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   "source": [
    "# Chapter 4 : The first law of thermodynamics"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.1 page no : 111"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "change in pot. energy per unit mass is 225.630000 m^2/s^2\n",
      "The total change in potential energy is 2256.300000 J\n"
     ]
    }
   ],
   "source": [
    "#calculate change in pot. energy per unit mass and total change in pot. energy\n",
    "\n",
    "# variables\n",
    "g=9.81;                 #m/s^2 acc. due to gravity\n",
    "dh=23.;                 #m change in height\n",
    "\n",
    "# calculation and result\n",
    "dpe=g*dh                #m^2/s^2 change in pot energy per unit mass\n",
    "print \"change in pot. energy per unit mass is %f m^2/s^2\"%dpe\n",
    "\n",
    "m=10.;                  #kg\n",
    "dPE=m*dpe               #kgm^2/s^2 or J change in pot. energy \n",
    "print \"The total change in potential energy is %f J\"%dPE"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.2 page no : 112"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the kinetic energy of bullet is 842.236025 J\n"
     ]
    }
   ],
   "source": [
    "#calculate the kinetic energy of bullet\n",
    "\n",
    "# variables\n",
    "m=0.01;                     #lbm mass of bullet\n",
    "v=2000.;                    #ft/s\n",
    "\n",
    "# calculation\n",
    "KE=(m*v**2/2)*(1.356/32.2)  #J\n",
    "\n",
    "# result\n",
    "print \"the kinetic energy of bullet is %f J\"%KE"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.3 page no : 112"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the kinetic energy of bullet fired from a airplane is 0.021056 J\n"
     ]
    }
   ],
   "source": [
    "#Calculate the kinetic energy of bullet fired from a airplane\n",
    "\n",
    "# variables\n",
    "v_bp=2000.;                    #ft/s vel of bullet wrt plane\n",
    "v_p=-1990.;                    #ft/s\n",
    "v_b=v_bp+v_p                   #ft/s vel of bullet wrt ground\n",
    "m=0.01;                        #lbm\n",
    "\n",
    "# calculation\n",
    "KE=(m*v_b**2/2)*(1.356/32.2)   #J\n",
    "\n",
    "# result\n",
    "print \"the kinetic energy of bullet fired from a airplane is %f J\"%KE"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.4 page no : 119"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the change in internal energy of the system is -44.720823 Btu\n"
     ]
    }
   ],
   "source": [
    "#calculate the change in internal energy of the system\n",
    "\n",
    "# variables\n",
    "p=14.7;                 #lbf/in^2 atmospheric pressure\n",
    "dV=1.;                  #ft^3 change in volume\n",
    "\n",
    "# calculation\n",
    "dW=p*dV*(144/778.)      #Btu work done\n",
    "dQ=-42.;                #Btu heat removed from the system\n",
    "dU=dQ-dW                #Btu change in internal energy of the system\n",
    "\n",
    "# result\n",
    "print \"the change in internal energy of the system is %f Btu\"%dU"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.5 page no : 120"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The work done is 101.650000 J/Kg\n"
     ]
    }
   ],
   "source": [
    "#calculate the work done\n",
    "#work done=change in pot. energy + change in kinetic energy considering steady flow and adiabatic conditions\n",
    "\n",
    "# variables\n",
    "v_in=3.;                        #m/s\n",
    "v_out=10.;                      #m/s\n",
    "dke=(v_in**2-v_out**2)/2.0;     #m^2/s^2\n",
    "g=9.81;                         #m/s^2\n",
    "dh=15.;                         #m change in height in inlet and outlet\n",
    "\n",
    "# calculation\n",
    "dpe=g*dh;                       #m^2/s^2\n",
    "W=dpe+dke                       #J/kg\n",
    "\n",
    "# result\n",
    "print \"The work done is %f J/Kg\"%W"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Example 4.6 page no : 124"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the mass consumed in a nuclear reactor per unit time is 2.222222 * 10^(-8) kg/s\n"
     ]
    }
   ],
   "source": [
    "#calculate the mass consumed in a nuclear reactor per unit time\n",
    "\n",
    "# variables\n",
    "#let D=d/dt\n",
    "DQ=-13.*10**8;                  #J/s\n",
    "DW=7*10.**8;                    #J/s\n",
    "\n",
    "#Dm=(DQ-DW)/c^2 where c is velocity of light sice E=mc^2\n",
    "c=3.0*10**8;                    #m/s\n",
    "c1=3.0;                         #velocity of light without power\n",
    "p=8.                            #power of 10 in speed of light \n",
    "\n",
    "# calculation\n",
    "Dm=float(DW-DQ)/c/c1            #kg/s\n",
    "\n",
    "# result\n",
    "print \"the mass consumed in a nuclear reactor per unit time is %f * 10^(-%d) kg/s\"%(Dm,p)"
   ]
  }
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