Added(A)/Deleted(D) following books

 A  Heat_Transfer_Principles_And_Applications_by_Dutta/README.txt
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch10.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch11.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch2.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch4.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch5.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch6.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch7.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch8.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/ch9.ipynb
A  Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/10.png
A  Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/5.png
A  Heat_Transfer_Principles_And_Applications_by_Dutta/screenshots/51.png
A  Heat_Transfer_in_SI_units_by_Holman/Chapter1.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter11.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter2.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter3.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter4.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter5.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter6.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter7.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter8.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/Chapter9.ipynb
A  Heat_Transfer_in_SI_units_by_Holman/README.txt
A  Heat_Transfer_in_SI_units_by_Holman/screenshots/9.1.png
A  Heat_Transfer_in_SI_units_by_Holman/screenshots/9.2.png
A  Heat_Transfer_in_SI_units_by_Holman/screenshots/9.4.png
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter1.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter2.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter3.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter4.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter5.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter6.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter7.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter8.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/Chapter9.ipynb
A  Power_Electronics_Principles_and_Applications_by_Jacob/README.txt
A  Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/4.png
A  Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/5.png
A  Power_Electronics_Principles_and_Applications_by_Jacob/screenshots/6.png
A  sample_notebooks/AviralYadav/Chapter5.ipynb

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diff --git a/Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb b/Heat_Transfer_in_SI_units_by_Holman/Chapter10.ipynb
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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 10 Heat Exchangers "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "hi is: 1409.0\n",
+      "% reduction because of fouling factor is  28.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.3\n",
+    "# influence of fouling factor\n",
+    "\n",
+    "#Variable declaration\n",
+    "\t\n",
+    "\t\n",
+    "Rf = 0.0002 \n",
+    "\t# using Rf=(1/hi-1/h_clean)\n",
+    "h_clean = 1961.0 \t\t\t# [W/square meter degree celsius]\n",
+    "\t# we obtain \n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "hi = 1/(Rf+(1/h_clean)) \t\t# [W/square meter degree celsius]\n",
+    "\n",
+    "#Result\n",
+    "\n",
