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-rwxr-xr-xsample_notebooks/Jaya Sravya/Chapter10.ipynb694
-rwxr-xr-xsample_notebooks/KumarRaju/Chapter12.ipynb647
-rwxr-xr-xsample_notebooks/MohdAkhlak/ch3.ipynb256
-rwxr-xr-xsample_notebooks/PradeepChoudari/Chapter01.ipynb326
-rwxr-xr-xsample_notebooks/RajbeeVakkalagadda/Chapter04.ipynb476
-rwxr-xr-xsample_notebooks/ajinkyakhair/chapter2.ipynb240
-rwxr-xr-xsample_notebooks/kartiksankhla/Chapter2.ipynb159
-rwxr-xr-xsample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer).ipynb278
-rwxr-xr-xsample_notebooks/kumargugloth/Chapter1.ipynb348
-rwxr-xr-xsample_notebooks/nemichand /Chapter1_1.ipynb227
-rwxr-xr-xsample_notebooks/saikiranm/CHAPTER4.ipynb196
-rwxr-xr-xsample_notebooks/saikomalchanagam/AKmaini_(1).ipynb293
-rwxr-xr-xsample_notebooks/vinodkumar/Chapter1.ipynb339
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+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7afbb46d48c2bdf700c3dc220becf18da8796860629c8c3b8505b530a4a07b18"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10:Rotational Mechanics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 10.1\n",
+ "#calculation of the number of revolutions made\n",
+ "#given data\n",
+ "import math \n",
+ "wzero=100.*2.*math.pi/60.#initial angular velocity(in rad/s) of the motor\n",
+ "w=0#final angular velocity(in rad/s) of the motor\n",
+ "t=15.#time interval(in s)\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(w-wzero)/t#equation of angular motion\n",
+ "theta=(wzero*t)+(alpha*t*t/2.)#equation of angular motion\n",
+ "\n",
+ "print '%s %.2f' %(\"the number of revolutions the motor makes before coming to rest is\",theta/(2*math.pi))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the number of revolutions the motor makes before coming to rest is 12.50\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1w - Pg 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.1w\n",
+ "#calculation of the number of revolutions made by the wheel \n",
+ "\n",
+ "#given data\n",
+ "wzero=0#initial angular velocity(in rad/s) of the wheel \n",
+ "alpha=2.#angular acceleration(in rad/s**2)\n",
+ "t=10#time(in s) interval\n",
+ "\n",
+ "#calculation\n",
+ "theta=(wzero*t)+(alpha*t*t/2.)#equation of angular motion\n",
+ "n=round(theta/(2.*math.pi))#number of revolutions\n",
+ "\n",
+ "print '%s %.2f' %(\"the number of revolutions made by the wheel is\",n)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the number of revolutions made by the wheel is 16.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.2\n",
+ "#calculation of the time taken by the fan to attain half of the maximum speed\n",
+ "\n",
+ "#given data\n",
+ "wzero=0#initial angular velocity(in rad/s) of the fan\n",
+ "w=400.*(2.*math.pi/60.)#final angular velocity(in rad/s) of the fan\n",
+ "t=5#tiem(in s) taken\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(w-wzero)/t#equation of angular motion\n",
+ "wdash=w/2.#half of maximum speed\n",
+ "t1=(wdash-wzero)/alpha#equation of angular motion\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the time taken by the fan to attain half of the maximum speed is\",t1,\"s\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the time taken by the fan to attain half of the maximum speed is 2.50 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2w - Pg 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.2w\n",
+ "#calculation of the angle rotated during the next second\n",
+ "\n",
+ "#given data\n",
+ "theta=2.5#angular displacement(in rad) of the wheel\n",
+ "t=1.#time(in s) required\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(theta*2.)/(t*t)#equation of angular motion\n",
+ "theta1=(alpha*(t+1.)*(t+1.)/2.)#angle rotated during first two seconds\n",
+ "thetar=theta1-theta#angle rotated during next second\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the angle rotated during the next second is\",thetar,\"rad\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angle rotated during the next second is 7.50 rad\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 10.3\n",
+ "#calculation of the angular velocity and angular acceleration of the pulley \n",
+ "\n",
+ "#given data\n",
+ "v=20.#linear speed(in cm/s) of the bucket\n",
+ "r=10.#radius(in cm) of the pulley\n",
+ "a=4.*10.**2.#linear acceleration(in cm/s**2) of the pulley\n",
+ "\n",
+ "#calculation\n",
+ "w=v/r#formula of angular velocity\n",
+ "alpha=a/r#formula of angular acceleration\n",
+ "\n",
+ "print '%s %.2f %s %s %.2f %s' %(\"the angular velocity of the pulley is\",w,\"rad/s\",\"and angular acceleration of the pulley is\",alpha,\"rad/s**2\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular velocity of the pulley is 2.00 rad/s and angular acceleration of the pulley is 40.00 rad/s**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3w - Pg 28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.3w\n",
+ "#calculation of the torque required to stop the wheel in one minute\n",
+ "\n",
+ "#given data\n",
+ "wzero=50.*(2.*math.pi/60.)#initial angular velocity(in rad/s) of the wheel\n",
+ "w=0#final angular velocity(in rad/s) of the wheel\n",
+ "t=60.#time(in s) taken to stop the wheel\n",
+ "I=2.#moment of inertia(in kg-m**2) of the wheel\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(w-wzero)/t#equation of angular motion\n",
+ "tau=I*abs(alpha)#torque\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the torque required to stop the wheel in one minute is\",tau,\"N-m\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the torque required to stop the wheel in one minute is 0.17 N-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4w - Pg 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.4w\n",
+ "#calculation of the angular velocity of the wheel\n",
+ "\n",
+ "#given data\n",
+ "F=20.#force(in N) of pull applied\n",
+ "I=.2#moment of inertia(in kg-m**2)\n",
+ "r=20.*10.**-2.#radius(in m) of the wheel\n",
+ "t=5.#time(in s) interval\n",
+ "wzero=0#initial angular velocity(in rad/s) of the wheel \n",
+ "\n",
+ "#calculation\n",
+ "tau=F*r#torque applied to the wheel\n",
+ "alpha=tau/I#angular acceleration\n",
+ "w=wzero+(alpha*t)#equation of angular motion\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the angular velocity of the wheel after 5 s is\",w,\"rad/s\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the angular velocity of the wheel after 5 s is 100.00 rad/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.5\n",
+ "#calculation of the moment of inertia of the wheel\n",
+ "\n",
+ "#given data\n",
+ "r=10.*10.**-2.#radius(in m) of the wheel\n",
+ "F=5.#force(in N) of pulling\n",
+ "aplha=2.#angular acceleration(in rad/s**2) of the wheel\n",
+ "\n",
+ "#calculation\n",
+ "tau=F*r#net torque\n",
+ "I=tau/aplha#moment of inertia\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the moment of inertia of the wheel is\",I,\"kg-m**2\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the moment of inertia of the wheel is 0.25 kg-m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7w - Pg 31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.7w\n",
+ "#calculation of the position of second kid on a balanced seesaw\n",
+ "\n",
+ "#given data\n",
+ "ma=10.#mass(in kg) of kid A\n",
+ "mb=15.#mass(in kg) of kid B\n",
+ "l=5.#length(in m) of the seesaw\n",
+ "la=(l/2.)#distance of A kid from fulcrum as he is sitting at an end\n",
+ "\n",
+ "#calculation\n",
+ "#taking torque about fulcrum...........(mb*g*x) = (ma*g*)\n",
+ "x=(ma*la)/mb\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the second kid should sit at a distance of\",x,\"m from the centre\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the second kid should sit at a distance of 1.67 m from the centre\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8w - Pg 31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 10.8w\n",
+ "#calculation of the normal force and the frictional force that the floor exerts on the ladder\n",
+ "\n",
+ "#given data\n",
+ "m=10.#mass(in kg) of the ladder\n",
+ "theta=53.#angle(in degree) made by the ladder against the vertical wall\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "#taking horizontal and vertical components\n",
+ "#N1 = f........................(1)\n",
+ "#N2 = W........................(2)\n",
+ "#taking torque about B\n",
+ "W=m*g\n",
+ "N2=W#from equation (2)\n",
+ "f=(W*math.sin(theta)*57.3/2.)/(math.cos(theta)*57.3)#from equation (1)\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the normal force that the floor exerts on the ladder is\",N2,\"N\\n\")\n",
+ "print '%s %.2f %s' %(\"the frictional force that the floor exerts on the ladder is\",f,\"N\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the normal force that the floor exerts on the ladder is 98.00 N\n",
+ "\n",
+ "the frictional force that the floor exerts on the ladder is -21.13 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9w - Pg 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.9w\n",
+ "#calculation of the contact force exerted by the floor on each leg of ladder\n",
+ "\n",
+ "#given data\n",
+ "theta=60.#angle(in degree) between the two legs\n",
+ "m=80.#mass(in kg) of the person\n",
+ "g=9.8#gravitational acceleration(in m/s**2) of the earth\n",
+ "\n",
+ "#calculation\n",
+ "N=m*g/2.\n",
+ "T=(N*2.*math.tan(90-theta)*57.3)/1.\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the contact force exerted by the floor on each leg of ladder\",N,\"N\\n\")\n",
+ "print '%s %.2f %s' %(\"the tension in the crossbar is\",T,\"N\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the contact force exerted by the floor on each leg of ladder 392.00 N\n",
+ "\n",
+ "the tension in the crossbar is -287747.97 N\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.12\n",
+ "#calculation of the kinetic energy of the sphere\n",
+ "\n",
+ "#given data\n",
+ "M=200.*10.**-3.#mass(in kg) of the sphere\n",
+ "vcm=2.*10.**-2.#speed(in m/s) of the sphere\n",
+ "\n",
+ "#calculation\n",
+ "#kinetic energy is K = (Icm*w*w/2) + (M*vcm*vcm/2)\n",
+ "#taking Icm = (2*M*r*r*w*w/5) and w=vcm/r\n",
+ "K=(M*vcm*vcm/5.)+(M*vcm*vcm/2.)#kinetic energy\n",
+ "\n",
+ "print '%s %.5f %s' %(\"the kinetic energy of the sphere is\",K,\"J\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the kinetic energy of the sphere is 0.00006 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E13w - Pg 37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.13w\n",
+ "#calculation of the kinetic energy and angular momentum of the disc\n",
+ "\n",
+ "#given data\n",
+ "M=200.*10.**-3.#mass(in kg) of the disc\n",
+ "r=4.*10.**-2.#radius(in m) of the disc\n",
+ "w=10.#angular velocity(in rad/s) \n",
+ "\n",
+ "#calculation\n",
+ "I=(M*r*r)/4.#moment of inertia\n",
+ "K=(I*w*w/2.)#kinetic energy\n",
+ "L=I*w#angular momentum\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the kinetic energy of the disc is\",K,\"J\")\n",
+ "print '%s %.2f %s' %(\"\\nthe angular momentum of the disc is\",L,\"J-s\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the kinetic energy of the disc is 0.00 J\n",
+ "\n",
+ "the angular momentum of the disc is 0.00 J-s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14w - Pg 38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 10.14w\n",
