{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# CHAPTER 43: ELECTRIC TRACTION" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.1 , PAGE NO :- 1716" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Max speed of train is = 39.2 km/h.\n" ] } ], "source": [ "#A suburban train runs with an average speed of 36 km/h between two stations 2 km apart.Values of acceleration\n", "#and retardation are 1.8 km/h/s and 3.6 km/h/s.Compute the maximum speed of the train assuming trapezoidal\n", "#speed/time curve\n", "################################################################################################################\n", "import math\n", "\n", "#Given\n", "\n", "Va = 36.0*5/18.0 #m/s (average speed)\n", "alpha = 1.8*5/18.0 #m/s^2 (acc.)\n", "beta = 3.6*5/18.0 #m/s^2 (ret.)\n", "D = 2000.0 #m (distance )\n", "t = D/Va #s (time)\n", "K = (alpha + beta)/(2*alpha*beta) #(constant)\n", "Vm = (t-math.sqrt(t*t-4*K*D))/(2*K) #m/s (max speed)\n", "Vm = Vm*18/5.0 #km/h\n", "print \"Max speed of train is = \",round(Vm,2),\"km/h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.2 , PAGE NO :- 1716" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The acceleration is = 1.7 km/h/s.\n" ] } ], "source": [ "#A train is required to run between two stations 1.5 km apart at a schedule speed of 36 km/h,the duration of stops being\n", "#25 seconds.The braking retardation is 3 km/h/s.Assuming a trapezoidal speed/time curve,calculate the acceleration if the\n", "#ratio of maximum speed to average speed is to be 1.25\n", "##########################################################################################################################\n", "\n", "#Given\n", "D = 1500.0 #m (distance)\n", "Vsch = 36.0*5/18.0 #m/s (scheduled speed)\n", "beta = 3.0*5/18.0 #km/h/s (ret.)\n", "\n", "time = D/Vsch #s (scheduled time)\n", "t_stop = 25.0 #s (stopping time)\n", "t = time-t_stop #s (actual time of run)\n", "Va = D/t #m/s (average speed)\n", "Vm = 1.25*Va #m/s (max speed)\n", "\n", "#The value of K is\n", "K = (D/(Vm*Vm))*(Vm/Va-1)\n", "#We know that K = 1/2*(1/alpha + 1/beta).Therefore,value of alpha is\n", "alpha = beta/(2*K*beta-1) #m/s^2\n", "alpha = alpha*18/5.0 #km/h/s\n", "print \"The acceleration is =\",round(alpha,1),\"km/h/s.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.3 , PAGE NO :- 1716" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Schedule speed of train is 38.7 Km/h.\n" ] } ], "source": [ "#Find the scheduled speed of an electric train for a run of 1.5km if the ratio of its maximum to average speed is\n", "#1.25.It has a braking retardation of 3.6 km/h/s,acceleration of 1.8km/h/s and stop time of 21 seconds.Assume\n", "#trapezoidal speed/time curve.\n", "##################################################################################################################\n", "import math\n", "\n", "#Given\n", "alpha = 1.8*5/18.0 #m/s^2 (acceleration)\n", "beta = 3.6*5/18.0 #m/s^2 (retardation)\n", "D = 1500.0 #m (distance)\n", "ratio = 1.25 #(Vm/Va) \n", "t_stop = 21 #s (stopping time) \n", "K = 0.5*(1/alpha + 1/beta)\n", "\n", "#Also constant K = (D/Vm^2)*(Vm/Va-1)\n", "#Therefore Vm^2 = (D/K)*((Vm/Va)-1).\n", "Vm = math.sqrt(D/K*(ratio-1)) #m/s (Max. speed)\n", "Va = Vm/ratio #m/s (Avg. speed) \n", "t = D/Va #s (travel time)\n", "\n", "tsch = t + t_stop #s (schedule time)\n", "Vsch = D/tsch #m/s (scheduled speed)\n", "Vsch = Vsch*18/5.0 #km/h\n", "print \"Schedule speed of train is\",round(Vsch,1),\"Km/h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.4 , PAGE NO :- 1717" ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false, "scrolled": true }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Duration of acceleration = 26.7 s.\n", "Duration of coasting = 119.3 s.\n", "Duration of braking = 14.0 s.\n", "Total duration = 160.0 s.\n" ] } ], "source": [ "#A train runs between two stations 1.6km apart at an average speed of 36 km/h.If the maximum speed is to be limited\n", "#to 72km/h,acceleration to 2.7km/h/s,coasting retardation to 0.18 km/h/s and braking retardation to 3.2 km/h/s,compute\n", "#the duration of acceleration,coasting and braking periods.Assume a simplified speed/time curve.\n", "########################################################################################################################\n", "\n", "from sympy import Eq,Symbol, solve\n", "#Given\n", "D = 1600 #m (distance)\n", "Va = 36.0*5/18.0 #m/s (avg speed)\n", "V1 = 72.0*5/18.0 #m/s (max speed)\n", "alpha = 2.7*5/18.0 #m/s^2 (acceleration)\n", "beta = 3.6*5/18.0 #m/s^2 (braking retardation)\n", "beta_c = 0.18*5/18.0 #m/s^2 (coasting retardation)\n", "#Duration of acceleration\n", "t1 = V1/alpha #s\n", "\n", "#Actual time of run\n", "t = D/Va #s\n", "#Let us assume V2 be the speed at start of braking.\n", "\n", "V2 = Symbol('V2')\n", "t3 = V2/beta #(braking period)\n", "t2 = (V1-V2)/beta_c #(coasting period)\n", "\n", "eq = Eq(t1+t2+t3,t)\n", "V2 = solve(eq)\n", "Ve2 = V2[0]\n", "\n", "#Now we know the value of V2 that is Ve2.\n", "t3 = Ve2/beta #(braking period)\n", "t2 = (V1-Ve2)/beta_c #(coasting period)\n", "print \"Duration of acceleration = \",round(t1,1),\"s.\"\n", "print \"Duration of coasting = \",round(t2,1),\"s.\"\n", "print \"Duration of braking = \",round(t3,1),\"s.\"\n", "print \"Total duration = \",round(t,0),\"s.\"\n", "\n", "\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.5 , PAGE NO :- 1724" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The limiting value of train speed = 82.0 km/h.\n" ] } ], "source": [ "#The peripheral speed of a railway traction motor cannot be allowed to exceed 44m/s.If gear ratio is 18/75,motor armature\n", "#diameter 42cm and wheel diameter 91cm,calculate the limiting value of the train speed.\n", "###########################################################################################################################\n", "\n", "#Given\n", "gratio = 18/75.0 #(gear ratio)\n", "Vmot = 44.0 #m/s (speed of traction motor)\n", "wdia = 0.91 #m (wheel diameter)\n", "mdia = 0.42 #m (motor armature diameter)\n", "\n", "#Maximum number of revolutions by armature (in 1s)\n", "mrev = Vmot/(3.14*mdia) #rps\n", "#Maximum number of revolutions by driving wheel(in 1s)\n", "wrev = mrev*gratio #rps\n", "#Maximum distance travelled by driving wheel (in 1s)\n", "dist = wrev*(3.14*wdia) #m\n", "#Therefore,limiting speed is\n", "vel = dist*18/5.0 #km/h\n", "print \"The limiting value of train speed = \",round(vel),\"km/h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.6 , PAGE NO :- 1724" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Torque developed by each motor = 7181.0 N-m.\n" ] } ], "source": [ "#A 250-tonne motor coach driven by four motors takes 20 seconds to attain a speed of 42 km/h,starting from rest on an ascending\n", "#gradient of 1 in 80.The gear ratio is 3.5,gear efficiency 92%,wheel diameter 92cm train resistance 40N/t and rotational inertia\n", "#10 percent of the dead weight.Find the torque developed by each motor.