path: root/Fundamental_of_Thermodynamics_by_Moran_and_Shapiro/Chapter_6.ipynb

{
 "metadata": {
  "name": "Chapter 6"
 }, 
 "nbformat": 2, 
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown", 
     "source": [
      "# Chapter 6 :- Using Entropy"
     ]
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>  Example 6.1 Page no- 219"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "T = 373.15                                # temperature in kelvin", 
      "", 
      "# From table A-2", 
      "", 
      "p = 1.014*(10**5)                         # pressure in pascal", 
      "vg = 1.673", 
      "vf = 1.0435e-3", 
      "sg = 7.3549", 
      "sf = 1.3069", 
      "", 
      "# Calculations", 
      "w = p*(vg-vf)*(10**(-3))", 
      "Q = T*(sg-sf)", 
      "", 
      "# Results ", 
      "print '-> The work per unit mass is',round(w,3),'KJ/kg.'", 
      "print '-> The heat transfer per unit mass is ',round(Q,2),'KJ/kg.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The work per unit mass is 169.536 KJ/kg.", 
        "-> The heat transfer per unit mass is  2256.81 KJ/kg."
       ]
      }
     ], 
     "prompt_number": 1
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>  Example 6.2 Page no- 224"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# Assumptions:", 
      "\"\"\"", 
      "1. The water in the piston\u2013cylinder assembly is a closed system.", 
      "", 
      "2. There is no heat transfer with the surroundings.", 
      "", 
      "3. The system is at an equilibrium state initially and finally. There is no", 
      "change in kinetic or potential energy between these two states.\"\"\"", 
      "", 
      "# From table A-2 at 100 degree celcius", 
      "ug = 2506.5                                  # in kj/kg", 
      "uf = 418.94                                  # in kj/kg", 
      "sg = 7.3549", 
      "sf = 1.3069", 
      "", 
      "", 
      "# Calculations:-", 
      "# From energy balance", 
      "W = -(ug-uf)", 
      "# From entropy balance", 
      "sigmabym = (sg-sf)", 
      "", 
      "# Results", 
      "print '-> The net work per unit mass is ',round(W,2),'KJ/kg.'", 
      "print '-> The amount of entropy produced per unit mass is ',round(sigmabym,2),'kJ/kg.k.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The net work per unit mass is  -2087.56 KJ/kg.", 
        "-> The amount of entropy produced per unit mass is  6.05 kJ/kg.k."
       ]
      }
     ], 
     "prompt_number": 2
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>  Example 6.3 Page no-225"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "T1 = 273.0                                # initial temperature of saturated vapor in kelvin", 
      "P2 = 0.7*(10**6)                          # final pressure in pascal", 
      "", 
      "# From table A-10,", 
      "u1 = 227.06                               # in kj/kg", 
      "", 
      "# minimum theoretical work corresponds to state of isentropic compression", 
      "# From table A-12,", 
      "u2s = 244.32                              # in kj/kg", 
      " ", 
      "# Calculations ", 
      "Wmin = u2s-u1", 
      "", 
      "# Results", 
      "print '-> The minimum theoretical work input required per unit mass of refrigerant is: ',round(Wmin,2),'kJ/kg.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The minimum theoretical work input required per unit mass of refrigerant is:  17.26 kJ/kg."
       ]
      }
     ], 
     "prompt_number": 3
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 6.4 Page no- 227"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given :-", 
      "Qdot = -1.2                          # in kilo watt", 
      "Tb = 300.0                           # in kelvin", 
      "Tf = 293.0                           # in kelvin", 
      "# Calculations", 
      "", 
      "# Part (a)", 
      "# From entropy balance ", 
      "sigmadot = -Qdot/Tb", 
      "", 
      "# Part(b)", 
      "# From entropy balance ", 
      "sigmadt = -Qdot/Tf", 
      "", 
      "# Results", 
      "print '-> The rate of entropy production with gearbox as system is ',round(sigmadot,5),'kw/k.'", 
      "print '-> The rate of entropy production with gearbox + sorrounding as system is',round(sigmadt,5),'kw/k.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The rate of entropy production with gearbox as system is  0.004 kw/k.", 
        "-> The rate of entropy production with gearbox + sorrounding as system is 0.0041 kw/k."
