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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# CHAPTER 45 : RATING AND SERVICE CAPACITY"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.1 , PAGE NO :- 1796"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Size of motor = 66.0 KW.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''An electric motor operates at full-load of 100KW for 10 minutes,at 3/4 full load for the next 10 minutes and at 1/2 load for\n",
+ "next 20 minutes,no-load for the next 20 minutes and this cycle repeats continously.Find the continous rating of the suitable \n",
+ "motor.'''\n",
+ "\n",
+ "import math as m\n",
+ "#Loads\n",
+ "l1 = 100.0 #kW (load 1)\n",
+ "l2 = 0.75*l1 #kW (load 2)\n",
+ "l3 = 0.5*l1 #kW (load 3)\n",
+ "l4 = 0.0 #kW (no-load)\n",
+ "\n",
+ "#coresponding time\n",
+ "t1 = 10.0 #minutes\n",
+ "t2 = 10.0 #minutes\n",
+ "t3 = 20.0 #minutes\n",
+ "t4 = 20.0 #minutes\n",
+ "\n",
+ "#size of motor required\n",
+ "size = m.sqrt((l1*l1*t1 + l2*l2*t2 + l3*l3*t3 + l4*l4*t4)/(t1+t2+t3+t4/3)) #kW\n",
+ "print \"Size of motor =\",round(size),\"KW.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.2 , PAGE NO :- 1797"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Size of motor = 141.0 KW.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''An electric motor has to be selected for a load which rises uniformly from zero to 200KW in 10 minutes after which it remains \n",
+ "constant at 200KW for the next 10 minutes,followed by a no-load period of 15 minutes before the cycle repeats itself.Estimate a \n",
+ "suitable size of continuosly rated motor.'''\n",
+ "\n",
+ "import math as m\n",
+ "#Loads\n",
+ "l1 = 200.0/2 #kW (load 1)\n",
+ "l2 = 200.0 #kW (load 2)\n",
+ "l3 = 0.0 #kW (no-load)\n",
+ "\n",
+ "#coresponding time\n",
+ "t1 = 10.0 #minutes\n",
+ "t2 = 10.0 #minutes\n",
+ "t3 = 15.0 #minutes\n",
+ "\n",
+ "\n",
+ "#size of motor required\n",
+ "size = m.sqrt((l1*l1*t1 + l2*l2*t2 + l3*l3*t3)/(t1+t2+t3/3)) #kW\n",
+ "print \"Size of motor =\",round(size),\"KW.\"\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.3 , PAGE NO :- 1797 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Size of motor = 70.0 KW.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A certain motor has to perform the following duty cycle:\n",
+ "(a) 100KW for 10 minutes (No-load for 5 minutes)\n",
+ "(b) 50KW for 8 minutes (No-load for 4 minutes)\n",
+ "The duty cycle is repeated indefinitely.Draw the curve for the load cycle.Assuming that the heating is propotional to the square of\n",
+ "the load,determine suitable size of a continuosly-rated motor.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "#Loads\n",
+ "l1 = 100.0 #KW (load 1)\n",
+ "l3 = 50.0 #KW (load 2)\n",
+ "\n",
+ "#Time\n",
+ "t1 = 10.0 #minutes\n",
+ "t2 = 5.0 #minutes\n",
+ "t3 = 8.0 #minutes\n",
+ "t4 = 4.0 #minutes\n",
+ "\n",
+ "#Size of the motor is\n",
+ "size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4)) #KW\n",
+ "print \"Size of motor = \",round(size,-1),\"KW.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.4 , PAGE NO :- 1799"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Size of motor = 69.07 H.P.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A motor has to perform the following duty cycle :-\n",
+ "(i) 100 H.P (10 mins) (ii) No Load (5 mins)\n",
+ "(iii)60 H.P (8 mins) (iv) No Load (4 mins)\n",
+ "which is repeated infinitely.Determine the suitable size of continuosly rated motor.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "#Loads\n",
+ "l1 = 100.0 #H.P (load 1)\n",
+ "l3 = 60.0 #H.P (load 2)\n",
+ "\n",
+ "#Time\n",
+ "t1 = 10.0 #minutes\n",
+ "t2 = 5.0 #minutes\n",
+ "t3 = 8.0 #minutes\n",
+ "t4 = 4.0 #minutes\n",
+ "\n",
+ "#Size of the motor is\n",
+ "size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4)) #H.P\n",