+ "#calculation of the work done by the torque in first two seconds\n",
+ "#given data\n",
+ "wzero=20.#initial angular velocity(in rad/s) of the motor\n",
+ "w=0#final angular velocity(in rad/s) of the motor\n",
+ "t=4.#time(in s) taken to attain rest position\n",
+ "I=.20#moment of inertia(in kg-m**2) of the disc about axis of rotation\n",
+ "t1=2.#time(in s)\n",
+ "\n",
+ "#calculation\n",
+ "alpha=(wzero-w)/t#equation of angular motion in case of deceleration\n",
+ "tau=I*alpha#torque\n",
+ "theta=(wzero*t1)-(alpha*t1*t1/2)#equation of angular motion\n",
+ "W=tau*theta#work done by the torque\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the work done by the torque in first two seconds is\",W,\"J\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the work done by the torque in first two seconds is 30.00 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E19w - Pg 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#example 10.19w\n",
+ "#calculation of the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point\n",
+ "\n",
+ "#given data\n",
+ "m=1.2#mass(in kg) of the sphere\n",
+ "R=10.*10.**-2.#radius(in cm) of the sphere\n",
+ "sep=50.*10.**-2.#separation(in m) between the two spheres\n",
+ "\n",
+ "#calculation\n",
+ "d=sep/2.#distance of each sphere from centre\n",
+ "Icm=(2.*m*R*R)/5.#moment of inertia about diameter\n",
+ "I=Icm+(m*d*d)#by parallel axis theorem,moment of inertia about given axis \n",
+ "#since second sphere has same moment of inertia\n",
+ "Isys=2.*I#moment of inertia of the system\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point is\",Isys,\"kg-m**2\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the moment of inertia of the system about the axis perpendicular to the rod passing through its middle point is 0.16 kg-m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E22w - Pg 42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#example 10.22w\n",
+ "#calculation of the number of revolutions made by the wheel per second\n",
+ "\n",
+ "#given data\n",
+ "p=220.*10.**-2.#perimeter(in cm) of the wheel\n",
+ "v=9.*10.**3./(60.*60.)#linear speed(in m/s) of wheel on the road\n",
+ "\n",
+ "#calculation\n",
+ "r=p/(2.*math.pi)#radius of the wheel\n",
+ "w=v/r#angular speed\n",
+ "n=w/(2.*math.pi)#number of revolutions\n",
+ "\n",
+ "print '%s %.2f %s' %(\"the number of revolutions made by the wheel per second is\",n,\"rev/s\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the number of revolutions made by the wheel per second is 1.14 rev/s\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/KumarRaju/Chapter12.ipynb b/sample_notebooks/KumarRaju/Chapter12.ipynb
new file mode 100755
index 00000000..d309457b
--- /dev/null
+++ b/sample_notebooks/KumarRaju/Chapter12.ipynb
@@ -0,0 +1,647 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2259086c8cee958c316a471676c823626713e37889548ae1aff2070252fca121"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter12:Power System Transients"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "rc=.5e-2;\n",
+ "rs=1.5e-2;\n",
+ "u=4\n",
+ "\n",
+ "L=2e-7 * math.log(rs/rc);\n",
+ "print '%s %.2f %s' %(\"\\nL=\",L*1e7,\"e-7H/m\")\n",
+ "C=u*1e-9/(18 * math.log(rs/rc))\n",
+ "print '%s %.2f %s' %(\"\\nC=\", C*1e9,\"e-9F/m\")\n",
+ "v=1/math.sqrt(L*C);\n",
+ "print '%s %.2f %s' %(\"\\nv=\", v*1e-8,\"e8m/s\")\n",
+ "Zc=math.sqrt(L/C)\n",
+ "print '%s %.2f %s' %(\"\\nZc=\",Zc,\"ohm\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "L= 2.20 e-7H/m\n",
+ "\n",
+ "C= 0.20 e-9F/m\n",
+ "\n",
+ "v= 1.50 e8m/s\n",
+ "\n",
+ "Zc= 32.96 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "ef=100.;\n",
+ "Zc=400.;\n",
+ "Z=50.;\n",
+ "et=2.*ef*Z/(Z+Zc)\n",
+ "print '%s %.2f %s' %(\"Surge transmitted=\",et,\"kV\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Surge transmitted= 22.22 kV\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "ef=200.;\n",
+ "Zc=400.;\n",
+ "Z1=500.;\n",
+ "Z2=300.;\n",
+ "et=2.*ef*(Z1*Z2/(Z1+Z2))/((Z1*Z2/(Z1+Z2))+Zc)\n",
+ "print '%s %.2f %s' %(\"\\nSurge Voltage transmitted=\",et,\"kV\\n\")\n",
+ "it1=et/Z1;\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current transmitted=\",it1,\"kA\\n\")\n",
+ "it2=et/Z2;\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current transmitted=\",(it2*100.)/100.,\"kA\\n\")\n",
+ "er=et-ef;\n",
+ "print '%s %.2f %s' %(\"\\nSurge Voltage Reflected=\",er,\"kV\\n\")\n",
+ "ir=it1+it2-(ef/Zc)\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current Reflected=\",ir,\"kA\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Surge Voltage transmitted= 127.66 kV\n",
+ "\n",
+ "\n",
+ "Surge Current transmitted= 0.26 kA\n",
+ "\n",
+ "\n",
+ "Surge Current transmitted= 0.43 kA\n",
+ "\n",
+ "\n",
+ "Surge Voltage Reflected= -72.34 kV\n",
+ "\n",
+ "\n",
+ "Surge Current Reflected= 0.18 kA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "E=100.\n",
+ "Zc=400.\n",
+ "L=4000.\n",
+ "\n",
+ "print '%s %.2f %s %.2f' %(\"et=\",2*E,\"math.exp( - %.1f t) KV\\n\",Zc/L)\n",
+ "print '%s %.2f %s %.2f' %(\"er=\",E,\"(2*math.exp( - %.1f t) -1) KV\\n\",Zc/L)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "et= 200.00 math.exp( - %.1f t) KV\n",
+ " 0.10\n",
+ "er= 100.00 (2*math.exp( - %.1f t) -1) KV\n",
+ " 0.10\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E7 - Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "V=300e3\n",
+ "R=400.\n",
+ "k=1.5e-27\n",
+ "\n",
+ "E=10.\n",
+ "x=1.\n",
+ "e=1e-5\n",
+ "while (E>e):\n",
+ " f=(k*R*x**6.) +x -(2.*V)\n",
+ " df=(6.* k*R*x**5.) +1.\n",
+ " x1=x-(f/df)\n",
+ " E=abs(x1-x)\n",
+ " x=x1\n",
+ "eA=round(x)\n",
+ "IA=k*eA**6.\n",
+ "\n",
+ "print '%s %.2f %s %.2f' %(\"eA=\",eA,\"Ia=\",IA)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "eA= 97101.00 Ia= 1257.28\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E8 - Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "V=300e3\n",
+ "R1=400.\n",
+ "R2=50.\n",
+ "R=1.+(400./50.)\n",
+ "k=1.5e-27\n",
+ "\n",
+ "E=10.\n",
+ "x=1.\n",
+ "e=1e-5\n",
+ "while (E>e) :\n",
+ " f=(k*R1*x**6.) +(R*x) -(2.*V)\n",
+ " df=(6.* k*R1*x**5.) +R\n",
+ " x1=x-(f/df)\n",
+ " E=abs(x1-x)\n",
+ " x=x1\n",
+ "eA=round(x)\n",
+ "IA=k*eA**6.\n",
+ "\n",
+ "print '%s %.2f %s %.2f' %(\"eA=\",eA,\"Ia=\",IA)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "eA= 62640.00 Ia= 90.62\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E9 - Pg 45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "\n",
+ "ef=3000.;\n",
+ "Zc=300.;\n",
+ "ea=1700.;\n",
+ "iF=ef/Zc\n",
+ "print '%s %.2f %s' %(\"\\nCurrent in line=\",iF,\"kA\\n\")\n",
+ "Ia=((2.*ef)-ea)/Zc\n",
+ "print '%s %.2f %s' %(\"\\nCurrent through Arrester=\",Ia,\"kA\\n\")\n",
+ "Ia=round(Ia *1000.)/1000.\n",
+ "R=ea/Ia\n",
+ "print '%s %.2f %s' %(\"\\nresistance of arrester=\",R,\"ohm\\n\")\n",
+ "er=ea-ef;\n",
+ "print '%s %.2f %s' %(\"\\nSurge Voltage Reflected=\",er,\"kV\\n\")\n",
+ "Cr=er/ef;\n",
+ "CR=ea/ef;\n",
+ "print '%s %.2f %s %.2f' %(\"\\nCoeff of Reflection =\",Cr,\"Coeff of Refraction=\",CR)\n",
+ "Cr=(R-Zc)/(R+Zc);\n",
+ "CR=(R*2)/(R+Zc);\n",
+ "print '%s %.2f %s %.2f' %(\"\\nVerification: Coeff of Reflection =\",Cr,\"Coeff of Refraction=\",CR)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Current in line= 10.00 kA\n",
+ "\n",
+ "\n",
+ "Current through Arrester= 14.33 kA\n",
+ "\n",
+ "\n",
+ "resistance of arrester= 118.61 ohm\n",
+ "\n",
+ "\n",
+ "Surge Voltage Reflected= -1300.00 kV\n",
+ "\n",
+ "\n",
+ "Coeff of Reflection = -0.43 Coeff of Refraction= 0.57\n",
+ "\n",
+ "Verification: Coeff of Reflection = -0.43 Coeff of Refraction= 0.57\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E10 - Pg 45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "ef=10000.;\n",
+ "Zc=400.;\n",
+ "iF=ef/Zc\n",
+ "print '%s %.2f' %(\"\\n(a)\\nIncident Wave magnitude= A\",iF)\n",
+ "\n",
+ "R=1000\n",
+ "et=ef*(R*2)/(R+Zc);\n",
+ "it=et/R;\n",
+ "er=et-ef;\n",
+ "print '%s %.2f' %(\"\\n(b)\\nSurge Voltage Reflected= KV\",er/1000)\n",
+ "ir=-1*er/Zc\n",
+ "print '%s %.2f' %(\"\\nSurge Current Reflected= A\",ir)\n",
+ "edr=et*it;\n",
+ "print '%s %.2f' %(\"\\nRate of dissipation of energy= KW\",edr/1000)\n",
+ "err=er*-ir;\n",
+ "print '%s %.2f' %(\"\\nRate of reflection of energy= KW\",err/1000)\n",
+ "\n",
+ "print '%s %.2f' %(\"\\n(c)\\nfor complete dissipation, R=Zc= ohm\",Zc);\n",
+ "\n",
+ "R=50\n",
+ "et=ef*(R*2)/(R+Zc);\n",
+ "print '%s %.2f' %(\"\\n(d)\\nSurge Voltage Transmitted= KV\",et/1000)\n",
+ "it=et/R;\n",
+ "print '%s %.2f' %(\"\\nSurge Current Transmitted= A\",it)\n",
+ "er=et-ef;\n",
+ "print '%s %.2f' %(\"\\nSurge Voltage Reflected= kV\",er/1000)\n",
+ "ir=-1*er/Zc\n",
+ "print '%s %.2f' %(\"\\nSurge Current Reflected= A\",ir)\n",
+ "edr=et*it;\n",
+ "print '%s %.2f' %(\"\\nRate of dissipation of energy= KW\",edr/1000)\n",
+ "err=er*-ir;\n",
+ "print '%s %.2f' %(\"\\nRate of reflection of energy= KW\",err/1000)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "(a)\n",
+ "Incident Wave magnitude= A 25.00\n",
+ "\n",
+ "(b)\n",
+ "Surge Voltage Reflected= KV 4.29\n",
+ "\n",
+ "Surge Current Reflected= A -10.71\n",
+ "\n",
+ "Rate of dissipation of energy= KW 204.08\n",
+ "\n",
+ "Rate of reflection of energy= KW 45.92\n",
+ "\n",
+ "(c)\n",
+ "for complete dissipation, R=Zc= ohm 400.00\n",
+ "\n",
+ "(d)\n",
+ "Surge Voltage Transmitted= KV 2.22\n",
+ "\n",
+ "Surge Current Transmitted= A 44.44\n",
+ "\n",
+ "Surge Voltage Reflected= kV -7.78\n",
+ "\n",
+ "Surge Current Reflected= A 19.44\n",
+ "\n",
+ "Rate of dissipation of energy= KW 98.77\n",
+ "\n",
+ "Rate of reflection of energy= KW 151.23\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E11 - Pg 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "Zc=400.\n",
+ "ef=20.\n",
+ "z1=150.;\n",
+ "z2=200.\n",
+ "z=round((z1*z2/(z1+z2))*100.)/100.\n",
+ "\n",
+ "et=2*ef*z/(Zc+z)\n",
+ "print '%s %.2f %s' %(\"\\nSurge Voltage Transmitted=\",et,\"kV\\n\")\n",
+ "\n",
+ "it1=et*1000./z1;\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current Transmitted in line 1=\",it1,\"A\\n\")\n",
+ "\n",
+ "it2=et*1000./z2;\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current Transmitted in line 2=\",it2,\"A\\n\")\n",
+ "\n",
+ "er=et-ef\n",
+ "print '%s %.2f %s' %(\"\\nSurge Voltage Reflected=\",er,\"kV\\n\")\n",
+ "ir=-1*er*1000./Zc\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current Reflected=\",ir,\"A\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Surge Voltage Transmitted= 7.06 kV\n",
+ "\n",
+ "\n",
+ "Surge Current Transmitted in line 1= 47.06 A\n",
+ "\n",
+ "\n",
+ "Surge Current Transmitted in line 2= 35.29 A\n",
+ "\n",
+ "\n",
+ "Surge Voltage Reflected= -12.94 kV\n",
+ "\n",
+ "\n",
+ "Surge Current Reflected= 32.35 A\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E12 - Pg 49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "ef=100.\n",
+ "Zc=400.\n",
+ "z=50.\n",
+ "R=z+Zc;\n",
+ "E=(2.*ef/(Zc+z+R))**2. *R\n",
+ "E=round(E*100.)/100.\n",
+ "print '%s %.2f %s %s %.2f %s' %(\"\\n(a)Energy transfer max when R=\",R,\"ohm\\n\",\"energy=\",E,\"KW\\n\");\n",
+ "etB=2.*ef*z/(z+Zc+R);\n",
+ "etB=round(etB*100.)/100.\n",