+    "print \"hi is:\",round(hi)\n",
+    "print \"% reduction because of fouling factor is \",round((h_clean-hi)*100/h_clean) \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area of heat-exchanger is 15.81 square meter\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.4\n",
+    "# calculation of heat exchanger size from known temperatures\n",
+    "\n",
+    "#Variable declaration\n",
+    "import math\n",
+    "m_dot = 68.0 \t\t\t# [kg/min] water flow rate \n",
+    "U = 320.0 \t\t\t# [W/sq m degree C] overall heat transfer coefficient\n",
+    "T1 = 35.0 \t\t\t# [degree celsius] initial temperature \n",
+    "T2 = 75.0 \t\t\t# [degree celsius] final temperature\n",
+    "Toe = 110.0 \t\t\t# [degree celsius] oil entering temperature \n",
+    "Tol = 75.0 \t\t\t# [degree celsius] oil leaving temperature\n",
+    "Cw = 4180.0 \t\t\t# [J/kg degree celsius] water specific heat capacity\n",
+    "\t# the total heat transfer is determined from the energy absorbed by the water:\n",
+    "\n",
+    "#Claculation\n",
+    "\n",
+    "q = m_dot*Cw*(T2-T1) \t\t# [J/min]\n",
+    "q = q/60 \t\t\t# [W]\n",
+    "\t# since all the fluid temperatures are known, the LMTD can be calculated by \t\tusing the temperature scheme in figure 10-7b\n",
+    "dT_m = ((Toe-Tol)-(T2-T1))/(math.log((Toe-Tol)/(T2-T1))) \t# [degree celsius]\n",
+    "\t\t\t\t# then, since  q = U*A*dT_m\n",
+    "A = q/(U*dT_m) \t\t\t# [square meter] area of heat-exchanger\n",
+    "\n",
+    "#Result\n",
+    "\n",
+    "print \"Area of heat-exchanger is\",round(A,2),\"square meter\" "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area required for this exchanger is 19.53 square meter\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.5\n",
+    "# shell-and-tube heat exchanger\n",
+    "\n",
+    "# Variable declaration\n",
+    "\t\n",
+    "\t# determine a correction factor from figure 10-8 to be used \n",
+    "\t# the parameters according to figure 10-8(page no.-532) are \n",
+    "T1 = 35 \t\t\t# [degree celsius]\n",
+    "T2 = 75 \t\t\t# [degree celsius]\n",
+    "t1 = 110 \t\t\t# [degree celsius]\n",
+    "t2 = 75 \t\t\t# [degree celsius]\n",
+    "P = (t2-t1)/(T1-t1) \n",
+    "R = (T1-T2)/(t2-t1) \n",
+    "\t# so the correction factor is \n",
+    "F = 0.81 \t\t\t# from figure 10-10(page no.-534)\n",
+    "\t# and the heat transfer is q = U*A*F*dT_m\n",
+    "\t# so that. from example 10-4 we have \n",
+    "U = 320 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n",
+    "q = 189493.33 \t\t\t# [W]\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "dT_m = 37.44 \t\t\t# [degree celsius]\n",
+    "A = q/(U*F*dT_m) \t\t# [square meter]\n",
+    "\n",
+    "#Result\n",
+    "\n",
+    "print \"Area required for this exchanger is\",round(A,2),\"square meter\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Number of tubes per pass 37.0\n",
+      "Number of passes = 2\n",
+      "Length of tube per pass = 1.708 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.6\n",
+    "# design of shell-and-tube heat exchanger\n",
+    "\n",
+    "# Variable declaration\n",
+    "\n",
+    "import math\n",
+    "m_dot_c = 3.8 \t\t\t# [kg/s] water flow rate\n",
+    "Ti = 38 \t\t\t# [degree celsius] initial temperature of water\n",
+    "Tf = 55 \t\t\t# [degree celsius] final temperature of water\n",
+    "m_dot_h = 1.9 \t\t\t# [kg/s] water flow rate entering the exchanger\n",