+ "print \"Size of motor = \",round(size,2),\"H.P.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.5 , PAGE nO :- 1800"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Size of motor = 48.11 H.P .\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A motor working in a coal mine has to exert power starting from zero and rising uniformly to 100 H.P in 5 min\n",
+ "after which it works at a constant rate of 50 H.P for 10 min.Then, a no load period of 3 min.The cycle is repeated\n",
+ "indefinitely,estimate suitable size of motor.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "#Load\n",
+ "l1 = 100.0 #H.P (load 1)\n",
+ "l2 = 50.0 #H.P (load 2)\n",
+ "l3 = 0.0 #H.P (no-load)\n",
+ "\n",
+ "#Time\n",
+ "t1 = 5.0 #min (time 1) \n",
+ "t2 = 10.0 #min (time 2)\n",
+ "t3 = 3.0 #min (time 3)\n",
+ "\n",
+ "#Using Simpson's one-third rule of Integration\n",
+ "rating = m.sqrt((1.0/3*l1*l1*t1 + l2*l2*t2)/(t1 + t2 + t3) ) #H.P\n",
+ "\n",
+ "print \"Size of motor =\",round(rating,2),\"H.P .\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.6 , PAGE NO :- 1800"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Size of motor = 263.19 H.P .\n",
+ "Therefore, suitable size of motor is 300.0 H.P\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A motor has following duty cycle\n",
+ "Load rising from 200 to 400 H.P - 4 min.\n",
+ "Uniform load 300 H.P - 2 min.\n",
+ "Regenerative braking (50 H.P to 0) - 1 min.\n",
+ "Idle - 1 min.\n",
+ "Estimate suitable H.P rating of the motor that can be used.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "#Loads\n",
+ "l1 = 200.0 #H.P (load 1)\n",
+ "l2 = 400.0 #H.P (load 2)\n",
+ "l3 = 300.0 #H.P (load 3)\n",
+ "l4 = 50.0 #H.P (load 4)\n",
+ "\n",
+ "#Time\n",
+ "t1 = 4.0 #min (time 1)\n",
+ "t2 = 2.0 #min (time 2)\n",
+ "t3 = 1.0 #min (time 3)\n",
+ "t4 = 1.0 #min (idle time)\n",
+ "\n",
+ "rating = m.sqrt((1.0/3*(l1*l1 + l1*l2 + l2*l2)*t1 + l3*l3*t2 + 1.0/3*l4*l4*t3)/(t1 + t2 + t3 + t4)) #H.P\n",
+ "\n",
+ "print \"Size of motor = \",round(rating,2),\"H.P .\"\n",
+ "print \"Therefore, suitable size of motor is\",round(rating,-2),\"H.P\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.7 , PAGE NO :- 1802"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Suitable motor size = 30.0 H.P .\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''The load cycle of a motor for 15 min in driving some equipment is as follows :\n",
+ "0 - 5 min - 30 H.P\n",
+ "5 - 9 min - No load\n",
+ "9 - 12 min - 45 H.P\n",
+ "12 - 15 min - No load\n",
+ "The load cycle is repeated indefinitely.Suggest a suitable size of continuosly rated motor.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "#Loads\n",
+ "l1 = 30.0 #H.P\n",
+ "l3 = 45.0 #H.P\n",
+ "\n",
+ "#Time\n",
+ "t1 = 5.0 #min\n",
+ "t2 = 4.0 #min\n",
+ "t3 = 3.0 #min\n",
+ "t4 = 3.0 #min\n",
+ "\n",
+ "#Size of motor is\n",
+ "Size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4)) #H.P\n",
+ "print \"Suitable motor size =\",round(Size,-1),\"H.P .\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.8 , PAGE NO :- 1802"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Size of continously rated motor = 505.0 H.P .\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A motor driving a colliery winder has the following acceleration period :\n",
+ " load cycle 0-15 sec : Load rises uniformly from 0-1000 H.P .\n",
+ " Full speed period : 15-85 sec. Load constant at 600 H.P .\n",
+ " Decceleration period : 85-95 sec. Regenerative braking the H.P returned uniformly from 200 to 0 H.P.\n",
+ " 95 - 120 sec : Motor stationary.\n",
+ "Estimate the size of continuosly rated motor.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "#Loads\n",
+ "l1 = 1000.0 #H.P (load 1)\n",
+ "l2 = 600.0 #H.P (load 2)\n",
+ "l3 = 200.0 #H.P (load 3)\n",
+ "\n",
+ "#Time\n",
+ "t1 = 15.0 #s\n",
+ "t2 = 70.0 #s\n",
+ "t3 = 10.0 #s\n",
+ "t4 = 25.0 #s\n",
+ "\n",
+ "#Size of motor is\n",
+ "\n",
+ "size = m.sqrt((l1*l1*t1/3 + l2*l2*t2 + l3*l3*t3/3)/(t1+t2+t3+t4)) #H.P\n",
+ "\n",
+ "while(round(size)%5!=0):\n",