       ]
      }
     ], 
     "prompt_number": 4
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 6.5 Page no-229"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# GIven:-", 
      "Tmi = 1200.0                                   # initial temperature of metal in kelvin", 
      "cm = 0.42                                      # specific heat of metal in KJ/kg.k", 
      "mm = 0.3                                       # mass of metal in kg", 
      "Twi = 300.0                                    # initial temperature of water in kelvin", 
      "cw = 4.2                                       # specific heat of water in KJ/Kg.k", 
      "mw = 9.0                                       # mass of water in kg", 
      "", 
      "# Calculations", 
      "import math", 
      "# Part(a)", 
      "# Solving energy balance equation yields", 
      "Tf = (mw*(cw/cm)*Twi+mm*Tmi)/(mw*(cw/cm)+mm)", 
      "", 
      "# Part (b)", 
      "# Solving entropy balance equation yields", 
      "sigma = mw*cw*math.log(Tf/Twi)+mm*cm*math.log(Tf/Tmi)", 
      "", 
      "# Results", 
      "print '-> The final equilibrium temperature of the metal bar and the water is',round(Tf,2),'  kelvin.'", 
      "print '-> The amount of entropy produced is:',round(sigma,2),'kJ/k.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The final equilibrium temperature of the metal bar and the water is 302.99   kelvin.", 
        "-> The amount of entropy produced is: 0.2 kJ/k."
       ]
      }
     ], 
     "prompt_number": 5
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 6.6 Page no- 234"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "P1 = 30.0                                            # pressure of steam entering the turbine in bar", 
      "T1 = 400.0                                           # temperature of steam entering the turbine in degree celcius", 
      "V1 = 160.0                                           # velocity of steam entering the turbine in m/s", 
      "T2 = 100.0                                           # temperature of steam exiting in degree celcius", 
      "V2 = 100.0                                           # velocity of steam exiting in m/s", 
      "Wcvdot = 540.0                                       # work produced by turbine in kJ/kg of steam", 
      "Tb = 350.0                                           # temperature of the boundary in kelvin", 
      "", 
      "# From table A-4 and table A-2", 
      "h1 = 3230.9                                          # specific enthalpy at entry in Kj/kg", 
      "h2 = 2676.1                                          # specific enthalpy at exit in kj/kg", 
      "", 
      "# Calculations", 
      "", 
      "# Reduction in mass and energy balance equations results in ", 
      "Qcvdot = Wcvdot + (h2 - h1)+ (V2**2-V1**2)/(2*(10**3))  # heat transfer rate", 
      "", 
      "# From table A-2", 
      "s2 = 7.3549                                             # in kj/kg.k", 
      "# From table A-4", 
      "s1 = 6.9212                                             # in kj/kg.k", 
      "", 
      "# From entropy and mass balance equations", 
      "sigmadot = -(Qcvdot/Tb) + (s2-s1)", 
      "", 
      "# Results", 
      "print '-> The rate at which entropy is produced within the turbine per kg of steam  flowing is',round(sigmadot,2),'kJ/kg.k.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The rate at which entropy is produced within the turbine per kg of steam  flowing is 0.5 kJ/kg.k."
