for maximum efficiency(in watt)=',x)\n",
- "eff_max=(x)/(x+P_i+P_i)*100.\n",
- "print '%s %.1f' %('Maximum efficiency(in%)=',eff_max)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Efficiency of transformer at half load(in %)= 96.2\n",
- "(b)Load for maximum efficiency(in watt)= 17320.5\n",
- "Maximum efficiency(in%)= 96.7\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E3 - Pg 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)% resistance (b)Regulation for power factors- unity, 0.8 lagging and 0.8 leading\n",
- "#Exa:2_3\n",
- "import math \n",
- "from math import acos,sin\n",
- "L_o=1.#Ohmic loss(%)\n",
- "X=6.#Reactance(in %)\n",
- "pf_1=0.8#lagging power factor\n",
- "pf_2=0.8#leading power factor\n",
- "R=L_o\n",
- "print '%s %.f %s' %('% resistance(in %)=',R,'\\n')\n",
- "Re_1=L_o\n",
- "print '%s %.f' %('(a) Regulation at unity power factor(in%)=',Re_1)\n",
- "theta=(acos(pf_1)*57.3)\n",
- "a=sin(theta)*57.3\n",
- "#Re_2=L_o*pf_1+X*a\n",
- "Re_2=4.4\n",
- "print '%s %.1f' %('(b) Regulation at 0.8 lagging power factor(in%)=',Re_2)\n",
- "#Re_3=L_o*pf_2-X*a\n",
- "Re_3=-2.8\n",
- "print '%s %.1f' %('(c) Regulation at 0.8 leading power factor(in%)=',Re_3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "% resistance(in %)= 1 \n",
- "\n",
- "(a) Regulation at unity power factor(in%)= 1\n",
- "(b) Regulation at 0.8 lagging power factor(in%)= 4.4\n",
- "(c) Regulation at 0.8 leading power factor(in%)= -2.8\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E4 - Pg 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find Regulation on full load at 0.8 power factor lagging\n",
- "#Exa:2_4\n",
- "import math \n",
- "from math import acos,sin\n",
- "P_s=500000.#Power supplied(in VA)\n",
- "V_1=2200.#Voltage on primary side(in volt)\n",
- "V_2=500.#Voltage on secondary side(in volt)\n",
- "f=50.#frequency(in hertz)\n",
- "Eff=97.#Efficiency of transformer(in %)\n",
- "Eff_m=75.#Maximum efficiency(in %) of its full load\n",
- "Z_1=10.#Impedance(in %)\n",
- "pf_1=1.#Power factor for maximum efficiency\n",
- "pf_2=0.8#Power factor lagging\n",
- "I_fl=P_s/V_2\n",
- "I=(Eff_m*I_fl)/100.\n",
- "Losses=(100.-Eff)/100.\n",
- "Cu=Losses/2.\n",
- "Fe=Losses/2.\n",
- "C=(Cu*P_s*Eff_m)/100.\n",
- "R=C/(I**2.)\n",
- "V=(Z_1*V_2)/100.\n",
- "Z=V/I_fl\n",
- "X=math.sqrt(Z**2.-R**2.)\n",
- "theta=(acos(pf_2)*57.3)\n",
- "#Re=(I_fl*R*pf_2)+(I_fl*X*sin(theta)*57.3)\n",
- "Re=37.4\n",
- "print '%s %.f' %('Regulation on full load at 0.8 power factor lagging(in volt)=',Re)\n",
- "#Reg=(Re/V_2)*100\n",
- "Reg=7.48\n",
- "print '%s %.2f' %('Percentage Regulation(in%)=',Reg)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Regulation on full load at 0.8 power factor lagging(in volt)= 37\n",
- "Percentage Regulation(in%)= 7.48\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E5 - Pg 27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find:(a) R_o (b)X_o (c)Resistance reffered to l.v side (d)Reactance reffered to l.v side\n",
- "#Exa:2.5\n",
- "import math \n",
- "from math import sin,acos\n",
- "P_s=5000.#Power Supplied(in VA)\n",
- "V_1=220.#Primary side voltage(in volt)\n",
- "V_2=440.#Secondary side voltage(in volt)\n",
- "f=50.#frequency(in hertz)\n",
- "I_o=0.75#Open circuit test current(in A)\n",
- "P_o=75.#Open circuit test power(in watt)\n",
- "V_s=16.#Short circuit test voltage(in volt)\n",
- "P_c=80.#Short circuit test power(in watt)\n",
- "pf=(P_o)/(V_1*I_o)\n",
- "a=sin(acos(pf)*57.3)*57.3\n",
- "R_o=(V_1)/(I_o*pf)\n",
- "print '%s %.f' %('(a)R_o(in ohms)=',R_o)\n",
- "#X_o=(V_1)/(I_o*a)\n",
- "X_o=328\n",
- "print '%s %.f' %('(b)X_o(in ohms)=',X_o)\n",
- "I_l=P_s/V_2\n",
- "Z=V_s/I_l\n",
- "R=(P_c)/(I_l**2.)\n",
- "X=math.sqrt(Z**2.-R**2.)\n",
- "n=V_2/V_1\n",
- "r=(R)/(n**2.)\n",
- "print '%s %.3f' %('(c)resistance reffered to low voltage side(in ohms)=',r)\n",
- "x=(X)/(n**2.)\n",
- "print '%s %.2f' %('(d)reactane reffered to low voltage side(in ohms)=',x)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)R_o(in ohms)= 645\n",
- "(b)X_o(in ohms)= 328\n",
- "(c)resistance reffered to low voltage side(in ohms)= 0.155\n",
- "(d)reactane reffered to low voltage side(in ohms)= 0.32\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E6 - Pg 28"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find voltage for h.v voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts\n",
- "#Exa:2.6\n",
- "import math\n",
- "from math import sin,acos\n",
- "P_s=100000.#Supplied power(in VA)\n",
- "V_1=6600.#Primary side voltage(in volt)\n",
- "V_2=240.#Secondary side voltage(in volt)\n",
- "f=50.#frequency(in hertz)\n",
- "I_sh=5.#short circuit test current(in A)\n",
- "P_sh=109.#short circuit test power(in watt)\n",
- "V_sh=50.#short circuit test voltage(in volt)\n",
- "pf=0.8#Power factor\n",
- "Z=V_sh/I_sh\n",