+ "print '%s %.2f %s' %(\"\\n(b)Surge Voltage Transmitted=\",etB,\"kV\\n\")\n",
+ "it=etB*1000./z;\n",
+ "it=round(it*100.)/100.\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current Transmitted =\",it,\"A\\n\")\n",
+ "etA=2.*ef*(z+R)/(z+Zc+R);\n",
+ "etA=round(etA*100.)/100.\n",
+ "erA=etA-ef\n",
+ "print '%s %.2f %s' %(\"\\n(c)Surge Voltage Reflected=\",erA,\"kV\\n\")\n",
+ "irA=-1.*erA*1000./Zc\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current Reflected=\",irA,\"A\\n\")\n",
+ "iF=ef*1000./Zc\n",
+ "Pi=ef*iF\n",
+ "print '%s %.2f %s' %(\"\\n(d)Power Incident=\",Pi,\"kW\\n\")\n",
+ "Pr=erA*-irA\n",
+ "print '%s %.2f %s' %(\"\\nPower Reflected=\",Pr,\"kW\\n\")\n",
+ "Pt=erA*it\n",
+ "print '%s %.2f %s' %(\"\\nPower Transmitted=\",Pt,\"kW\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "(a)Energy transfer max when R= 450.00 ohm\n",
+ " energy= 22.22 KW\n",
+ "\n",
+ "\n",
+ "(b)Surge Voltage Transmitted= 11.11 kV\n",
+ "\n",
+ "\n",
+ "Surge Current Transmitted = 222.20 A\n",
+ "\n",
+ "\n",
+ "(c)Surge Voltage Reflected= 11.11 kV\n",
+ "\n",
+ "\n",
+ "Surge Current Reflected= -27.77 A\n",
+ "\n",
+ "\n",
+ "(d)Power Incident= 25000.00 kW\n",
+ "\n",
+ "\n",
+ "Power Reflected= 308.58 kW\n",
+ "\n",
+ "\n",
+ "Power Transmitted= 2468.64 kW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E14 - Pg 50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "I=5.;\n",
+ "z1=400.\n",
+ "z2=50.\n",
+ "V=I * z1* z2/(z2+z1)\n",
+ "print '%s %.2f %s' %(\"\\nSurge Voltage Transmitted=\",V,\"kV\\n\")\n",
+ "ic=V/z2\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current Transmitted in cable=\",ic,\"kA\\n\")\n",
+ "\n",
+ "io=-V/z1;\n",
+ "print '%s %.2f %s' %(\"\\nSurge Current Transmitted in OH line=\",io,\"kA\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Surge Voltage Transmitted= 222.22 kV\n",
+ "\n",
+ "\n",
+ "Surge Current Transmitted in cable= 4.44 kA\n",
+ "\n",
+ "\n",
+ "Surge Current Transmitted in OH line= -0.56 kA\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E16 - Pg 51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "i=100.\n",
+ "L=4e-3\n",
+ "C=300e-12\n",
+ "E=i* math.sqrt(L/C)\n",
+ "T=1./ math.sqrt(L*C)\n",
+ "print '%s %.2f %s %.2f' %(\"e=\",E/1e3,\"*1e3 sin( %.3f *1e6 t) kV\",T/1e6)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "e= 365.15 *1e3 sin( %.3f *1e6 t) kV 0.91\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/MohdAkhlak/ch3.ipynb b/sample_notebooks/MohdAkhlak/ch3.ipynb
new file mode 100755
index 00000000..82035c0d
--- /dev/null
+++ b/sample_notebooks/MohdAkhlak/ch3.ipynb
@@ -0,0 +1,256 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:931ccf820f91dc3bd263fe2430c73233b66e02ab90eedbd3236bc42b5727ead5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3, Waveform generators"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex 1 - page : 145"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Given data \n",
+ "\n",
+ "R2 = 100.0 # ohm\n",
+ "R1 = 50.0*10**3 # ohm\n",
+ "Vref = 0 # V\n",
+ "vi = 1 # V\n",
+ "Vsat = 14.0 # V\n",
+ "# Solution \n",
+ "\n",
+ "VUT = (R2*Vsat)/(R1+R2) # V\n",
+ "VLT = (R2*-Vsat)/(R1+R2) # V\n",
+ "\n",
+ "# Displaying the values\n",
+ "\n",
+ "print \"The value of Upper threshold voltage = \",int(round(VUT*10**3,0)),\"mV\"\n",
+ "print \"The value of Lower threshold voltage = \",int(round(VLT*10**3,0)),\"mV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Upper threshold voltage = 28 mV\n",
+ "The value of Lower threshold voltage = -28 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex 2 - page : 148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import numpy as np \n",
+ "# Given data\n",
+ "\n",
+ "VUT = 0\n",
+ "VH = 0.3\n",
+ "f = 10.0**3\n",
+ "# Solution \n",
+ "\n",
+ "VLT = VUT - VH\n",
+ "theta = round(np.arcsin(VH/2),1)\n",
+ "T = 1/f\n",
+ "Ttheta = theta/(2*np.pi*f) \n",
+ "T1 = (T/2) + Ttheta\n",
+ "T2 = (T/2) - Ttheta \n",
+ "\n",
+ "# Displayig the values \n",
+ "\n",
+ "print \"The value of Lower threshold voltage = \",VLT,\"V\"\n",
+ "print \"The value of angle theta = \",theta,\"Radian\"\n",
+ "print \"The value of T1 = \",round(T1*10**3,3),\"ms\"\n",
+ "print \"The value of T2 = \",round(T2*10**3,3),\"ms\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Lower threshold voltage = -0.3 V\n",
+ "The value of angle theta = 0.2 Radian\n",
+ "The value of T1 = 0.532 ms\n",
+ "The value of T2 = 0.468 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex 3 - page : 152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# Given data \n",
+ "\n",
+ "f = 100\n",
+ "C = 0.1*10**-6\n",
+ "\n",
+ "# Solution \n",
+ "\n",
+ "R = 1/(math.sqrt(6)*2*math.pi*(10**-7)*100)\n",
+ "R1 = 10*R\n",
+ "Rf = 29*R1\n",
+ "\n",
+ "# Displaying the solutions \n",
+ "\n",
+ "print \"The value of R1 =\",int(round(R1*10**-3,0)),\"Kilo Ohms\"\n",
+ "print \"The value of Rf =\",int(round(Rf*10**-3,0)),\"Kilo Ohms\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R1 = 65 Kilo Ohms\n",
+ "The value of Rf = 1884 Kilo Ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex 4 - page 158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# Given data \n",
+ "\n",
+ "f = 50\n",
+ "C = 0.2*10**-6\n",
+ "\n",
+ "# Solution \n",
+ "\n",
+ "R = 1/(math.sqrt(6)*2*math.pi*(10**-7)*100)\n",
+ "R1 = 10*R\n",
+ "Rf = 29*R1\n",
+ "\n",
+ "# Displaying the solutions \n",
+ "\n",
+ "print \"The value of R1 is %0.2f kohms\"%(R1*10**-3)\n",
+ "print \"The value of Rf is %0.2f kohms\"%(Rf*10**-3)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of R1 is 64.97 kohms\n",
+ "The value of Rf is 1884.27 kohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex 5 - page 162"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from numpy import pi, arcsin \n",
+ "# Given data\n",
+ "\n",
+ "VUT = 0\n",
+ "VH = 0.5\n",
+ "f = 1000\n",
+ "# Solution \n",
+ "\n",
+ "VLT = VUT - VH\n",
+ "theta = round(arcsin(VH/2),1)\n",
+ "T = 1/f\n",
+ "Ttheta = theta/(2*pi*f) \n",
+ "T1 = (T/2) + Ttheta\n",
+ "T2 = (T/2) - Ttheta \n",
+ "\n",
+ "# Displayig the values \n",
+ "\n",
+ "print \"The value of Lower threshold voltage = \",VLT,\"V\"\n",
+ "print \"The value of angle theta = \",theta,\"Radian\"\n",
+ "print \"The value of T1 = \",round(T1*10**3,3),\"ms\"\n",
+ "print \"The value of T2 = \",round(T2*10**3,3),\"ms\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Lower threshold voltage = -0.5 V\n",
+ "The value of angle theta = 0.3 Radian\n",
+ "The value of T1 = 0.048 ms\n",
+ "The value of T2 = -0.048 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/PradeepChoudari/Chapter01.ipynb b/sample_notebooks/PradeepChoudari/Chapter01.ipynb
new file mode 100755
index 00000000..00a54c26
--- /dev/null
+++ b/sample_notebooks/PradeepChoudari/Chapter01.ipynb
@@ -0,0 +1,326 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:18faacea8ac62a8f633d38b4dcf07dd31cdf79c796296b846bc3a06dfe56da57"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter01:Single phase transformer Principle and Constructions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 23"
+ ]
+ },
+ {
+ "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 %.2f' %('(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 %.2f' %('(c)load impedance on low voltage side for full load (in ohm)=',Z_2)\n",
+ "Z_1=Z_2*(n**2)\n",
+ "print '%s %.2f' %('(d)impedance referred to high voltage side(in ohm)=',Z_1)"
+ ],
+ "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)= 13.09\n",
+ "(c)load impedance on low voltage side for full load (in ohm)= 2.30\n",
+ "(d)impedance referred to high voltage side(in ohm)= 145.92\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 24"
+ ]
+ },
+ {
+ "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 %.2f' %('(a)Power factor on no load=',Pf)\n",
+ "I_c=I_0*Pf\n",
+ "print '%s %.2f' %('(b)Active current(in A)=',I_c)\n",
+ "theta=(acos(Pf)*57.3)\n",
+ "I_m=I_0*sin(theta)\n",
+ "print '%s %.2f' %('(c)magnetising current is(in A)=',I_m)\n",
+ "Culoss=(I_0**2)*R\n",
+ "print '%s %.2f' %('(d)copper loss in the primary winding is(in watt)=',Culoss)\n",
+ "Coreloss=P-Culoss\n",
+ "print '%s %.2f' %('(e)core loss(in watt)=',Coreloss)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)Power factor on no load= 0.12\n",
+ "(b)Active current(in A)= 0.25\n",
+ "(c)magnetising current is(in A)= 1.82\n",
+ "(d)copper loss in the primary winding is(in watt)= 3.20\n",
+ "(e)core loss(in watt)= 56.80\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 26"
+ ]
+ },
+ {
+ "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 %.2f' %('(a)number of turns on low voltage side is=',T_2)\n",
+ "T_1=T_2*V_1/V_2\n",
+ "print '%s %.2f' %('(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= 14.00\n",
+ "(b)number of turns on high voltage side is= 192.55\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 29"
+ ]
+ },
+ {
+ "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 %.2f' %('(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 %.2f' %('\\nequivalent reactance of low voltage side in terms of high voltage side=',X_2)\n",
+ "R_1=r_1/(n**2)\n",
+ "print '%s %.2f' %('\\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 %.2f' %('\\ntotal reactance of transformer in terms of high voltage side=',X_t)\n",
+ "R_e=r_2+R_1\n",
+ "print '%s %.2f' %('\\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= 4.00\n",
+ "\n",
+ "equivalent reactance of low voltage side in terms of high voltage side= 15.00\n",
+ "\n",
+ "(b)equivalent resistance of high voltage side in terms of low voltage side = 0.01\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= 19.00\n",
+ "\n",
+ "(d)total resistance of transformer in terms of low voltage side= 0.05\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 29"
+ ]
+ },
+ {
+ "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 %.2f' %('(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 %.2f' %('(c)efficiency of transformer is(in %)=',eff)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)terminal voltage on load(in volt)= 193.00\n",
+ "(b)voltage on load at hgih voltage terminals(in volt)= 1930.00\n",
+ "(c)efficiency of transformer is(in %)= 93.58\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 - Pg 30"
+ ]
+ },
+ {
+ "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 %.2f' %('(b) Resistance(in %)=',R)\n",
+ "x=math.sqrt(Z**2-r**2)\n",
+ "X=(x*I/V_2)*100\n",
+ "print '%s %.2f' %('Reactance(in %)=',X)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)Impedance(in ohms)= 0.05\n",
+ "(b) Resistance(in %)= 2.00\n",
+ "Reactance(in %)= 9.80\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/RajbeeVakkalagadda/Chapter04.ipynb b/sample_notebooks/RajbeeVakkalagadda/Chapter04.ipynb
new file mode 100755
index 00000000..1b360b5e
--- /dev/null
+++ b/sample_notebooks/RajbeeVakkalagadda/Chapter04.ipynb
@@ -0,0 +1,476 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:36d59fdfc24a2deb7849a2ba98aa4b099fd70d413ba4136a658e9181b78b3321"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter04:The Transistor At Low Frequency"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.1\n",
+ "import math\n",
+ "# Given data\n",
+ "R1 = 100.*10.**3.;# in ohm\n",
+ "R2 = 10.*10.**3.;# in ohm\n",
+ "h_fe = 50.;\n",