+    "Te = 93 \t\t\t# [degree celsius] entering water temperature\n",
+    "U = 1419 \t\t\t# [W/sq m degree C] overall heat transfer coefficient\n",
+    "d = 0.019 \t\t\t# [m] diameter of tube\n",
+    "v_avg = 0.366 \t\t\t# [m/s] average water velocity in exchanger\n",
+    "Cc = 4180 \t\t\t# [] specific heat of water\n",
+    "Ch = Cc \t\t\t# [] specific heat \n",
+    "rho = 1000 \t\t\t# [kg/cubic meter] density of water\n",
+    "\t# we first assume one tube pass and check to see if it satisfies the \t\t\tconditions of this problem. the exit temperature of the hot water is \t\t\tcalculated from\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "dTh = m_dot_c*Cc*(Tf-Ti)/(m_dot_h*Ch) \t# [degree celsius]\n",
+    "Th_exit = Te-dTh \t\t\t# [degree celsius]\n",
+    "\t# the total required heat transfer is obtained for the cold fluid is \n",
+    "q = m_dot_c*Cc*(Tf-Ti) \t\t\t# [W]\n",
+    "\t# for a counterflow exchanger, with the required temperature \n",
+    "LMTD = ((Te-Tf)-(Th_exit-Ti))/math.log((Te-Tf)/(Th_exit-Ti)) \t# [degree celsius]\n",
+    "dTm = LMTD \t\t\t\t# [degree celsius]\n",
+    "A = q/(U*dTm) \t\t\t\t# [square meter]\n",
+    "\n",
+    "\n",
+    "\t#calculate the total area with\n",
+    "A1 = m_dot_c/(rho*v_avg) \t\t# [square meter]\n",
+    "\t# this area is the product of number of tubes and the flow area per tube:\n",
+    "n = A1*4/(math.pi*d**(2)) \t\t# no. of tubes\n",
+    "n = round(n) \t\n",
+    "\t# rounding of value of n because no. of pipe is an integer value\n",
+    "\t# the surface area per tube per meter of length is \n",
+    "S = math.pi*d \t\t\t\t# [square meter/tube meter]\n",
+    "\t# total surface area required for a one tube pass exchanger\t\t  was \t\tcalculated above .\n",
+    "\t# we may thus compute the length of tube for this type of exchanger from \n",
+    "L = A/(S*n) \t\t\t\t# [m]\n",
+    "\t# this length is greater than the allowable 2.438 m, so we must use more than \t\tone tube pass.\n",
+    "\t\n",
+    "\t# we next try two tube passes. from figure 10-8(page no.-532) \n",
+    "F = 0.88 \n",
+    "A_total = q/(U*F*dTm) \t\t\t# [square meter]\n",
+    "\t# the number of tubes per pass is still 37 because of the velocity \t\t\trequirement. for the two pass exchanger the total surface area is now related \t\tto the length by\n",
+    "L1 = A_total/(2*S*n) \t\t\t# [m]\n",
+    "\t# this length is within the 2.438 m requirement, so the final design choice is \n",
+    "\n",
+    "#Result\n",
+    "\n",
+    "print \"Number of tubes per pass\",n \n",
+    "print \"Number of passes = 2\" \n",
+    "print \"Length of tube per pass =\",round(L1,3),\"m\" \n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Surface area of heat exchanger is  10.84 square meter\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.7\n",
+    "# cross flow exchanger with one fluid mixed \n",
+    "\n",
+    "# Variable declaration\n",
+    "import math\n",
+    "\n",
+    "m_dot = 5.2 \t\t\t# [kg/s] mass flow rate\n",
+    "T1 = 130.0 \t\t\t# [degree celsius] temperature of entering steam\n",
+    "T2 = 110.0 \t\t\t# [degree celsius] temperature of leaving steam\n",
+    "t1 = 15.0 \t\t\t# [degree celsius] temperature of entering oil\n",
+    "t2 = 85.0 \t\t\t# [degree celsius] temperature of leaving oil\n",
+    "c_oil = 1900.0 \t\t\t# [J/kg degree celsius] heat capacity of oil\n",