- "R=P_sh/(I_sh**2.)\n",
- "X=math.sqrt(Z**2.-R**2.)\n",
- "I_l=P_s/V_1\n",
- "Re=(I_l*R*pf)+(I_l*X*sin(acos(pf))*57.3)*57.3\n",
- "#V_r=Re+V_1m\n",
- "V_r=6735\n",
- "print '%s %.f' %('Voltage for high voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts(in volt)=',V_r)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage for high voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts(in volt)= 6735\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E7 - Pg 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate (a)Z,X,R reffered to h.v side (b)Regulaton on full load at 0.8 power factor lagging (c)Terminal voltage on l.v side on full load at pf=0.8 lagging (d)Efficiency of transformer when current=250A,pf=0.8 lagging is load connected to l.v side and voltage at h.v side is 11000 volts\n",
- "#Exa:2.7\n",
- "import math\n",
- "from math import sin,acos\n",
- "P_s=220000.#Supplied power (in VA)\n",
- "V_1=11000.#Primary side voltage(in volt)\n",
- "V_2=440.#Secondary side voltage(in volt)\n",
- "P_i=2200.#Iron losses(in watt)\n",
- "V=500.#voltage applied to high voltage side for open circuit test(in volt)\n",
- "P=2000.#Wattmeter reading for open circuit test(in watt)\n",
- "pf=0.8#Power factor\n",
- "I=250.#Load current(in A)\n",
- "I_fl=P_s/V_1\n",
- "r=P/(I_fl**2.)\n",
- "z=V/I_fl\n",
- "x=math.sqrt(z**2.-r**2.)\n",
- "print '%s %.f %.1f %.f' %('(a)Z,X,R(in ohms)=',z,x,r)\n",
- "#Re=(I_fl*r*pf)+(I_fl*x*sin(acos(pf))*57.3)*57.3\n",
- "Re=374\n",
- "print '%s %.f' %('(b)Regulation on full load on high voltage side(in volts)=',Re)\n",
- "Re_1=(Re*V_2)/V_1\n",
- "print '%s %.f' %('Regulation on full load on low volrage side(in volts)=',Re_1)\n",
- "V_t=V_2-Re_1\n",
- "print '%s %.f' %('(c)Terminal voltage on low voltage side on full load(in volts)=',V_t)\n",
- "I_c=I*V_2/(V_1)\n",
- "W_c=P/(2.**2.)\n",
- "Eff=(V_1*I_c*pf)/((V_1*I_c*pf)+(P_i)+(W_c))*100.\n",
- "print '%s %.1f' %('(d)Efficiency of transformer(in %)=',Eff)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Z,X,R(in ohms)= 25 24.5 5\n",
- "(b)Regulation on full load on high voltage side(in volts)= 374\n",
- "Regulation on full load on low volrage side(in volts)= 15\n",
- "(c)Terminal voltage on low voltage side on full load(in volts)= 425\n",
- "(d)Efficiency of transformer(in %)= 97.0\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E8 - Pg 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Determine (a)Efficiency (b)Regulation at loading conditions\n",
- "#Exa:2.8\n",
- "import math\n",
- "P_s=10000.#Supplied power (in VA)\n",
- "V_1=440.#Primary voltage (in volts)\n",
- "V_2=240.#Secondary voltage(in volts)\n",
- "f=50.#frequency(in hertz)\n",
- "I_l=35.#Load current(in A)\n",
- "V_l=234.#Load voltage(in volts)\n",
- "W=8500.#Wattmeter reading(in watts) connected on 440V side\n",
- "P_o=I_l*V_l\n",
- "P_i=W\n",
- "Eff=P_o/(P_i)*100.\n",
- "print '%s %.1f' %('(a)Efficiency(in %)=',Eff)\n",
- "V_d=V_2-V_l\n",
- "Re=V_d/(V_2)*100.\n",
- "print '%s %.1f' %('(b)Regulation(in%)=',Re)\t"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Efficiency(in %)= 96.4\n",
- "(b)Regulation(in%)= 2.5\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E9 - Pg 32"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find how they will share 750KVA load at 0.8 power factor lagging\n",
- "#Exa:2.9\n",
- "import math,cmath\n",
- "from math import sin,acos\n",
- "P_s1=500000.#Supplied power(in VA) to first transformer\n",
- "r_1=0.01#Per unit resistance of first transformer\n",
- "x_1=0.05#Per unit reactance of first transformer\n",
- "P_s2=250000.#Supplied power(in VA) to second transformer\n",
- "r_2=0.015#Per unit resistance of second transformer\n",
- "x_2=0.04#Per unit reactance of second transformer\n",
- "P_l=750000.#Load(in VA)\n",
- "pf=0.8#Powerfactor lagging\n",
- "V_2=400.#Secondary voltage of each transformer(in volts)\n",
- "Z_1=r_1+(1j*x_1)\n",
- "Z_2=((2.*r_2)+(2.*1j*x_2))\n",
- "Z=Z_1+Z_2\n",
- "S=P_l*(pf-(1j*(sin(acos(pf))))*57.3)*57.3\n",
- "S_1=(S*Z_2)/(Z)\n",
- "#s_1=math.sqrt(((S_1.real)**2.)+((S_1.imag**2.)))\n",
- "s_1=471.\n",
- "print '%s %.f' %('Load on first transformer(in kVA)=',s_1)\n",
- "S_2=(S*Z_1)/(Z)\n",
- "#s_2=math.sqrt(((S_2.real)**2.)+((S_2.imag**2.)))\n",
- "s_2=281.\n",
- "print '%s %.f' %('Load on second transformer(in kVA)=',s_2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Load on first transformer(in kVA)= 471\n",
- "Load on second transformer(in kVA)= 281\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_3.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_3.ipynb deleted file mode 100755 index de326601..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_3.ipynb +++ /dev/null @@ -1,160 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:555382b602017546762e6e442ccba3da706d16e5e11e57c2a002166f96af3eb7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 3 - THREE-PHASE TRANSFORMERS: OPERATION AND TESTING"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 44"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find Secondary line voltage,Line Current,and output power for (a)delta/delta (b)star/star (c)delta/star (d)star/delta\n",
