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| author | hardythe1 | 2015-04-07 15:58:05 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-04-07 15:58:05 +0530 |
| commit | 92cca121f959c6616e3da431c1e2d23c4fa5e886 (patch) | |
| tree | 205e68d0ce598ac5caca7de839a2934d746cce86 /Fundamentals_Of_Thermodynamics/Chapter5_6.ipynb | |
| parent | b14c13fcc6bb6d01c468805d612acb353ec168ac (diff) | |
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diff --git a/Fundamentals_Of_Thermodynamics/Chapter5_6.ipynb b/Fundamentals_Of_Thermodynamics/Chapter5_6.ipynb new file mode 100755 index 00000000..5a9734bc --- /dev/null +++ b/Fundamentals_Of_Thermodynamics/Chapter5_6.ipynb @@ -0,0 +1,635 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:bc9fbdf4ecee4bce9770355a576adf487ef4df6d34bad14ada42e8383ea01d91"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5:THE FIRST LAW OF THERMODYNAMICS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.1:pg-131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1\n",
+ "#calculating height\n",
+ "\n",
+ "m=1100 #mass of car in kg\n",
+ "ke=400 #kinetic energy of car in kJ\n",
+ "V=(2*ke*1000/m)**0.5 #velocity of car in m/s\n",
+ "g=9.807 #acc. due to gravity in m/s^2\n",
+ "H=ke*1000/(m*g) #height to which the car should be lifted so that its potential energy equals its kinetic energy\n",
+ "print\"hence,the car should be raised to a height is\",round(H,1),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hence,the car should be raised to a height is 37.1 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.1E:pg-132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1E\n",
+ "#calculating height\n",
+ "\n",
+ "m=2400 #mass of car in lbm\n",
+ "ke=400 #kinetic energy of car in Btu\n",
+ "V=(2*ke*778.17*32.174/m)**0.5 #velocity of car in ft/s\n",
+ "g=32.174 #acc. due to gravity in ft/s^2\n",
+ "H=ke*778.17*32.174/(m*g) #height to which the car should be lifted so that its potential energy equals its kinetic energy\n",
+ "print\"hence,the car should be raised to a height is\",round(H,1),\"ft\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hence,the car should be raised to a height is 129.7 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.2:pg-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2\n",
+ "#change in internal energy\n",
+ "\n",
+ "W=-5090 #work input to paddle wheel in kJ\n",
+ "Q=-1500 #heat transfer from tank in kJ\n",
+ "dU=Q-W #change in internal energy in kJ\n",
+ "print\"hence,change in internal energy is\",round(dU),\"kj\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hence,change in internal energy is 3590.0 kj\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.3:pg-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3\n",
+ "#analysis of energy transfer\n",
+ "\n",
+ "g=9.806 #acceleration due to gravity in m/s^2\n",
+ "m=10 #mass of stone in kg\n",
+ "H1=10.2 #initial height of stone above water in metres\n",
+ "H2=0 #final height in metres\n",
+ "dKE1=-m*g*(H2-H1) #change in kinetic energy when stone enters state 2 in J\n",
+ "dPE1=-1 #change in potential energy when stone enters state 2 in J\n",
+ "print\"\\n hence,when stone is \",round(dKE1),\"J\"\n",
+ "print\"\\n and change in potential energy is \",round(dPE1),\"J\"\n",
+ "dPE2=0 #change in potential energy when stone enters state 3 in JQ2=0 //no heat transfer when stone enters state 3 in J\n",
+ "W2=0 #no work done when stone enters state 3 in J\n",
+ "dKE2=-1 #change in kinetic energy when stone enters state 3\n",
+ "dU2=-dKE2 #change in internal energy when stone enters state 3 in J\n",
+ "print\"\\n hence,when stone has just come to rest in the bucket is \",round(dKE2),\"J\" \n",
+ "print\"\\n and dU is\",round(dU2),\"J\" \n",
+ "dKE3=0 #change in kinetic energy when stone enters state 4\n",