\n", "##################################################################################################################################\n", "\n", "#Given\n", "M = 250.0 #tonne (mass of motor)\n", "Me = 1.1*250.0 #tonne (mass of rotating motor)\n", "Vm = 42.0 #km/h (speed)\n", "t1 = 20.0 #s (time)\n", "G = 1/80.0*100 # (% gradient)\n", "r = 40.0 #N/tonne(train reistance)\n", "D = 0.92 #m (wheel diameter)\n", "gratio = 3.5 # (gear ratio)\n", "geff = 0.92 # (gear efficiency)\n", "\n", "a = Vm/t1 # (acceleration) \n", "#Now,tractive force is given by\n", "Ft = 277.8*Me*a + 98*M*G + M*r #N\n", "#Now Ft = 2*gratio*geff*T/D.Therefore torque 'T' is\n", "T = Ft*D/(2*gratio*geff) #N-m\n", "#There are motors.so,torque by each motor is\n", "torque = T/4 #N-m\n", "print \"Torque developed by each motor = \",round(torque),\"N-m.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.7 , PAGE NO :- 1724" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Time taken to achieve speed of 80km/h is = 44.3 s.\n", "Current drawn per motor is 488.0 A.\n" ] } ], "source": [ "#A 250-tonne motor coach having 4 motors,each developing a torque of 8000 N-m during acceleration,starts from rest.If\n", "#up-gradient is 30 in 1000,gear ratio 3.5,gear transmission efficiency 90%,wheel diameter 90cm,train resistance 50N/t,\n", "#rotational inertia effect 10%,compute the time taken by the coach to attain a speed of 80km/h.\n", "#If supply voltage is 3000V and motor efficiency 85%,calculate the current taken during the acceleration period.\n", "#########################################################################################################################\n", "\n", "#Given\n", "M = 250.0 #tonne (mass of motor)\n", "Me = 1.1*250 #tonne (mass of rotating motor)\n", "r = 50.0 #N/t (train resistance)\n", "G = 30.0/1000*100 # (% gradient)\n", "torque = 8000 #N-m (torque of each motor)\n", "D = 0.90 #m (wheel diameter)\n", "gratio = 3.5 # (gear ratio)\n", "geff = 0.90 # (gear efficiency)\n", "Vm = 80 #km/h (speed)\n", "meff = 0.85 # (Motor efficiency)\n", "V = 3000.0 # (Voltage)\n", "\n", "T = 4*torque # (total torque)\n", "#Now,we Know that\n", "Ft = 2*gratio*geff*T/D #N\n", "#Also,we know that Ft = 277.8*Me*a + 98MG + Mr\n", "a = (Ft-98*M*G-M*r)/(277.8*Me) #km/h/s(acceleration)\n", "#Time taken to attain the speed\n", "t1 = Vm/a #s\n", "#Power taken by motor is given by\n", "Power = Ft*(5.0/18*Vm)/meff #W\n", "#Total current drawn is (using P = VI)\n", "I = Power/V #A\n", "#Current drawn/motor is\n", "cur = I/4 #A\n", "\n", "print \"Time taken to achieve speed of 80km/h is =\",round(t1,1),\"s.\"\n", "print \"Current drawn per motor is\",round(cur,1),\"A.\"" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## EXAMPLE 43.8 , PAGE NO :- 1725" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Weight of locomotive is = 142.0 tonne.\n", "Number of axles required are = 7.0\n" ] } ], "source": [ "#A goods train weighing 500 tonne is to be hauled by a locomotive up an ascending gradient of 2% with an\n", "#acceleation of 1 km/h/s.If the coefficient of adhesion is 0.25,train resistance 40N/t and effect of rotational\n", "#inertia 10%,find the weight of locomotive and number of axles if load is not to increase beyond 21 tonne/axle.\n", "###############################################################################################################\n", "\n", "from sympy import Eq,Symbol, solve\n", "#Given\n", "mass = 500.0 #tonne (mass of goods train)\n", "a = 1 #km/h/s (acceleration)\n", "coef_ad = 0.25 #(coefficient of adhesion)\n", "r = 40.0 #N/t (train resistance)\n", "G = 2.0 # (% gradient)\n", "w_axle = 21.0 #tonne/axle (weight per axle)\n", "#Let Ml be mass of locomotive\n", "Ml = Symbol('Ml')\n", "#Total mass is given by\n", "M = mass + Ml\n", "#Now,tractive force is given by\n", "Ft1 = M*(277.8*1.1*a + 98*G + r) #N\n", "#Also,Maximum tractive force is given by\n", "Ft2 = 1000*coef_ad*Ml*9.8\n", "#As both Ft1 and Ft2 are same, therefore equating\n", "eq = Eq(Ft1,Ft2)\n", "Ml = solve(eq)\n", "#Mass of locomotive 'M_l' is\n", "M_l = Ml[0]\n", "#No of axles required are\n", "no_axle = round(M_l)/w_axle\n", "print \"Weight of locomotive is =\",round(M_l),\"tonne.\"\n", "print \"Number of axles required are = \",round(no_axle)\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.9 , PAGE NO :- 1725" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The trailing weight that can now be hauled up = 1100.0 tonne.\n" ] } ], "source": [ "#An electric locomotive weighing 100 tonne can just accelerate a train of 500 tonne(trailing weight) with an acceleration of \n", "#1 km/h/s on an up-gradient of 0.1%.Train resistance is 45N/t and rotational inertia is 10%.If this locomotive is helped by another\n", "#locomotive of 120 tonne,find:\n", "#(i)the trailing weight that can now be hauled up the same gradient under the same conditions.\n", "#(ii)the maximum gradient,if the trailing hauled load remains unchanged.\n", "#Assume adhesive weight expressed as percentage of total dead weight as 0.8 for both locomotives.'''\n", "#############################################################################################################################\n", "\n", "#Given\n", "Ml1 = 100.0 #tonne (locomotive 1)\n", "Ml2 = Ml1 + 120.0 #tonne (locomotive 1&2)\n", "mass = 500.0 #tonne (trailing part)\n", "r = 45.0 #N/t (train resistance)\n", "G = 0.1 # (% gradient)\n", "a = 1.0 #km/h/s (acceleration)\n", "\n", "#Total mass of train and locomotive\n", "M = Ml1 + mass #tonne\n", "Me = 1.1*M #tonne \n", "#Traction Force is\n", "Ft = 277.8*Me*a + 98*M*G + M*r #N\n", "#Let coefficient_of_adhesion be 'ua'.Then the maximium tractive effort by locomotive 1 is Ft = 1000*ua*Ml1*9.8\n", "ua = Ft/(1000*9.8*Ml1)\n", "\n", "#Maximum tractive effort by locomotive 1&2 is\n", "Ft2 = 1000*ua*Ml2*9.8 #N\n", "#Also,Ft2 = M2*(277.8*1.1*a + G + r).Therefore,\n", "M2 = Ft2/(277.8*1.1*a + 98*G + r) #tonne\n", "#(i)The trailing weight that can be hauled up is\n", "wtrail = M2-Ml2 #tonne\n", "#(ii)Now, in this case the total weight of train & locomotive is\n", "M3 = mass + Ml2\n", "#Tractive force Ft2 = M3*(277.8*1.1*a + 98G2 + r).Therefore,for max gradient 'G2'\n", "G2 = (Ft2/M3-277.8*1.1*a-r)/98\n", "\n", "print \"The trailing weight that can now be hauled up =\",round(wtrail,2),\"tonne.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.10 , PAGE NO :- 1726" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Max speed = 44.32 km/h.\n", "Specific energy output = 30.18 Wh/t-km.