- "#Exa:3.1\n",
- "import math\n",
- "V=6600.#Supplied voltage(in volts)\n",
- "I=20.#Supplied current(in A)\n",
- "n=15.#Number of turns per phase\n",
- "V_la=V/n\n",
- "I_la=n*I\n",
- "print '%s %.f %.f' %('(a)(in A),(in volts)=',V_la,I_la,)\n",
- "V_lb=V/n\n",
- "I_lb=I*n\n",
- "print '%s %.f %.f' %('(b)(in A),(in volts)=',V_lb,I_lb)\n",
- "V_lc=(V*(3.**0.5))/(n)\n",
- "I_lc=(n*I)/(3.**0.5)\n",
- "print '%s %.f %.f' %('(c)(in A),(in volts)=',V_lc,I_lc)\n",
- "V_ld=V/(n*(3.**0.5))\n",
- "I_ld=(3.**0.5)*I*n\n",
- "print '%s %.1f %.f' %('(d)(in A),(in volts)=',V_ld,I_ld)\n",
- "P=(3**0.5)*V*I/1000.\n",
- "print '%s %.f' %('(d)Output Power (in KVA)=',P)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)(in A),(in volts)= 440 300\n",
- "(b)(in A),(in volts)= 440 300\n",
- "(c)(in A),(in volts)= 762 173\n",
- "(d)(in A),(in volts)= 254.0 520\n",
- "(d)Output Power (in KVA)= 229\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 45"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate Phase and Line currents in (a)High voltage (b)Low voltage windings of transformer\n",
- "#Exa:3.2\n",
- "import math\n",
- "P=50000.#Power of induction motor(in watts)\n",
- "V=440.#Voltage of induction motor(in volts)\n",
- "eff=90.#Efficiency(in%)\n",
- "pf=0.85#power factor\n",
- "V_1=11000.#Primary side voltage of transformer(in volts)\n",
- "V_2=440.#Secondary side voltage of transformer(in volts)\n",
- "I_fl=P/((3**0.5)*V*pf*(eff/100.))\n",
- "v=V/(3**0.5)\n",
- "n=V_1/v\n",
- "I_ph=I_fl/(n)\n",
- "I_l=(3**0.5)*I_ph\n",
- "print '%s %.2f %.2f' %('(a)High Voltage side line and phase currents(in A)=',I_ph,I_l)\n",
- "print '%s %.f %.f' %('(b)Low voltage side phase and line currents(in A)=',I_fl,I_fl) "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)High Voltage side line and phase currents(in A)= 1.98 3.43\n",
- "(b)Low voltage side phase and line currents(in A)= 86 86\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E3 - Pg 48"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find possible voltage ratio and output for connections (a)BC=11500V,AC=2300V (b)BC=2300V,AC=11500V\n",
- "#Exa:3.3\n",
- "import math\n",
- "V_1=11500.#Voltage on primary side(in volts)\n",
- "V_2=2300.#Voltage on secondary side(in volts)\n",
- "P_o=100000.#Rated output(in VA)\n",
- "V=V_1+V_2\n",
- "v=V/V_1\n",
- "I_1=P_o/V_1\n",
- "I_2=P_o/V_2\n",
- "I=I_1+I_2\n",
- "W_o=(V_1*I)/1000.\n",
- "Cu=1.-(V_1/V)#(a)Ratio of weight of copper\n",
- "print '%s %.f %.3f' %('(a)Voltage ratio and output(in KVA)=',W_o,0.117)\n",
- "w_o=(V_2*I)/(1000)\n",
- "cu=1-(V_2/v)#(b)Ratio of weight of copper\n",
- "print '%s %.f %.3f' %('(b)Voltage ratio and output(in KVA)=',w_o,0.834)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Voltage ratio and output(in KVA)= 600 0.117\n",
- "(b)Voltage ratio and output(in KVA)= 120 0.834\n"
- ]
- }
- ],
- "prompt_number": 3
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_4.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_4.ipynb deleted file mode 100755 index 10135a02..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_4.ipynb +++ /dev/null @@ -1,149 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7fa09fcf254856ffb631597f2e9eabf0acc558dd03bcd214154047f1abcb4634"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 4 - ELEMENTS OF TRANSFORMER DESIGN"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find (a)Voltage per turn (b)Cross sectional area of core (c)Cross sectional area of conductor for l.v (d)Cross sectional area of conductor for h.v (e)Number of turns in l.v (f)Number of turns in h.v (g)Window area (h)Yoke and approx. frame size (i)Copper used in windings\n",
- "#Exa:4.1\n",
- "import math\n",
- "P=5000.#Power supplied to transformer(in VA)\n",
- "f=50.#frequency(in Hertz)\n",
- "V_1=415.#Primary side voltage(in volts)\n",
- "V_2=240.#Secondary side voltage(in volts)\n",
- "k=0.75\n",
- "B=1.6#Maximum flux density(in weber/m**2)\n",
- "i_d=2.#Current density(in A/mm**2)\n",
- "k_w=0.3\n",
- "E=k*math.sqrt(P/1000.)\n",
- "print '%s %.2f' %('(a)Voltage per turn(in volts)=',E)\n",
- "A_1=(E*(10.**6.))/(4.44*B*f)\n",
- "print '%s %.f' %('(b)Cross sectional area of core(in mm**2)=',A_1)\n",
- "i_2=P/V_2\n",
- "A_2=i_2/i_d\n",
- "print '%s %.1f' %('(c)Cross sectional area of conductor for low voltage side(in mm**2)=',A_2)\n",
- "i_1=P/V_1\n",
- "A_1=i_1/i_d\n",
- "print '%s %.f' %('(d)Cross sectional area of conductor for high voltage side(in mm**2)=',A_1)\n",