+ "h_oe = 1./40.;# in ohm\n",
+ "R_L = 5.*10.**3.;# in ohm\n",
+ "R_S= 5.*10.**3;# in ohm\n",
+ "h_ie = 1.1*10.**3.;# in ohm\n",
+ "h_re = 2.5*10.**-4.;\n",
+ "R_B = (R1*R2/(R1+R2));# in ohm\n",
+ "A_I = (-h_fe)/(1 + h_oe*R_L);\n",
+ "print '%s %.2f %s' %(\"The internal current gain is\",A_I,\"\\n\");\n",
+ "#Internal input impedance, Zi = Vbe/Ib or \n",
+ "Zi = (h_ie + h_re*A_I*R_L);# in ohm\n",
+ "Zi= Zi*10.**-3.;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The internal input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "Zi= Zi*10.**3.;# in ohm\n",
+ "#Internal voltage gain, Av = Vce/Vbe or \n",
+ "Av = (A_I*R_L)/Zi;\n",
+ "print '%s %.2f %s' %(\"The internal voltage gain is\",Av,\"\\n\");\n",
+ "Ri =round(R_B*Zi/(R_B+Zi));# in ohm\n",
+ "Ri= Ri*10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Ri,\"\\n\");\n",
+ "Ri= Ri*10**3;# in ohm\n",
+ "# V_S= I_i*R_S+v_be or\n",
+ "VS_by_vbe= Ri/(Ri+R_S);\n",
+ "Avs= Av*VS_by_vbe;\n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is : \",Avs,\"\\n\")\n",
+ "# R_B*(I_i-I_b)= Zi*I_b or\n",
+ "I_bBYI_i= R_B/(R_B+Zi);\n",
+ "A_IS= A_I*I_bBYI_i;\n",
+ "print '%s %.2f %s' %(\"The overall current gain is : \",A_IS,\"\\n\")\n",
+ "Rdesh_S= R_B*R_S/(R_B+R_S);# in ohm\n",
+ "Rdesh_S= 3220\n",
+ "I_bByVce= -h_re/(h_ie+Rdesh_S);\n",
+ "Yo= h_oe-h_fe*h_re/(h_ie+Rdesh_S)*10**3;\n",
+ "Zo= 1/Yo;\n",
+ "print '%s %.2f %s' %(\"The Output impedance in ohm is : \",Zo,\"\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal current gain is -0.40 \n",
+ "\n",
+ "The internal input impedance in k ohm is 1.10 \n",
+ "\n",
+ "The internal voltage gain is -1.80 \n",
+ "\n",
+ "The overall input impedance in k ohm is 0.98 \n",
+ "\n",
+ "The overall voltage gain is : -0.30 \n",
+ "\n",
+ "The overall current gain is : -0.35 \n",
+ "\n",
+ "The Output impedance in ohm is : 45.24 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - Pg 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.2\n",
+ "import math\n",
+ "# Given data\n",
+ "V_CC = 15.;# in V\n",
+ "R_L = 10.;# in k ohm\n",
+ "Rf = 200.;# in k ohm\n",
+ "R_S = 5.;# in k ohm\n",
+ "Rf2 = Rf;# in k ohm\n",
+ "h_fe = 50.;\n",
+ "V_S= 10.*10.**-3.;# in V\n",
+ "h_oe = 1./40.;# in k ohm\n",
+ "R_L = (R_L*Rf2)/(R_L+Rf2);# in k ohm\n",
+ "Ai = -h_fe/(1.+h_oe*R_L);\n",
+ "print '%s %.2f %s' %(\"The internal current gain is\",Ai,\"\\n\");\n",
+ "#Zi = Vbe/Ib = h_ie +Ai*h_re*R_L;\n",
+ "h_ie = 1.1;# in k ohm\n",
+ "h_re = 2.5*10.**-4.;\n",
+ "Zi = h_ie +Ai*h_re*R_L;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The internal input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "#A_V = Vce/Vbe = (Ai*R_L)/Zi;\n",
+ "A_V = (Ai*R_L)/Zi;\n",
+ "print '%s %.2f %s' %(\"The internal voltage gain is\",A_V,\"\\n\");\n",
+ "Rf1= Rf/(1-A_V)\n",
+ "# Rf1 = Rf/(1-A_V);# in k ohm\n",
+ "#Ri = Vi/Ii = Vbe/Ii = (Rf1*Zi)/(Rf1+Zi);\n",
+ "Ri = (Rf1*Zi)/(Rf1+Zi);# in k ohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Ri,\"\\n\");\n",
+ "#A_VS = Vo/V_S or \n",
+ "A_VS = A_V*(Ri/(R_S+Ri));\n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is\",A_VS,\"\\n\");\n",
+ "#A_IS = I_L/Ii or\n",
+ "A_IS = (Rf2/(Rf2+R_L))*Ai*(Rf1/(Rf1+Zi));\n",
+ "print '%s %.2f %s' %(\"The overall current gain is\",A_IS,\"\\n\");\n",
+ "Rdesh_S= Rf1*R_S/(Rf1+R_S);# in k ohm\n",
+ "Yo= h_oe-h_re*h_fe/(h_ie+Rdesh_S);# in mho\n",
+ "Zo= 1/Yo;# in ohm\n",
+ "print '%s %.2f %s' %(\"The output impedance in ohm is : \",Zo,\"\\n\")\n",
+ "Zdesh_o= Rf2*Zo/(Rf2+Zo);# in ohm\n",
+ "print '%s %.2f %s' %(\"The overall output impedance in ohm is : \",Zdesh_o,\"\\n\");\n",
+ "Vo= V_S*abs(A_VS);# in V\n",
+ "Vo= Vo*10**3;# in mV\n",
+ "print '%s %.2f %s' %(\"The magnitude of output voltage in mV is : \",Vo,\"\\n\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The internal current gain is -40.38 \n",
+ "\n",
+ "The internal input impedance in k ohm is 1.00 \n",
+ "\n",
+ "The internal voltage gain is -383.14 \n",
+ "\n",
+ "The overall input impedance in k ohm is 0.34 \n",
+ "\n",
+ "The overall voltage gain is -24.58 \n",
+ "\n",
+ "The overall current gain is -13.17 \n",
+ "\n",
+ "The output impedance in ohm is : 58.66 \n",
+ "\n",
+ "The overall output impedance in ohm is : 45.36 \n",
+ "\n",
+ "The magnitude of output voltage in mV is : 245.85 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.3\n",
+ "import math \n",
+ "# Given data\n",
+ "h_ic = 2.;# in k ohm\n",
+ "h_fc = -51.;\n",
+ "h_oc = 25.*10.**-6.;# in ohm\n",
+ "h_rc= 1.;\n",
+ "V_CC = 20.;# in V\n",
+ "R1 = 10.;# in k ohm\n",
+ "R2 = 10.;# in k ohm\n",
+ "R_S = 1.;# in k ohm\n",
+ "R_E = 5.;# in k ohm\n",
+ "R_B= 5.;# in k ohm\n",
+ "R_L= 5.;# in k ohm\n",
+ "# (i) Current Gain\n",
+ "Ai = (-h_fc)/(1.+h_oc*R_E*10.**3.);\n",
+ "print '%s %.2f %s' %(\"The current gain is\",Ai,\"\\n\");\n",
+ "# (ii) Input impedance\n",
+ "Zi = h_ic*10**3 + h_rc*Ai*R_E*10**3;# in ohm\n",
+ "Zi = Zi * 10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "# (iii) Voltage Gain\n",
+ "A_V = (Ai*R_L*10**3)/(Zi*10**3);\n",
+ "A_V = 1;# (approx)\n",
+ "print '%s %.2f %s' %(\"The voltage gain is\",A_V,\"\\n\");\n",
+ "# (iv) Overall Input Impedance\n",
+ "Z_IS = (R_B*Zi)/(R_B+Zi);# in k ohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Z_IS,\"\\n\");\n",
+ "# (v) Overall voltage gain\n",
+ "A_VS = (A_V*Zi)/(Zi+R_S); \n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is\",A_VS,\"\\n\");\n",
+ "# (vi) Overall current gain\n",
+ "A_IS =Ai*(R_B/(R_B+Zi));\n",
+ "print '%s %.2f %s' %(\"The overall current gain is\",A_IS,\"\\n\");\n",
+ "# (vii) Output impedance\n",
+ "RdasS = (R_S*R_B)/(R_S+R_B);# in k ohm\n",
+ "Yo = h_oc - ( (h_fc*h_rc)/(h_ic*10.**3.+RdasS*10.**3.) );# in mho \n",
+ "Zo = 1./Yo;# in ohm\n",
+ "print '%s %.2f %s' %(\"The output impedance in ohm is\",Zo,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current gain is 45.33 \n",
+ "\n",
+ "The input impedance in k ohm is 228.67 \n",
+ "\n",
+ "The voltage gain is 1.00 \n",
+ "\n",
+ "The overall input impedance in k ohm is 4.89 \n",
+ "\n",
+ "The overall voltage gain is 1.00 \n",
+ "\n",
+ "The overall current gain is 0.97 \n",
+ "\n",
+ "The output impedance in ohm is 55.48 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E44 - Pg 44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.4\n",
+ "# Given data\n",
+ "h_ie = 1.1;# in k ohm\n",
+ "h_re = 2.5*10.**-4.;\n",
+ "h_fe = 50.;\n",
+ "h_oe = 25.*10.**-6.;# in A\n",
+ "V_CC = 15.;# in V\n",
+ "R1 = 20.;# in k ohm\n",
+ "R_C = 2.;# in k ohm\n",
+ "R2 = 10.;# in k ohm\n",
+ "R_S = 1.;# in k ohm\n",
+ "R_E = 1.;# in k ohm\n",
+ "# (i) Current Gain\n",
+ "Ai = -h_fe/(1. + h_oe*R_C*10.**3.);\n",
+ "print '%s %.2f %s' %(\"The current gain is\",Ai,\"\\n\");\n",
+ "# (ii) Input impedance\n",
+ "Zi = (h_ie*10**3) + (h_re*Ai*R_C*10**3);#in ohm\n",
+ "Zi = Zi * 10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "# (iii) Voltage gain\n",
+ "A_V = (Ai*R_C)/Zi;\n",
+ "print '%s %.2f %s' %(\"The voltage gain is\",A_V,\"\\n\");\n",
+ "# (iv) Overall input impedance\n",
+ "R_B = (R1*R2)/(R1+R2);# in k ohm\n",
+ "Z_IS = (Zi*R_B)/(Zi+R_B);# in k ohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Z_IS,\"\\n\");\n",
+ "# (v) Overall voltage gain\n",
+ "A_VS = A_V * (Z_IS/(Z_IS+R_S));\n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is\",A_VS,\"\\n\");\n",
+ "# (vi) Overall current gain\n",
+ "A_IS =Ai*(R_B/(R_B+Zi));\n",
+ "print '%s %.2f %s' %(\"The overall current gain is\",A_IS,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current gain is -47.62 \n",
+ "\n",
+ "The input impedance in k ohm is 1.08 \n",
+ "\n",
+ "The voltage gain is -88.50 \n",
+ "\n",
+ "The overall input impedance in k ohm is 0.93 \n",
+ "\n",
+ "The overall voltage gain is -42.56 \n",
+ "\n",
+ "The overall current gain is -41.00 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.5\n",
+ "# Given data\n",
+ "h_ie = 1.1;# in k ohm\n",
+ "h_oe = 25.;# in A/V\n",
+ "h_oe = h_oe * 10.**-6.;# in A/V\n",
+ "h_fe = 50.;\n",
+ "h_re = 2.5*10.**-4.;\n",
+ "R_L = 1.6;# in ohm\n",
+ "R_S = 1.;# in k ohm\n",
+ "V_CC = 15.;# in V\n",
+ "# (i) Current Gain\n",
+ "Ai = -h_fe/(1. + (h_oe*R_L));\n",
+ "print '%s %.2f %s' %(\"The current gain is\",Ai,\"\\n\");\n",
+ "# (ii) Input impedance\n",
+ "Zi = (h_ie*10**3) + (h_re*Ai*R_L);# in ohm\n",
+ "Zi= Zi*10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "Zi= Zi*10**3;# in ohm\n",
+ "# (iii) Voltage gain\n",
+ "A_V = Ai*R_L/Zi;\n",
+ "print '%s %.2f %s' %(\"The voltage gain is\",A_V,\"\\n\");\n",
+ "# (iv) Power gain\n",
+ "A_P = Ai*A_V;\n",
+ "print '%s %.2f %s' %(\"The power gain is\",A_P,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current gain is -50.00 \n",
+ "\n",
+ "The input impedance in k ohm is 1.10 \n",
+ "\n",
+ "The voltage gain is -0.07 \n",
+ "\n",
+ "The power gain is 3.64 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 - Pg 50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa 4.6\n",
+ "# Given data\n",
+ "h_fe = 150.;\n",
+ "Beta_dc = h_fe;\n",
+ "h_ie = 1.*10.**3.;# in ohm\n",
+ "h_re = 0;\n",
+ "h_oe = 0;\n",
+ "V_CC = 18.;# in V\n",
+ "V_BE= 0.7;# in V\n",
+ "R1 = 100.*10.**3.;# in ohm\n",
+ "R2 = 50.*10.**3.;# in ohm\n",
+ "R_C = 1.*10.**3.;# in ohm\n",
+ "R_E = 0.5*10.**3.;# in ohm\n",
+ "V_Th = (V_CC/(R1+R2))*R2;# in V\n",
+ "R_Th =(R1*R2)/(R1+R2);# in ohm\n",
+ "# V_Th - I_B*R_Th - V_BE - (1+Beta)*-I_B*R_E = 0;\n",
+ "I_B = (V_Th-V_BE)/( R_Th + (1+Beta_dc)*R_E);# in A\n",
+ "#I_C = I_CQ = Beta*I_B;\n",
+ "I_C = Beta_dc*I_B;# in A\n",
+ "I_CQ = I_C;# in A\n",
+ "I_CQ= I_CQ*10.**3.;# in mA\n",
+ "print '%s %.2f %s' %(\"The value of I_CQ in mA is\",I_CQ,\"\\n\");\n",
+ "I_E = (1+Beta_dc)*I_B;# in mA\n",
+ "# V_CC - I_C*R_C - V_CE - I_E*R_E = 0;\n",
+ "V_CE = V_CC - (I_C*R_C) - (I_E*R_E);# in V\n",
+ "print '%s %.2f %s' %(\"The value of V_CE in V is\",V_CE,\"\\n\");\n",
+ "R_L =R_C;# in ohm\n",
+ "Ai = -h_fe/(1+(h_oe*R_L));\n",
+ "print '%s %.2f %s' %(\"The current gain is \",Ai,\"\\n\");\n",
+ "Zi = h_ie + h_re*Ai*R_L;# in ohm\n",
+ "Zi= Zi*10**-3;# in k ohm\n",
+ "print '%s %.2f %s' %(\"The input impedance in k ohm is\",Zi,\"\\n\");\n",
+ "Zi= Zi*10**3;# in ohm\n",
+ "A_V = Ai*(R_L/Zi);\n",
+ "print '%s %.2f %s' %(\"The voltage gain is\",A_V,\"\\n\");\n",
+ "R_B= (R1*R2)/(R1+R2);# in ohm\n",
+ "Z_IS =(Zi*R_B)/(Zi+R_B);# in ohm\n",
+ "Z_IS= Z_IS*10**-3;# in kohm\n",
+ "print '%s %.2f %s' %(\"The overall input impedance in k ohm is\",Z_IS,\"\\n\");\n",
+ "Z_IS= Z_IS*10**3;# in ohm\n",
+ "A_VS =A_V*(Z_IS/Z_IS);\n",
+ "print '%s %.2f %s' %(\"The overall voltage gain is\",A_VS,\"\\n\");\n",
+ "A_IS =Ai * (R_B/(R_B+Zi));\n",
+ "print '%s %.2f %s' %(\"The overall current gain is\",A_IS,\"\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of I_CQ in mA is 7.30 \n",
+ "\n",
+ "The value of V_CE in V is 7.02 \n",
+ "\n",
+ "The current gain is -150.00 \n",
+ "\n",
+ "The input impedance in k ohm is 1.00 \n",
+ "\n",
+ "The voltage gain is -150.00 \n",
+ "\n",
+ "The overall input impedance in k ohm is 0.97 \n",
+ "\n",
+ "The overall voltage gain is -150.00 \n",
+ "\n",
+ "The overall current gain is -145.63 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/ajinkyakhair/chapter2.ipynb b/sample_notebooks/ajinkyakhair/chapter2.ipynb