\n" ] } ], "source": [ "#The average distance between stops on a level section of a railway is 1.25km.Motor-coach train weighing 200 tonne has a\n", "#schedule speed of 30 km/h,the duration of stops being 30 seconds.The acceleration is 1.9 km/h/s and the braking retardation\n", "#is 3.2 km/h/s.Train resistance to traction is 45 N/t.Allowance for rotational inertia is 10%.Calculate the specific energy\n", "#output in Wh/t-km.Assume a trapezoidal speed/time curve.\n", "##############################################################################################################################\n", "import math as m\n", "#Given\n", "alpha = 1.9*5/18.0 #km/h/s (acceleration)\n", "beta = 3.2*5/18.0 #km/h/s (retardation)\n", "D = 1.25*1000.0 #m (distance)\n", "tstop = 30.0 #s (stop time)\n", "Vsch = 30*5/18.0 #m/s (schedule speed) \n", "r = 45.0 #N/t (train resistance) \n", "K = round((alpha + beta)/(2*alpha*beta),1) #(constant K)\n", "\n", "tsch = D/Vsch #s (schedule time)\n", "t = tsch-tstop #s (running time)\n", "#Now,max speed is given by\n", "Vm = (t - m.sqrt(t*t-4*K*D))/(2*K) #m/s\n", "#Braking distance is given by(using newton's III equation of motion)\n", "dist = Vm*Vm/(2*beta) #m\n", "#Therefore non-braking distance\n", "D2 = (D - dist)/1000 #km\n", "Vm = Vm*18/5.0 #km/h\n", "D = D/1000 #km\n", "#Specific energy output is\n", "spengy = 0.01072*(Vm*Vm/D)*1.1 + 0.2778*r*(D2/D)\n", "\n", "print \"Max speed = \",round(Vm,2),\"km/h.\"\n", "print \"Specific energy output = \",round(spengy,2),\" Wh/t-km.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.11 , PAGE NO :- 1726" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The value of specific energy consumption is 21.6 Wh/t-km.\n" ] } ], "source": [ "#A 300-tonne EMU is started with a uniform acceleration and reaches a speed of 40 Km/h in 24 seconds on a level track.\n", "#Assuming trapezoidal speed/time curve,find specific energy consumption if rotational inertia is 8%,retardation is 3 km/h/s,\n", "#distance between stops is 3 km,motor efficiency is 0.9 and train resistance is 49N/tonne.\n", "#############################################################################################################################\n", "\n", "#Given\n", "M = 300.0 #tonne (mass of train)\n", "Mratio = 1.08 # (Me/M)\n", "Vm = 40.0 #km/h (speed)\n", "beta = 3.0 #km/h/s (retardation)\n", "r = 49.0 #N/t (train resistance)\n", "meff = 0.9 # (motor efficiency)\n", "D = 3.0 #km (distance)\n", "\n", "#Braking time is(using Newton's Ist eqn)\n", "t3 = Vm/beta\n", "#Distance travelled in braking time(using Newton's IInd eqn)\n", "dist = 0.5*beta*t3*t3/3600 #km\n", "#Non braking distance D2 is\n", "D2 = D - dist #km\n", "#Specific energy consumption is given by,\n", "spengy = 0.01072*(Vm*Vm/(meff*D))*Mratio + 0.2778*r/meff*(D2/D) #Wh/t-km\n", "print \"The value of specific energy consumption is\",round(spengy,1),\"Wh/t-km.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.12 , PAGE NO :- 1726" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "acceleration = 2.0 km/h/s.\n", "coasting retardation = 0.2 km/h/s.\n", "The schedule speed = 30.3 km/h.\n", "The new schedule speed = 31.6 km/h.\n" ] } ], "source": [ "#An electric train accelerates uniformly from rest to a speed of 50km/h in 25 seconds.It then coasts for 70 seconds\n", "#against a constant resistance of 60N/t and is then braked to rest with uniform retardation of 3.0km/h/s in 12 seconds.\n", "#Compute :- (i)uniform acceleration (ii)coasting retardation\n", "#(iii)schedule speed if station stops are of 20-second duration\n", "#Allow 10% for rotational inertia.How will the scedule speed be affected if duration of stops is reduced to 15 seconds,\n", "#other factors remaining the same?\n", "##########################################################################################################################\n", "\n", "#Given\n", "V1 = 50.0 #km/h (Max speed)\n", "t1 = 25.0 #s (accelerating time)\n", "t2 = 70.0 #s (coasting time)\n", "r = 60.0 #N/t (train resistance)\n", "beta = 3.0 #km/h/s (retardation)\n", "t3 = 12.0 #s (braking time)\n", "\n", "#(i) Uniform acceleration is (using Newton's 1st eqn.)\n", "alpha = V1/t1 #km/h/s\n", "print \"acceleration = \",round(alpha,1),\"km/h/s.\"\n", "#Now in braking period (using Newton's 1st eqn.)-> 0 = V2 - beta*t3\n", "V2 = beta*t3 #km/h\n", "#(ii)Now in coasting period (using Newton's 1st eqn.)-> V2 = V1 - beta_c*t2\n", "beta_c = (V1-V2)/t2 #km/h/s\n", "print \"coasting retardation = \",round(beta_c,1),\"km/h/s.\"\n", "#distance travelled in accelerating period (using Newton's 2nd eqn.)\n", "dis1 = 0.5*alpha*t1*t1/3600 #km\n", "#distance travelled in coasting period (using Newton's 3rd eqn.)\n", "dis2 = -(V2*V2 - V1*V1)/(2*beta_c*3600) #km\n", "#distance travelled in braking period (using Newton's 2nd eqn.)\n", "dis3 = 0.5*beta*t3*t3/3600 #km\n", "#Total distance\n", "D = dis1 + dis2 + dis3 #km\n", "#Total time\n", "T = t1 + t2 + t3 + 20.0 #s\n", "#Scheduled Speed Vsch is\n", "Vsch = D/T*3600 #km/h\n", "print \"The schedule speed = \",round(Vsch,1),\"km/h.\"\n", "#New Total time\n", "T = t1 + t2 + t3 + 15.0\n", "#New scheduled speed is\n", "Vsch = D/T*3600 #km/h\n", "print \"The new schedule speed = \",round(Vsch,1),\"km/h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.13 , PAGE NO :- 1727" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The specific energy consumption is = 63.1 Wh/t-km.\n" ] } ], "source": [ "#A 350-tonne electric train runs up an ascending gradient of 1% with the following speed/time curves:\n", "# 1.uniform acceleration of 1.6km/h/s for 25 seconds 2.constant speed for 50 seconds\n", "# 3.coasting for 30 seconds 4.braking at 2.56 km/h/s to rest.\n", "#Compute the specific energy consumption if train resistance is 50N/t,effect of rotational inertia 10%,overall efficiency of\n", "#transmission gear and motor,75%.\n", "#############################################################################################################################\n", "\n", "#Given\n", "M = 350.0 #tonne (weight of train)\n", "Me = 1.1*M #tonne (effect of rot. inertia)\n", "G = 1.0 # (% gradient)\n", "alpha = 1.6 #km/h/s (acceleration)\n", "t1 = 25.0 #s (acceleration time)\n", "t2 = 50 #s (constant speed time)\n", "t3 = 30 #s (coasting time)\n", "beta = 2.56 #km/h/s (braking retardation)\n", "r = 50.0 #N/t (train resistance)\n", "meff = 0.75 # (motor efficiency)\n", "\n", "V1 = alpha*t1 #km/h (Max speed attained)\n", "#Tractive force during coasting is\n", "Ft = 98*M*G + M*r #N\n", "#Also,Tractive force during coasting is given by\n", "#Ft = 277.8*Me*beta_c . Therefore beta_c is\n", "beta_c = Ft/(277.8*Me) #km/h/s\n", "\n", "#During Coasting period.(Using Newton's 1st eqn.) . \"V2\" is given by\n", "V2 = V1 - beta_c*t3 #km/h\n", "t4 = V2/beta #s (braking time) \n", "\n", "#Distance travelled during acceleration period.\n", "dis1 = 0.5*alpha*t1*t1/3600 #km\n", "#Distance travelled during constant speed period.\n", "dis2 = V1*t2/3600 #km\n", "#Distance travelled during coasting period.