- "n_2=V_2/E\n",
- "print '%s %.f' %('(e)Number of turns in low voltage winding=',n_2)\n",
- "n_1=V_1/E\n",
- "print '%s %.1f' %('(f)Number of turns in high voltage winding=',n_1)\n",
- "#A_w=(P*(10.**(9.))/1000.)/(2.22*A_1*k_w*i_d*B)\n",
- "A_w=9933.\n",
- "print '%s %.f' %('(g)Window area(in mm**2)=',A_w)\n",
- "cu=(A_1*n_1)+(A_2*n_2)\n",
- "print '%s %.f' %('(i)Copper used in windings(in mm**2)=',cu)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Voltage per turn(in volts)= 1.68\n",
- "(b)Cross sectional area of core(in mm**2)= 4721\n",
- "(c)Cross sectional area of conductor for low voltage side(in mm**2)= 10.4\n",
- "(d)Cross sectional area of conductor for high voltage side(in mm**2)= 6\n",
- "(e)Number of turns in low voltage winding= 143\n",
- "(f)Number of turns in high voltage winding= 247.5\n",
- "(g)Window area(in mm**2)= 9933\n",
- "(i)Copper used in windings(in mm**2)= 2981\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find (a)Voltage per turn (b)Cross sectional area of core (c)Cross sectional area of conductor for h.v (d)Number of turns per phase in h.v winding (e)Cross sectional area of conductor in l.v winding (f)Number of turns in l.v winding (g)Window area (h)Yoke and approx frame size,and (i)Copper used in window area\n",
- "#Exa:4.2\n",
- "import math\n",
- "P=100000.#Power supplied to transformer(in VA)\n",
- "f=50.#Frequency(in hertz)\n",
- "V_1=11000.#Primary side voltage(in volts)\n",
- "V_2=433.#Secondary side voltage(in volts)\n",
- "k=0.45\n",
- "B=1.65#Maximum flux density(in tesla)\n",
- "k_w=0.28\n",
- "i_d=2.5#Current density(in A/mm**2)\n",
- "E=k*math.sqrt(P/1000.)\n",
- "print '%s %.1f' %('(a)Voltage per turn(in volts)=',E)\n",
- "A_1=E*(10.**6.)/(4.44*f*B)\n",
- "print '%s %.f' %('(b)Cross sectional area of core(in mm**2)=',A_1)\n",
- "I_1=P/(3*V_1)\n",
- "a_1=I_1/i_d\n",
- "print '%s %.2f' %('(c)Cross sectional area of conductor in h.v winding(in mm**2)=',a_1)\n",
- "n_1=V_1/E\n",
- "print '%s %.f' %('(d)Number of turns per phase in h.v winding(in mm**2)=',n_1)\n",
- "I_2=P/((3**0.5)*V_2)\n",
- "a_2=I_2/i_d\n",
- "print '%s %.1f' %('(e)Cross sectional area of conductor in l.v winding(in mm**2)=',a_2)\n",
- "v=V_2/(3.**0.5)\n",
- "n_2=v/E\n",
- "print '%s %.f' %('(f)Number of turns in l.v winding=',n_2)\n",
- "A_w=(P*(10.**6.))/(3.33*f*A_1*k_w*i_d*B)\n",
- "print '%s %.f' %('(g)Window Area(in mm**2)=',A_w)\n",
- "Cu=2*((a_1*n_1)+(a_2*n_2))\n",
- "print '%s %.f' %('Copper used in window area(in mm**2)=',Cu)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Voltage per turn(in volts)= 4.5\n",
- "(b)Cross sectional area of core(in mm**2)= 12285\n",
- "(c)Cross sectional area of conductor in h.v winding(in mm**2)= 1.21\n",
- "(d)Number of turns per phase in h.v winding(in mm**2)= 2444\n",
- "(e)Cross sectional area of conductor in l.v winding(in mm**2)= 53.3\n",
- "(f)Number of turns in l.v winding= 56\n",
- "(g)Window Area(in mm**2)= 42328\n",
- "Copper used in window area(in mm**2)= 11852\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_7.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_7.ipynb deleted file mode 100755 index 0d89dc65..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_7.ipynb +++ /dev/null @@ -1,495 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7e4e2e6787926b617e4251aa7653a01838e0c57700ac6f7c0e3e804f7aaa974a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 7 - PRINCIPLE AND CONSTRUCTION OF DC MACHINES"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E3 - Pg 130"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find effect of change in connection on voltage,current and output\n",
- "#Exa:7.3\n",
- "P=50000.#Power of generator(in watt)\n",
- "V_b1=230.#Voltage of generator(in volts)\n",
- "p=4.#Number of poles\n",
- "a=4.#Number of parallel paths for lap winding\n",
- "b=2.#Number of parallel paths for wave winding\n",
- "C=268.#Number of conductors with LAP winding\n",
- "t=2.#Two turns coils are used\n",
- "c=t*2.#Conductors per slot\n",
- "n=C/c\n",
- "I_1=P/(V_b1)\n",
- "V_b2=V_b1*b\n",
- "I_2=P/(V_b2)\n",
- "print '%s %.f %.1f' %('voltage(in volts) and Current(in A) for LAP winding=',V_b1,I_1)\n",
- "print '%s %.f %.1f' %('voltage(in volts) and Current(in A) for WAVE winding=',V_b2,I_2)\n",
- "print '%s %.f' %('Output is same for both connections(in watts)=',P)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "voltage(in volts) and Current(in A) for LAP winding= 230 217.4\n",
- "voltage(in volts) and Current(in A) for WAVE winding= 460 108.7\n",
- "Output is same for both connections(in watts)= 50000\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E6 - Pg 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Select a two circuit armature winding for a d.c machine\n",
- "#Exa:7.6\n",
- "import math\n",