new file mode 100755
index 00000000..8b221e49
--- /dev/null
+++ b/sample_notebooks/ajinkyakhair/chapter2.ipynb
@@ -0,0 +1,240 @@
+{
+ "metadata": {
+ "celltoolbar": "Raw Cell Format",
+ "name": "",
+ "signature": "sha256:4fe36e3e0da1a77ee9793bbcdad9ed8d44455b05327e70b42ad389ca8fb3e239"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2: Semiconductor Physics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.21.1,Page number 2-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given Data:\n",
+ "\n",
+ "ro=1.72*10**-8 #resistivity of Cu\n",
+ "s=1/ro #conductivity of Cu\n",
+ "n=10.41*10**28 #no of electron per unit volume\n",
+ "e=1.6*10**-19 #charge on electron\n",
+ "\n",
+ "u=s/(n*e)\n",
+ "print\"mobility of electron in Cu =\",round(u,4),\"m**2/volt-sec\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "mobility of electron in Cu = 0.0035 m**2/volt-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.21.2,Page number 2-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given Data:\n",
+ "\n",
+ "m=63.5 #atomic weight\n",
+ "u=43.3 #mobility of electron\n",
+ "e=1.6*10**-19 #charge on electron\n",
+ "N=6.02*10**23 #Avogadro's number\n",
+ "d=8.96 #density\n",
+ "\n",
+ "Ad=N*d/m #Atomic density\n",
+ "n=1*Ad\n",
+ "\n",
+ "ro=1/(n*e*u)\n",
+ "\n",
+ "print\"Resistivity of Cu =\",\"{0:.3e}\".format(ro),\"ohm-cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistivity of Cu = 1.699e-06 ohm-cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.21.3,Page number 2-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given Data:\n",
+ "\n",
+ "e=1.6*10**-19 #charge on electron\n",
+ "ne=2.5*10**19 #density of carriers\n",
+ "nh=ne #for intrinsic semiconductor\n",
+ "ue=0.39 #mobility of electron\n",
+ "uh=0.19 #mobility of hole\n",
+ "\n",
+ "s=ne*e*ue+nh*e*uh #conductivity of Ge\n",
+ "ro=1/s #resistivity of Ge\n",
+ "\n",
+ "print\"Resistivity of Ge =\",round(ro,4),\"ohm-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistivity of Ge = 0.431 ohm-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.21.6,Page number 2-49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given Data:\n",
+ "\n",
+ "c=5*10**28 #concentration of Si atoms\n",
+ "e=1.6*10**-19 #charge on electron\n",
+ "u=0.048 #mobility of hole\n",
+ "s=4.4*10**-4 #conductivity of Si\n",
+ "\n",
+ "#since millionth Si atom is replaced by an indium atom\n",
+ "\n",
+ "n=c*10**-6\n",
+ "sp=u*e*n #conductivity of resultant\n",
+ "\n",
+ "print\"conductivity =\",sp,\"mho/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "conductivity = 384.0 mho/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.21.7,Page number 2-49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Given Data:\n",
+ "\n",
+ "m=28.1 #atomic weight of Si\n",
+ "e=1.6*10**-19 #charge on electron\n",
+ "N=6.02*10**26 #Avogadro's number\n",
+ "d=2.4*10**3 #density of Si\n",
+ "p=0.25 #resistivity\n",
+ "\n",
+ "#no. of Si atom/m**3\n",
+ "Ad=N*d/m #Atomic density\n",
+ "\n",
+ "#impurity level is 0.01 ppm i.e. 1 atom in every 10**8 atoms of Si\n",
+ "n=Ad/10**8 #no of impurity atoms\n",
+ "\n",
+ "#since each impurity produce 1 hole\n",
+ "nh=n\n",
+ "print\"1) hole concentration =\",\"{0:.3e}\".format(n),\"holes/m**3\"\n",
+ "up=1/(e*p*nh)\n",
+ "print\"2) mobility =\",round(up,4),\"m**2/volt.sec\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "1) hole concentration = 5.142e+20 holes/m**3\n",
+ "2) mobility = 0.0486 m**2/volt.sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/kartiksankhla/Chapter2.ipynb b/sample_notebooks/kartiksankhla/Chapter2.ipynb
new file mode 100755
index 00000000..21f2d4c4
--- /dev/null
+++ b/sample_notebooks/kartiksankhla/Chapter2.ipynb
@@ -0,0 +1,159 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e984fee9b841dd6e9b7eedf1533b0a0d297cd9f484c047f051ce48a09b156826"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Nuclear Engineering"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.1\n",
+ "import math\n",
+ "#determine atoms in deuterium\n",
+ "## Given data\n",
+ "atom_h = 6.6*10**24; ## Number of atoms in Hydrogen\n",
+ "## Using the data given in Table II.2, Appendix II for isotropic abundance of deuterium\n",
+ "isoab_H2 = 0.015; ## Isotropic abundance of deuterium\n",
+ "## Calculation\n",
+ "totatom_d=(isoab_H2*atom_h)/100.;\n",
+ "## Result\n",
+ "print\"%s %.2e %s \"%('\\n Number of deuterium atoms = ',totatom_d,'');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Number of deuterium atoms = 9.90e+20 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.2\n",
+ "import math\n",
+ "#determine atomic weight of oxygen\n",
+ "## Given data \n",
+ "## Using the data given in the example 2.2\n",
+ "atwt_O16 = 15.99492; ## Atomic weight of O-16 isotope\n",
+ "isoab_O16 = 99.759; ## Abundance of O-16 isotope\n",
+ "atwt_O17 = 16.99913; ## Atomic weight of O-17 isotope\n",
+ "isoab_O17 = 0.037; ## Abundance of O-17 isotope\n",
+ "atwt_O18 = 17.99916; ## Atomic weight of O-18 isotope\n",
+ "isoab_O18 = 0.204; ## Abundance of O-18 isotope\n",
+ "## Calculation\n",
+ "atwt_O=(isoab_O16*atwt_O16 + isoab_O17*atwt_O17 + isoab_O18*atwt_O18)/100.;\n",
+ "## Result\n",
+ "print\"%s %.2f %s \"%('\\n Atomic Weight of Oxygen = ',atwt_O,'');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Atomic Weight of Oxygen = 16.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.3\n",
+ "import math\n",
+ "#determine rest mass energy of electron\n",
+ "## Given data\n",
+ "me = 9.1095*10**(-28); ## Mass of electron in grams\n",
+ "c = 2.9979*10**10; ## Speed of light in vacuum in cm/sec\n",
+ "## Calculation\n",
+ "rest_mass = me*c**2;\n",
+ "## Result\n",
+ "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass,' ergs\\n');\n",
+ "print('Expressing the result in joules')\n",
+ "## 1 Joule = 10^(-7)ergs\n",
+ "rest_mass_j = rest_mass*10**(-7);\n",
+ "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass_j,' joules\\n');\n",
+ "print('Expressing the result in MeV')\n",
+ "## 1 MeV = 1.6022*10^(-13)joules\n",
+ "rest_mass_mev = rest_mass_j/(1.6022*10**(-13));\n",
+ "print\"%s %.2f %s \"%('\\n Rest mass energy of electron = ',rest_mass_mev,' MeV\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Rest mass energy of electron = 8.19e-07 ergs\n",
+ " \n",
+ "Expressing the result in joules\n",
+ "\n",
+ " Rest mass energy of electron = 8.19e-14 joules\n",
+ " \n",
+ "Expressing the result in MeV\n",
+ "\n",
+ " Rest mass energy of electron = 0.51 MeV\n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer).ipynb b/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer).ipynb
new file mode 100755
index 00000000..b114e915
--- /dev/null
+++ b/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer).ipynb
@@ -0,0 +1,278 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "#Chapter:2 CONDUCTION"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.1"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " heat is Btu/hr 69120.0\n",
+ "\t approximate values are mentioned in the book \n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "#given\n",
+ "Tavg=900; # average temperature of the wall,F\n",
+ "k=0.15; # Thermal conductivity at 932 F,Btu/(hr)(ft^2)(F/ft)\n",
+ "T1=1500; # hot side temperature,F\n",
+ "T2=300; # cold side temperature,F\n",
+ "A=192; # surface area,ft^2\n",
+ "L=0.5; # thickness,ft\n",
+ "#solution\n",
+ "Q=(k)*(A)*(T1-T2)/L; # formula for heat,Btu/hr\n",
+ "print \" heat is Btu/hr \",Q\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#end\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example2.2"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t resistance offered by firebrick : (hr)*(F)/Btu 0.97\n",
+ "\t resistance offered by insulating brick : (hr)*(F)/Btu 2.2\n",
+ "\t resistance offered by buildingbrick : (hr)*(F)/Btu 1.25\n",
+ "\t total resistance offered by three walls : (hr)*(F)/Btu 4.42\n",
+ "\t heat loss/ft^2 : Btu/hr 331.0\n",
+ "\t delta is : F 322.0\n",
+ "\t temperature at interface of firebrick and insulating brick F 1278.0\n",
+ "\t deltb is : F 729.0\n",
+ "\t temperature at interface of insulating brick and building brick F 549.0\n",
+ "\t approximate values are mentioned in the book \n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "#given\n",
+ "La=0.66; # Thickness of firebrick wall,ft\n",
+ "Lb=0.33; # Thickness of insulating brick wall,ft\n",
+ "Lc=0.5; # Thickness of building brick wall,ft\n",
+ "Ka=0.68; # themal conductivity of firebrick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Kb=0.15; # themal conductivity of insulating brick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Kc=0.40; # themal conductivity of building brick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "A=1.; # surface area,ft^2\n",
+ "Ta=1600.; # temperature of inner wall,F\n",
+ "Tb=125.; # temperature of outer wall.F\n",
+ "#solution\n",
+ "Ra=La/(Ka)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by firebrick : (hr)*(F)/Btu \",round(Ra,2)\n",
+ "Rb=Lb/(Kb)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by insulating brick : (hr)*(F)/Btu \",round(Rb,2)\n",
+ "Rc=Lc/(Kc)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by buildingbrick : (hr)*(F)/Btu \",round(Rc,2)\n",
+ "R=Ra+Rb+Rc; # total resistance offered by three walls,(hr)*(F)/Btu\n",
+ "print\"\\t total resistance offered by three walls : (hr)*(F)/Btu \",round(R,2)\n",
+ "Q=(1600-125)/4.45; # using formula for heat loss/ft^2,Btu/hr\n",
+ "print\"\\t heat loss/ft^2 : Btu/hr \",round(Q,0)\n",
+ "# T1,T2 are temperatures at interface of firebrick and insulating brick, and insulating brick and building brick respectively,F\n",
+ "delta=(Q)*(Ra); # formula for temperature difference,F\n",
+ "print\"\\t delta is : F \",round(delta,0)\n",
+ "T1=Ta-((Q)*(Ra)); # temperature at interface of firebrick and insulating brick,F\n",
+ "print\"\\t temperature at interface of firebrick and insulating brick F \",round(T1,0)\n",
+ "deltb=Q*(Rb);\n",
+ "print\"\\t deltb is : F \",round(deltb,0)\n",
+ "T2=T1-((Q)*(Rb)); #temperature at interface of insulating brick and building brick,F\n",
+ "print\"\\t temperature at interface of insulating brick and building brick F \",round(T2,0)\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#end\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t approximate values are mentioned in the book \n",
+ "\n",
+ "\t resistance offered by air film (hr)(F)/Btu 0.79\n",
+ "\t total resistance (hr)(F)/Btu 5.24\n",
+ "\t heat loss Btu/hr 282.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#given\n",
+ "Lair=0.25/12; # thickness of air film,ft\n",
+ "Kair=0.0265; # thermal conductivity of air at 572F,Btu/(hr)*(ft^2)(F/ft)\n",
+ "A=1; # surface area,ft^2\n",
+ "#solution\n",
+ "Rair=Lair/(Kair*(A)); # resistance offered by air film, (hr)(F)/Btu\n",
+ "print\"\\t resistance offered by air film (hr)(F)/Btu \",round(Rair,2)\n",
+ "R=4.45; # resistance from previous example 2.2,(hr)(F)/Btu\n",
+ "Rt=(R)+Rair; # total resistance,(hr)(F)/Btu\n",
+ "print\"\\t total resistance (hr)(F)/Btu \",round(Rt,2)\n",
+ "Ta=1600; # temperature of inner wall,F\n",
+ "Tb=125; # temperature of outer wall,F\n",
+ "Q=(1600-125)/Rt; # heat loss, Btu/hr\n",