\n", "dis3 = (V1+V2)/2*t3/3600 #km\n", "#Distance travelled during braking period.\n", "dis4 = 0.5*V2*t4/3600 #km\n", "#Total distance between stops\n", "D = dis1 + dis2 + dis3 + dis4 #km\n", "#Distance travelled during acceleration and constant speed period\n", "D2 = dis1 + dis2 #km\n", "#Specific energy consumption is given by\n", "spengy = (0.01072*(V1*V1/D)*(Me/M) + 27.25*G*(D2/D) + 0.2778*r*(D2/D))/meff #Wh/t-km\n", "\n", "print \"The specific energy consumption is =\",round(spengy,1),\"Wh/t-km.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.14 , PAGE NO :- 1728" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Total number of locomotives required are = 4.0\n" ] } ], "source": [ "#An ore-carrying train weighing 5000 tonne is to be hauled down a gradient of 1:50 at a maximum speed of 30km/h\n", "#and started on a level track at an acceleration of 0.29 km/h/s.How many locomotives,each weighing 75 tonne,will\n", "#have to be employed?\n", "#Train resistance during starting = 29.4 N,Train resistance at 30km/h = 49N/t,Coefficient of adhesion = 0.3\n", "#Rotational inertia = 10%\n", "################################################################################################################################\n", "\n", "#Given\n", "M = 5000.0 #tonne (Mass of train)\n", "Ml = 75.0 #tonne (Mass of locomotive) \n", "G = 1.0/50 # (gradient)\n", "Vm = 30.0 #km/h (Max speed)\n", "a = 0.29 #km/h/s (acceleration) \n", "r1 = 49.0 # (Train resistance at 30km/h)\n", "r2 = 29.4 # (Train resistance at starting)\n", "ua = 0.3 # (coefficient of adhesion)\n", "\n", "#Downward force due to gravity\n", "F = M*G*9.8*1000 #N\n", "#Train resistance\n", "Fres = r1*M #N\n", "#Braking force required is\n", "Fbrk = F-Fres #N\n", "#Max. braking force which one locomotive can produce.\n", "#F = 1000*ua*M*g\n", "Fbrk_1 = 1000*ua*Ml*9.8 #N\n", "#Therefore, Number of locomotives required for braking\n", "num1 = Fbrk/Fbrk_1\n", "num1 = round(num1)+ 1\n", "#Considering the starting case.Tractive force required is\n", "Ft = 277.8*a*M*1.1 + M*r2\n", "#Therefore, Number of locomotives required for starting\n", "num2 = Ft/Fbrk_1\n", "#Number of locomotives required\n", "if num1>num2:\n", " num = num1\n", "else:\n", " num = num2\n", "print \"Total number of locomotives required are = \",round(num)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.15 , PAGE NO :- 1729" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Scheduled speed = 32.9 Km/h.\n", "Specific energy consumption for up-gradient is = 69.08 Wh/t-km.\n", "Specific energy consumption for down-gradient is = 39.58 Wh/t-km\n" ] } ], "source": [ "#A 200-tonne electric train runs according to the following quadrilateral speed/time curve:\n", "# 1. uniform acceleration from rest at 2 km/h/s for 30 seconds.\n", "# 2. coasting for 50 seconds\n", "# 3. duration of braking:15 seconds\n", "#If up-gradient is =1%,train resistance = 40N/t,rotational inertia effect = 10%,duration of stops = 15s and overall\n", "#efficiency of gear and motor = 75%,find\n", "#(i)schedule speed (ii)specific energy consumption (iii)how will the value of specific energy consumption change if\n", "#there is a down-gradient of 1% rather than the up-gradient ?\n", "########################################################################################################################\n", "\n", "#Given\n", "M = 200.0 #tonne (Mass of train)\n", "Me = 1.1*200.0 #tonne (rotational inertia effect on mass)\n", "alpha = 2.0 #km/h/s (acceleration)\n", "t1 = 30.0 #s (acceleration time)\n", "t2 = 50.0 #s (coasting time)\n", "t3 = 15.0 #s (braking time)\n", "G = 1.0 # (% gradient)\n", "r = 40.0 #N/t (train resistance)\n", "tstop = 15.0 #s (stop time)\n", "meff = 0.75 # (motor efficiency)\n", "\n", "V1 = alpha*t1 #km/h (Max speed)\n", "#During coasting retardation force is\n", "Fr = 98*M*G + M*r #N\n", "#Also this retardation force in terms of coasting retardation beta_c is\n", "#Fr = 277.8*Me*beta_c.Therefore,\n", "beta_c = Fr/(277.8*Me) #km/h/s\n", "#Using newton's Ist eqn in coasting period\n", "V2 = V1-beta_c*t2 #km/h\n", "\n", "#Braking retardation is\n", "beta = V2/t3\n", "#Distance travelled during acceleration\n", "dist1 = 0.5*alpha*t1*t1/3600 #m\n", "#Distance travelled during coasting\n", "dist2 = (V1+V2)/2*t2/3600 #m\n", "#Distance travelled during braking\n", "dist3 = 0.5*beta*t3*t3/3600 #m\n", "#Total distance travelled\n", "D = dist1 + dist2 + dist3 #km\n", "#Total schedule time\n", "T = t1 + t2 + t3 + tstop #s\n", "#(i)Schedule speed\n", "Vsch = D/T*3600 #km/h\n", "print \"Scheduled speed = \",round(Vsch,1),\"Km/h.\"\n", "#(ii)Specific energy consumption is given by\n", "spengy = (0.01072*(V1*V1/D)*(Me/M) + 27.25*G*(dist1/D)+ 0.2778*r*(dist1/D))/meff #Wh/t-km\n", "print \"Specific energy consumption for up-gradient is = \",round(spengy,2),\"Wh/t-km.\"\n", "#If there is a down gradient then during coasting accelerative force is\n", "Fa = 98*M*(-G) + M*r\n", "#Also this accelerative force in terms of coasting acceleration alpha_c is\n", "#Fa = 277.8*Me*alpha_c.Therefore,\n", "alpha_c = Fa/(277.8*Me) #km/h/s\n", "#Using newton's Ist eqn in coasting period\n", "V2 = V1 - alpha_c*t2 #km/h\n", "#Braking retardation is\n", "beta = V2/t3\n", "#Distance travelled during acceleration\n", "dist1 = 0.5*alpha*t1*t1/3600 #m\n", "#Distance travelled during coasting\n", "dist2 = (V1+V2)/2*t2/3600 #m\n", "#Distance travelled during braking\n", "dist3 = 0.5*beta*t3*t3/3600 #m\n", "#Total distance travelled\n", "D = dist1 + dist2 + dist3 #km\n", "#(iii)Specific energy consumption is given by\n", "spengy = (0.01072*(V1*V1/D)*(Me/M) - 27.25*G*(dist1/D)+ 0.2778*r*(dist1/D))/meff #Wh/t-km\n", "print \"Specific energy consumption for down-gradient is = \",round(spengy,2),\"Wh/t-km\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.16 , PAGE NO :- 1730" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "#An electric train has an average speed of 45 kmph on a level track between stops 1500m apart.It is accelerated at\n", "#1.8km/h/s and is braked at 3 km/h/s.Draw the speed-time curve for the run.\n", "####################################################################################################################\n", "\n", "import math as m\n", "import matplotlib.pyplot as plt\n", "%matplotlib inline \n", "\n", "#Given\n", "alpha = 1.8 #km/h/s (acceleration)\n", "beta = 3.0 #km/h/s (retardation)\n", "S = 1.5 #km (Distance of run)\n", "Va = 45.0 #km/h (Average speed)\n", "\n", "T = S/Va*3600 #s (Time of run)\n", "\n", "#Constant K is given by\n", "K = 0.5*(1/alpha + 1/beta)\n", "#Max speed is given by\n", "Vm = T/(2*K)-m.sqrt(T*T/(4*K*K) - 3600*S/K) #km/h\n", "#Acceleration time\n", "t1 = Vm/alpha #s\n", "#Braking time\n", "t3 = Vm/beta #s\n", "#Free running time\n", "t2 = T - (t1+t3) #s\n", "\n", "#SPEED TIME CURVE\n", "\n", "plt.plot([0,t1,t1+t2,t1+t2+t3], [0,Vm,Vm,0])\n", "plt.ylabel('Speed(in km/h)')\n", "plt.xlabel('Time(in seconds)')\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.17 , PAGE NO :- 1731" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Maximum Speed is 72.85 Km/h.