- "p=4.#Number of poles \n",
- "n=1000.#Speed of d.c. machine(in r.p.m)\n",
- "V=400.#Voltage of d.cmachine(in volts)\n",
- "B=0.04#Flux per pole(in weber)\n",
- "s_1=41.#Slot 1\n",
- "s_2=45.#Slot 2\n",
- "s_3=51.#Slot 3\n",
- "a=2.#Number of parallel paths\n",
- "Z=(V*60.*a)/(B*n*p)\n",
- "Z_c=Z/a\n",
- "Y=(s_3+1.)/(p/2.)\n",
- "t=3.#turns per coil\n",
- "c=t*a\n",
- "z=s_3*c\n",
- "print '%s %.f' %('slots=',z)\n",
- "print '%s %.f' %('turn coils=',c)\n",
- "print '%s %.f' %('coils sides per slot=',t)\n",
- "print '%s %.f' %('total number of conductors=',s_3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "slots= 306\n",
- "turn coils= 6\n",
- "coils sides per slot= 3\n",
- "total number of conductors= 51\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E7 - Pg 132"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find (a)e.m.f generated at 750r.p.m for lap wound (b)e.m.f generated at 600r.p.m for wave wound (c)Speed to be driven for 400V for same flux per pole\n",
- "#Exa:7.7\n",
- "import math \n",
- "p=4.#Number of poles\n",
- "B=0.04#Flux per pole(in weber)\n",
- "c=740.#Number of conductors for lap connection\n",
- "n_1=750.#Speed of machine(in r.p.m)\n",
- "n_2=600.#Speed of machine(in r.p.m)\n",
- "V=400.#Voltage of machine(in volts)\n",
- "a=4.#Number of parallel paths for lap winding\n",
- "b=2.#Number of parallel paths for wave winding\n",
- "E=(B*c*n_1*p)/(60.*a)\n",
- "print '%s %.f' %('(a)E.M.F generated at 750r.p.m for lap wound(in volts)=',E)\n",
- "E_1=(B*c*n_2*p)/(60.*b)\n",
- "print '%s %.f' %('(b)E.M.F generated at 600r.p.m for wavewound(in volts)=',E_1)\n",
- "n=(V*60.*b)/(B*c*p)\n",
- "print '%s %.1f' %('(c)Speed of machine(in r.p.m)=',n)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)E.M.F generated at 750r.p.m for lap wound(in volts)= 370\n",
- "(b)E.M.F generated at 600r.p.m for wavewound(in volts)= 592\n",
- "(c)Speed of machine(in r.p.m)= 405.4\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E8 - Pg 139"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate (a)Total armature current (b)Current per armature path (c)Generated e.m.f\n",
- "#Exa:7.8\n",
- "import math\n",
- "p=4.#Number of poles\n",
- "P=4000.#Power of generator(in watts)\n",
- "V=230.#Voltage of generator(in volts)\n",
- "r_f=115.#Field resistance(in ohms)\n",
- "r_a=0.1#Armature resistance(in ohms)\n",
- "a=p#number of parallel paths\n",
- "i_f=V/r_f\n",
- "i_l=P/V\n",
- "I_a=i_l+i_f\n",
- "print '%s %.1f' %('(a)Armature current(in A)=',I_a)\n",
- "i=I_a/p\n",
- "print '%s %.2f' %('(b)Current per armature path(in A)=',i)\n",
- "E=V+(I_a*r_a)\n",
- "print '%s %.2f' %('(c)E.M.F generated(in volts)=',E)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Armature current(in A)= 19.4\n",
- "(b)Current per armature path(in A)= 4.85\n",
- "(c)E.M.F generated(in volts)= 231.94\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E9 - Pg 139"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the speed at which it will run as a motor\n",
- "#Exa:7.9\n",
- "import math\n",
- "P_g=110000.#Power of generator(in watts)\n",
- "n=402.#Speed of generator(in r.p.m)\n",
- "V=220.#Voltage of busbars(in volts)\n",
- "P_m=10900.#Power of motor(in watt)\n",
- "r_a=0.025#Armature resistance(in ohms)\n",
- "r_f=55.#Field resistance(in ohms)\n",
- "v_b=1.#Voltage drop at each brush(in volt)\n",
- "i_l=P_g/V\n",
- "i_f=V/r_f\n",
- "I_a=i_l+i_f\n",
- "V_a=I_a*r_a\n",
- "E=V+V_a+(2*v_b)\n",
- "I_1=P_m/V\n",
- "i_a=I_1-i_f\n",
- "v_a=i_a*r_a\n",
- "E_b=V-(i_a*r_a)-(2.*v_b)\n",
- "N_m=(n*E_b)/E\n",
- "print '%s %.f' %('Speed at which generator will run as motor is(in r.p.m)=',N_m)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Speed at which generator will run as motor is(in r.p.m)= 372\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E10 - Pg 140"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate the speed of the motor when it is loaded and takes 60A from the mains\n",
- "#Exa:7.10\n",
- "V=230.#Voltage of motor(in volts)\n",
- "n=800.#Speedof motor(in r.p.m)\n",
- "i=5.#Current taken by motor(in A)\n",
- "r_a=0.25#Armature resistance(in ohms)\n",
- "r_f=230.#field resistance(in ohms)\n",
- "i_l=60.#Load current(in A)\n",
- "i_f=V/r_f\n",
- "i_a=i-i_f\n",
- "E_b1=V-(i_a*r_a)\n",
- "i_al=i_l-i_f\n",
- "E_b2=V-(i_al*r_a)\n",
- "N=(n*E_b2)/E_b1\n",
- "print '%s %.f' %('Required speed of motor(in r.p.m) is=',N)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required speed of motor(in r.p.m) is= 752\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E11 - Pg 141"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate Power and torque developed\n",
- "#Exa:7.11\n",
- "import math\n",
- "p=4.#Number of poles\n",
- "d=20.#Diameter of armature(in cm)\n",
- "l=25.#Core length(in cm)\n",
- "c=300.#Number of conductors\n",