+    "c_steam = 1860.0\t\t# [J/kg degree celsius] heat capacity of steam\n",
+    "U = 275 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n",
+    "\t#the total heat transfer may be obtained from an energy balance on the steam \n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "q = m_dot*c_steam*(T1-T2) \t\t\t\t# [W]\n",
+    "\t# we can solve for the area from equation (10-13). the value of dT_m is \t\tcalculated as if the exchanger were counterflow double pipe,thus\n",
+    "dT_m = ((T1-t2)-(T2-t1))/math.log((T1-t2)/(T2-t1)) \t# [degree celsius]\n",
+    "\n",
+    "\t# t1,t2 is representing the unmixed fluid(oil) and T1,T2 is representing the \t\tmixed fluid(steam) so that:\n",
+    "\t# we calculate \n",
+    "\n",
+    "R = (T1-T2)/(t2-t1) \n",
+    "P = (t2-t1)/(T1-t1) \n",
+    "\t# consulting figure 10-11(page no.-534) we find \n",
+    "F = 0.97 \n",
+    "\t# so the area is calculated from \n",
+    "A = q/(U*F*dT_m) \t\t\t\t\t# [square meter]\n",
+    "\n",
+    "#Result\n",
+    "print \"Surface area of heat exchanger is \",round(A,2),\"square meter\" \n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "the objective of this example is to show that an iterative procedure is required when the inlet and outlet temperatures are not known or easily calculated\n",
+      "there is no need to go through this iteration because it can be avoided by using the techniques described in section 10-6\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.8\n",
+    "# effects of off-design flow rates for exchanger in example 10-7 \n",
+    "# Variable declaration\n",
+    "\n",
+    "# we did not calculate the oil flow in example 10-7 but can do so now from \n",
+    "q = 193 # [kW]\n",
+    "c_oil = 1.9 # [J/kg degree celsius] heat capacity of oil\n",
+    "t1 = 15 # [degree celsius] temperature of entering oil\n",
+    "t2 = 85 # [degree celsius] temperature of leaving oil\n",
+    "m_dot_o = q/(c_oil*(t2-t1)) # [kg/s]\n",
+    "# the new flow rate will be half this value \n",
+    "m_dot_o = m_dot_o/2 # [kg/s]\n",
+    "# we are assuming the inlet temperatures remain the same at 130 degree celsius for the steam and 15 degree celsius for the oil.\n",
+    "# the new relation for the heat transfer is q = m_dot_o*c_oil*(Teo-15) = m_dot_s*cp*(130-Tes)                       (a)\n",
+    "# but the exit temperatures, Teo and Tes are unknown. furthermore, dT_m is unknown without these temperatures, as are the values of R and P from figure  10-11(page no.-535). this means we must use an iterative procedure to solve for the exit temperatures using equation (a) and   q = U*A*F*dT_m          (b)\n",
+    "# the general procedure is to assume values of the exit temperatures until the q's agree between equations(a) and (b).\n",
+    "print \"the objective of this example is to show that an iterative procedure is required when the inlet and outlet temperatures are not known or easily calculated\" \n",
+    "print \"there is no need to go through this iteration because it can be avoided by using the techniques described in section 10-6\" \n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "A Reduction in the oil flow rate of 50 % causes a reduction in heat transfer of only  34.0 percent\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.9\n",
+    "# off-design calculation using E-NTU method \n",
+    " \n",