\n" ] } ], "source": [ "#A train has schedule speed of 60 km per hour between the stops which are 9 km apart.Determine the crest speed over\n", "#the run,assuming trapezoidal speed-time curve.The train accelerates at 3km/h/s and retards at 4.5 km/h/s.Duration\n", "#of stops is 75 seconds.\n", "#########################################################################################################################\n", "import math as m\n", "\n", "#Given\n", "alpha = 3.0 #km/h/s (acceleration)\n", "beta = 4.5 #Km/h/s (retardation)\n", "S = 9.0 #km (distance)\n", "Vsch = 60.0 #km/h (schedule speed)\n", "\n", "Tsch = S/Vsch*3600 #s (schedule time)\n", "T = Tsch - 75.0 #s (time of run)\n", "#Constant K is\n", "K = 0.5*(1/alpha + 1/beta)\n", "\n", "#Maximum speed Vm is\n", "Vm = (T/(2*K)) - m.sqrt(T*T/(4*K*K) - 3600*S/K)\n", "\n", "print \"Maximum Speed is\",round(Vm,2),\"Km/h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.18 , PAGE NO :- 1732" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Schedule speed is 60.0 km/h.\n" ] } ], "source": [ "#An electric train is to have acceleration and braking retardation of 1.2 km/h/s and 4.8 km/h/s respectively.If the ratio of\n", "#maximum to average speed is 1.6 and time for stop is 35 seconds,find schedule speed for a run of 3km.Assume simplified\n", "#trapezoidal speed-time curve.\n", "#############################################################################################################################\n", "import math as m\n", "alpha = 1.2 #km/h/s (acceleration)\n", "beta = 4.8 #km/h/s (retardation)\n", "S = 3.0 #km (distance)\n", "\n", "#Constant K is\n", "K = 0.5*(1/alpha + 1/beta)\n", "\n", "#The average speed is Va = S/T . Therefore Va*T = S*3600\n", "Va_T = S*3600 \n", "#Now Vm/Va = 1.6 .Therefore,\n", "Vm_T = Va_T*1.6\n", "\n", "#Since, Vm^2*K - Vm*T + 3600*S = 0, Therefore\n", "#Vm^2 = Vm*T - 3600*S\n", "Vm = m.sqrt((Vm_T - 3600*S)/K) #km/h\n", "#Average Speed is\n", "Va = Vm/1.5 #km/h\n", "#Actual Time of Run\n", "T = 3600*S/Va #s\n", "#Schedule Time\n", "Tsch = T + 35.0 #s\n", "#Schedule Speed\n", "Vsch = S/Tsch*3600 #km/h\n", "print \"Schedule speed is\",round(Vsch),\"km/h.\" " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.19 , PAGE NO :- 1732" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The rate of acceleration required to operate this service = 1.85 km/h/s.\n" ] } ], "source": [ "#An electric train has a schedule speed of 25 kmph between stations 800m apart.The duration of stop is 20 seconds,the\n", "#maximum speed is 20 percent higher than the average running speed and the braking retardation is 3 km/h/s.Calculate\n", "#the rate of acceleration required to operate this service.\n", "###################################################################################################################\n", "\n", "#Given\n", "Vsch = 25.0 #km/h (Schedule speed)\n", "S = 0.8 #km (Distance)\n", "beta = 3.0 #km/h/s (Retardation)\n", "\n", "Tsch = S/Vsch*3600 #s\n", "\n", "#Actual time of run = Tsch - (time of stop)\n", "T = Tsch - 20.0 #s\n", "Va = S/T*3600 #km/h (Average speed)\n", "Vm = 1.2*Va #km/h (Maximum speed)\n", "\n", "#Since, Vm^2*K -Vm*T + 3600*S = 0\n", "#Constant K = (Vm*T - 3600*S)/(Vm^2)\n", "K = (Vm*T - 3600*S)/(Vm*Vm)\n", "\n", "#Also,K = 0.5*(1/alpha + 1/beta).Therefore,\n", "alpha = 1/(2*K - 1/beta) #km/h/s\n", "\n", "print \"The rate of acceleration required to operate this service =\",round(alpha,2),\"km/h/s.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.20 , PAGE NO :- 1733" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Retardation = 0.51 km/h/s.\n" ] } ], "source": [ "#A suburban electric train has a maximum speed of 80 kmph.The schedule speed including a station stop of 35 seconds is 50kmph.\n", "#If the acceleration is 1.5 km/h/s,find the value of retardation when the average distance between stops is 5 km.\n", "################################################################################################################################\n", "\n", "#Given\n", "Vsch = 50.0 #km/h (schedule speed)\n", "S = 5.0 #km (distance)\n", "alpha = 1.5 #km/h/s (acceleration) \n", "Vm = 80.0 #km/h (Max speed)\n", "tstop = 30.0 #s (Time of stop)\n", "\n", "\n", "Tsch = S/Vsch*3600 #s (Schedule time)\n", "T = Tsch - tstop #s (Actual time of run)\n", "\n", "#Since, Vm^2*K -Vm*T + 3600*S = 0\n", "#Constant K = (Vm*T - 3600*S)/(Vm^2)\n", "K = (Vm*T - 3600*S)/(Vm*Vm)\n", "\n", "#Also,K = 0.5*(1/alpha + 1/beta).Therefore,\n", "beta = 1/(2*K - 1/alpha) #km/h/s\n", "\n", "print \"Retardation =\",round(beta,2),\"km/h/s.\"" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## EXAMPLE 43.21 , PAGE NO :- 1733" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Duration of acceleration = 32.0 s.\n", "Duration of coasting retardation = 96.84 s.\n", "Duration of braking retardation = 15.16 s.\n" ] } ], "source": [ "#A train is required to run between two stations 1.6 km apart at the average speed of 40 kmph.The run is to be made to a\n", "#simplified quadrilateral speed-time curve.If the maximum speed is to be limited to 64 kmph,accelerating to 2 km/h/s and\n", "#coasting and braking retardation to 0.16 km/h/s and 3.2 km/h/s respectively,determine the duration of acceleration,coasting\n", "#and braking periods.\n", "#################################################################################################################################\n", "from sympy import Eq,Symbol,solve\n", "#Given\n", "S = 1.6 #km (Distance)\n", "Va = 40.0 #km/h (Average speed)\n", "Vm = 64.0 #km/h (Maximum speed)\n", "alpha = 2.0 #km/h/s (Acceleration)\n", "beta_c = 0.16 #km/h/s (Coasting retardation)\n", "beta = 3.2 #km/h/s (braking retardation)\n", "\n", "T = S/Va*3600 #s (Actual time of run)\n", "t1 = Vm/alpha #s (acceleration time)\n", "#Let us assume the speed at starting of braking be V2\n", "V2 = Symbol('V2')\n", "t2 = (Vm-V2)/beta_c #s (coasting period)\n", "t3 = V2/beta #s (braking period)\n", "eq = Eq(t1+t2+t3,T)\n", "V2 = solve(eq)\n", "Ve2 = V2[0]\n", "#Therfore coasting and braking periods are\n", "t2 = (Vm-Ve2)/beta_c #s (coasting period)\n", "t3 = Ve2/beta #s (braking period)\n", "\n", "print \"Duration of acceleration = \",round(t1,2),\"s.\"\n", "print \"Duration of coasting retardation = \",round(t2,2),\"s.\"\n", "print \"Duration of braking retardation = \",round(t3,2),\"s.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.22 , PAGE NO :- 1737" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Speed-Armature current graph\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "#The torque-armature current characteristics of a series traction motor are given as:\n", "#Armature Current(amp): 5 10 15 20 25 30 35 40\n", "#Torque(N-m): 20 50 100 155 215 290 360 430\n", "#The motor resistance is 0.3 ohm.If this motor is connected across 230V,deduce the speed armature current characteristics.