       ]
      }
     ], 
     "prompt_number": 6
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>  Example 6.7 Page no- 235"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "T1 = 294.0                            # entry temperature of air in kelvin", 
      "P1 = 5.1                              # entry pressure of air in bars", 
      "T2 = 352.0                            # exit temperature of hot stream in kelvin", 
      "P2 = 1.0                              # exit pressure of hot stream in bars", 
      "T3 = 255.0                            # exit temperature of cold stream in kelvin", 
      "P3 = 1.0                              # exit pressure of cold stream in bars", 
      "cp = 1.0                              # in kj/kg.k", 
      "", 
      "# Calculations", 
      "import math", 
      "R = 8.314/28.97", 
      "se = 0.4*(cp*math.log((T2)/(T1))-R*math.log(P2/P1)) + 0.6*(cp*math.log((T3)/(T1))-R*math.log(P3/P1))", 
      "                                      # specific entropy in kj/kg.k", 
      "", 
      "", 
      "# Results", 
      "print '-> Specific entropy in kj/kg.k =  ',round(se,3),' KJ/kg.'", 
      "print '-> Since se > 0, the claim of the writer is true'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> Specific entropy in kj/kg.k =   0.454  KJ/kg.", 
        "-> Since se > 0, the claim of the writer is true"
       ]
      }
     ], 
     "prompt_number": 7
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 6.8 Page no- 237"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "P1 = 3.5                                          # pressure of refrigerant entering the compressor in bars", 
      "T1 = 268.0                                        # temperature of refrigerant entering the compressor in kelvin", 
      "P2 = 14.0                                         # pressure of refrigerant entering the condenser in bars", 
      "T2 = 348.0                                        # temperature of refrigerant entering the condenser in kelvin", 
      "P3 = 14.0                                         # pressure of refrigerant exiting the condenser in bars", 
      "T3 = 301.0                                        # temperature of refrigerant exiting the condenser in kelvin", 
      "P4 = 3.5                                          # pressure of refrigerant after passing through expansion valve in bars", 
      "P5 = 1.0                                          # pressure of indoor return air entering the condenser in bars", 
      "T5 = 293.0                                        # temperature of indoor return air entering the condenser in kelvin", 
      "AV5 = 0.42                                        # volumetric flow rate of indoor return air entering the condenser in m^3/s", 
      "P6 =  1.0                                         # pressure of return air exiting the condenser in bar", 
      "T6 = 323.0                                        # temperature of return air exiting the condenser in kelvin", 
      "", 
      "# Part(a)", 
      "", 
      "# From table A-9", 
      "s1 = 0.9572                                       # in kj/kg.k", 
      "# Interpolating in table A-9", 
      "s2 = 0.98225                                      # in kj/kg.k", 
      "h2 = 294.17                                       # in kj/kg", 
      "# From table A-7", 
      "s3 = 0.2936                                       # in kj/kg.k", 
      "h3 = 79.05                                        # in kj/kg", 
      "", 
      "h4 = h3                                           # since expansion through valve is throttling process", 
      "", 
      "# From table A-8", 
      "hf4 = 33.09                                       # in kj/kg", 
      "hg4 = 246.00                                      # in kj/kg", 
      "sf4 = 0.1328                                      # in kj/kg.k", 
      "sg4 = 0.9431                                      # in kj/kg.k", 
      "cp = 1.005                                        # in kj/kg.k", 
      "", 
      "# Calculations", 
      "import math", 
      "x4 = (h4-hf4)/(hg4-hf4)                           # quality at state 4", 
      "s4 = sf4 + x4*(sg4-sf4)                           # specific entropy at state 4", 
      "", 
      "# CONDENSER!!", 
      "v5 = ((8314/28.97)*T5)/(P5*(10**5))               # specific volume at state 5", 
      "mairdot = AV5/v5                            ", 
      "h6 = cp*T6", 
      "h5 = cp*T5", 
      "mrefdot = mairdot*(h6-h5)/(h2-h3)", 
      "deltaS65 = cp*math.log(T6/T5)-(8.314/28.97)*math.log(P6/P5) # change in specific entropy", 
      "sigmacond = (mrefdot*(s3-s2)) + (mairdot*(deltaS65))", 
      "", 
      "# COMPRESSOR!!", 
      "sigmacomp = mrefdot*(s2-s1)", 
      "", 
      "# VALVE!!", 
      "sigmavalve = mrefdot *(s4-s3)", 
      "", 
      "# Results", 
      "print '-> The rates of entropy production for control volume enclosing the condenser  is ',sigmacond,'kW/K.'", 
      "print '-> The rates of entropy production for control volume enclosing the compressor  is ',sigmacomp,'kW/K.'", 
      "print '-> The rates of entropy production for control volume enclosing the expansion valve  is  ',sigmavalve,'kW/K.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The rates of entropy production for control volume enclosing the condenser  is  0.000724165354577 kW/K.", 
        "-> The rates of entropy production for control volume enclosing the compressor  is  0.00175361560919 kW/K.", 
        "-> The rates of entropy production for control volume enclosing the expansion valve  is   0.000988192525533 kW/K."
