Revised list of TBCs

1 files changed, 1 insertions, 108 deletions

diff --git a/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5.ipynb b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5.ipynb
index dfdf53ef..6e2c62ae 100755..100644
--- a/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5.ipynb
+++ b/basic_electrical_engineering_by_nagsarkar_and_sukhija/Chapter5.ipynb
@@ -34,41 +34,26 @@
    ],
    "source": [
     "import math\n",
-    "\n",
     "#given\n",
     "vl=400 #line voltage\n",
-    "\n",
     "va=vl/math.sqrt(3)\n",
     "vb=230.94  #angle(-120)\n",
     "vc=230.94  #angle(-240)\n",
-    "\n",
     "#case a\n",
-    "\n",
     "#the line currents are given by\n",
-    "\n",
     "ia=12000/230.94  #with angle 0\n",
-    "\n",
     "ib=10000/230.94  #with angle 120\n",
-    "\n",
     "ic=8000/230.94   #with angle 240\n",
-    "\n",
     "print\"ia=\",round(ia,3),\"A\"\n",
     "print \"ib=\",round(ib,5),\"A\"\n",
     "print \"ic=\",round(ic,5),\"A\"\n",
-    "\n",
     "#case b\n",
-    "\n",
     "#IN=ia+ib+ic\n",
-    "\n",
     "#ia,ib and ic are phase currents hence contain with angles they are in the form sin(angle)+icos(angle)\n",
-    "\n",
     "#IN=51.96*(sin(0)+i*cos(0))+43.3*(sin(120)+i*cos(120))+34.64*(sin(240)+i*cos(240))\n",
-    "\n",
     "#IN=51.96+(-21.65+i*37.5)+34.64*(-0.5-i*0.866)\n",
-    "\n",
     "#12.99+i*7.5 on which the sin+icos=sin**2+cos**2 operation is performed\n",
     "#therefore  \n",
-    "\n",
     "IN=15 #at angle 30\n",
     "print \"IN=\",round(IN,10),\"A\"\n",
     "\n"
@@ -103,43 +88,27 @@
    ],
    "source": [
     "import math\n",
-    "\n",
     "#case a\n",
-    "\n",
     "vab=400 #phase angle of 0\n",
     "vbc=400 #phase angle of 120\n",
     "vca=400 #phase angle of 240\n",
-    "\n",
     "#the phase currents are given by iab,ibc,ica\n",
-    "\n",
     "iab=400/150 #from the diagram \n",
-    "\n",
     "print \"iab=\",round(iab,5),\"A\"\n",
     "#ibc=(400*314*50)/10**6  numerator with an angle of -120 and denominator angle of -90 which amounts to -30 in numerator\n",
     "#this leads to simplifying with the formula as the value obtained for ibc after simplification from above mutiplied by values of cos(-30)+jsin(-30)\n",
     "#therefore print as below\n",
-    "\n",
     "print\"ibc=5.4414-j3.1416\",\"A\"\n",
-    "\n",
     "#same method for ica\n",
-    "\n",
-    "\n",
     "print \"ica=3.1463+j4.2056\",\"A\"\n",
-    "\n",
     "#case b\n",
-    "\n",
     "#ia=iab-ica\n",
-    "\n",
     "#ia=2.667-(3.1463+j4.2056)\n",
-    "\n",
     "#leads to 4.2328 with an angle of -96.51\n",
     "#angle calculated using tan formula\n",
     "print \"ia=4.2328 with an angle of -96.51\",\"A\"\n",
-    "\n",
     "#same for ib and ic\n",
-    "\n",
     "print \"ib=4.1915 with angle of -48.55\",\"A\"\n",
-    "\n",
     "print \"ic=7.6973 with an angle of 107.35\",\"A\""
    ]
   },
@@ -170,40 +139,25 @@
    ],
    "source": [
     "import math\n",
-    "\n",
     "#case a\n",
-    "\n",
     "#given\n",
     "zl=5 #load impedanc with an angle of 36.87 degrees\n",
     "vl=400 #line voltage\n",
     "il=46.19\n",
     "va=400/3**0.5 #phase voltage\n",
-    "\n",
     "ia=va/zl  #line current with an angle of -36.87 degrees\n",
-    "\n",
     "#ib and ic are also the same values with changes in in their angles\n",
-    "\n",
     "#case b\n",
     "#cos(-36.87)=0.8 lagging\n",
-    "\n",
     "print \"power factor =0.8\"\n",
-    "\n",
     "#case c\n",
-    "\n",
     "p=3**0.5*vl*il*0.8 #power where 0.8 is power factor\n",
-    "\n",
     "print\"p=\",round(p,2),\"KW\"\n",
-    "\n",
     "#case d\n",
-    "\n",
     "q=3**0.5*vl*il*0.6 #where 0.6 is sin(36.87) and q is reactive volt ampere\n",
-    "\n",
     "print\"q=\",round(q,2),\"Kvar\"\n",
-    "\n",
     "#case e\n",
-    "\n",
     "t=3**0.5*vl*il #total volt ampere\n",
-    "\n",
     "print \"t=\",round(t,0),\"KVA\"\n",
     "\n"
    ]
@@ -238,39 +192,25 @@
    ],
    "source": [
     "import math\n",
-    "\n",
     "#given\n",
-    "\n",
     "za=50\n",
     "zb=15  #j15\n",
     "zc=-15 #-j15\n",
-    "\n",
     "vl=440\n",
-    "\n",
     "vab=440 #with an angle of 0\n",
-    "\n",
     "vbc=440 #with an angle of -120\n",
-    "\n",
     "vca=440 #with an angle of -240\n",