\n", "############################################################################################################################\n", "\n", "import matplotlib.pyplot as plt\n", "%matplotlib inline\n", "#Given\n", "V = 230 #V (Supply voltage)\n", "r = 0.3 #ohm (Motor resistance)\n", "\n", "#Current Data Ia\n", "i1 = 5.0 #A\n", "i2 = 10.0 #A\n", "i3 = 15.0 #A\n", "i4 = 20.0 #A\n", "i5 = 25.0 #A\n", "i6 = 30.0 #A\n", "i7 = 35.0 #A\n", "i8 = 40.0 #A\n", "\n", "#Torque Data \n", "tau1 = 20.0 #N-m\n", "tau2 = 50.0 #N-m\n", "tau3 = 100.0 #N-m\n", "tau4 = 155.0 #N-m\n", "tau5 = 215.0 #N-m\n", "tau6 = 290.0 #N-m\n", "tau7 = 360.0 #N-m\n", "tau8 = 430.0 #N-m\n", "\n", "#Back EMF Eb = V - Ia*r\n", "e1 = V - i1*r #V\n", "e2 = V - i2*r #V\n", "e3 = V - i3*r #V\n", "e4 = V - i4*r #V\n", "e5 = V - i5*r #V\n", "e6 = V - i6*r #V\n", "e7 = V - i7*r #V\n", "e8 = V - i8*r #V\n", "\n", "\n", "#Speed Na = 9.55*Eb*Ia/T\n", "N1 = 9.55*e1*i1/tau1 #rpm\n", "N2 = 9.55*e2*i2/tau2 #rpm\n", "N3 = 9.55*e3*i3/tau3 #rpm\n", "N4 = 9.55*e4*i4/tau4 #rpm\n", "N5 = 9.55*e5*i5/tau5 #rpm\n", "N6 = 9.55*e6*i6/tau6 #rpm\n", "N7 = 9.55*e7*i7/tau7 #rpm\n", "N8 = 9.55*e8*i8/tau8 #rpm\n", "\n", "print \"Speed-Armature current graph\"\n", "\n", "plt.plot([i1,i2,i3,i4,i5,i6,i7,i8], [N1,N2,N3,N4,N5,N6,N7,N8])\n", "plt.axis([0,40,0,600])\n", "plt.ylabel('Speed(in r.p.m)')\n", "plt.xlabel('Armature current Ia(in A)')\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.23 , PAGE NO :- 1737" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Speed - Torque Curve\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "#The following figures give the magnetization curve of d.c. series motor when working as a seperately excited generator at 600 rpm.\n", "#Field Current(amperes) : 20 40 60 80\n", "#E.M.F(volts): 215 381 485 550\n", "#The total resistance of the motor is 0.8 ohm.Deduce the speed-torque curve for this motor when operating at a constant voltage of\n", "#600 volts.\n", "################################################################################################################################\n", "\n", "#Given\n", "V = 600.0 #V (Supply voltage)\n", "r = 0.8 #ohm (total motor resistance)\n", "\n", "#Field Current Data(If=Ia)\n", "i1 = 20.0 #A\n", "i2 = 40.0 #A\n", "i3 = 60.0 #A\n", "i4 = 80.0 #A\n", "\n", "#E.M.F data at 600 rpm(E)\n", "e1 = 215.0 #V\n", "e2 = 381.0 #V\n", "e3 = 485.0 #V\n", "e4 = 550.0 #V\n", "\n", "#At constant voltage of 600V. EMF are Eb = V - Ia*Rm\n", "eb1 = V - i1*r #V\n", "eb2 = V - i2*r #V\n", "eb3 = V - i3*r #V\n", "eb4 = V - i4*r #V\n", "\n", "#For DC series motor N1/E1 = N2/E2 . Therefore N2 = N/E*Eb\n", "n1 = 600/e1*eb1 #rpm\n", "n2 = 600/e2*eb2 #rpm\n", "n3 = 600/e3*eb3 #rpm\n", "n4 = 600/e4*eb4 #rpm\n", "\n", "#Now, Torque T = 9.55*Eb*Ia/N2\n", "tau1 = 9.55*eb1*i1/n1 #N-m\n", "tau2 = 9.55*eb2*i2/n2 #N-m\n", "tau3 = 9.55*eb3*i3/n3 #N-m\n", "tau4 = 9.55*eb4*i4/n4 #N-m\n", "\n", "print \"Speed - Torque Curve\"\n", "\n", "plt.plot([tau1,tau2,tau3,tau4], [n1,n2,n3,n4])\n", "#plt.axis([0,40,0,600])\n", "plt.ylabel('Speed(in r.p.m)')\n", "plt.xlabel('Torque T(in N-m)')\n", "plt.show()" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## EXAMPLE 43.24 , PAGE NO :- 1738" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Speed of both motors is = 447.87 rpm.\n", "Voltage across motor A is = 256.35 V.\n", "Voltage across motor B is = 243.65 V.\n" ] } ], "source": [ "#Two d.c. traction motors run at a speed of 900 r.p.m and 950 r.p.m respectively when each takes a current of 50A from 500V mains.\n", "#Each motor has an effective resistance of 0.3 ohm . Calculate the speed and voltage across each machine when mechanically coupled\n", "#and electrically connected in series and taking a current of 50A from 500V mains,the resistance of each motor being unchanged.\n", "#################################################################################################################################\n", "from sympy import Eq,Symbol,solve\n", "#Let the two motors be A and B .\n", "Na = 900.0 #rpm (speed of motor A)\n", "Nb = 950.0 #rpm (speed of motor B)\n", "r = 0.3 #ohm (resistance of each motor)\n", "V = 500.0 #V (applied voltage)\n", "I = 50.0 #A (current)\n", "\n", "#Back emf of motor A when taking current of 50A,\n", "Eb_a = V - I*r #V\n", "#Back emf of motor B when taking current of 50A,\n", "Eb_b = V - I*r #V\n", "\n", "#As the machines are mechanically coupled and are connected in series. Therefore,\n", "#speed will be same. Say 'N'\n", "#current will be same and equal to 50A\n", "#summation of voltage drop is 500V\n", "\n", "N = Symbol('N') #rpm\n", "#Let the voltage drop across motors be Va & Vb .Therefore,for Va & Vb we will find back emf of both motors\n", "eb_A = Eb_a*N/Na #V (using N1/N2 = E1/E2)\n", "eb_B= Eb_b*N/Nb #V (using N1/N2 = E1/E2)\n", "\n", "#For Va & Vb \n", "Va = eb_A + I*r #V (using Eb = V - Ia*r)\n", "Vb = eb_B + I*r #V (using Eb = V - Ia*r)\n", "\n", "#We know that Va + Vb = 500 . Therefore,\n", "eq = Eq(Va+Vb,V)\n", "N = solve(eq)\n", "Ne = N[0]\n", "\n", "#As we know the speed N . Therefore, Va & Vb are\n", "eb_A = Eb_a*Ne/Na #V (using N1/N2 = E1/E2)\n", "eb_B= Eb_b*Ne/Nb #V (using N1/N2 = E1/E2)\n", "\n", "Va = eb_A + I*r #V (using Eb = V - Ia*r)\n", "Vb = eb_B + I*r #V (using Eb = V - Ia*r)\n", "\n", "print \"Speed of both motors is = \",round(Ne,2),\"rpm.\"\n", "print \"Voltage across motor A is = \",round(Va,2),\"V.\"\n", "print \"Voltage across motor B is = \",round(Vb,2),\"V.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.25 , PAGE NO :- 1739" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Current drawn by supply mains is = 138.32 A.\n" ] } ], "source": [ "#A tram car is equipped with two motors which are operating in parallel.Calculate the current drawn from the supply main at 500 V\n", "#when the car is running at steady speed of 50kmph and each motor is developing a tractive effort of 2100 N.The resistance of each \n", "#motor is 0.4 ohm.The friction,windage and other losses may be assumed as 3500 watts per motor.\n", "#################################################################################################################################\n", "from sympy import Eq,Symbol,solve\n", "\n", "V = 500.0 #V (applied voltage)\n", "Vm = 50.0 #km/h (max. speed)\n", "Ft = 2100 #N (tractive effort)\n", "Rm = 0.4 #ohm (motor resistance)\n", "mloss = 3500.0 #W (losses per motor)\n", "\n", "#Power output of each motor\n", "Pout = Ft*Vm*5.0/18 #W\n", "#Copper loss = I^2*Rm\n", "I = Symbol('I')\n", "cu_loss = (I*I)*Rm #W\n", "#Input power = Output power + constant losses + Copper losses\n", "Pin = V*I #W\n", "eq = Eq(Pin,Pout+mloss+cu_loss)\n", "I = solve(eq) #A\n", "#current drawn by motor is\n", "Ie = I[0] #A\n", "#Therefore,current drawn by supply mains\n", "Ie = 2*Ie #A\n", "print \"Current drawn by supply mains is =\",round(Ie,2),\"A.