+ " size = size + 1\n",
+ " \n",
+ "print \"Size of continously rated motor = \",round(size),\"H.P .\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.9 , PAGE NO :- 1807"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " 1/2 hour rating = 75.13 KW.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A 40KW motor when run continuosly on full load,attains a temperature of 35C , above the surrounding air.Its heating time \n",
+ "constant is 90 min.What would be the 1/2 hour rating of the motor for this temperature rise?Assume that the machine cools down \n",
+ "completely between each load period and that the losses are propotional to square of the load.'''\n",
+ "\n",
+ "\n",
+ "from sympy import Symbol,Eq,solve\n",
+ "import math as m\n",
+ "# Let 'P' KW be 1/2 hour rating of the motor\n",
+ "# theta1 - final temp rise at P KW\n",
+ "# theta2 - final temp rise at 40 KW\n",
+ "#Losses at P KW is directlt propotional to P^2\n",
+ "\n",
+ "theta2 = 35.0 # *C\n",
+ "tau = 1.5 #hr (time constant)\n",
+ "t = 0.5 #hr (motor running time)\n",
+ "\n",
+ "#Now, (theta1/theta2) = loss at P KW/loss at 40KW = (P/40)^2\n",
+ "P = Symbol('P')\n",
+ "theta1 = theta2*(P/40)*(P/40) #*C\n",
+ "\n",
+ "#Now, theta2 = theta1*(1 - exp(-t/tau))\n",
+ "\n",
+ "theta2a = theta1*(1-m.exp(-t/tau)) #*C\n",
+ "eq = Eq(theta2,theta2a)\n",
+ "P = solve(eq)\n",
+ "P1 = P[1] #KW\n",
+ "\n",
+ "print \"1/2 hour rating = \",round(P1,2),\"KW.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.10 , PAGE NO :- 1807"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "1 hour rating = 24.0 H.P.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''Determine the one-hour rating of a 15 H.P motor having heating time contant of 2 hours.The motor attains the temperature of\n",
+ "40*C on continuos run at full load.Assume that the losses are propotional to square of the load and the motor is allowed to cool\n",
+ "down to the ambient temperature before being loaded again.'''\n",
+ "\n",
+ "from sympy import Symbol,Eq,solve\n",
+ "import math as m\n",
+ "# Let 'P' H.P be 1 hour rating of the motor\n",
+ "# theta2 - final temp rise at P H.P\n",
+ "# theta1 - final temp rise at 15 H.P\n",
+ "#Losses at P H.P is directlt propotional to P^2\n",
+ "\n",
+ "theta1 = 40.0 # *C\n",
+ "tau = 2.0 #hr (time constant)\n",
+ "t = 1.0 #hr (motor running time)\n",
+ "\n",
+ "#Now, (theta2/theta1) = loss at P H.P/loss at 15 H.P = (P/15)^2\n",
+ "P = Symbol('P')\n",
+ "theta2 = theta1*(P/15)*(P/15) #*C\n",
+ "\n",
+ "#Now, theta1 = theta2*(1 - exp(-t/tau))\n",
+ "\n",
+ "theta1a = theta2*(1-m.exp(-t/tau)) #*C\n",
+ "eq = Eq(theta1,theta1a)\n",
+ "P = solve(eq)\n",
+ "P1 = P[1] #H.P\n",
+ "\n",
+ "print \"1 hour rating = \",round(P1),\"H.P.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.11 , PAGE NO :- 1808"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Temperature rise of motor = 35.6 *C .\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''The heating and cooling time constants of a motor are 1 hour and 2 hours respectively.Final temperature rise attained is \n",
+ "100*C.This motor runs at full load for 30 minutes and then kept idle for 12 min and the cycle is repeated indefinitely.Determine \n",
+ "the temperature rise of motor after one cycle.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "theta2 = 100.0 #*C (Final temperature rise)\n",
+ "tau_h = 1.0 #hr (heating time constant)\n",
+ "tau_c = 2.0 #hr (cooling time constant)\n",
+ "t1 = 30.0/60 #hr (motor running time)\n",
+ "t2 = 12.0/60 #hr (motor idle time)\n",
+ "\n",
+ "#Heating cycle\n",
+ "theta1 = theta2*(1 - m.exp(-t1/tau_h))\n",
+ "\n",
+ "#Cooling cycle\n",
+ "thetac = theta1*m.exp(-t2/tau_c)\n",
+ "\n",
+ "print \"Temperature rise of motor = \",round(thetac,2),\"*C .\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.12 , PAGE NO :- 1808"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 68,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maximum overload that can be carried by motor = 25.82 KW.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''Calculate the maximum overload that can be carried by a 20KW output motor,if the temperature rise is not to exceed 50*C after\n",
+ "one hour on overload .The temperature rise on full load,after 1 hour is 30*C and after 2 hour is 40*C . Assume losses propotional\n",
+ "to square of load.'''\n",
+ "\n",
+ "from sympy import solve,Symbol,Eq\n",
+ "import math as m\n",
+ "\n",
+ "# As theta = thetaf*(1 - exp(-t/T))\n",
+ "\n",
+ "theta1 = 30.0 #*C (temperature rise in time1)\n",
+ "t1 = 1 #hr (time 1)\n",
+ "theta2 = 40.0 #*C (temperature rise in time2)\n",
+ "t2 = 2 #hr (time 2)\n",
+ "#Now theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))\n",
+ "#Let us assume that x = exp(-1/T).Therefore\n",
+ "x = Symbol('x')\n",
+ "ratio1 = (1 - x**t1)/(1-x**t2) #(theta1/theta2)\n",
+ "ratio2 = theta1/theta2\n",
+ "\n",
+ "#As ratio1 = ratio2\n",
+ "eq = Eq(ratio1,ratio2)\n",
+ "x1 = solve(eq)\n",
+ "x = x1[0] #variable \n",
+ "\n",
+ "#x = exp(-1/T) . Therefore,\n",
+ "T = -1/m.log(x)\n",
+ "\n",
+ "#Now, theta1 = thetaf1*(1 - exp(-t1/T))\n",
+ "\n",
+ "thetaf1 = theta1/(1-x**t1) #*C\n",
+ "\n",
+ "#Also theta3 = thetaf3*(1 - exp(-t3/T))\n",
+ "theta3 = 50.0 #*C (max temp)\n",
+ "t3 = 1 #hr (time 3) \n",
+ "thetaf3 = theta3/(1-x**t3) #*C\n",
+ "\n",
+ "#Given that temp is directly propotional to square of power output i.e thetaf1/thetaf3 = (Power1/Power3)^2\n",
+ "P1 = 20.0 #KW\n",
+ "P3 = m.sqrt(thetaf3/thetaf1)*P1 #KW\n",
+ "\n",
+ "print \"Maximum overload that can be carried by motor = \",round(P3,2),\"KW.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.13 , PAGE NO :- 1809"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 67,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Final steady temperature rise = 50.0 *C.\n",
+ "Cooling time constant = 0.93 hr\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''In a transformer the temperature rise is 25*C after 1 hour and 37.5*C after 2 hours,starting from cold conditions.Calculate \n",
+ "its final steady temperature rise and the heating time constant.If the transformer temerature falls from the final steady state\n",
+ "value to 40*C in 1.5 hours when disconnected,calculate its cooling time constant.Ambient temperature is 30*C.'''\n",
+ "\n",
+ "from sympy import solve,Symbol,Eq\n",
+ "import math as m\n",
+ "\n",
+ "# As theta = thetaf*(1 - exp(-t/T))\n",
+ "\n",
+ "theta1 = 25.0 #*C (temperature rise in time1)\n",
+ "t1 = 1 #hr (time 1)\n",
+ "theta2 = 37.5 #*C (temperature rise in time2)\n",
+ "t2 = 2 #hr (time 2)\n",
+ "#Now theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))\n",
+ "#Let us assume that x = exp(-1/T).Therefore\n",
+ "x = Symbol('x')\n",
+ "ratio1 = (1 - x**t1)/(1-x**t2) #(theta1/theta2)\n",
+ "ratio2 = theta1/theta2\n",
+ "\n",
+ "#As ratio1 = ratio2\n",
+ "eq = Eq(ratio1,ratio2)\n",
+ "x1 = solve(eq)\n",
+ "x = x1[0] #variable \n",
+ "\n",
+ "#x = exp(-1/T) . Therefore,\n",
+ "T = -1/m.log(x)\n",
+ "\n",
+ "#As theta1 = thetaf1*(1 - exp(-t1/T))\n",
+ "thetaf1 = theta1/(1-x**t1) #*C\n",
+ "print \"Final steady temperature rise =\",round(thetaf1,2),\"*C.\"\n",
+ "\n",
+ "#Cooling conditions\n",
+ "theta_rise = 40.0 - 30.0 #*C (temp rise above ambient conditions)\n",
+ "t3 = 1.5 #hr (time taken)\n",
+ "\n",
+ "#Now, theta_rise = thetaf1*exp(-t3/T)\n",
+ "T = -t3/m.log(theta_rise/thetaf1) #hr\n",
+ "print \"Cooling time constant =\",round(T,2),\"hr\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.14 , PAGE NO :- 1809"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 66,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Temperature of machine = 70.58 *C.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''The initial temperature of machine is 20*C.Calculate the temperature of machine after 1.2 hours,if its final steady \n",
+ "temperature rise is 85*C and the heating time constant is 2.4 hours.Ambient temperature is 25*C.''' \n",