+    "#Variable declaration\t\n",
+    "import math\n",
+    "m_dot_o = 0.725 \t\t\t# [kg/s] oil flow rate\n",
+    "m_dot_s = 5.2 \t\t\t\t# [kg/s] steam flow rate\n",
+    "t1 = 15 \t\t\t\t# [degree celsius] temperature of entering oil\n",
+    "T1 = 130 \t\t\t\t# [deg C] temperature of entering steam\n",
+    "c_oil = 1900 \t\t\t\t# [J/kg degree celsius] heat capacity of oil\n",
+    "c_steam = 1860 \t\t\t\t# [J/kg degree celsius] heat capacity of steam\n",
+    "\t# for the steam \n",
+    "Cs = m_dot_s*c_steam \t\t\t# [W/degree celsius]\n",
+    "\t# for the oil\n",
+    "Co = m_dot_o*c_oil \t\t\t# [W/degree celsius]\n",
+    "\t# so the oil is minium fluid. we thus have\n",
+    "C_min_by_C_max = Co/Cs \n",
+    "U = 275 \t\t\t\t# [W/sq m deg C] overall heat transfer \t\t\t\t\t\t\t  coefficient\n",
+    "A = 10.83 \t\t\t\t# [sq meter] surface area of heat exchanger\n",
+    "NTU = U*A/Co \n",
+    "\t# we choose to use the table and note that Co(minimum) is unmixed and \t\t\tCs(maximum) is mixed so that the first relation in the table 10-3 applies.\n",
+    "\t# we therfore calculate E(effectiveness) as \n",
+    "\n",
+    "E = (1/C_min_by_C_max)*(1-math.exp(-C_min_by_C_max*(1-math.exp(-NTU)))) \n",
+    "\t# if we were using figure 10-14(page no.-544) we would have to evaluate \n",
+    "C_mixed_by_C_unmixed = Cs/Co \n",
+    "\t# and would still determine \n",
+    "E = 0.8 # approximately\n",
+    "\t# now, using the effectiveness we can determine the temperature difference of \t\tthe minimum fluid(oil as)\n",
+    "dT_o = E*(T1-t1) \t\t\t# [degree celsius]\n",
+    "\t# so that heat transfer is \n",
+    "q = m_dot_o*c_oil*(dT_o) \t\t# [W]\n",
+    "q_initial = 193440 \t\t\t# [W] heat transfer when oilrate is 100 %\n",
+    "print \"A Reduction in the oil flow rate of 50 % causes a reduction in heat transfer of only \",round((q_initial-q)*100/q_initial),\"percent\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Exit water temperature is 90.8 degree celcius\n",
+      "The total heat transfer under the new flow conditions is 155.5 kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.10\n",
+    "# off-design calculation of exchanger in example 10-4  \n",
+    "\n",
+    "# Variable declaration\n",
+    "\n",
+    "m_dot_c = 68 \t\t\t# [kg/min] water flow rate\n",
+    "\n",
+    "T1 = 35 \t\t\t# [degree celsius] initial temperature \n",
+    "T2 = 75 \t\t\t# [degree celsius] final temperature\n",
+    "Toe = 110 \t\t\t# [degree celsius] oil entering temperature \n",
+    "Tol = 75 \t\t\t# [degree celsius] oil leaving temperature\n",
+    "Cc = 4180 \t\t\t# [J/kg degree celsius] water specific heat capacity\n",
+    "Ch = 1900 \t\t\t# [J/kg degree celsius] heat capacity of oil\n",
+    "U = 320 \t\t\t# [W/squ m deg C] overall heat transfer coefficient\n",
+    "A = 15.814568 \t\t\t# [sq m] area of heat exchanger (from example 10-4)\n",
+    "\t# the flow rate of oil is calculated from the energy balance for the original \t\tproblem:\n",
+    "\n",
+    "#Calculation\n",
+    "m_dot_h = m_dot_c*Cc*(T2-T1)/(Ch*(Toe-Tol)) \t# [kg/min]\n",
+    "\t# the capacity rates for the new conditions are calculated as \n",
+    "C_h = m_dot_h*Ch/60 \t\t\t\t# [W/degree celsius]\n",
+    "C_c = m_dot_c*Cc/60 \t\t\t\t# [W/degree celsius]\n",
+    "\t# so that the water (cold fluid) is the minimum fluid, and \n",
+    "C_min_by_C_max = C_c/C_h \n",