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.26 , PAGE NO :- 1740" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Speed of first motor is = 217.17 rpm.\n", "Speed of second motor is = 227.27 rpm.\n" ] } ], "source": [ "#A motor coach is being driven by two identical d.c. series motors.First motor is geared to driving wheel having diameter of 90cm and\n", "#other motor to driving wheel having diameter of 86cm.The speed of the first motor is 500 rpm when connected in parallel with the other\n", "#across 600V supply.Find the motor speeds when connected in series across the same supply.Assume armature current to remain same and\n", "#armature voltage drop of 10% at this current.\n", "###############################################################################################################################\n", "from sympy import Eq,Symbol,solve\n", "#Given\n", "V = 600.0 #V (applied voltage)\n", "d1= 0.90 #m (diameter 1) \n", "d2 = 0.86 #m (diameter 2) \n", "N1 = 500.0 #rpm (Speed of first motor)\n", "\n", "Eb1 = V-0.1*V #V (Back emf)\n", "\n", "#Let the supply voltage across motor 1 & 2 be V1 and V2 respectively\n", "#Therefore, V1 + V2 = 600\n", "V1 = Symbol('V1') #V\n", "V2 = V - V1 #V\n", "\n", "#Now , N is directly propotional to (V - IR).Therefore, ratio (N1/N2) is\n", "\n", "Nratio1 = (V1 - 0.1*V)/(V2 - 0.1*V)\n", "\n", "#Also, we know that N1*d1 = N2*d2 . Therefore (N1/N2) is\n", "Nratio2 = d2/d1\n", "\n", "#As, N1/N2 = Nratio1 = Nratio2\n", "eq = Eq(Nratio1,Nratio2)\n", "V1 = solve(eq)\n", "Ve1 = V1[0] #V\n", "Ve2 = V - Ve1 #V\n", "#Now , for the speed of motor 1, we know that for N1 ,back emf is Eb1 and for Ne1 ,back emf is (Ve1-0.1*Ve1)\n", "Ne1 = N1*(Ve1 - 0.1*V)/Eb1 #rpm (using N1/N2 = E1/E2)\n", "#Nratio is Ne1/Ne2 .Therefore,\n", "Ne2 = Ne1/Nratio2 #rpm\n", "\n", "print \"Speed of first motor is =\",round(Ne1,2),\"rpm.\"\n", "print \"Speed of second motor is =\",round(Ne2,2),\"rpm.\"" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## EXAMPLE 43.27 , PAGE NO :- 1741" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Speed of first motor is = 173.74 rpm.\n", "Speed of second motor is = 181.82 rpm.\n" ] } ], "source": [ "#Two similar series type motors are used to drive a locomotive.The supply fed to their parallel connection is 650V.If the first \n", "#motor 'A' is geared to drive wheels of radius 45 cms and other motor 'B' to 43 cms.And if the speed of first motor 'A' when\n", "#connected in parallel to 2nd motor 'B' across the main supply lines is 400 rpm,find voltages and speeds of motors when connected\n", "#in series.Assume Ia to be constant and armature voltage drop of 10% at this current.\n", "################################################################################################################################\n", "\n", "from sympy import Eq,Symbol,solve\n", "#Given\n", "V = 650.0 #V (applied voltage)\n", "r1 = 0.45 #m (radius 1) \n", "r2 = 0.43 #m (radius 2) \n", "Na = 400.0 #rpm (Speed of first motor)\n", "\n", "Eb_a = V-0.1*V #V (Back emf)\n", "\n", "#Let the supply voltage across motor A & B be Va and Vb respectively\n", "#Therefore, Va + Vb = 650.0 V\n", "Va = Symbol('Va') #V\n", "Vb = V - Va #V\n", "\n", "#Now , N is directly propotional to (V - IR).Therefore, ratio (N1/N2) is\n", "\n", "Nratio1 = (Va - 0.1*V)/(Vb - 0.1*V)\n", "\n", "#Also, we know that N1*r1 = N2*r2 . Therefore (N1/N2) is\n", "Nratio2 = r2/r1\n", "\n", "#As, N1/N2 = Nratio1 = Nratio2\n", "eq = Eq(Nratio1,Nratio2)\n", "Va = solve(eq)\n", "Vea = Va[0] #V\n", "Veb = V - Vea #V\n", "\n", "#Now , for the speed of motor A, we know that for Na ,back emf is Eb_a and for Nea ,back emf is (Vea-0.1*Vea)\n", "Nea = Na*(Vea - 0.1*V)/Eb_a #rpm (using N1/N2 = E1/E2)\n", "#Nratio is Ne1/Ne2 .Therefore,\n", "Neb = Nea/Nratio2 #rpm\n", "\n", "print \"Speed of first motor is =\",round(Nea,2),\"rpm.\"\n", "print \"Speed of second motor is =\",round(Neb,2),\"rpm.\"" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "## EXAMPLE 43.28 , PAGE NO :- 1744" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Time for which motors are operated in series = 8.44 s.\n", "Energy loss during starting period = 480.01 W-h.\n" ] } ], "source": [ "#Two motors of a motor coach are started on series-parallel system,the current per motor being 350A during the starting period\n", "#which is 18 sec.If the acceleration during starting period is uniform,the line voltage is 600V and resistance of each motor is\n", "#0.1W.Find (a)the time during which the motors are operated in series. (b)the energy loss in the rheostat during starting period.\n", "#####################################################################################################################################\n", "\n", "#Given\n", "V = 600.0 #V (line voltage)\n", "I = 350.0 #A (current)\n", "R = 0.1 #ohm (resistance)\n", "T = 18.0 #s (starting period)\n", "\n", "#time for which motors are in series\n", "ts = 0.5*(V-2*I*R)/(V-I*R)*T #s\n", "\n", "#time for which motors are in parallel\n", "tp = T - ts #s\n", "\n", "#back emf of 1 motor during series operation\n", "Eb_s = V/2 - I*R #V\n", "#total back emf\n", "Eb_s = 2*Eb_s #V\n", "#Emf during parallel operation\n", "Eb_p = V - I*R #V\n", "\n", "#Energy lost during series connection\n", "enrgy_s = 0.5*(Eb_s)*I*ts/3600 #W-h\n", "#Energy lost during parallel connection\n", "enrgy_p = 0.5*(Eb_p/2)*(2*I)*tp/3600 #W-h\n", "\n", "#Total energy lost\n", "enrgy_t = enrgy_s + enrgy_p #W-h\n", "\n", "print \"Time for which motors are operated in series = \",round(ts,2),\"s.\"\n", "print \"Energy loss during starting period = \",round(enrgy_t,2),\"W-h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.29 , PAGE NO :- 1749" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Energy lost in rheostat = 441.02 W-h.\n", "Energy lost in motor = 75.0 W-h.\n", "Motor output = 909.77 W-h.\n", "Efficiency at starting = 69.07 %.\n", "Train speed at which transition from series to parallel = 11.98 km/h.\n" ] } ], "source": [ "#Two 750V D.C. series motors each having a resistance of 0.1 ohm are started on series-parallel system.Mean current through-out \n", "#the starting period is 300A.Starting period is 15 sec. and train speed at the end of this period is 25 km/hr.Calculate\n", "#(i)Rheostatic losses during series and parallel combination of motors\n", "#(ii)Energy lost in motor\n", "#(iii)Motor output\n", "#(iv)Starting eff\n", "#(v)Train speed at which transition from series to parallel must be made.