+ "print\"\\t heat loss Btu/hr \",round(Q,0)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "#given\n",
+ "k=0.63; # thermal conductivity of pipe, Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Do=6. # in\n",
+ "Di=5. # in\n",
+ "Ti=200.;# inner side temperature,F\n",
+ "To=175.; # outer side temperature,F\n",
+ "#solution\n",
+ "import math\n",
+ "from math import log\n",
+ "q=(2*(3.14)*(k)*(Ti-To))/(log (Do/Di)); # formula for heat flow,Btu/(hr)*(ft)\n",
+ "print\"\\t heat flow is : Btu/(hr)*(ft) \",round(q,0)\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "# caculation mistake in book\n",
+ "# end\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t approximate values are mentioned in the book \n",
+ "\n",
+ "\t heat loss for linear foot is : Btu/(hr)*(lin ft) 104.4\n",
+ "\t Check between ts and t1, since delt/R = deltc/Rc \n",
+ "\t t1 is : F 122.300238658\n",
+ "\t heat loss for linear foot is : Btu/(hr)*(lin ft) 102.9\n",
+ "\t Check between ts and t1, since delt/R = deltc/Rc \n",
+ "\t t1 is : F \n",
+ "125.4\n"
+ ]
+ }
+ ],
+ "source": [
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#given\n",
+ "t1=150; # assume temperature of outer surface of rockwool,F\n",
+ "ta=70; # temperature of surrounding air,F\n",
+ "ha=2.23; # surface coefficient,Btu/(hr)*(ft^2)*(F)\n",
+ "#solution\n",
+ "import math\n",
+ "from math import log\n",
+ "q=(3.14)*(300-70)/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.23)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft), calculation mistake\n",
+ "print\"\\t heat loss for linear foot is : Btu/(hr)*(lin ft) \",round(q,1)\n",
+ "print\"\\t Check between ts and t1, since delt/R = deltc/Rc \"\n",
+ "t1=300-(((104.8)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F\n",
+ "print\"\\t t1 is : F \",t1\n",
+ "t1=125; # assume temperature of outer surface of rockwool,F\n",
+ "ha=2.10; # surface coefficient,Btu/(hr)*(ft^2)*(F)\n",
+ "q=((3.14)*(300-70))/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.10)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft)\n",
+ "print\"\\t heat loss for linear foot is : Btu/(hr)*(lin ft)\",round(q,1)\n",
+ "print\"\\t Check between ts and t1, since delt/R = deltc/Rc \"\n",
+ "t1=300-(((103)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F\n",
+ "print\"\\t t1 is : F \\n\",round(t1,1)\n",
+ "# end \n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/kumargugloth/Chapter1.ipynb b/sample_notebooks/kumargugloth/Chapter1.ipynb
new file mode 100755
index 00000000..df9ba4d0
--- /dev/null
+++ b/sample_notebooks/kumargugloth/Chapter1.ipynb
@@ -0,0 +1,348 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:84e452258bd05b64c16351467c4970051f4494cb47d7a832df03bdce07abddb8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Introduction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math\n",
+ " #determine\n",
+ "##This numerical is Ex 1_1E,page 9.\n",
+ "Pso=20.5\n",
+ "Psc=20.5*550##converting hp to fps system\n",
+ "Qo=385.\n",
+ "Qc=385./449.##converting gpm to ft^3/s\n",
+ "E=0.83\n",
+ "dp=E*Psc/(Qc*144.)\n",
+ "print\"%s %.2f %s \"%('The pressure rise is ',dp,' psi')\n",
+ "print(\"After rounding off,pressure rise is 75.8 psi\")\n",
+ "dpr=75.8\n",
+ "dHw=75.8*144/62.4##62.4 is accelaration due to gravity in fps system\n",
+ "print\"%s %.2f %s \"%(' The head of water is ',dHw,' ft of water')\n",
+ "print(\"After rounding off the value of head of water the answer is 175 ft of water.\")\n",
+ "dhwr=175##rounded off value of head of water\n",
+ "sg=0.72##specific gravity of oil\n",
+ "dHo=dhwr/sg\n",
+ "print\"%s %.2f %s \"%(' The head of oil is ',dHo,' ft of oil')\n",
+ "print(\"After rounding off the value of head of oil the answer is 243 ft of oil.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pressure rise is 75.79 psi \n",
+ "After rounding off,pressure rise is 75.8 psi\n",
+ " The head of water is 174.92 ft of water \n",
+ "After rounding off the value of head of water the answer is 175 ft of water.\n",
+ " The head of oil is 243.06 ft of oil \n",
+ "After rounding off the value of head of oil the answer is 243 ft of oil.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math \n",
+ "#determine\n",
+ "##This numerical is Ex 1_1S,page 10.\n",
+ "E=0.83##efficiency\n",
+ "Ps=15300.\n",
+ "Q=87.4\n",
+ "Qs=87.4/3600.##flow rate in meter cube per sec\n",
+ "rho=998.\n",
+ "g=9.81\n",
+ "sg=0.72\n",
+ "dp=E*Ps/Qs\n",
+ "print\"%s %.2f %s \"%('\\n The change in pressure (dp)is ',dp,'')\n",
+ "dpr=523000##rounded value of dp\n",
+ "print(\"The rounded off value of dp is 523kPa.\")\n",
+ "dHw=dpr/(rho*g)\n",
+ "print\"%s %.2f %s \"%(' dHw is equal to ',dHw,' m of water')\n",
+ "print(\"The rounded off value of dHw is 53.4 m of water.\")\n",
+ "dHwr=53.4##rounded off value of dHw\n",
+ "print(\"Thus we can determine head of oil.\")\n",
+ "dHoil=dHwr/sg\n",
+ "print\"%s %.2f %s \"%(' dHoil is given by ',dHoil,' m of oil')\n",
+ "print(\"The rounded off value of dHoil is 74.2 m of oil.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The change in pressure (dp)is 523070.94 \n",
+ "The rounded off value of dp is 523kPa.\n",
+ " dHw is equal to 53.42 m of water \n",
+ "The rounded off value of dHw is 53.4 m of water.\n",
+ "Thus we can determine head of oil.\n",
+ " dHoil is given by 74.17 m of oil \n",
+ "The rounded off value of dHoil is 74.2 m of oil.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine\n",
+ "##This numerical is Ex 1_2E,page 10.\n",
+ "Q=12000.\n",
+ "A=3.5\n",
+ "rho_a=0.0762\n",
+ "E=0.85\n",
+ "r=2.5##resistance of duct system\n",
+ "V=Q/(60.*A)\n",
+ "print\"%s %.2f %s \"%('The air flow velocity at discharge is ',V,' ft/s')\n",
+ "KE=(rho_a*(V**2))/(32.2*2)\n",
+ "print\"%s %.2f %s \"%('\\n The product is ',KE,' lb/ft^2')\n",
+ "##PE=KE\n",
+ "Hv=KE/62.4\n",
+ "print\"%s %.2f %s \"%('\\n The dynamic head is ',Hv,' ft')\n",
+ "print(\"The value of dynamic head in inches of water is 0.74.\")\n",
+ "Hvi=0.74##Head in inches\n",
+ "Ht=r+Hvi\n",
+ "print\"%s %.2f %s \"%('\\n The total head is ',Ht,' inches of water')\n",
+ "p_tot=Ht*62.4\n",
+ "Ps=Q*p_tot/(60.*12.*E)\n",
+ "print\"%s %.2f %s \"%('\\n The shaft power is ',Ps,' ft-lb/s')\n",
+ "print(\"The shaft power is 7.2 hp.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The air flow velocity at discharge is 57.14 ft/s \n",
+ "\n",
+ " The product is 3.86 lb/ft^2 \n",
+ "\n",
+ " The dynamic head is 0.06 ft \n",
+ "The value of dynamic head in inches of water is 0.74.\n",
+ "\n",
+ " The total head is 3.24 inches of water \n",
+ "\n",
+ " The shaft power is 3964.24 ft-lb/s \n",
+ "The shaft power is 7.2 hp.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##This numerical is Ex 1_2S,page 11.\n",
+ "Q=340.\n",
+ "A=0.325\n",
+ "V=Q/(60.*A)\n",
+ "print\"%s %.2f %s \"%('The air flow velocity at discharge is ',V,' m/s')\n",
+ "rho_a=1.22\n",
+ "Vr=17.4\n",
+ "Hd=(rho_a*(Vr**2))/2.\n",
+ "print\"%s %.2f %s \"%('\\n The dynamic pressure head is ',Hd,' Pa')\n",
+ "Hdr=184.7##rounded off value of Hd\n",
+ "rho_w=998.##density of water=rhow\n",
+ "g=9.81\n",
+ "H=0.0635\n",
+ "dp=rho_w*g*H##static pressure head\n",
+ "print\"%s %.2f %s \"%('\\n The static pressure head is ',dp,' Pa')\n",
+ "dpr=621.7\n",
+ "p_tot=Hdr+dpr\n",
+ "print\"%s %.2f %s \"%('\\n The total pressure head is ',p_tot,' Pa')\n",
+ "p_tot=806.4\n",
+ "E=0.85##efficiency\n",
+ "Ps=Q*p_tot/(60*E)\n",
+ "print\"%s %.2f %s \"%('\\n The shaft power is',Ps, 'W')\n",
+ "print(\"The shaft power is 5.376 kW.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The air flow velocity at discharge is 17.44 m/s \n",
+ "\n",
+ " The dynamic pressure head is 184.68 Pa \n",
+ "\n",
+ " The static pressure head is 621.69 Pa \n",
+ "\n",
+ " The total pressure head is 806.40 Pa \n",
+ "\n",
+ " The shaft power is 5376.00 W \n",
+ "The shaft power is 5.376 kW.\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine \n",
+ "import math\n",
+ "##This numerical is Ex 1_3E,page 11.\n",
+ "H=295.##net head in ft\n",
+ "Q=148.##water flow rate\n",
+ "n=1800.##rpm\n",
+ "E=0.87##efficiency\n",
+ "a=62.4##product of density and accelaration due to gravity\n",
+ "omega=(n*2.*math.pi)/60.\n",
+ "dp=a*H\n",
+ "print\"%s %.2f %s \"%('The pressure is ',dp,' lb/ft^2')\n",
+ "Ps=E*Q*dp\n",
+ "print\"%s %.2f %s \"%('\\n Output power is equal to ',Ps,' lb-ft/s')\n",
+ "print(\"The output output power can also be written as 2.37*10^6 lb-ft/s\")\n",
+ "print(\"Output power in terms of horsepower is given by 4309hp.\")\n",
+ "Psr=2370000##rounded off value of Ps\n",
+ "Torque=Psr/omega\n",
+ "print\"%s %.2f %s \"%(' The output torque is ',Torque,' lb-ft.')\n",
+ "print(\"The output torque can also be written as 12.57*10^3 lb-ft\")\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pressure is 18408.00 lb/ft^2 \n",
+ "\n",
+ " Output power is equal to 2370214.08 lb-ft/s \n",
+ "The output output power can also be written as 2.37*10^6 lb-ft/s\n",
+ "Output power in terms of horsepower is given by 4309hp.\n",
+ " The output torque is 12573.24 lb-ft. \n",
+ "The output torque can also be written as 12.57*10^3 lb-ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine c\n",
+ "import math\n",
+ "##This numerical is Ex 1_3S,page 12.\n",
+ "H=90.\n",
+ "Q=4.2##water flow rate(in m^3/s)\n",
+ "n=1800.\n",
+ "E=0.87##efficiency\n",
+ "rho=998.\n",
+ "g=9.81\n",
+ "omega=(n*2.*math.pi)/60.\n",
+ "dp=rho*g*H\n",
+ "print\"%s %.2f %s \"%('The pressure is ',dp,' N/m^2')\n",
+ "Ps=E*Q*dp\n",
+ "print\"%s %.2f %s \"%('\\n Output power is equal to ',Ps,' N-m/s')\n",
+ "print(\"After rounding off the value of output power is 3220 kW.\")\n",
+ "Psr=3220000.##rounded off value of Ps\n",
+ "Torque=Psr/omega\n",
+ "print\"%s %.2f %s \"%(' The output torque is ',Torque,' N-m.')\n",
+ "print(\"After rounding off the output torque comes out to be 17.1*10^3 N-m.\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pressure is 881134.20 N/m^2 \n",
+ "\n",
+ " Output power is equal to 3219664.37 N-m/s \n",
+ "After rounding off the value of output power is 3220 kW.\n",
+ " The output torque is 17082.63 N-m. \n",
+ "After rounding off the output torque comes out to be 17.1*10^3 N-m.\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/nemichand /Chapter1_1.ipynb b/sample_notebooks/nemichand /Chapter1_1.ipynb
new file mode 100755
index 00000000..43f5bce4
--- /dev/null
+++ b/sample_notebooks/nemichand /Chapter1_1.ipynb
@@ -0,0 +1,227 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:60b1203d60983bbbb28528cd720bc31b7ac71ec9fc83b7d2e5e78e90b9f2b472"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Introduction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math\n",
+ "\n",
+ "#calculate the steady state\n",
+ "\n",
+ "\n",
+ "##The thickness of the slab(L) is 80mm or .08m\n",
+ "##The thermal conductivity(k)of the material is .20 W/(m*K)\n",
+ "T1=40.;\n",
+ "T2=20.;\n",
+ "L=.08;\n",
+ "k=.20;\n",
+ "##The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L\n",
+ "print(\"The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m^2 \")\n",
+ "q=k*(T1-T2)/L\n",
+ "print(q)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m^2 \n",
+ "50.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math\n",
+ "#calculate the thickness of masonry wall\n",