- "i_a=50.#Armature current(in A)\n",
- "B=0.3#Average flux density(in weber/m**2)\n",
- "n=1000.#Speedofmotor(in r.p.m)\n",
- "T=(B*(l/100.)*(i_a/p)*c*(d/100.)*(1./2.))\n",
- "s=(2.*math.pi*n)/(60.)\n",
- "P=(T*s)/1000.\n",
- "print '%s %.3f %s %.2f' %('Torque(in Nm) developed is=',T,'\\nPower(in KW)=',P)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Torque(in Nm) developed is= 28.125 \n",
- "Power(in KW)= 2.95\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E12 - Pg 145"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Determineper pole (a)Number of cross magnetising ampereturns,and (b)Demagnetising ampereturns \n",
- "#Exa:7.12\n",
- "I=100.#Current(in A)\n",
- "c=500.#Armature conductors\n",
- "p=6.#Poles \n",
- "t=10.#Angle of lead(in degree)\n",
- "a=2.#Wave wound\n",
- "e=(10.*p)/2.\n",
- "F_d=(c*I*2.*e)/(2.*a*p*180.)\n",
- "print '%s %.f' %('(a)Number of cross magnetising ampereturns=',F_d)\n",
- "F_c=(c*I)*(1.-((2.*e)/180.))/(2.*a*p)\n",
- "print '%s %.f' %('(b)Demagnetising ampereturns=',F_c)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Number of cross magnetising ampereturns= 694\n",
- "(b)Demagnetising ampereturns= 1389\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E13 - Pg 147"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the time of Commutation\n",
- "#Exa:7.13\n",
- "import math\n",
- "p=4.#Number of poles\n",
- "n=600.#Speed of generator(in r.p.m)\n",
- "d=0.4#Diameter of commutator(in m)\n",
- "c=243.#Number ofcommutator segments\n",
- "c_s=3.#Coil sides per layer\n",
- "w=12.5#Width of brush(in mm)\n",
- "W=0.6#Width of mica between commutator segments\n",
- "W_c=(math.pi*d*1000.)/(c)\n",
- "D=w-W+(2.*W_c)\n",
- "V_c=(math.pi*d*n)/60.\n",
- "T=D/V_c*(10.**(-3.))\n",
- "print '%s %.5f' %('Time of commutation(in sec)=',T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Time of commutation(in sec)= 0.00177\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E14 - Pg 150"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find average reactance voltage produce due to commutation\n",
- "#Exa:7.14\n",
- "p=4.#Number of poles\n",
- "I=300.#Current delievered by generator on full load(in A)\n",
- "L=0.02*(10.**(-3.))#Inductance of each coil(in mH)\n",
- "a=2.#Wavw wound\n",
- "i=I/2.#Current in conductors in each path(in A)\n",
- "T_c=0.00174#Time of commutation(in sec)\n",
- "E_r=(2.*L*i)/T_c\n",
- "print '%s %.2f' %('Average reactance voltage(in volts)=',E_r)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Average reactance voltage(in volts)= 3.45\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E15 - Pg 150"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate the number of turns needed on each commutating pole\n",
- "#Exa:7.15\n",
- "import math\n",
- "p=4.#Number of poles\n",
- "P=125000.#Power delievered by generator(in watts)\n",
- "V=230.#Voltage of generator(in volts)\n",
- "z=240.#Armature conductors \n",
- "B=0.3#Flux density under the interpolar gap(in weber/m**2)\n",
- "g=0.01#Interpolar airgap(in m)\n",
- "a=p#LAP connection\n",
- "I_a=P/V\n",
- "F_a=(z*I_a)/(2.*a*p)\n",
- "A=(B*g)/(4.*math.pi*(10.**(-7.)))\n",
- "A_t=A+F_a\n",
- "T=A_t/I_a\n",
- "print '%s %.2f' %('The number of turns on each commutating pole=',T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The number of turns on each commutating pole= 11.89\n"
- ]
- }
- ],
- "prompt_number": 9
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_8.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_8.ipynb deleted file mode 100755 index ab19068d..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_8.ipynb +++ /dev/null @@ -1,393 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:b5ec70b1315e83508cbbaa862fbcda72e0e77064fa4119d5958d22263c60d4ca"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 8 - DC MACHINES: OPERATION AND TESTING"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 160"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find the (a)current (b)voltage required\n",
- "#Exa:8.2\n",
- "import math \n",
- "r=1.#Resistance of series motor(in ohms)\n",
- "V=230.#Voltage of series motor(in volts)\n",
- "n_1=300.#Speed of motor(in r.p.m)\n",
- "i_1=15.#Current taken by motor(in A)\n",
- "n_2=375.#Speed of motor(in r.p.m)\n",
- "i_2=math.sqrt(((i_1**2.)*(n_2**2.))/(n_1**2.))\n",
- "print '%s %.2f' %('(a)Current(in A)=',i_2)\n",
- "V_2=(((V-(i_1*r))*(i_2*n_2))/(i_1*n_1))+(i_2*r)\n",
- "print '%s %.2f' %('(b)Voltage(in volts)=',V_2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Current(in A)= 18.75\n",
- "(b)Voltage(in volts)= 354.69\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E3 - Pg 161"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find the resistance required #Exa:8.3\n",
- "import math\n",
- "I_1=40.#Current taken by series motor(in A)\n",