+    "NTU_max = U*A/C_c \n",
+    "\t# from figure 10-13(page no.-542) or table 10-3(page no.-543) the \t\teffectiveness is \n",
+    "E = 0.744 \n",
+    "\t# and because the cold fluid is the minimum, we can write \n",
+    "dT_cold = E*(Toe-T1) \t\t\t\t# [degree celsius]\n",
+    "\t\t\t\t\t\t# and the exit water temperature is \n",
+    "Tw_exit = T1+dT_cold \t\t\t\t# [degree celsius]\n",
+    "\t# the total heat transfer under the new flow conditions is calculated  as \n",
+    "m_dot_c = 40 \t\t\t\t\t# [kg/min]\n",
+    "q = m_dot_c*Cc*dT_cold/60 \t\t\t# [W]\n",
+    "\n",
+    "#Result\n",
+    "\n",
+    "print \"Exit water temperature is\",Tw_exit,\"degree celcius\" \n",
+    "print \"The total heat transfer under the new flow conditions is\",round(q/1000,1),\"kW\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The exit water temperature is 21.1 degree celsius\n",
+      "the heat transfer is 40.33 kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.11\n",
+    "# cross-flow exchanger with both fluid unmixed \n",
+    " \n",
+    "# Variable declaration\n",
+    "\n",
+    "pa = 101325 \t\t\t# [Pa] pressure of air\n",
+    "Ti = 15.55 \t\t\t# [degree celsius] initial temperature of air\n",
+    "Tf = 29.44 \t\t\t# [degree celsius] final temperature of air\n",
+    "Thw = 82.22 \t\t\t# [degree celsius] hot water temperature\n",
+    "U = 227 \t\t\t# [W/sq m deg C] overall heat transfer coefficient\n",
+    "S = 9.29 \t\t\t# [square meter] total surface area of heat exchanger\n",
+    "R = 287 \t\t\t# [] universal gas constant\n",
+    "Cc = 1006 \t\t\t# [J/kg degree celsius] specific heat of air \n",
+    "Ch = 4180 \t\t\t# [J/kg degree celsius] specific heat of water\n",
+    "\t# the heat transfer is calculated from the energy balance on the air. first, \t\tthe inlet air density is \n",
+    "rho = pa/(R*(Ti+273.15)) \t# [kg/cubic meter]\n",
+    "\t# so the mass flow of air (the cold fluid) is \n",
+    "mdot_c = 2.36*rho \t\t# [kg/s]\n",
+    "\t# the heat transfer is then \n",
+    "q = mdot_c*Cc*(Tf-Ti) \t\t# [W]\n",
+    "\t# from the statement of the problem we do not know whether the air or water is \tthe minimum fluid. a trial and error procedur must be used \n",
+    "\t# we assume that the air is the minimum fluid and then check out our \t\tassumption.then\n",
+    "Cmin = mdot_c*Cc \t\t# [W/degree celsius]\n",
+    "NTU_max = U*S/Cmin \n",
+    "\t# and the effectiveness based on the air as the minimum fluid is \n",
+    "E = (Tf-Ti)/(Thw-Ti) \n",
+    "\t# we must assume values for the water flow rate until we are able to match the \tperformance as given by figure 10-15 or table 10-3. we first note that\n",
+    "Cmax = mdot_c*Cc \t\t# [W/degree celsius]           (a)\n",
+    "\t\t\t\t# NTU_max = U*S/Cmin                             (b)\n",
+    "\t\t\t\t# E = dT_h/(Thw-Ti)                             (c)\n",
+    "\t\t\t\t# dT_h = q/Cmin                                 (d)\n",
+    "\n",
+    "\t# now we assume different values for Cmin abd calculate different-different \t\tvalues for NTU_max, dT_h, and E\n",
+    "\n",
+    "\t# for \n",
+    "Cmin_by_Cmax1 = 0.5 \n",
+    "Cmin1 = Cmin_by_Cmax1*Cmax \t\t\t# [W/degree celsius]\n",
+    "NTU_max1 = U*S/Cmin1 \n",
+    "dT_h1 = q/Cmin1 \t\t\t\t# [degree celsius]\n",
+    "E1_c1 = dT_h1/(Thw-Ti) \t\t\t\t# calculated\n",
+    "E1_t1 = 0.65 \t\t\t\t\t# from table \n",