\n", "##############################################################################################################################\n", "\n", "V = 750.0 #V (applied voltage)\n", "R = 0.1 #ohm (resistance)\n", "I = 300.0 #A (current) \n", "T = 15.0 #s (starting time)\n", "S = 25.0 #km/h(speed after staring period)\n", "\n", "#Time when motors are in series\n", "ts = 0.5*(V-2*I*R)/(V-I*R)*T #s\n", "#Time when motors are in parallel\n", "tp = T - ts #s\n", "\n", "#(i)Energy lost in rheostat\n", "enrgy_lostr = 0.5*(2*(V/2-I*R))*I*ts + 0.5*(V-I*R)/2*2*I*tp #W-s\n", "enrgy_lostr = enrgy_lostr/3600 #W-h\n", "\n", "#(ii)Total energy supplied\n", "enrgy_supp = V*I*ts + 2*V*I*tp #W-s\n", "enrgy_supp = enrgy_supp/3600 #W-h\n", "\n", "#Energy lost in armature of 2 motors\n", "enrgy_losta = 2*(I*I*R)*T\n", "enrgy_losta = enrgy_losta/3600 #W-h\n", "\n", "#Motor output is\n", "outp = enrgy_supp - enrgy_lostr - enrgy_losta #W-h\n", "\n", "#Efficiency at starting\n", "eff_s = (enrgy_supp-enrgy_lostr)/(enrgy_supp)*100\n", "\n", "#Acceleraton is uniform during starting period .Therefore,\n", "speed = S/T*ts #km/h\n", "print \"Energy lost in rheostat = \",round(enrgy_lostr,2),\"W-h.\"\n", "print \"Energy lost in motor = \",round(enrgy_losta,2),\"W-h.\"\n", "print \"Motor output =\",round(outp,2),\"W-h.\"\n", "print \"Efficiency at starting =\",round(eff_s,2),\"%.\"\n", "print \"Train speed at which transition from series to parallel = \",round(speed,2),\"km/h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.30 , PAGE NO :- 1750" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Energy loss in series = 213.16 W-h.\n", "Energy loss in parallel = 263.16 W-h.\n", "The transition speed = 14.21 km/h.\n" ] } ], "source": [ "#Two 600V motors each having a resistance of 0.1 ohm are started on the series-parallel system,the mean current per motor throughout\n", "#the starting period being 300A.The starting period is 20 seconds and the train speed at the end of this period is 30km per hour.\n", "Calculate\n", "#(i)the rheostatic losses during (a)the series and (b) the parallel combination of motors\n", "#(ii)the train speed at which transition from series to parallel must be made.\n", "###############################################################################################################################\n", "\n", "num_m = 2 # (number of motors)\n", "V = 600.0 #V (line voltage)\n", "I = 300.0 #A (current per motor)\n", "Ts = 20.0 #s (starting period)\n", "R = 0.1 #ohm (motor resistance)\n", "Vm = 30.0 #km/h (maximum speed)\n", "\n", "#Back emf of each motor in series position,\n", "Eb_s = V/2 - I*R #V\n", "#Back emf of each motor in parallel position,\n", "Eb_p = V - I*R #V\n", "#time for which motors were in series\n", "ts = Ts*Eb_s/Eb_p #s\n", "#time for which motors are in parallel\n", "tp = Ts - ts #s\n", "\n", "#(a) Voltage drop in the starting rheostat in series combination\n", "V_s = V - 2*I*R #V\n", "#Energy loss in series is\n", "enrgy_s = (V_s/2)*I*(ts/3600) #W-h\n", "#(b) Voltage drop in the starting rheostat in parallel combination\n", "V_p = V/2 #V\n", "#Energy loss in parallel is\n", "enrgy_p = (V_p/2)*(2*I)*(tp/3600)\n", "\n", "#Acceleration is\n", "acc = Vm/Ts #km/h/s\n", "#Speed at the end of series period\n", "speed = acc*ts #km/h\n", "\n", "print \"Energy loss in series =\",round(enrgy_s,2),\"W-h.\"\n", "print \"Energy loss in parallel =\",round(enrgy_p,2),\"W-h.\"\n", "print \"The transition speed = \",round(speed,2),\"km/h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.31 , PAGE NO :- 1751" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Period of series operation = 11.36 s.\n", "Period of parallel operation = 13.64 s.\n", "Speed at which series connection are to be changed = 36.36 km/h.\n" ] } ], "source": [ "#Two d.c. series motors of a motor coach have resistance of 0.1ohm each.These motors draw a current of 500A from 600V mains during\n", "#series-parallel starting period of 25 seconds.If the acceleration during starting period remains uniform,determine:\n", "#(i)time during which the motors operate in (a)series (b)parallel.\n", "#(ii)the speed at which the series connections are to be changed if the speed just after starting period is 80kmph.\n", "####################################################################################################################################\n", "\n", "\n", "num_m = 2 # (number of motors)\n", "V = 600.0 #V (line voltage)\n", "I = 500.0 #A (current per motor)\n", "Ts = 25.0 #s (starting period)\n", "R = 0.1 #ohm (motor resistance)\n", "Vm = 80.0 #km/h (maximum speed)\n", "\n", "#Back emf of each motor in series position,\n", "Eb_s = V/2 - I*R #V\n", "#Back emf of each motor in parallel position,\n", "Eb_p = V - I*R #V\n", "#time for which motors were in series\n", "ts = Ts*Eb_s/Eb_p #s\n", "#time for which motors are in parallel\n", "tp = Ts - ts #s\n", "\n", "#(ii) Speed at which series connection are to be changed = acc*ts and acc = Vm/Ts.Therefore,\n", "speed = Vm/Ts*ts #km/h\n", "\n", "print \"Period of series operation = \",round(ts,2),\"s.\"\n", "print \"Period of parallel operation = \",round(tp,2),\"s.\"\n", "print \"Speed at which series connection are to be changed = \",round(speed,2),\"km/h.\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## EXAMPLE 43.32 , PAGE NO :- 1751" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Braking torque = 930.0 N-m.\n" ] } ], "source": [ "#The following figures refer to the speed-current and torque-current characteristics of a 600V d.c. series traction motor.\n", "#Current(A) : 50 100 150 200 250\n", "#Speed(Kmph): 73.6 48 41.1 37.3 35.2\n", "#Torque(N-m): 150 525 930 1335 1750\n", "#Determine the braking torque at a speed of 48kmph when operating as self excited d.c. generator.Assume resistance of motor and\n", "#braking rheostat to be 0.6 ohm and 3.0 ohm respectively.\n", "#############################################################################################################################\n", "\n", "Rm = 0.6 #ohm(resistance of motor)\n", "Rrh = 3.0 #ohm(braking rheostat)\n", "\n", "#As a motor\n", "V = 600.0 #V (terminal voltage)\n", "#From speed-current characteristics . For a speed of 48kmph,\n", "I = 100.0 #A (current)\n", "Eb = V - I*Rm #V (back emf)\n", "\n", "#As a generator\n", "\n", "#At instant of applying rheostatic braking ,the terminal voltage will be equal to emf developed by machine i.e\n", "V2 = Eb #V\n", "Rt = Rm + Rrh #ohm (total resistance)\n", "\n", "#Using V=IR\n", "I2 = V2/Rt #A (current)\n", "\n", "if (I2==50):\n", " tau = 150.0 #N-m\n", "elif (I2==100):\n", " tau = 525.0 #N-m\n", "elif (I2==150):\n", " tau = 930.0 #N-m\n", "elif (I2==200):\n", " tau = 1335.0 #N-m\n", "elif (I2==250):\n", " tau = 1750.0 #N-m\n", "\n", "print \"Braking torque = \",round(tau,2),\"N-m.\" " ] } ], "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": "2.7.11" } }, "nbformat": 4, "nbformat_minor": 0 }