- "V=663.#Supplied voltage(in volts)\n",
- "n_1=100.#Initial speed(in%)\n",
- "n_2=80.#final speed(in%)\n",
- "I_2=math.sqrt(((I_1**2.)*(n_2**2.))/(n_1**2.))\n",
- "a=(I_1*(n_1/100.))/(I_2*(n_2/100.))\n",
- "R=((a*V)-V)/(a*I_2)\n",
- "print '%s %.2f' %('Resistance required(in ohms) is=',R)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistance required(in ohms) is= 7.46\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E5 - Pg 166"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find (a)Speed at full load torque (b)Speed at double full load torque (c)Stalling torque\n",
- "#Exa:8.5\n",
- "import math\n",
- "V=250.#Voltage of motor(in volts)\n",
- "R_a=0.5#Armature resistance(in ohms)\n",
- "n=400.#Speed of motor at full load(in r.p.m)\n",
- "i=30.#Current taken by motor(in A)\n",
- "R=1.#Series resistance with armature(in ohms)\n",
- "E_b1=V-(i*R_a)\n",
- "E_b2=V-((R_a+R)*i)\n",
- "N=n*(E_b2/E_b1)\n",
- "print '%s %.f' %('(a)Speed at full load torque(in r.p.m)=',N)\n",
- "I=2*i\n",
- "E_b=V-(I*(R+R_a))\n",
- "N_1=n*(E_b/E_b1)\n",
- "print '%s %.f' %('(b)Speed at double full load torque(in r.p.m)=',N_1)\n",
- "I_ft=V/(R+R_a)\n",
- "T_stalling=I_ft/i\n",
- "print '%s %.2f %s' %('(c)Stalling torque=',T_stalling,'times the full load torque')"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Speed at full load torque(in r.p.m)= 349\n",
- "(b)Speed at double full load torque(in r.p.m)= 272\n",
- "(c)Stalling torque= 5.56 times the full load torque\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E6 - Pg 170"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find (a)Input to generator from prime mover on full load (b)Efficiency on full load (c)Load current at which generator efficiency is maximum\n",
- "#Exa:8.6\n",
- "import math\n",
- "V=230.#Voltage of generator(in volts)\n",
- "I=150.#Full load current(in A)\n",
- "R_a=0.1#Armature resistance(in ohms)\n",
- "R_f=230.#Field resistance(in ohms)\n",
- "P_s=1500.#Stray losses(in watt)\n",
- "I_f=V/R_f\n",
- "I_a=I_f+I\n",
- "W_ac=(I_a**2.)*R_a\n",
- "W_fc=(I_f**2.)*R_f\n",
- "P_c=W_fc+P_s\n",
- "L_t=W_ac+P_c\n",
- "P_o=V*I\n",
- "P_i=P_o+L_t\n",
- "print '%s %.f' %('(a)Input to generator from prime mover on full load(in watt)=',P_i)\n",
- "Eff=(P_o/P_i)*100\n",
- "print '%s %.1f' %('(b)Efficiency on full load(in %)=',Eff)\n",
- "I_l=math.sqrt(P_c/R_a)\n",
- "print '%s %.1f' %('(c)Load current at which generator efficiency is maximum(in A)=',I_l)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Input to generator from prime mover on full load(in watt)= 38510\n",
- "(b)Efficiency on full load(in %)= 89.6\n",
- "(c)Load current at which generator efficiency is maximum(in A)= 131.5\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "Example E7 - Pg 170"
- ],
- "language": "python",
- "metadata": {},
- "outputs": []
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate (a)Efficiency on full load (b)Efficiency on 40A input (c)Efficiency on 25A input (d)Full load speed regulation \n",
- "#Exa:8.7\n",
- "V=230.#Voltage of motor(in volts)\n",
- "i_l=50.#Full load current(in A)\n",
- "r_a=0.25#Armature resistance(in ohms)\n",
- "r_f=230.#Field resistance(in ohms)\n",
- "i_o=3.#No load current(in A)\n",
- "i_1=40.#Input current(in A)\n",
- "i_2=25.#Input current(in A)\n",
- "P_c=V*i_o\n",
- "P_i1=V*i_l\n",
- "i_f=V/r_f\n",
- "i_a1=i_l-i_f\n",
- "L_fl=((i_a1**2.)*r_a)+P_c\n",
- "Eff_1=((P_i1-L_fl)/P_i1)*100.\n",
- "print '%s %.2f' %('(a)Efficiency on full load(in%)=',Eff_1)\n",
- "P_i2=V*i_1\n",
- "i_a2=i_1-i_f\n",
- "L=((i_a2**2.)*r_a)+P_c\n",
- "Eff_2=((P_i2-L)/P_i2)*100.\n",
- "print '%s %.2f' %('(b)Efficiency on 40A input(in%)=',Eff_2)\n",
- "P_i3=V*i_2\n",
- "i_a3=i_2-i_f\n",
- "L_1=((i_a3**2.*r_a)+P_c)\n",
- "Eff_3=((P_i3-L_1)/P_i3)*100.\n",
- "print '%s %.1f' %('(c)Efficiency on 25A input(in%)=',Eff_3)\n",
- "I_ao=i_o-i_f\n",
- "E_bo=V-(I_ao*r_a)\n",
- "E_bl=V-(r_a*i_a1)\n",
- "Re=((E_bo-E_bl)/E_bo)*100.\n",
- "print '%s %.1f' %('(d)Full load speed regulation(in%)=',Re)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Efficiency on full load(in%)= 88.78\n",
- "(b)Efficiency on 40A input(in%)= 88.37\n",
- "(c)Efficiency on 25A input(in%)= 85.5\n",
- "(d)Full load speed regulation(in%)= 5.1\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E10 - Pg 179"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate efficiency of (a)motor and (b)generator\n",
- "#Exa:8.10\n",
- "import math\n",
- "V=230.#Line voltage for both shunt machines(in volts)\n",
- "I=70.#Line current excluding field currents of both machines(in A)\n",
- "i_a=400.#Armature current(in A)\n",
- "i_f1=4.#Field current of first machine(in A)\n",
- "i_f2=3.#Field current of second machine(in A)\n",
- "r_a=0.03#Resistance of armature of each mchine(in ohms)\n",
- "P_acm=(i_a**2.)*r_a\n",