+    "\n",
+    "\t# for \n",
+    "Cmin_by_Cmax2 = 0.25 \n",
+    "Cmin2 = Cmin_by_Cmax2*Cmax \t\t\t# [W/degree celsius]\n",
+    "NTU_max2 = U*S/Cmin2 \n",
+    "dT_h2 = q/Cmin2 \t\t\t\t# [degree celsius]\n",
+    "E1_c2 = dT_h2/(Thw-Ti) \t\t\t\t# calculated\n",
+    "E1_t2 = 0.89 \t\t\t\t\t# from table \n",
+    "\n",
+    "\t# for \n",
+    "Cmin_by_Cmax3 = 0.22 \n",
+    "Cmin3 = Cmin_by_Cmax3*Cmax \t\t\t# [W/degree celsius]\n",
+    "NTU_max3 = U*S/Cmin3 \n",
+    "dT_h3 = q/Cmin3 \t\t\t\t# [degree celsius]\n",
+    "E1_c3 = dT_h3/(Thw-Ti) \t\t\t\t# calculated\n",
+    "E1_t3 = 0.92 \t\t\t\t\t# from table \n",
+    "\n",
+    "\t# we estimate the water-flow rate as about\n",
+    "Cmin = 660 \t\t\t\t\t# [W/degree celsius]\n",
+    "mdot_h = Cmin/Ch \t\t\t\t# [kg/s]\n",
+    "\t# the exit water temperature is accordingly\n",
+    "Tw_exit = Thw-q/Cmin \t\t\t\t# [degree celsius]\n",
+    "\n",
+    "print \"The exit water temperature is\",round(Tw_exit,1),\"degree celsius\" \n",
+    "print \"the heat transfer is\",round(q/1000,2),\"kW\" "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area required for the heat exchanger is 20.09 square meter\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.13\n",
+    "# shell and tube exchangeras air heater\n",
+    "  \n",
+    "# Variable declaration\n",
+    "\n",
+    "import math\n",
+    "To = 100.0 \t\t\t\t# [degree celsius] temperature of hot oil\n",
+    "m_dot_a = 2.0 \t\t\t\t# [kg/s] flow rate of air\n",
+    "T1 = 20.0 \t\t\t\t# [degree celsius] initial temperature of air \n",
+    "T2 = 80.0 \t\t\t\t# [degree celsius] final temperature of air\n",
+    "Cp_o = 2100.0 \t\t\t\t# [J/kg deg C] specific heat of the oil\n",
+    "Cp_a = 1009.0 \t\t\t\t# [J/kg deg C] specific heat of the air\n",
+    "m_dot_o = 3 \t\t\t\t# [kg/s] flow rate of oil\n",
+    "U = 200.0 \t\t\t\t# [W/sq m] overall heat transfer coefficient\n",
+    "\t# the basic energy balance is m_dot_o*Cp_o*(To-Toe) = m_dot_a*Cp_a*(T2-T1)\n",
+    "#Calculation\n",
+    "Toe = To-m_dot_a*Cp_a*(T2-T1)/(m_dot_o*Cp_o) # [degree celsius]\n",
+    "\n",
+    "\t# we have\n",
+    "m_dot_h_into_Ch = m_dot_o*Cp_o \t\t# [W/degree celsius]\n",
+    "m_dot_c_into_Cc = m_dot_a*Cp_a \t\t# [W/degree celsius]\n",
+    "\t# so the air is minimum fluid\n",
+    "C = m_dot_c_into_Cc/m_dot_h_into_Ch \n",
+    "\t# the effectiveness is \n",
+    "E = (T2-T1)/(To-T1) \n",
+    "\t# now we may use figure 10-16  to obtain NTU. \n",
+    "NTU = -(1+C**(2))**(-1.0/2.0)*math.log((2/E-1-C-(1+C**2)**(1.0/2.0))/(2/E-1-C+(1+C**2)**(1.0/2.0))) \n",
+    "\t# now, we calcuate the area as \n",
+    "A = NTU*m_dot_c_into_Cc/U \t\t# [square meter]\n",
+    "\n",
+    "#Result\n",
+    "\n",
+    "print \"Area required for the heat exchanger is\",round(A,2),\"square meter\" "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.14"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Area to achieve a heat exchanger effectiveness of 60% with an exit water temperature of 40 degree celsius is 9.16 square meter\n",
+      "by reducing the flow rate we have lowered the heat transfer by 37.0  percent\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.14\n",
+    "# ammonia condenser  \n",
+    "\n",
+    "#Variable declaration\n",
+    "\n",
+    "Ta = 50 \t\t\t# [degree C] temperature of entering ammonia vapour\n",