+ "\n",
+ "import math as m\n",
+ "thetaf = 85.0 #*C (final temp. rise)\n",
+ "theta1 = 20.0 #*C (initial temp)\n",
+ "t1 = 1.2 #hr (time taken)\n",
+ "T = 2.4 #hr (heat time constant)\n",
+ "#Now, Temperature rise above coling medium is\n",
+ "theta = thetaf - (thetaf - theta1)*m.exp(-t1/T) #*C\n",
+ "\n",
+ "#Therefore, temp. of machine after t1 time is\n",
+ "temp = theta + 25.0\n",
+ "\n",
+ "print \"Temperature of machine =\",round(temp,2),\"*C.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.15 , PAGE NO :- 1809"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 69,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Final steady temperature rise = 45.0 *C .\n",
+ "Time constant = 2.47 hr.\n",
+ "The steady temperature rise = 30.0 *C .\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''The following rises were observed in a teperature rise test on a D.C machine at full loads. :-\n",
+ "After 1 hour - 15*C\n",
+ "After 2 hours - 25*C\n",
+ "Find out (i) Final steady temperature rise and time constant.\n",
+ " (ii)The steady temperature rise after 1 hour at 50% overload,from cold.\n",
+ "Assume that the final temperature rise on 50% overload is 90*C.'''\n",
+ "\n",
+ "\n",
+ "from sympy import solve,Symbol,Eq\n",
+ "import math as m\n",
+ "\n",
+ "# As theta = thetaf*(1 - exp(-t/T))\n",
+ "\n",
+ "theta1 = 15.0 #*C (temperature rise in time1)\n",
+ "t1 = 1 #hr (time 1)\n",
+ "theta2 = 25.0 #*C (temperature rise in time2)\n",
+ "t2 = 2 #hr (time 2)\n",
+ "#Now theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))\n",
+ "#Let us assume that x = exp(-1/T).Therefore\n",
+ "x = Symbol('x')\n",
+ "ratio1 = (1 - x**t1)/(1-x**t2) #(theta1/theta2)\n",
+ "ratio2 = theta1/theta2\n",
+ "\n",
+ "#As ratio1 = ratio2\n",
+ "eq = Eq(ratio1,ratio2)\n",
+ "x1 = solve(eq)\n",
+ "x = x1[0] #variable \n",
+ "\n",
+ "#x = exp(-1/T) . Therefore,\n",
+ "T = -1/m.log(x) #hr (time constant)\n",
+ "\n",
+ "#As theta1 = thetaf1*(1 - exp(-t/T))\n",
+ "thetaf1 = theta1/(1-x**t1) #*C (Final steady temp. rise)\n",
+ "print \"Final steady temperature rise =\",round(thetaf1,2),\"*C .\"\n",
+ "print \"Time constant =\",round(T,2),\"hr.\"\n",
+ "\n",
+ "#(ii) Now at 50% overload .Final temp rise is\n",
+ "thetaf3 = 90.0 #*C\n",
+ "t3 = 1 #hr (time taken)\n",
+ "#As , theta = thetaf*(1 - exp(-t/T))\n",
+ "theta3 = thetaf3*(1 - m.exp(-t3/T)) #*C\n",
+ "\n",
+ "print \"The steady temperature rise =\",round(theta3,2),\"*C .\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.16 , PAGE NO :- 1813"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Moment of Inertia = 2.947543e+06 kg-m^2\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''The following data refers to a 500 H.P rolling mill,induction motor equipped with a flywheel.\n",
+ " No load speed -> 40 rpm\n",
+ " Slip at full load(torque) -> 12%\n",
+ " Load torque during actual rolling -> 41500 kg-m\n",
+ " Duration of each rolling period -> 10 sec.\n",
+ "Determine inertia of flywheel required in the above case to limit motor torque to twice its full load value.Neglect no-load \n",
+ "losses and assume that the rolling mill torque falls to zero between each rolling period.Assume motor slip propotional to\n",
+ "full load torque.'''\n",
+ "\n",
+ "\n",
+ "import math as m\n",
+ "N = 40.0 #rpm (No load speed)\n",
+ "P = 500.0*(735.5) #W (Power)\n",
+ "w = 2*(3.14)*N/60 #rad/sec (angular speed)\n",
+ "T0 = 0 #kg-m (initial torque)\n",
+ "Tl = 41500.0 #kg-m (Torque load) \n",
+ "t = 10.0 #sec (time taken)\n",
+ "s = 0.12 # (slip)\n",
+ "g = 9.81 #m/s^2 \n",
+ "Tfull = P/(w*(1-s)) #N-m (full load torque)\n",
+ "Tfull = Tfull/g #kg-m\n",
+ "Tm = 2*Tfull #kg-m (Max torque)\n",
+ "S = 2*3.14*(0.12*40)/60\n",
+ "#Now, S = K*Tfl\n",
+ "K = S/Tfull #constant\n",
+ "#Also, Tm = Tl - (Tl-T0)*exp(-tg/IK) .Therefore I is\n",
+ "I =(-t*g)/(K*m.log((Tl-Tm)/(Tl-T0))) #kg-m^2\n",
+ "\n",