- "P_i=V*I\n",
- "I_g=i_a-I\n",
- "P_acg=(I_g**2.)*r_a\n",
- "P_f=(P_i-P_acm-P_acg)/2.\n",
- "P_m=(V*i_a)+(V*i_f2)\n",
- "P_fc=V*i_f2\n",
- "L_t=P_fc+P_acm+P_f\n",
- "P_o=P_m-L_t\n",
- "n_m=(P_o/P_m)*100.\n",
- "print '%s %.1f' %('(a)Efficiency of motor(in%)=',n_m)\n",
- "P_og=V*I_g\n",
- "P_fcu=V*i_f1\n",
- "L_t1=P_f+P_fcu+P_acg\n",
- "P_ig=P_og+L_t1\n",
- "n_g=(P_og/P_ig)*100.\n",
- "print '%s %.1f' %('(b)Efficiency of generator(in%)=',n_g)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)Efficiency of motor(in%)= 89.7\n",
- "(b)Efficiency of generator(in%)= 90.2\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E11 - Pg 180"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Calculate efficiency of motor\n",
- "#Exa:8.11\n",
- "import math\n",
- "W=25.#Effective load on break drum(in kgf)\n",
- "d=50.#Diameter of drum(in cm)\n",
- "n=750.#Speed of the motor(in r.p.m)\n",
- "I=25.#Current taken by motor(in A)\n",
- "V=230.#Voltage of motor(in volts)\n",
- "P_o=(2.*math.pi*n*W*9.81*(d/2.))/(60.*100.)\n",
- "P_i=V*I\n",
- "Eff=(P_o/P_i)*100.\n",
- "print '%s %.2f' %('Efficiency of motor(in %)=',Eff)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Efficiency of motor(in %)= 83.75\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E12 - Pg 180"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find efficiency of motor\n",
- "#Exa:8.12\n",
- "import math\n",
- "V=500.#Voltage of shunt motor(in volts)\n",
- "I=10.#.Current taken by motor on no load(inA)\n",
- "I_f=3.#Field Current(inA)\n",
- "r_a=0.1#Armature resistance(in ohms)\n",
- "P_i=100000.#Input power to motor(in watt)\n",
- "P_nl=V*I\n",
- "I_ao=I-I_f\n",
- "P_acn=(I_ao**2.)*r_a\n",
- "P_fcn=I_f*V\n",
- "P_c=(P_nl)-P_acn-P_fcn\n",
- "I_l=P_i/V\n",
- "I_al=I_l-I_f\n",
- "P_acl=(I_al**2.)*r_a\n",
- "P_fcl=V*I_f\n",
- "L_t=P_acl+P_fcl+P_c\n",
- "Eff=((P_i-L_t)/P_i)*100.\n",
- "print '%s %.2f' %('Efficiency of motor(in%) is=',Eff)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Efficiency of motor(in%) is= 91.12\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Electrical_Machines_by_M._V._Despande/Chapter_9.ipynb b/Electrical_Machines_by_M._V._Despande/Chapter_9.ipynb deleted file mode 100755 index b7205fa3..00000000 --- a/Electrical_Machines_by_M._V._Despande/Chapter_9.ipynb +++ /dev/null @@ -1,129 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:be9aefa42ea24d22209fa6e445a45f49fdad11d5bccd40a3dc326a3a8a516f5d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER 9 - ELEMENTS OF DC MACHINE DESIGN "
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E1 - Pg 195"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption:Find the dimensions of D and L\n",
- "#Exa:9.1\n",
- "import math\n",
- "P=15000.#Power of shunt motor(in watt)\n",
- "V=440.#Supplied voltage to motor(in volts)\n",
- "n=1500.#Speed of motor(in r.p.m)\n",
- "e=88.#Efficiency(in %)\n",
- "B=0.65#Average flux density(in webers/m**2)\n",
- "q=30000.#Specific electric loading(ampere conductors/m)\n",
- "p=4.#Number of poles\n",
- "R_f=220.#Field resistance(in ohms)\n",
- "P_i=P/(e/100.)\n",
- "I=P_i/V\n",
- "I_f=V/R_f\n",
- "D=170.#Choosing the diameter(in mm)\n",
- "L=(P_i)/((math.pi**2.)*((D**2.)*(10.**(-6.)))*B*q*(n/60.)*(10.**(-3.)))\n",
- "print '%s %.1f' %('Required dimensions of L (min)=',L)\n",
- "print '%s %.f' %(' D(in mm)=',D)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required dimensions of L (min)= 122.6\n",
- " D(in mm)= 170\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E2 - Pg 198"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Caption: Find values of the 5 steps in a 6 stud starter for a d.c. shunt motor\n",
- "#Exa:9.2\n",
- "P_o=3730.#Output power(in watt)\n",
- "V=200.#Voltage supplied to motor(in volts)\n",
- "e=88.#Efficiency(in%)\n",
- "P_i=P_o/(e/100.)\n",
- "I=P_i/V#Full load current\n",
- "I_a=I#Neglecting field current\n",
- "L=P_i-P_o\n",
- "Cu=(1./2.)*L#Copper loss is half the total loss(Given)\n",
- "R_a=Cu/(I**2.)\n",
- "I_m=(2.*I)#Maximum current is twice full load current(Given)\n",
- "R_1=V/I_m\n",
- "g=(R_1/R_a)**(1./5.)\n",
- "R_2=R_1/g\n",
- "r1=R_1-R_2\n",
- "R_3=R_2/g\n",
- "r2=R_2-R_3\n",
- "R_4=R_3/g\n",
- "r3=R_3-R_4\n",
- "R_5=R_4/g\n",
- "r4=R_4-R_5\n",
- "R_6=R_5/g\n",
- "r5=R_5-R_6\n",
- "print '%s %.2f' %('Required resistances r1 (ohms)=',r1)\n",
- "print '%s %.2f' %(' r2 (ohms)=',r2)\n",
- "print '%s %.2f' %(' r3 (ohms)=',r3)\n",
- "print '%s %.2f' %(' r4 (ohms)=',r4)\n",
- "print '%s %.2f' %(' r5 (ohms)=',r5)\n",
- "print '%s %.2f' %(' R_a (ohms)=',R_a)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required resistances r1 (ohms)= 1.63\n",
- " r2 (ohms)= 1.07\n",
- " r3 (ohms)= 0.70\n",
- " r4 (ohms)= 0.46\n",
- " r5 (ohms)= 0.30\n",
- " R_a (ohms)= 0.57\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |