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diff --git a/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter18.ipynb b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter18.ipynb new file mode 100644 index 00000000..97968f87 --- /dev/null +++ b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter18.ipynb @@ -0,0 +1,703 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 18:Elements of Heat Transfer" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.1:pg-757" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.1\n", + "\n", + "\n", + " The rate of heat removal is 486.40484238 W\n", + "\n", + " Temperature at inside surface of brick is 20.2812224957 degree celcius\n" + ] + } + ], + "source": [ + "\n", + "ho = 12.0 # Outside convective heat transfer coefficient in W/m**2K \n", + "x1 = 0.23# Thickness of brick in m\n", + "k1 = 0.98 # Thermal conductivity of brick in W/mK\n", + "x2 = 0.08 # Thickness of foam in m\n", + "k2 = 0.02# Thermal conductivity of foam in W/mK\n", + "x3 = 1.5# Thickness of wood in cm\n", + "k3 = 0.17# Thermal conductivity of wood in W/cmK\n", + "hi = 29.0# Inside convective heat transfer coefficient in W/m**2K \n", + "A = 90.0 # Total wall area in m**2\n", + "to = 22.0# outside air temperature in degree Celsius\n", + "ti = -2.0 # Inside air temperature in degree Celsius\n", + "print \"\\n Example 18.1\\n\"\n", + "U = (1/((1/ho)+(x1/k1)+(x2/k2)+(x3*1e-2/k3)+(1/hi)))# Overall heat transfer coefficient\n", + "Q = U*A*(to-ti) # Rate of heat transfer\n", + "R = (1/ho)+(x1/k1)\n", + "t2 = to-Q*R/A # Temperature at inside surface of brick\n", + "\n", + "print \"\\n The rate of heat removal is \",Q ,\" W\"\n", + "\n", + "print \"\\n Temperature at inside surface of brick is \",t2 ,\" degree celcius\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.2:pg-758" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.2\n", + "\n", + "\n", + " Heat transfer rate is 2.33519645654 kW\n" + ] + } + ], + "source": [ + "import math\n", + "r1 = 5.0 # Inner radius of steel pipe in cm\n", + "r2 = 10.0 # Extreme radius of inner insulation in cm\n", + "r3 = 13.0# Extreme radius of outer insulation in cm\n", + "K1 = 0.23 # Thermal conductivity of inner insulation in W/mK\n", + "K2 = 0.37 # Thermal conductivity of outer insulation in W/mK\n", + "hi = 58.0 # Inner heat transfer coefficient in W/m**2K\n", + "h0 = 12.0 # Inner heat transfer coefficient in W/m**2K\n", + "ti = 60.0 # Inner temperature in degree Celsius\n", + "to = 25.0 # Outer temperature in degree Celsius\n", + "L = 50.0 # Length of pipe in m\n", + "\n", + "print \"\\n Example 18.2\\n\"\n", + "Q =((2*math.pi*L*(ti-to))/((1/(hi*r1*1e-2))+(math.log(r2/r1)/(K1))+(math.log(r3/r2)/(K2))+(1/(h0*r3*1e-2))))\n", + "# Rate of heat transfer\n", + "print \"\\n Heat transfer rate is \",Q/1e3 ,\" kW\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.3:pg-759" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.3\n", + "\n", + "\n", + " Thermal conductivity of rod A is 57.4969670417 W/mK\n", + "\n", + " Thermal conductivity of rod B is 86.076212035 W/mK\n", + "\n", + " Thermal conductivity of rod C is 116.0 W/mK\n" + ] + } + ], + "source": [ + "\n", + "to = 20 # Environment temperature in degree