+ "print \"Moment of Inertia = %e kg-m^2\" %round(I,2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.17 , PAGE NO :- 1814 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Tm = 47.84 kg-m.\n",
+ "Actual Speed = 942.59 rpm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A 6 pole,50 Hz Induction Motor has a flywheel of 1200 kg-m^2 as moment of inertia.Load torque is 100 kg-m for 10 sec.No load\n",
+ "period is long enough for the flywheel,to regain its full speed.Motor has a slip of 6% at a torque of 50 kg-m.Calculate\n",
+ "(i)Maximum torque exerted by motor\n",
+ "(ii)Speed at the end of deacceleration period.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "Tl = 100.0 #kg-m (load torque)\n",
+ "t = 10.0 #s (time taken)\n",
+ "g = 9.81 #m/s^2 (gravitational acceleration)\n",
+ "I = 1200.0 #kg-m^2 (moment of inertia)\n",
+ "p = 6 # (poles)\n",
+ "f = 50.0 #Hz (frequency)\n",
+ "s = 0.06 # (slip)\n",
+ "Tfull = 50.0 #kg-m (full load torque)\n",
+ "Ns = 120*f/p\n",
+ "Nr = (1-s)*Ns\n",
+ "\n",
+ "#Now, S = K*T\n",
+ "S = 2*3.14*(Ns - Nr)/60 #rad/sec\n",
+ "K = S/Tfull #constant\n",
+ "\n",
+ "#As Tm = Tl*(1-exp(-t*g/I*K))\n",
+ "Tm = Tl*(1 - m.exp(-t*g/(I*K))) #kg-m\n",
+ "print \"Tm = \",round(Tm,2),\"kg-m.\"\n",
+ "\n",
+ "#(ii)Slip speed\n",
+ "S = K*Tm #rad/sec (slip speed)\n",
+ "S = S*(60/(2*3.14)) #rpm\n",
+ "N = Ns - S #rpm (actual speed)\n",
+ "print \"Actual Speed =\",round(N,2),\"rpm.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.18 , PAGE NO :- 1815"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Speed 1 = 691.75 rpm.\n",
+ "Speed 2 = 731.62 rpm.\n",
+ "Speed 1 = 683.89 rpm.\n",
+ "Speed 2 = 731.0 rpm.\n",
+ "Speed 1 = 683.62 rpm.\n",
+ "Speed 2 = 730.98 rpm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''An Induction Motor equipped with a flywheel is driving a rolling mill which requires a load torque of 1900 N-m for 10 sec\n",
+ "followed by 250 N-m for 30 sec.This cycle being repeated indefinitely.The synchronus speed of motor is 750 rpm and it has slip of\n",
+ "10% when delivering 1400 N-m torque.The total Moment of Inertia of the flywheel and other rotating parts is 2100 kg-m^2.Draw the \n",
+ "curves showing the torque exerted by the motor and the speed for five complete cycles,assuming the initial torque is zero.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "Tl1 = 1900.0 #N-m (load torque 1)\n",
+ "t1 = 10.0 #s (time 1)\n",
+ "Tl2 = 280.0 #N-m (load torque 2)\n",
+ "t2 = 30.0 #s (time 2)\n",
+ "s = 0.1 # (slip)\n",
+ "Ns = 750.0 #rpm (synchronus speed)\n",
+ "I = 2100.0 #kg-m^2 (moment of inertia)\n",
+ "Tm = 1400.0 #N-m\n",
+ "S = Ns*s #rpm (slip speed)\n",
+ "S = S*(2*3.14/60) #rad/sec \n",
+ "\n",
+ "K = S/Tm #constant\n",
+ "T0 = 0 #N-m\n",
+ "#(i) During First Cycle\n",
+ "Tm = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K)) #N-m\n",
+ "s2 = K*Tm*(60/(2*3.14)) #rpm\n",
+ "Speed1 = Ns - s2 #rpm\n",
+ "print \"Speed 1 = \",round(Speed1,2),\"rpm.\"\n",
+ "\n",
+ "# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))\n",
+ "Tmb = Tm #N-m\n",
+ "T0 = Tl2 #N-m (No Load Torque)\n",
+ "\n",
+ "Tm = T0 + (Tmb - T0)*m.exp(-t2/(I*K))\n",
+ "S2 = K*Tm*(60/(2*3.14)) #rpm\n",
+ "Speed2 = Ns - S2 #rpm\n",
+ "print \"Speed 2 = \",round(Speed2,2),\"rpm.\"\n",
+ "#################################################################\n",
+ "\n",
+ "#(ii) During Second cycle\n",
+ "T0 = Tm\n",
+ "Tm2 = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K)) #N-m\n",
+ "s2 = K*Tm2*(60/(2*3.14)) #rpm\n",
+ "Speed1 = Ns - s2 #rpm\n",
+ "print \"Speed 1 = \",round(Speed1,2),\"rpm.\"\n",
+ "\n",
+ "# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))\n",
+ "Tm2b = Tm2 #N-m\n",
+ "T0 = Tl2 #N-m (No Load Torque)\n",
+ "\n",
+ "Tm = T0 + (Tm2b - T0)*m.exp(-t2/(I*K))\n",
+ "S2 = K*Tm*(60/(2*3.14)) #rpm\n",
+ "Speed2 = Ns - S2 #rpm\n",
+ "print \"Speed 2 = \",round(Speed2,2),\"rpm.\"\n",
+ "###################################################################\n",
+ "\n",