       ]
      }
     ], 
     "prompt_number": 8
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>  Example 6.9 Page no-243"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "P1 = 1.00                                    # initial pressure in bar", 
      "T1 = 300.00                                  # initial temperature in kelvin", 
      "T2 = 650.00                                  # final temperature in kelvin", 
      "", 
      "# Part(a)", 
      "# From table A-22", 
      "pr2 = 21.86                               ", 
      "pr1 = 1.3860", 
      "k = 1.39                                     # From table A-20", 
      "", 
      "# Calculations", 
      "p2 = P1*(pr2/pr1)", 
      "p2a = P1*((T2/T1)**(k/(k-1)))", 
      "", 
      "# Results", 
      "print '-> P2  =  ',p2,'bar.'", 
      "print '-> Part(b) IT software problem'", 
      "print '-> P2a  = ',p2a,'bar.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> P2  =   15.772005772 bar.", 
        "-> Part(b) IT software problem", 
        "-> P2a  =  15.7324909817 bar."
       ]
      }
     ], 
     "prompt_number": 9
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 6.10 Page no- 244"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "m1 = 5.00                                           # initial mass in kg", 
      "P1 = 5.00                                           # initial pressure in bar", 
      "T1 = 500.00                                         # initial temperature in kelvin", 
      "P2 = 1.00                                           # final pressure in bar", 
      "", 
      "# From table A-22", 
      "pr1 = 8.411", 
      "", 
      "", 
      "", 
      "# Using this value of pr2 and interpolation in table A-22", 
      "T2 = 317.00                                         # in kelvin", 
      "", 
      "# Calculations ", 
      "pr2 = (P2/P1)*pr1", 
      "m2 = (P2/P1)*(T1/T2)*m1", 
      "", 
      "# Results", 
      "print '-> The amount of mass remaining in the tank  is ',m2,'kg.'", 
      "print 'and its temperature is ',T2,'kelvin.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The amount of mass remaining in the tank  is  1.57728706625 kg.", 
        "and its temperature is  317.0 kelvin."
       ]
      }
     ], 
     "prompt_number": 10
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>  Example 6.11 Page no- 249"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "P1 = 1.00                                              # inlet pressure in bar", 
      "T1 = 593.00                                            # inlet temperature in kelvin", 
      "P2 = 1.00                                              # exit pressure in bar", 
      "eta =0.75                                              # turbine efficiency", 
      "", 
      "# From table A-4", 
      "h1 = 3105.6                                            # in Kj/kg", 
      "s1 = 7.5308                                            # in kj/kg.k", 
      "# From table A-4 at 1 bar", 
      "h2s = 2743.00                                          # in kj/kg", 
      "", 
      "# Calculations", 
      "w = eta*(h1 - h2s)", 
      "", 
      "# Result", 
      "print '-> The work developed per unit mass of steam flowing through is ',w,'kJ/kg.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The work developed per unit mass of steam flowing through is  271.95 kJ/kg."
       ]
      }
     ], 
     "prompt_number": 11
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 6.12 Page no-250"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "P1 = 3.00                                  # pressure of air entering in bar", 
      "T1 = 390.00                                # temperature of air entering in kelvin", 
      "P2 = 1.00                                  # pressure of exit air", 
      "Wcvdot = 74.00                             # work developed in kj/kg", 
      "", 
      "# From table A-22,at 390k", 
      "h1 = 390.88                                # in kj/kg", 
      "pr1 = 3.481", 
      "", 
      "# From interpolation table A-22", 
      "h2s = 285.27                               # in kj/kg", 
      "", 
      "# calculations", 
      "pr2 = (P2/P1)*pr1", 
      "Wcvdots = h1 - h2s", 
      "eta = Wcvdot/Wcvdots", 
      "", 
      "# Result", 
      "print '-> The turbine efficiency is ',round(eta,4),'.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The turbine efficiency is  0.7007 ."