+    "Tw1 = 20 \t\t\t# [degree celsius] temperature of entering water\n",
+    "q = 200 \t\t\t# [kW] total heat transfer required\n",
+    "U = 1 \t\t\t\t# [kW/sq m deg C] overall heat transfer coefficient\n",
+    "Tw2 = 40 \t\t\t# [deg C] temperature of exiting water\n",
+    "Cw = 4.18 \t\t\t# [kJ/kg degree celsius] specific heat of water\n",
+    "\t# the mass flow can be calculated from the heat transfer with\n",
+    "m_dot_w = q/(Cw*(Tw2-Tw1)) \t# [kg/s]\n",
+    "\t# because this is the condenser the water is the minimum fluid and \n",
+    "C_min = m_dot_w*Cw # [kW/degree celsius]\n",
+    "\t# the value of NTU is obtained from the last entry of table 10-4\n",
+    "E = 0.6 \t\t\t# effectiveness\n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "import math\n",
+    "NTU = -math.log(1-E) \n",
+    "\n",
+    "\t# so that area is calculated as \n",
+    "A = C_min*NTU/U \t\t# [square meter]\n",
+    "\n",
+    "\t# when the flow rate is reduced in half the new value of NTU is \n",
+    "NTU1 = U*A/(C_min/2) \n",
+    "\n",
+    "\t# and the effectiveness is computed from the last entry of table 10-3\n",
+    "E1 = 1-math.exp(-NTU1) \n",
+    "\n",
+    "\t# the new water temperature difference is computed as \n",
+    "dT_w = E1*(Ta-Tw1) \t\t# [degree celsius]\n",
+    "\n",
+    "\t# so that the heat transfer is \n",
+    "q1 = C_min*dT_w/2 \t\t# [kW]\n",
+    "\n",
+    "#Result\n",
+    "\n",
+    "print \"Area to achieve a heat exchanger effectiveness of 60% with an exit water temperature of 40 degree celsius is\",round(A,2),\"square meter\" \n",
+    "print \"by reducing the flow rate we have lowered the heat transfer by\",(q-q1)*100/q,\" percent\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Exa 10.16"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Heat-transfer coefficient is 174.0 W/square meter degree celsius\n"
+     ]
+    }
+   ],
+   "source": [
+    "#Example Number 10.16 \n",
+    "# heat-transfer coefficient in compact exchanger   \n",
+    "\n",
+    "# Variable declaration\n",
+    "\n",
+    "p = 101325.0 \t\t\t# [Pa] pressure of air\n",
+    "T = 300.0\t\t\t# [K] temperature of entering air\n",
+    "u = 15.0 \t\t\t\t# [m/s] velocity of air\n",
+    "\t# we obtain the air properties from table A-5\n",
+    "rho = 1.774 \t\t\t# [kg/cubic meter] density of air\n",
+    "Cp = 1005.7 \t\t\t# [J/kg degree celsius] specific heat of air\n",
+    "mu = 1.983*10**(-5) \t\t# [kg/m s] viscosity of air\n",
+    "Pr = 0.708 \t\t\t# prandtl number\n",
+    "\t# from figure 10-19 we have\n",
+    "Ac_by_A = 0.697 \n",
+    "sigma = 0.697 \n",
+    "\n",
+    "#Calculation\n",
+    "\n",
+    "Dh = 3.597*10**(-3) \t\t# [m] \n",
+    "\n",
+    "\t\t\t\t# the mass velocity is thus \n",
+    "G = ((rho*u)/sigma) \t\t# [kg/square meter s]\n",
+    "\n",
+    "\n",
+    "\t# and the reynolds number is \n",
+    "Re = Dh*G/mu \n",
+    "\n",
+    "\t# from figure 10-19(page no.-557) we can read\n",
+    "\n",
+    "St_into_Pr_exp_2_by_3 = 0.0036\n",
+    " \n",
+    "\t# and the heat transfer coefficient is \n",
+    "\n",
+    "h = St_into_Pr_exp_2_by_3*G*Cp*(Pr)**(-2.0/3.0) \t# [W/sq m deg C]\n",
+    "\n",
+    "#Result\n",
+    "\n",
+    "print \"Heat-transfer coefficient is\",round(h),\"W/square meter degree celsius\""
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
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