+ "#(iii) During Third cycle\n",
+ "T0 = Tm\n",
+ "Tm3 = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K)) #N-m\n",
+ "s2 = K*Tm3*(60/(2*3.14)) #rpm\n",
+ "Speed1 = Ns - s2 #rpm\n",
+ "print \"Speed 1 = \",round(Speed1,2),\"rpm.\"\n",
+ "\n",
+ "# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))\n",
+ "Tm3b = Tm3 #N-m\n",
+ "T0 = Tl2 #N-m (No Load Torque)\n",
+ "\n",
+ "Tm = T0 + (Tm3b - T0)*m.exp(-t2/(I*K))\n",
+ "S2 = K*Tm*(60/(2*3.14)) #rpm\n",
+ "Speed2 = Ns - S2 #rpm\n",
+ "print \"Speed 2 = \",round(Speed2,2),\"rpm.\"\n",
+ "####################################################################"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## EXAMPLE 45.19 , PAGE NO :- 1817"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Moment of inertia = 2860.94 kg-m^2.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A motor fitted with a flywheel supplies a load torque of 150 kg-m for 15 sec.During the no-load period,the flywheel regains \n",
+ "its original speed.The motor torque is required to be limited to 85 kg-m.Determine moment of inertia of flywheel.\n",
+ "The no-load speed of motor is 500 rpm and it has slip of 10% on full load.'''\n",
+ "\n",
+ "from sympy import Symbol,solve,Eq,exp\n",
+ "import math as m\n",
+ "\n",
+ "Tm = 85.0 #kg-m (Max torque)\n",
+ "Tl = 150.0 #kg-m (load torque with flywheel)\n",
+ "T0 = 0 #kg-m (constant load torque)\n",
+ "t = 15.0 #s (time)\n",
+ "N = 500.0 #rpm (no load speed)\n",
+ "s = 0.1 # (slip)\n",
+ "g = 9.82 #m/s^2 \n",
+ "# s = K*T\n",
+ "K = 2*(3.14)*N*s/(60*Tm) #constant\n",
+ "\n",
+ "# As Tm = Tl*(1 - exp(-t*g/(I*K)))\n",
+ "\n",
+ "I =(-t*g)/(K*m.log(1 - Tm/Tl)) #kg-m^2 (Moment of inertia)\n",
+ "\n",
+ "print \"Moment of inertia =\",round(I,2),\"kg-m^2.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.20 , PAGE NO :- 1817"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 35,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Moment of inertia of flywheel = 1908.86 kg-m^2.\n",
+ "Time taken after removal of additional load = 8.88 s.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A 3-phase ,50 KW,6 pole,960 rpm induction motor has a constant load torque of 300 N-m and at wide intervals additional \n",
+ "torque of 1500 N-m for 10 sec.Calculate\n",
+ "(a)The moment of inertia of the flywheel used for load equalization,if the motor torque is not to exceed twice the rated torque.\n",
+ "(b)Time taken after removal of additional load,before the motor torque becomes 700 N-m.'''\n",
+ "\n",
+ "from sympy import Symbol,Eq,solve\n",
+ "import math as m\n",
+ "\n",
+ "P = 50.0e+3 #W (output power)\n",
+ "Nr = 960.0 #rpm (rotational speed)\n",
+ "p = 6.0 # (no. of poles)\n",
+ "t = 10.0 #s (time)\n",
+ "T0 = 300.0 #N-m (constant load torque)\n",
+ "Tl = T0 + 1500.0 #N-m (total load torque)\n",
+ "f = 50.0 #Hz (frequency) \n",
+ "\n",
+ "# Power = T*w (torque*ang_speed)\n",
+ "T = P/(2*3.14*Nr/60) #N-m (Full-load torque)\n",
+ "Tm = 2*T #N-m (Max torque)\n",
+ "\n",
+ "Ns = 120*f/p #rpm (synchronus speed)\n",
+ "\n",
+ "#Slip speed\n",
+ "sl = Ns-Nr #rpm\n",
+ "\n",
+ "#Now, s = K*T\n",
+ "K = 2*3.14*sl/(60*T) #constant\n",
+ "\n",
+ "#As Tm = Tl - (Tl - T0)*exp(-t/I*K)\n",
+ "\n",
+ "I = (-t)/(K*m.log((Tl - Tm)/(Tl - T0))) #kg-m^2 (moment of inertia)\n",
+ "\n",
+ "print \"Moment of inertia of flywheel =\",round(I,2),\"kg-m^2.\"\n",
+ "\n",
+ "#(b) Tm2 = T0 + (Tm-T0)*exp(-t/I*K)\n",
+ "Tm2 = 700.0 #N-m (Max torque - case 2)\n",
+ "\n",
+ "t1= (-I*K)*m.log((Tm2 - T0)/(Tm - T0)) #s (time after removal of load)\n",
+ "print \"Time taken after removal of additional load =\",round(t1,2),\"s.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "## EXAMPLE 45.21 , PAGE NO :- 1818"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 39,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Moment of Inertia = 551.35 kg-m^2.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A 3-phase,8 pole,50 cps.Induction Motor equipped with a flywheel supplies a constant load torque of 100 N-m and at wide \n",