       ]
      }
     ], 
     "prompt_number": 12
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2> Example 6.13 Page no-251"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "P1 = 1.00                                       # pressure of entering steam in Mpa", 
      "T1 = 593.00                                     # temperature of entering steam in kelvin", 
      "V1 = 30.00                                      # velocity of entering steam in m/s", 
      "P2 = 0.3                                        # pressure of exit steam in Mpa", 
      "T2 = 453.00                                     # temperature of exit steam in kelvin", 
      "", 
      "# From table A-4, at T1 = 593 kelvin and P1 = 1 Mpa;", 
      "# and at T2 = 453 kelvin and P2 = .3 Mpa", 
      "h1 = 3093.9                                     # in kj/kg", 
      "s1 = 7.1962                                     # in kj/kg.k", 
      "h2 = 2823.9                                     # in kj/kg", 
      "", 
      "", 
      "# Interpolating in table A-4", 
      "h2s = 2813.3                                    # in kj/kg", 
      "", 
      "# Calculations", 
      "V2squareby2 = h1 - h2 + (V1**2)/2000", 
      "V2squareby2s = h1 - h2s + (V1**2)/2000", 
      "eta = V2squareby2/V2squareby2s", 
      "", 
      "# Results", 
      "print '-> The nozzle efficiency is ',round(eta,4),'.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The nozzle efficiency is  0.9623 ."
       ]
      }
     ], 
     "prompt_number": 13
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 6.14 Page no- 252"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "# From table A-9", 
      "h1 = 249.75                                             # in kj/kg", 
      "h2 = 294.17                                             # in kj/kg", 
      "mdot = 0.07                                             # in kg/s", 
      "", 
      "# From table A-9", 
      "s1 = 0.9572                                             # in Kj/Kg.k", 
      "h2s = 285.58                                            # in kj/kg", 
      "", 
      "# Calculations", 
      "wcvdot = -(mdot*(h2-h1))", 
      "eta = (h2s-h1)/(h2-h1)        ", 
      "", 
      "# Results", 
      "print '-> The power in  is',wcvdot,'KW.'", 
      "print '-> The isentropic efficiency is ',round(eta,3),'.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The power in  is -3.1094 KW.", 
        "-> The isentropic efficiency is  0.807 ."
       ]
      }
     ], 
     "prompt_number": 14
    }, 
    {
     "cell_type": "markdown", 
     "source": [
      "##<heading2>Example 6.15 Page no-256"
     ]
    }, 
    {
     "cell_type": "code", 
     "collapsed": false, 
     "input": [
      "", 
      "# Given:-", 
      "P1 = 1.00                               # pressure of entering air in bar", 
      "T1 = 293.00                             # temperature of entering air in kelvin", 
      "P2 = 5.00                               # pressure of exit air in bar", 
      "n = 1.3", 
      "R = 8.314/28.97", 
      "", 
      "# From table A-22", 
      "h1 = 293.17                             # in kj/kg", 
      "h2 = 426.35                             #  in kj/kg", 
      "", 
      "# Calculations", 
      "T2 = T1*((P2/P1)**((n-1)/n))             # in kelvin", 
      "wcvdot=((n*R)/(n-1))*(T1-T2)            # in kj/kg", 
      "Qcvdot= wcvdot + (h2-h1)                # in kj/kg", 
      "", 
      "# Results", 
      "print '-> The work per unit mass passing through the device is',round(wcvdot,2),'kJ/kg.'", 
      "print '-> The heat transfer per unit mass is  ',round(Qcvdot,2),'kJ/kg.'"
     ], 
     "language": "python", 
     "outputs": [
      {
       "output_type": "stream", 
       "stream": "stdout", 
       "text": [
        "-> The work per unit mass passing through the device is -163.89 kJ/kg.", 
        "-> The heat transfer per unit mass is   -30.71 kJ/kg."
       ]
      }
     ], 
     "prompt_number": 15
    }
   ]
  }
 ]
}