+ "dPE=0 #change in potential energy when stone enters state 4 in J\n",
+ "W3=0 #no work done when stone enters state 4 in J\n",
+ "dU3=-1 #change in internal energy when stone enters state 4 in J\n",
+ "Q3=dU3 #heat transfer when stone enters state 4 in J\n",
+ "print\"\\n hence,when stone has entered state 4 is\",round(dU3),\"J\" \n",
+ "print\"\\n and Q3 is \",round(Q3),\"J\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " hence,when stone is 1000.0 J\n",
+ "\n",
+ " and change in potential energy is -1.0 J\n",
+ "\n",
+ " hence,when stone has just come to rest in the bucket is -1.0 J\n",
+ "\n",
+ " and dU is 1.0 J\n",
+ "\n",
+ " hence,when stone has entered state 4 is -1.0 J\n",
+ "\n",
+ " and Q3 is -1.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.4:pg-136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4\n",
+ "#Determinig the missing properties \n",
+ "\n",
+ "T1=300 #given temp. in Celsius\n",
+ "u1=2780 #given specific internal enrgy in kJ/kg\n",
+ "print\"From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\" #hiven pressure in kPa\n",
+ "u2=2000 #given specific intrernal energy in kJ/kg\n",
+ "print\"at P=2000 kPa\"\n",
+ "uf=906.4 #in kJ/kg\n",
+ "ug=2600.3 #in kJ/kg \n",
+ "x2=(u2-906.4)/(ug-uf) \n",
+ "print\"Also, under the given conditions\"\n",
+ "vf=0.001177 #in m^3/kg \n",
+ "vg=0.099627 #in m^3/kg\n",
+ "v2=vf+x2*(vg-vf)#Specific volume for water in m^3/kg\n",
+ "print\"\\n hence,specific volume for water is \",round(v2,5),\"m^3/kg\" \n",
+ "print\"\\n Therefore ,this state is \",round(x2,4),\"N\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "From steam table, at T=300 C,ug=2563.0 kJ/kg.So,u1>ug ,it means the state is in the superheated vapor region.So, by interplotation,we find P=1648 kPa and v=0.1542 m^3/kg\n",
+ "at P=2000 kPa\n",
+ "Also, under the given conditions\n",
+ "\n",
+ " hence,specific volume for water is 0.06474 m^3/kg\n",
+ "\n",
+ " Therefore ,this state is 0.6456 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.5:pg-138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5\n",
+ "#calculating heat transfer for the given process\n",
+ "\n",
+ "Vliq=0.05 #volume of saturated liquid in m^3\n",
+ "vf=0.001043 #in m^3/kg\n",
+ "Vvap=4.95 #volume of saturated water vapour in m^3\n",
+ "vg=1.6940 #in m^3/kg\n",
+ "m1liq=Vliq/vf #mass of liquid in kg\n",
+ "m1liq=round(m1liq,2)\n",
+ "m1vap=Vvap/vg #mass of vapors in kg\n",
+ "m1vap=round(m1vap,2)\n",
+ "u1liq=417.36 #specific internal energy of liquid in kJ/kg\n",
+ "u1vap=2506.1 #specific internal energy of vapors in kJ/kg\n",
+ "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in kJ\n",
+ "m=m1liq+m1vap #total mass in kg\n",
+ "V=5.0 #total volume in m^3\n",
+ "v2=V/m #final specific volume in m^3/kg\n",
+ "print\"by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\"\n",
+ "u2=2600.5 #specific internal energy at final state in kJ/kg\n",
+ "U2=m*u2 #internal energy at final state in kJ\n",
+ "Q=U2-U1 #heat transfer for the process in kJ\n",
+ "print\"\\n hence,heat transfer for the process is\",round(Q),\"kJ\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "by interplotation we find that for steam, if vg=0.09831 m^3/kg then pressure is 2.03 Mpa\n",
+ "\n",
+ " hence,heat transfer for the process is 104935.0 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.5E:pg-140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 5E\n",
+ "#calculating heat transfer for the given process\n",
+ "\n",
+ "Vliq=1 #volume of saturated liquid in ft^3\n",
+ "vf=0.01672 #in ft^3/lbm\n",
+ "Vvap=99 #volume of saturated water vapour in m^3\n",
+ "vg=26.80 #in ft^3/lbm\n",
+ "m1liq=Vliq/vf #mass of liquid in lbm\n",
+ "m1liq=round(m1liq,2)\n",
+ "m1vap=Vvap/vg #mass of vapors in lbm\n",