Celsius\n", + "t = 100# Temperature of steam path in degree Celsius\n", + "ta1 = 26.76 # Temperature at other end in degree Celsius for rod A \n", + "d = 10 # diameter of rod in mm\n", + "L = 0.25 # length of rod in m\n", + "h = 23 # heat transfer coefficient in W/m**2 K\n", + "tb1 = 32.00 # Temperature at other end in degree Celsius for rod B \n", + "tc1 = 36.93 # Temperature at other end in degree Celsius for rod C \n", + "\n", + "print \"\\n Example 18.3\\n\"\n", + "A = math.pi/4 * (d*1e-3)**2 #Area of rod\n", + "p = math.pi*d*1e-3 # perimeter of rod\n", + "# For rod A\n", + "a = (ta1-to)/(t-to) \n", + "ma = (math.acosh(1/a))/L\n", + "\n", + "Ka = (h*p)/(ma**2*A) # Thermal conductivity of rod A\n", + "print \"\\n Thermal conductivity of rod A is \",Ka ,\" W/mK\"\n", + "# For rod B\n", + "b = (tb1-to)/(t-to) \n", + "mb = (math.acosh(1/b))/L\n", + "\n", + "Kb = (h*p)/(mb**2*A) # Thermal conductivity of rod B\n", + "print \"\\n Thermal conductivity of rod B is \",Kb ,\" W/mK\"\n", + "c = (tc1-to)/(t-to) \n", + "mc = (math.acosh(1/c))/L\n", + "\n", + "Kc = (h*p)/(mc**2*A) # Thermal conductivity of rod A\n", + "print \"\\n Thermal conductivity of rod C is \",math. ceil(Kc) ,\" W/mK\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.4:pg-760" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.4\n", + "\n", + "\n", + " Midway temperature of rod is 88.7138777413 degree Celcius\n", + "\n", + " Heat loss rate is 88.0331604603 W\n" + ] + } + ], + "source": [ + "h = 17.4 # Convective heat transfer coefficient in W/m**2K\n", + "K = 52.2 # Thermal conductivity in W/mK\n", + "t = 120 # Heat reservoir wall temperature in degree celcius\n", + "t0 = 35 # Ambient temperature in degree celcius\n", + "L = 0.4 # Lenght of rod in m\n", + "b = .050 # width of rod in mm\n", + "H = .050 # Heigth of rod in mm\n", + "\n", + "print \"\\n Example 18.4\\n\"\n", + "l= L/2\n", + "A = b*H\n", + "m = math.sqrt(4*h*b/(K*b*H))\n", + "t1 = (t-t0)/math.cosh(m*l) + t0 # Midway temperature of rod\n", + "Q1 = 2*5.12*K*A*(t-t0)*math.tanh(m*l) # Heat loss rate \n", + "print \"\\n Midway temperature of rod is \",t1 ,\" degree Celcius\"\n", + "print \"\\n Heat loss rate is \",Q1 ,\"W\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.5:pg-760" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.5\n", + "\n", + "\n", + " Time to cool down to 2 degree celcius is 30.5933342864 min\n", + "\n", + " Temperature of peas after 10 minutes is 13.1714792663 degree celcius\n", + "\n", + " Temperature of peas after 30 minutes is 1.0393274697 degree celcius\n" + ] + } + ], + "source": [ + "\n", + "d = 8.0 # Average diameter in mm\n", + "r = 750.0 # Density in Kg/m**3\n", + "t = 2.0 # Intermediate temperature in degree celcius\n", + "t_inf = 1.0 # Ambient temperature in degree celcius\n", + "t0 = 25.0 # Initial temperature in degree celcius\n", + "c = 3.35 # Specific heat in kJ/KgK\n", + "h = 5.8 # Heat transfer coeeficient in W/m**2K\n", + "T1 = 10.0 # time period in minutes\n", + "T2 = 30.0 # time period in minutes \n", + "t1 = 5.0 # Intermediate temperature in degree celcius\n", + "print \"\\n Example 18.5\\n\"\n", + "tau1 = c*1e3*math.log((t0-t_inf)/(t-t_inf))/(h*60) # Time to cool down to 2 degree celcius\n", + "tau2 = (t0-t_inf)*(math.exp(-(c*T1*60)/(c*1e3))) # Temperature of peas after 10 minutes\n", + "Y = math.exp(-1*(c*T2*60)/(c*1e3))\n", + "tau3 = (t0*Y-t1)/(Y-1)\n", + "\n", + "print \"\\n Time to cool down to 2 degree celcius is \",tau1 ,\" min\"\n", + "print \"\\n Temperature of peas after 10 minutes is \",tau2 ,\" degree celcius\"\n", + "print \"\\n Temperature of peas after 30 minutes is \",tau3 ,\" degree celcius\"\n", + "#The answers given in book are incorrect\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.6:pg-761" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.6\n", + "\n", + "\n", + " Surface area of heat exchanger is 53.1155468795 m**2\n" + ] + } + ], + "source": [ + "\n", + "mh = 1000 # mass flow rate of hot fluid in Kg/h\n", + "mc = 1000 # mass flow rate of cold fluid in Kg/h\n", + "ch = 2.09 # Specific heat capacity of hot fluid in kJ/kgK\n", + "cc = 4.187 #Specific heat capacity of cold fluid in kJ/kgK \n", + "th1 = 80# Inlet temperature of hot fluid in degree celcius\n", + "th2 = 40 # Exit temperature of hot fluid in degree Celsius\n", + "tc1 = 30 # Inlet temperature of cold fluid in degree Celsius\n", + "U = 24 # heat transfer coefficient in W/m**2K\n", + "\n", + "print \"\\n Example 18.6\\n\"\n", + "Q = mh*ch*(th1-th2)\n", + "tc2 = Q/(mc*cc) + tc1# outlet temperature of cold fluid\n", + "te = th2-tc1 # Exit end temperature difference in degree Celsius\n", + "ti = th1 - tc2 # Inlet end temperature difference in degree Celsius\n", + "t_lm = (ti-te)/(math.log(ti/te))\n", + "A = Q / (U*t_lm*3.6) # Surface are of heat exchanger\n", + "\n", + "print \"\\n Surface area of heat exchanger is \",A ,\" m**2\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.7:pg-762" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.7\n", + "\n", + "\n", + " Surface area of heat exchanger is 3.52948841744 m**2\n" + ] + } + ], + "source": [ + "\n", + "Hfg = 2257.0 # Latent heat at 100 degree Celsius\n", + "\n", + "ma = 500.0 # mass flow rate of air in Kg/h\n", + "ch = 1.005 # Specific heat capacity of hot air in kJ/kgK\n", + "ta1 = 260.0 # Inlet temperature of hot air in degree Celsius\n", + "ta2 = 150.0 # Inlet temperature of cold air in degree Celsius\n", + "tc1 = 100.0 # Inlet temperature of steam\n", + "tc2 = tc1 # Exit temperature of steam\n", + "U = 46.0 # heat transfer coefficient in W/m**2K\n", + "\n", + "print \"\\n Example 18.7\\n\"\n", + "Q = ma*ch*(ta1-ta2)\n", + "m = Q/Hfg # mass flow rate of steam\n", + "te = ta2-tc1 # Exit end temperature difference in degree Celsius\n", + "ti = ta1 - tc2 # Inlet end temperature difference in degree Celsius\n", + "t_lm = (ti-te)/(math.log(ti/te))\n", + "A = Q / (U*t_lm*3.6) # Surface are of heat exchanger\n", + "\n", + "print \"\\n Surface area of heat exchanger is \",A ,\" m**2\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.8:pg-763" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.8\n", + "\n", + "\n", + " Exit temperature of oil is 90.1251029717 degree celcius\n", + "\n", + " Rate of heat transfer is 1302.7384927 kW\n" + ] + } + ], + "source": [ + "import math\n", + "mh = 20.15 # mass flow rate of hot fluid in Kg/s\n", + "mc = 5.04 # mass flow rate of cold fluid in Kg/h\n", + "ch = 2.094 # Specific heat