+ "intervals an additional load torque of 300 N-m for 6 sec.The motor runs at 735 rpm at 100 N-m torque.Find moment of inertia of \n",
+ "the flywheel,if the motor torque is not to exceed 250 N-m.'''\n",
+ "\n",
+ "from sympy import Symbol,Eq,solve\n",
+ "import math as m\n",
+ "\n",
+ "T0 = 100.0 #N-m (constant load torque)\n",
+ "Tl = T0 + 300.0 #N-m (Total load torque)\n",
+ "f = 50.0 #Hz (frequency)\n",
+ "P = 8.0 # (poles)\n",
+ "Tm = 250.0 #N-m (Max torque) \n",
+ "Ns = 120*f/P #rpm (Synchronus speed)\n",
+ "sl = Ns - 735.0 #rpm (Slip speed)\n",
+ "t = 6.0 #s (time)\n",
+ "#Now, s = K*T0\n",
+ "K = 2*3.14*sl/(60*T0) #constant\n",
+ "\n",
+ "#Also, Tm = Tl - (Tl-T0)*exp(-t/I*K)\n",
+ "\n",
+ "I = -t/(K*m.log((Tl - Tm)/(Tl-T0))) #kg-m^2 (moment of inertia)\n",
+ "\n",
+ "print \"Moment of Inertia =\",round(I,2),\"kg-m^2.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.22 , PAGE NO :- 1818"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 42,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maximum torque developed by motor = 615.35 N-m.\n",
+ "Speed at the end of deacceleration period = 938.47 rpm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A 6 pole,50 Hz,3-phase wound rotor Induction Motor has a flywheel coupled to its shaft.The total moment of inertia is \n",
+ "1000kg-m^2.Load torque is 1000 N-m for 10 sec followed by a no-load period which is long enough for the motor to reach its \n",
+ "no-load speed.Motor has a slip of 5% at a torque of 500 N-m.Find\n",
+ "(a)Maximum torque developed by motor\n",
+ "(b)Speed at the end of deacceleration period.'''\n",
+ "\n",
+ "import math as m\n",
+ "\n",
+ "P = 6.0 # (No of poles)\n",
+ "I = 1000.0 #kg-m^2 (Moment of Inertia)\n",
+ "Tl = 1000.0 #N-m (Load torque with flywheel)\n",
+ "t = 10.0 #s (time)\n",
+ "s = 0.05 # (slip)\n",
+ "Tfl = 500.0 #N-m (full load Torque)\n",
+ "f = 50.0 #Hz (frequency)\n",
+ "\n",
+ "Ns = 120*f/P #rpm (Synchronus speed)\n",
+ "\n",
+ "#Now, s = K*Tfl\n",
+ "K = 2*3.14*(Ns*s)/(60*Tfl) #constant\n",
+ "\n",
+ "#K = 6.2e-3 #(considered value) \n",
+ "\n",
+ "#Also Tm = Tl*(1-exp(-t/I*K)\n",
+ "Tm = Tl*(1 - m.exp(-t/(I*K))) #N-m\n",
+ "print \"Maximum torque developed by motor = \",round(Tm,2),\"N-m.\"\n",
+ " \n",
+ "#(b) s = K*Tfl where s = 2*3.14*(Ns - N)/60\n",
+ "N = Ns - (60/(2*3.14))*K*Tm #rpm\n",
+ "print \"Speed at the end of deacceleration period =\",round(N,2),\"rpm.\" "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## EXAMPLE 45.23 , PAGE NO :- 1819"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 53,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Moment of inertia of flywheel = 275.67 kg-m^2.\n"
+ ]
+ }
+ ],
+ "source": [
+ "'''A motor fitted with a flywheel supplies a load torque of 1000 N-m for 2 seconds.During no load period,the flywheel regains its\n",
+ "original speed.The motor torque is to be limited to 500 N-m.Find moment of inertia of the flywheel.No load speed of the motor is \n",
+ "500 rpm and its full load slip is 10%.'''\n",
+ "\n",
+ "from sympy import solve,Eq,Symbol\n",
+ "import math as m\n",
+ "N = 500.0 #rpm (No load speed)\n",
+ "s = 0.1 # (slip)\n",
+ "Tfl = 500.0 #N-m (full load torque)\n",
+ "Tl = 1000.0 #N-m (load torque with flywheel)\n",
+ "t = 2.0 #s (time) \n",
+ "#Now, s = K*Tfl\n",
+ "K = (2*3.14*(N*s))/(Tfl*60) #constant\n",
+ "\n",
+ "\n",
+ "#Also, Tm = Tl*(1 - exp(-t/I*K))\n",
+ "I =-t/(K*m.log(1 - Tfl/Tl)) #(moment of inertia)\n",
+ "\n",
+ "print \"Moment of inertia of flywheel =\",round(I,2),\"kg-m^2.\""
+ ]
+ },
+ {
+ "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",
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+ "pygments_lexer": "ipython2",
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+ }
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+ "nbformat_minor": 0
+}
generated by cgit (git 2.34.1) at 2024-12-20 18:41:21 +0000