+ "m1vap=round(m1vap,2)\n",
+ "u1liq=180.1 #specific internal energy of liquid in Btu/lbm\n",
+ "u1vap=1077.6 #specific internal energy of vapors in Btu/lbm\n",
+ "U1=m1liq*u1liq + m1vap*u1vap #total internal energy in Btu\n",
+ "U1=round(U1)\n",
+ "m=m1liq+m1vap #total mass in lbm\n",
+ "V=100.0 #total volume in ft^3\n",
+ "v2=V/m #final specific volume in ft^3/lbm\n",
+ "print\"by interplotation we find that for steam, if vg=1.575 ft^3/lbm then pressure is 294 lbf/in^2\"\n",
+ "u2=1117.0 #specific internal energy at final state in Btu/lbm\n",
+ "U2=m*u2 #internal energy at final state in Btu\n",
+ "U2=round(U2)\n",
+ "Q=U2-U1 #heat transfer for the process in Btu\n",
+ "print\"\\n hence,heat transfer for the process is\",round(Q),\"Btu\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "by interplotation we find that for steam, if vg=1.575 ft^3/lbm then pressure is 294 lbf/in^2\n",
+ "\n",
+ " hence,heat transfer for the process is 56182.0 Btu\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.6:pg-143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6\n",
+ "#calculating work and heat transfer for the process\n",
+ "\n",
+ "V1=0.1 #volume of cylinder in m^3\n",
+ "m=0.5 #mass of steam in kg\n",
+ "v1=V1/m #specific volume of steam in m^3/kg\n",
+ "vf=0.001084 #m^3/kg\n",
+ "vfg=0.4614 #m^3/kg\n",
+ "x1=(v1-vf)/vfg #quality\n",
+ "hf=604.74 #kJ/kg\n",
+ "hfg=2133.8#kJ/kg\n",
+ "h2=3066.8 #final specific heat enthalpy in kJ/kg\n",
+ "h1=hf+x1*hfg #initial specific enthalpy in kJ/kg\n",
+ "Q=m*(h2-h1) #heat transfer for this process in kJ\n",
+ "P=400 #pressure inside cylinder in kPa\n",
+ "v2=0.6548 #specific enthalpy in m^3/kg\n",
+ "W=m*P*(v2-v1) #work done for the process in kJ\n",
+ "print\"\\n hence, work done for the process is\",round(W),\"kJ\" \n",
+ "print\"\\n and heat transfer is \",round(Q,1),\"kJ\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " hence, work done for the process is 91.0 kJ\n",
+ "\n",
+ " and heat transfer is 771.1 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.7:pg-144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 7\n",
+ "#calculating work and heat transfer for the process\n",
+ "V1=0.010 #volume of cylinder in m^3\n",
+ "V2=0.040 # m^3\n",
+ "P1=573 #kPa\n",
+ "v1=0.03606 #specific volume of steam in m^3/kg\n",
+ "u1=389.2 #kJ/kg\n",
+ "m=V1/v1#mass of steam in kg\n",
+ "v2=v1*(V2/V1)\n",
+ "P2=163 #kPa\n",
+ "u2=395.8 # kJ/kg\n",
+ "W=8 # kJ\n",
+ "Q2=m*(u2-u1)+W\n",
+ "u3=383.4#kJ/kg\n",
+ "W2=m*(u1-u3)#kJ\n",
+ "print\"\\n and heat transfer is \",round(Q2,2),\"kJ\" \n",
+ "print\"\\n hence, work done for the process is\",round(W2,1),\"kJ\" \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " and heat transfer is 9.83 kJ\n",
+ "\n",
+ " hence, work done for the process is 1.6 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.8:pg-151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8\n",
+ "#calculating change in enthalpy\n",
+ "import math\n",
+ "\n",
+ "h1=273.2 #specific heat enthalpy for oxygen at 300 K\n",
+ "h2=1540.2 #specific heat enthalpy for oxygen at 1500 K\n",
+ "T1=300 #initial temperature in K\n",
+ "T2=1500 #final temparature in K\n",
+ "\n",
+ "dh1=h2-h1 #this change in specific heat enthalpy is calculated using ideal gas tables \n",
+ "dh3=0.922*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 300K\n",
+ "dh4=1.0767*(T2-T1) #it is claculated if we assume specific heat enthalpy to be constant and uses its value at 900K i.e mean of initial and final temperature\n",
+ "print\"\\n Hence,change in specific heat enthalpy if ideal gas tables are used is \",round(dh1,1),\"kJ/kg\"\n",
+ "print\"\\n if specific heat is assumed to be constant and using its value at T1 is\",round(dh3,1),\"kJ/kg\"\n",