+ "print(\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 2\")\n",
+ "##The thermal conductivity(km)of masonry wall is .8 W/(mK)\n",
+ "##The thermal conductivity(kc)of composite wall is .2 W/(mK)\n",
+ "##The thickness of composite wall(Lc) is 100 mm or .1 m\n",
+ "km=.8;\n",
+ "kc=.2;\n",
+ "Lc=.1;\n",
+ "##The thickness of masonry wall(Lm) is to be found. \n",
+ "##The steady state heat flow(qm)through masonry wall is km(T1-T2)/L\n",
+ "## The steady state heat flow(qc)through composite wall is kc(T1-T2)/L\n",
+ "##As the steady rate of heat flow through masonry wall is 80% that through composite wall and both the wall have same surface area and same temp. difference so qm/qc=0.8=(km/kc)*(Lc/Lm)\n",
+ "##The thickness of masonry wall is Lm.\n",
+ "print (\"The thickness of masonry wall is Lm in m\")\n",
+ "Lm=(km/kc)*(Lc/(0.8))\n",
+ "print(Lm)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Introduction to heat transfer by S.K.Som, Chapter 1, Example 2\n",
+ "The thickness of masonry wall is Lm in m\n",
+ "0.5\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "## printlay warning for floating point exception\n",
+ "print(\"Introduction to heat transfer by S.K.Som Chapter 1 Example 4\")\n",
+ "hbr=200.;\n",
+ "Tinf=100.;\n",
+ "Ts=20.;\n",
+ "##The rate of heat transfer per unit area is q\n",
+ "print (\"The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m^2\")\n",
+ "q=hbr*(Tinf-Ts)\n",
+ "print(q)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Introduction to heat transfer by S.K.Som Chapter 1 Example 4\n",
+ "The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m^2\n",
+ "16000.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "import math\n",
+ "#calculat e surface area\n",
+ "print(\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 5\")\n",
+ "\n",
+ "hbr=800.;\n",
+ "deltaT=(75.-25.);\n",
+ "Q=20.;\n",
+ "print(\"The heat exchanger surface area(A)in m^2 required for 20 MJ/h of heating is \")\n",
+ "A = (Q*10**6.)/(3600.*hbr*deltaT)\n",
+ "print round(A,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Introduction to heat transfer by S.K.Som, Chapter 1, Example 5\n",
+ "The heat exchanger surface area(A)in m^2 required for 20 MJ/h of heating is \n",
+ "0.14\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##The ambient temprature (Tinf) \n",
+ "print(\"Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\")\n",
+ "Ts=225.;\n",
+ "Tinf=25.;\n",
+ "## |because it is modulus function and it converts negative values to positive value.\n",
+ "X=0.02;\n",
+ "A=.1;\n",
+ "m=4;\n",
+ "cp=2.8;\n",
+ "hbr=(m*cp*10**3*X)/(A*(Ts-Tinf))\n",
+ "print round(hbr,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Introduction to heat transfer by S.K.Som, Chapter 1, Example 6\n",
+ "11.2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/saikiranm/CHAPTER4.ipynb b/sample_notebooks/saikiranm/CHAPTER4.ipynb
new file mode 100755
index 00000000..951201ff
--- /dev/null
+++ b/sample_notebooks/saikiranm/CHAPTER4.ipynb
@@ -0,0 +1,196 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:77734ce79918b96bed43d7baa269682912f0f51f73bf77fadd993778504d6908"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 4 - Characteristics of AC motor"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 4.1 - PG NO:72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.1, Page 72\n",
+ "import numpy\n",
+ "p1=numpy.poly1d([1, -3, 1])#Polynomial equation\n",
+ "print('Part a')\n",
+ "print('roots of the equation when slip at max torque')\n",
+ "x=numpy.roots(p1)\n",
+ "print(x)\n",
+ "\n",
+ "\n",
+ "p2=numpy.poly1d([1, -1.719,0.146])#polynomial equation in scilab function\n",
+ "y=numpy.roots(p2)\n",
+ "print('Part b')\n",
+ "print('roots of the equation when slip at max load')\n",
+ "print(y)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part a\n",
+ "roots of the equation when slip at max torque\n",
+ "[ 2.61803399 0.38196601]\n",
+ "Part b\n",
+ "roots of the equation when slip at max load\n",
+ "[ 1.62939626 0.08960374]\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 4.3 - PG NO:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.3, Page 76\n",
+ "import numpy\n",
+ "p1=numpy.poly1d([0.04, -0.0266,.0016 ])#Polynomial equation\n",
+ "print('Part a')\n",
+ "print('roots of the equation that slip will run is')\n",
+ "x=numpy.roots(p1)\n",
+ "print(x)\n",
+ "#answers after calculation are accurate than textbook answers due to approximations"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part a\n",
+ "roots of the equation that slip will run is\n",
+ "[ 0.59812426 0.06687574]\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 4.5 - PG NO:81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.5, page 81\n",
+ "import math\n",
+ "pole=24.\n",
+ "Ns=245.#in rpm\n",
+ "N=(120.*50.)/pole#synchronous speed in rpm\n",
+ "f=(N-Ns)/N\n",
+ "p=110.#in kw\n",
+ "T=(p*1000.*60.)/(2.*math.pi*Ns)\n",
+ "v1=440./math.sqrt(3)#in v\n",
+ "ws=(2*math.pi*250)/60\n",
+ "s=0.02\n",
+ "R=0.03125#in ohm\n",
+ "x=math.sqrt(((3*R*v1**2)/(T*ws*s))-(R/s)**2)#by rearranging formula\n",
+ "print'%s %.5f %s' %('Stator resistance per phase is=',x,'ohm')\n",
+ "#calculating original resistance\n",
+ "\n",
+ "#Example 4.1, Page 72\n",
+ "import numpy\n",
+ "p1=numpy.poly1d([3190,-3235,72.78 ])#Polynomial equation\n",
+ "print('Part a')\n",
+ "print('The value of original resistance is')\n",
+ "x=numpy.roots(p1)\n",
+ "print(x)\n",
+ "\n",
+ "#Taking r=0.99108\n",
+ "r=(0.99108-R)/1.25**2\n",
+ "\n",
+ "print'%s %.4f %s' %('The value of resistance to be added is =',r,' ohm')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stator resistance per phase is= 0.50358 ohm\n",
+ "Part a\n",
+ "The value of original resistance is\n",
+ "[ 0.99108634 0.02302024]\n",
+ "The value of resistance to be added is = 0.6143 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 4.6 - PG NO:92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.6, Page no 92\n",
+ "import math\n",
+ "print(\"Part ii\")\n",
+ "new_sin_delta=math.sin(math.pi/4)/.95\n",
+ "delta=math.asin(new_sin_delta)\n",
+ "x=math.degrees(delta)\n",
+ "print'%s %.2f %s' %('The value of delta is =',x,'degree')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Part ii\n",
+ "The value of delta is = 48.10 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/saikomalchanagam/AKmaini_(1).ipynb b/sample_notebooks/saikomalchanagam/AKmaini_(1).ipynb
new file mode 100755
index 00000000..71b9c040
--- /dev/null
+++ b/sample_notebooks/saikomalchanagam/AKmaini_(1).ipynb
@@ -0,0 +1,293 @@
+{
+ "metadata": {
+ "name": "AKmaini"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": "Chapter 11: Satellites and Satellite Communications\n"
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 1, Pg No: 567"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math\n\n# Variable Declaration\nh = 150; # height of satellite from earth in km\nG = 6.67*10**-11; # Gravitational constant\nM = 5.98*10**24; # mass of the earth in kg\nRe = 6370; # radius of earth in km\n\n# Calculations\nu = G*M\nV = math.sqrt(u/((Re + h)*10**3)) # orbital velocity\nV1 = V/1000; # orbital velocity in km/s\n\n# Result\nprint 'Orbital velocity = %3.3f'%V1,'km/s';",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Orbital velocity = 7.821 km/s\n"
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 2, Pg No: 568"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Declaration\nAp_Pe_diff = 30000; # difference between apogee and perigee in Km\na = 16000; # semi major axis of orbit\n\n# Calculations\ne = Ap_Pe_diff/float(2*a); # Eccentricity\n\n# Result\nprint 'Eccentricity = %3.2f'%e;",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Eccentricity = 0.94\n"
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 3, Pg No:568"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Decalaration\na1 = 18000; # semi major axis of the elliptical orbits of satellite 1\na2 = 24000; # semi major axis of the elliptical orbits of satellite 2\n\n# Calculations\n#T = 2*%pi*sqrt(a^3/u);\n#let K = T2/T1;\nK = (float(a2)/a1)**(3/float(2)); # Ratio of orbital periods\n\n# Result\nprint 'The orbital period of satellite-2 is %3.2f' %K,' times the orbital period of satellite-1';\n",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "The orbital period of satellite-2 is 1.54 times the orbital period of satellite-1\n"
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 4, Pg No:569"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Declaration\n\nh = 35800; # height of satellite orbit from earth in km\nG = 6.67*10**-11; # Gravitational constant\nM = 5.98*10**24; # mass of the earth in kg\nRe = 6364; # radius of earth in km\ni = 2; # inclination angle\n\n# Calculations\nu = G*M\nr = Re+h\nVi = math.sqrt(u/r*10**3)* math.tan(i*math.pi/180); # magnitude of velocity impulse\nV = Vi/1000; # magnitude of velocity impulse in m/s\n\n#Result\nprint 'Magnitude of velocity impulse = %d' %V,' m/s';",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Magnitude of velocity impulse = 107 m/s\n"
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 5, Pg No: 571"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Declaration\nh = float(13622); # ht of circular orbit from earth's surface\nRe = 6378; # Radius of earth in km\n\n# Calculations\nR = Re+h; # Radius of circular orbit\npimax = 180 - (2*math.acos(Re/R))*(180/math.pi); # Maximum shadow angle\neclipmax_time = (pimax/360)*24; # maximum daily eclipse duration\n\n# Result\nprint ' Maximum shadow angle = %3.1f\u00b0' %pimax\nprint ' Maximum daily eclipse duration = %3.2f'%eclipmax_time,' hours';",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": " Maximum shadow angle = 37.2\u00b0\n Maximum daily eclipse duration = 2.48 hours\n"
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 6, Pg No:572"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math\n\n# Variable Declaration\n\nh = 35786; # ht of geo.stationary orbit above earth surface\nT = 365; # time in days\nr = 6378 # radius of earth in km\n\n# ie(t) = 23.4*sin(2*%pi*t/T)\n# for a circular orbit of 20000 km radius ,phi = 37.4\u00b0 ,Therefore, the time from first day of eclipse to equinox is given by substituting ie(t) = 37.4/2 = 18.7\u00b0\nphi = 37.4\nie = (phi/2)*(math.pi/180)\nk = 23.4*(math.pi/180)\nt = (365/(2*math.pi))*math.asin((ie/k)) \n# for geostationary orbit\nphimax = 180 - 2*(math.acos(r/(r+h)))*(180/math.pi)\nt_geo = (365/(2*math.pi))*math.asin((8.7*math.pi/180)/k)\n\n# Result\nprint 'Total time from first day of eclipse to last day of eclipse = %3.1f' %t,' days';\nprint 'Total time from first day of eclipse to last day of eclipse for geostationary orbit = %3.2f' %t_geo, 'days'",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Total time from first day of eclipse to last day of eclipse = 53.8 days\nTotal time from first day of eclipse to last day of eclipse for geostationary orbit = 22.13 days\n"
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 7, Pg No:600"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Declaration\nm = 100; # mass of satellite\nV = 8000; # orbital velocity in m/s\nRe = 6370; # radius of earth in Km\nH = 200; # satellite height above earth surface\n\n# Calculations\nCF = (m*V**2)/((Re+H)*10**3); #centrifugal force\n\n# Result\nprint 'Centrifugal Force = %d' %CF,' Newtons';",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Centrifugal Force = 974 Newtons\n"