-    "\n",
     "#applying kvl to meshes as in the diagram we get the following equations\n",
-    "\n",
     "#50i1+j15(i1-i2)-440(angle 0)=0,j15(i2-i1)+(-j15)i2-440(angle 120)=0\n",
-    "\n",
     "#solving the above 2 eqns we get the values of ia,ib and ic as follows\n",
-    "\n",
     "print \"ia=29.33A\" #at angle -30\n",
     "print \"ib=73.83A\" #at angle -131.45\n",
     "print \"ic=73.82A\" #at angle 71.5\n",
-    "\n",
     "#the voltage drops across vr,vl and vc which are voltages across resistance ,inducctance and capacitance are given as follows\n",
-    "\n",
     "print \"vr=1466.5V\" #at angle -30\n",
     "print \"vl=73.83V\"  #at angle -41.45\n",
     "print \"vc=73.83V\"  #at angle -18.5\n",
-    "\n",
     "#the potential of neutral point\n",
-    "\n",
     "print \"vn=1212.45V\" #at angle 150\n"
    ]
   },
@@ -299,24 +239,16 @@
    ],
    "source": [
     "import math\n",
-    "\n",
     "#given\n",
-    "\n",
     "v=440 #voltage\n",
     "o=25000 #output power\n",
     "e=0.9 #efficiency\n",
     "p=0.85 #poer factor\n",
-    "\n",
     "#case a\n",
-    "\n",
     "il=o/(3**0.5*v*p*e) #line current\n",
-    "\n",
     "print \"il=\",round(il,5),\"A\"\n",
-    "\n",
     "#case b\n",
-    "\n",
     "ip=o/(3*v*e*p) #phase current for delta current winding\n",
-    "\n",
     "print \"ip=\",round(ip,5),\"A\"\n"
    ]
   },
@@ -353,55 +285,32 @@
    ],
    "source": [
     "import math\n",
-    "\n",
     "#given\n",
-    "\n",
     "#25kW at power factor 1 for branch AB\n",
     "#40KVA at power factor 0.85 for branch BC\n",
     "#30KVA at power factor 0.6 for branch CA\n",
-    "\n",
     "#line voltages with vab as reference phasor\n",
-    "\n",
     "vab=415 #at angle 0\n",
     "vbc=415 #at angle -120\n",
     "vca=415 #at angle -240\n",
-    "\n",
     "#phase currents are given with x+jy form of an imaginary number and vary according to angles.The values below are only the values of the currents without conversion into imaginary form\n",
-    "\n",
     "iab=(25*10**3)/(3**0.5*415*1)\n",
-    "\n",
     "print \"iab=\",round(iab,3),\"A\"\n",
-    "\n",
     "ibc=(40*10**3)/(3**0.5*415)\n",
-    "\n",
     "print \"ibc=\",round(ibc,3),\"A\"\n",
-    "\n",
     "ica=(30*10**3)/(3**0.5*415)\n",
-    "\n",
     "print \"ica=\",round(ica,3),\"A\"\n",
-    "\n",
     "#the line currents are as below.The following values can also be converted to x+iy form where x is real and y is imaginary\n",
-    "\n",
     "#ia=iab-ibc and subtraction is done of x+iy forms where the value of the term varies as obtained by sqrt(x**2+y**2)\n",
-    "\n",
     "print \"ia=76.38A\" #at angle -3.75\n",
-    "\n",
     "#ib=ibc-iab\n",
-    "\n",
     "print \"ib=87.85A\"\n",
-    "\n",
     "#ic=ica-ibc\n",
-    "\n",
     "print \"ic=32.21A\"\n",
-    "\n",
     "#wattmeter readings on phase A\n",
-    "\n",
     "#w1=vab*ia*cos(-3.35) where the cos angle is given by phase angle between ia and vab\n",
-    "\n",
     "print \"w1=31.63KW\"\n",
-    "\n",
     "#same formula for wattmeter readings in phase c where the angle is 16.35\n",
-    "\n",
     "print \"w2=12.827KW\""
    ]
   },
@@ -432,40 +341,24 @@
    ],
    "source": [
     "import math\n",
-    "\n",
     "#given\n",
-    "\n",
     "w1=500\n",
     "w2=200\n",
     "w=w1+w2\n",
-    "\n",
     "#case a\n",
-    "\n",
     "print \"the total input power=\",round(w,0),\"KW\"\n",
-    "\n",
     "#case b\n",
-    "\n",
     "#tan(angle)=3**0.5*(w1-w2)/(w1+w2) where the angle=36.58 and cos(36.58)=0.803 which is the power factor\n",
-    "\n",
     "print \"power factor=0.803\"\n",
-    "\n",
     "#case c\n",
-    "\n",
     "#given\n",
-    "\n",
     "vl=2200\n",
-    "\n",
     "il=w/(3**0.5*vl*0.803)  #0.803 is the value of the cos angle and il is the line current\n",
-    "\n",
     "print \"il=\",round(il,5),\"A\"\n",
-    "\n",
     "#case d\n",
-    "\n",
     "#efficiency=o/i #i is input and o is output\n",
-    "\n",
     "hp=746 #horse power\n",
     "o=0.9*w/hp #0.9 is efficiency\n",
-    "\n",
     "print \"output=\",round(o,3),\"hp\"\n",
     "\n",
     "\n"
@@ -497,7 +390,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython2",
-   "version": "2.7.9"
+   "version": "2.7.5"
   }
  },
  "nbformat": 4,