+ "print\"\\n if specific heat is assumed to be constant at its value at (T1+T2)/2 is\",round(dh4,1),\"kJ/kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Hence,change in specific heat enthalpy if ideal gas tables are used is 1267.0 kJ/kg\n",
+ "\n",
+ " if specific heat is assumed to be constant and using its value at T1 is 1106.4 kJ/kg\n",
+ "\n",
+ " if specific heat is assumed to be constant at its value at (T1+T2)/2 is 1292.0 kJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.9:pg-152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 9\n",
+ "#determining amount of heat transfer\n",
+ "\n",
+ "P=150 #pressure of nitrogen in cylinder in kPa\n",
+ "V=0.1 #initial volume of cylinder in m^3\n",
+ "T1=25 #initial temperature of nitrogen in celsius\n",
+ "T2=150 #final tempareture of nitrogen in celsius\n",
+ "R=0.2968 #in kJ/kg-K\n",
+ "m=P*V/(R*(T1+273)) #mass of nitrogen in kg\n",
+ "Cv=0.745 #constant volume specific heat for nitrogen in kJ/kg-K\n",
+ "W=-20 #work done on nitrogen gas in kJ\n",
+ "Q=m*Cv*(T2-T1)+W #heat transfer during the process in kJ\n",
+ "print\"\\n hence,the heat transfer for the above process is\",round(Q,1),\"kJ\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " hence,the heat transfer for the above process is -4.2 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.9E:pg-153"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 9E\n",
+ "#determining amount of heat transfer\n",
+ "\n",
+ "P=20 #pressure of nitrogen in cylinder in lbf/in^2\n",
+ "V=2 #initial volume of cylinder in ft^3\n",
+ "T1=80 #initial temperature of nitrogen in F\n",
+ "T2=300 #final tempareture of nitrogen in F\n",
+ "R=55.15 #in ft*lbf/lbm-R\n",
+ "m=P*V*144/(R*(540)) #mass of nitrogen in lbm\n",
+ "Cv=0.177 #constant volume specific heat for nitrogen in Btu/lbm-R\n",
+ "W=-9.15 #work done on nitrogen gas in Btu\n",
+ "Q=m*Cv*(T2-T1)+W #heat transfer during the process in Btu\n",
+ "print\"\\n hence,the heat transfer for the above process is\",round(Q,2),\"Btu\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " hence,the heat transfer for the above process is -1.62 Btu\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.10:pg-155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 10\n",
+ "#calculating rate of increase of internal energy\n",
+ "\n",
+ "W=-12.8*20 #power consumed in J/s\n",
+ "Q=-10 #heat transfer rate from battery in J/s\n",
+ "r=Q-W #rate of increase of internal energy\n",
+ "print\"\\n hence,the rate of increase of internal energy is\",round(r),\"J/s\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " hence,the rate of increase of internal energy is 246.0 J/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.11:pg-155"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 11\n",
+ "#rate of change of temperature\n",
+ "\n",
+ "Q=1500.0 #power produced by burning wood in J/s\n",
+ "mair=1 #mass of air in kg\n",
+ "mwood=5 #mass of soft pine wood in kg \n",
+ "miron=25 #mass of cast iron in kg\n",
+ "Cvair=0.717 #constant volume specific heat for air in kJ/kg\n",
+ "Cwood=1.38 #constant volume specific heat for wood in kJ/kg\n",
+ "Ciron=0.42 #constant volume specific heat for iron in kJ/kg\n",
+ "dT=75-20 #increase in temperature in Celsius\n",
+ "T=(Q/1000)/(mair*Cvair+mwood*Cwood+miron*Ciron) #rate of change of temperature in K/s\n",
+ "dt=(dT/T)/60 #in minutes\n",
+ "print\" hence,the rate of change of temperature is\",round(T,4),\"k/s\" \n",
+ "print\" and time taken to reach a temperature of T is\",round(dt),\"min\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " hence,the rate of change of temperature is 0.0828 k/s\n",
+ " and time taken to reach a temperature of T is 11.0 min\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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