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 8, Pg No:601"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Declaration\n\nApogee = 30000; # Apogee pt of satellite elliptical orbit\nPerige = 1000; # perigee pt of satellite elliptical orbit\n\n# Calculations\na = (Apogee + Perige)/2; # semi major axis\n\n# Result\nprint 'Semi-major axis = %d' %a,' Km';",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Semi-major axis = 15500 Km\n"
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 9, Pg No:603"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math\n\n# Variable Declaration\nfarth = 30000; # farthest point in satellite elliptic eccentric orbit\nclosest = 200; # closest point in satellite elliptic eccentric orbit\nRe = float(6370); # Radius of earth in km\n\n# Calculations\nApogee = farth + Re; # Apogee in km\nPerigee = closest + Re; # perigee in km\na = (Apogee + Perigee)/(2); # semi-major axis\ne = (Apogee - Perigee)/(2*a); # orbit eccentricity\n\n# Result\nprint 'Apogee = %d' %Apogee,' km';\nprint 'Perigee = %d' %Perigee,' km';\nprint 'Orbit eccentricity = %3.3f' %e;",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Apogee = 36370 km\nPerigee = 6570 km\nOrbit eccentricity = 0.694\n"
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 10, Pg No:604"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Declaration\ne = 0.5; # orbit eccentricity\nae = 14000; # from fig. the distance from center of ellipse to the centre of earth\n\n# Calculations\na = ae/(e); # semi major axis\napogee = a*(1 + e); # Apogee in km\nperige = a*(1 - e); # perigee in km\n\n# Result\nprint 'Apogee = %d' %apogee,' km'\nprint 'Perigee = %d' %perige,' km'",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Apogee = 42000 km\nPerigee = 14000 km\n"
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 11, Pg No:604"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Declaration\nG = 6.67*10**-11; # Gravitational constant\nM = 5.98*10**24; # mass of the earth in kg\nRe = 6370*10**3; # radius of earth in m\n\n# Calculations\nu = G*M\nVesc = math.sqrt(2*u/Re);\nVes = Vesc/1000; # escape velocity in km/s\n\n# Result\nprint 'Escape velocity = %3.1f' %Ves,' km/s';",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Escape velocity = 11.2 km/s\n"
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 12, Pg No:605"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# variable Declartion\na = 25000*10**3; # semimajor axis in m from fig\nG = 6.67*10**-11; # Gravitational constant\nM = 5.98*10**24; # mass of the earth in kg\nh = 0\n\n# Calculations\nu = G*M;\nT = 2*math.pi*math.sqrt((a**3)/u)\nhr = T/3600 # conv. from sec to hrs and min\nt = T%3600 # conv. from sec to hrs and min\nmi = t/60 # conv. from sec to hrs and min\n\n# Result\nprint 'Orbital time period = %d' %hr,' Hours',' %d'%mi, 'minutes'",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Orbital time period = 10 Hours 55 minutes\n"
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": "Example 13, Pg No:605"
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": "import math;\n\n# Variable Declaration\napogee = float(35000); # farthest point in kms\nperigee = 500; # closest point in kms\nr = float(6360); # radius of earth in kms\nG = 6.67*10**-11 # gravitational constant\nM = 5.98*10**24; # mass of earth in kgs\n\n# calculations\n#funcprot(0)\napogee_dist = apogee + r # apogee distance in kms\nperigee_dist= perigee+r ; # perigee distance in kms\na = (apogee_dist + perigee_dist)/2; # semi-major axis of elliptical orbit\nT = (2*math.pi)*math.sqrt((a*10**3)**3/(G*M)); # orbital time period\nhr = T/3600 # conv. from sec to hrs and min\nt = (T%3600) # conv. from sec to hrs and min\nmi = t/60 # conv. from sec to hrs and min\nu = G*M\nVapogee = math.sqrt(u*((2/(apogee_dist*10**3)) - (1/(a*10**3))))/1000; # velocity at apogee point\nVperigee = math.sqrt((G*M)*((2/(perigee_dist*10**3)-(1/(a*10**3)))))/1000 # velocity at perigee point\n\n#Result\nprint 'Orbital Time Period = %d'%hr,' Hrs'' %d'%mi,' min'\nprint 'Velocity at apogee = %3.3f' %Vapogee,' Km/s'\nprint'Velocity at perigee = %3.3f' %Vperigee,' Km/s'\nprint'Note: Calculation mistake in textbook in finding velocity at apogee point'\n",
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": "Orbital Time Period = 10 Hrs 20 min\nVelocity at apogee = 1.656 Km/s\nVelocity at perigee = 9.987 Km/s\nNote: Calculation mistake in textbook in finding velocity at apogee point\n"
+ }
+ ],
+ "prompt_number": 57
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/sample_notebooks/vinodkumar/Chapter1.ipynb b/sample_notebooks/vinodkumar/Chapter1.ipynb
new file mode 100755
index 00000000..0225739c
--- /dev/null
+++ b/sample_notebooks/vinodkumar/Chapter1.ipynb
@@ -0,0 +1,339 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:cd92732b89aba9dec0c4f42de79fa75bac0a4c551a65b1e08ee3d7196bca88ee"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Basic Circuit Concepts"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Basic Circuit Concepts\n",
+ "##page no-1.9\n",
+ "##example1.1\n",
+ "print(\"Current through 15Ohm resistor is given by:\");\n",
+ "print(\"I1=30/15\");\n",
+ "I1=30/15\n",
+ "print\"%s %.2f %s \"%(\"current through 15Ohm resistor = \",I1,\" Ampere\")\n",
+ "print(\"Current through 5Ohm resistor is given by:\")\n",
+ "print(\"I2=5+2\");\n",
+ "I2=5+2\n",
+ "print\"%s %.2f %s \"%(\"current through 5ohm resistor = \",I2,\" Ampere\")\n",
+ "print(\"R=100-30-5*I2/I1\");\n",
+ "R=(100-30-5*I2)/I1\n",
+ "print\"%s %.2f %s \"%(\"R = \",R,\" Ohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through 15Ohm resistor is given by:\n",
+ "I1=30/15\n",
+ "current through 15Ohm resistor = 2.00 Ampere \n",
+ "Current through 5Ohm resistor is given by:\n",
+ "I2=5+2\n",
+ "current through 5ohm resistor = 7.00 Ampere \n",
+ "R=100-30-5*I2/I1\n",
+ "R = 17.00 Ohm \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Basic Circuit Concepts\n",
+ "##page no-1.10\n",
+ "##example1.2\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"from the given fig:\")\n",
+ "print(\"I2-I3=13\");\n",
+ "print(\"-20*I1+8*I2=0\");\n",
+ "print(\"-12*I1-16*I3=0\");\n",
+ "##solving these equations in the matrix form\n",
+ "A=numpy.matrix([[0, 1 ,-1],[-20, 8, 0],[-12 ,0 ,-16]])\n",
+ "B=numpy.matrix([[13], [0] ,[0]])\n",
+ "print(\"A=\")\n",
+ "print[A]\n",
+ "print(\"B=\")\n",
+ "print[B]\n",
+ "X=numpy.dot(numpy.linalg.inv(A),B)\n",
+ "print(\"X=\")\n",
+ "print[X]\n",
+ "print(\"I1 = 4Ampere\")\n",
+ "print(\"I2 = 10Ampere\")\n",
+ "print(\"I3 = -3Ampere\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "from the given fig:\n",
+ "I2-I3=13\n",
+ "-20*I1+8*I2=0\n",
+ "-12*I1-16*I3=0\n",
+ "A=\n",
+ "[matrix([[ 0, 1, -1],\n",
+ " [-20, 8, 0],\n",
+ " [-12, 0, -16]])]\n",
+ "B=\n",
+ "[matrix([[13],\n",
+ " [ 0],\n",
+ " [ 0]])]\n",
+ "X=\n",
+ "[matrix([[ 4.],\n",
+ " [ 10.],\n",
+ " [ -3.]])]\n",
+ "I1 = 4Ampere\n",
+ "I2 = 10Ampere\n",
+ "I3 = -3Ampere\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Basic Circuit Concepts\n",
+ "##pg no-1.11\n",
+ "##example 1.3\n",
+ "print(\"Iaf=x\")\n",
+ "print(\"Ife=x-30\")\n",
+ "print(\"Ied=x+40\")\n",
+ "print(\"Idc=x-80\")\n",
+ "print(\"Icb=x-20\")\n",
+ "print(\"Iba=x-80\")\n",
+ "print(\"Applying KVL to the closed path AFEDCBA:\")##Applying KVL to the path AFEDCBA\n",
+ "print(\"x=4.1/0.1\")\n",
+ "x=4.1/0.1;\n",
+ "Iaf=x;\n",
+ "print\"%s %.2f %s \"%(\"\\nIaf = \",Iaf,\" Ampere\");\n",
+ "Ife=x-30.\n",
+ "print\"%s %.2f %s \"%(\"\\nIfe = \",Ife,\" Ampere\");\n",
+ "Ied=x+40.;\n",
+ "print\"%s %.2f %s \"%(\"\\nIed = \",Ied,\" Ampere\");\n",
+ "Idc=x-80;\n",
+ "print\"%s %.2f %s \"%(\"\\nIdc = \",Idc,\" Ampere\");\n",
+ "Icb=x-20.;\n",
+ "print\"%s %.2f %s \"%(\"\\nIcb = \",Icb,\" Ampere\");\n",
+ "Iba=x-80.;\n",
+ "print\"%s %.2f %s \"%(\"\\nIba = \",Iba,\" Ampere\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Iaf=x\n",
+ "Ife=x-30\n",
+ "Ied=x+40\n",
+ "Idc=x-80\n",
+ "Icb=x-20\n",
+ "Iba=x-80\n",
+ "Applying KVL to the closed path AFEDCBA:\n",
+ "x=4.1/0.1\n",
+ "\n",
+ "Iaf = 41.00 Ampere \n",
+ "\n",
+ "Ife = 11.00 Ampere \n",
+ "\n",
+ "Ied = 81.00 Ampere \n",
+ "\n",
+ "Idc = -39.00 Ampere \n",
+ "\n",
+ "Icb = 21.00 Ampere \n",
+ "\n",
+ "Iba = -39.00 Ampere \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Basic Circuit Concepts\n",
+ "##pg no- 1.12\n",
+ "##example 1.4\n",
+ "import math\n",
+ "import numpy\n",
+ "print(\"Applying KVL to the closed path OBAO\");##Applying KVL to the closed path OBAO\n",
+ "print(\"3*x-3*y=2\");\n",
+ "print(\"Applying KVL to the closed path ABCA\");##Applying KVL to the closed path ABCA\n",
+ "print(\"9*x+12*y=4\");\n",
+ "a=numpy.matrix([[3, -3],[9, 12]]);\n",
+ "b=([[2] ,[4]])\n",
+ "print(\"a=\")\n",
+ "print[a]\n",
+ "print(\"b=\")\n",
+ "print[b]\n",
+ "X=numpy.dot(numpy.linalg.inv(a),b)\n",
+ "\n",
+ "print(X)\n",
+ "print(\"x=0.5714286 Ampere\");\n",
+ "print(\"y=-0.095238 Ampere\");\n",
+ "print(\"Ioa=0.57A\")\n",
+ "print(\"Iob=1-0.57\")\n",
+ "Iob=1-0.57;\n",
+ "print\"%s %.2f %s \"%(\"\\nIob = \",Iob,\" A\");\n",
+ "print(\"Iab = 0.095\");\n",
+ "Iac=0.57-0.095;\n",
+ "print\"%s %.2f %s \"%(\"\\nIac =\",Iac,\" A\");\n",
+ "print(\"Iab=1-0.57 + 0.095\")\n",
+ "Iab=1-0.57 + 0.095;\n",
+ "print\"%s %.2f %s \"%(\"\\nIob = \",Iab,\" A\") "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Applying KVL to the closed path OBAO\n",
+ "3*x-3*y=2\n",
+ "Applying KVL to the closed path ABCA\n",
+ "9*x+12*y=4\n",
+ "a=\n",
+ "[matrix([[ 3, -3],\n",
+ " [ 9, 12]])]\n",
+ "b=\n",
+ "[[[2], [4]]]\n",
+ "[[ 0.57142857]\n",
+ " [-0.0952381 ]]\n",
+ "x=0.5714286 Ampere\n",
+ "y=-0.095238 Ampere\n",
+ "Ioa=0.57A\n",
+ "Iob=1-0.57\n",
+ "\n",
+ "Iob = 0.43 A \n",
+ "Iab = 0.095\n",
+ "\n",
+ "Iac = 0.47 A \n",
+ "Iab=1-0.57 + 0.095\n",
+ "\n",
+ "Iob = 0.53 A \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Basic Circuit Concepts\n",
+ "##pg no-1.12\n",
+ "##example 1.5\n",
+ "I1=2./5.;\n",
+ "print\"%s %.2f %s \"%(\"I1=2/5= \",I1,\" Ampere\")\n",
+ "I2=4./8.;\n",
+ "print\"%s %.2f %s \"%(\"\\nI2=4/8= \",I2,\" Ampere\")\n",
+ "print(\"\\nPotential difference between points x and y = Vxy = Vx-Vy\")\n",
+ "print(\"\\nWriting KVL equations for the path x to y\")##Writing KVL equation from x to y\n",
+ "print(\"\\nVs+3*I1+4-3*I2-Vy=0\")\n",
+ "print(\"\\nVs+3*(0.4) + 4- 3*(0.5) -Vy = 0\")\n",
+ "print(\"\\nVs+3*I1+4-3*I2-Vy = 0\")\n",
+ "print(\"\\nVx-Vy = -3.7\")\n",
+ "print(\"\\nVxy = -3.7V\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I1=2/5= 0.40 Ampere \n",
+ "\n",
+ "I2=4/8= 0.50 Ampere \n",
+ "\n",
+ "Potential difference between points x and y = Vxy = Vx-Vy\n",
+ "\n",
+ "Writing KVL equations for the path x to y\n",
+ "\n",
+ "Vs+3*I1+4-3*I2-Vy=0\n",
+ "\n",
+ "Vs+3*(0.4) + 4- 3*(0.5) -Vy = 0\n",
+ "\n",
+ "Vs+3*I1+4-3*I2-Vy = 0\n",
+ "\n",
+ "Vx-Vy = -3.7\n",
+ "\n",
+ "Vxy = -3.7V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
generated by cgit (git 2.34.1) at 2024-12-20 07:00:49 +0000