capacity of hot fluid in kJ/kgK\n", + "cc = 4.2 #Specific heat capacity of cold fluid in kJ/kgK \n", + "th1 = 121# Inlet temperature of hot fluid in degree Celsius\n", + "th2 = 40 # Exit temperature of hot fluid in degree Celsius\n", + "tc1 = 10 # Inlet temperature of cold fluid in degree Celsius\n", + "U = 0.34 # heat transfer coefficient in kW/m**2K\n", + "n = 200 # total number of tubes\n", + "l = 4.87 # length of tube in m\n", + "d = 1.97 # Outer diameter in cm\n", + "print \"\\n Example 18.8\\n\"\n", + "A = math.pi*n*d*1e-2*l # Total surface area\n", + "mc_oil = mh*ch\n", + "mc_water = mc*cc\n", + "c_min = mc_water\n", + "c_max =mc_oil\n", + " \n", + "if (mc_oil<mc_water):\n", + " c_min = mc_oil\n", + " c_max =mc_water\n", + "\n", + "R = c_min/c_max\n", + "NTU = U*A/c_min\n", + "e = (1-math.exp(-1*NTU*(1-R)))/(1-R*math.exp(-1*NTU*(1-R)))\n", + "t_larger = e*(th1-tc1)\n", + "t_water = t_larger \n", + "t_oil = t_water*mc_water/mc_oil\n", + "th2 = th1 - t_oil # Exit temperature of oil\n", + "Q = mh*ch*(th1-th2) # Rate of heat transfer\n", + "\n", + "print \"\\n Exit temperature of oil is \",th2 ,\" degree celcius\"\n", + "print \"\\n Rate of heat transfer is \",Q ,\" kW\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.9:pg-763" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.9\n", + "\n", + "\n", + " Heat transfer coefficient is 4074.68413756 W/m**2K\n", + "\n", + " Rate of heat transfer is 38.4029932568 kW\n" + ] + } + ], + "source": [ + "u_m = 0.8 # mean velocity in m/s\n", + "D = 5 # Diameter in cm\n", + "v = 4.78e-7 # dynamic coefficient of viscosity\n", + "Pr = 2.98 # Prantl number\n", + "K = 0.66 # Thermal conductivity in W/mK\n", + "l = 3 # length of pipe in m\n", + "tw = 70 # Wall temperature\n", + "tf = 50 # mean water temperature\n", + "print \"\\n Example 18.9\\n\"\n", + "Re = u_m*D*1e-2/v # Reynold number\n", + "Nu = 0.023*(Re**0.8)*(Pr**0.4)\n", + "h = K*Nu/(D*1e-2) # Heat transfer coefficient\n", + "A = math.pi*D*1e-2*l # Surface area\n", + "Q = h*A*(tw-tf) # Rate of heat transfer\n", + "print \"\\n Heat transfer coefficient is \",h ,\" W/m**2K\"\n", + "print \"\\n Rate of heat transfer is \",Q/1e3 ,\" kW\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.10:pg-764" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.10\n", + "\n", + "\n", + " Rate of heat dissipation is 31.392 W\n" + ] + } + ], + "source": [ + "\n", + "b = 10 # width of plate in cm\n", + "h = 15 # Height of plate in cm\n", + "hr = 8.72 # Radiative heat transfer coefficient in W/m**2K\n", + "tw = 140 # temperature of wall in degree Celsius\n", + "tf = 20 # Atmospheric temperature in degree Celsius\n", + "v = 2.109e-5 # Coefficient of dynamic viscosity in m**2/s\n", + "Pr = 0.692 # Prantl number\n", + "K = 0.0305 # Thermal conductivity in W/mK\n", + "L = 0.15 # characteristic length in m\n", + "g = 9.81 # Gravitational acceleration in m/s**2\n", + "\n", + "print \"\\n Example 18.10\\n\"\n", + "A = 2*b*1e-2*h*1e-2 # total area of plate\n", + "t_mean = (tw+tf)/2 +273\n", + "B = 1/t_mean\n", + "del_t = tw-tf\n", + "Gr = g*B*del_t*L**3/v**2 # Grashoff number\n", + "x = Gr*Pr\n", + "Nu = 0.59*(Gr*Pr)**0.25\n", + "hc = Nu*K/L\n", + "Q = (hc+hr)*A*del_t # Rate of heat dissipation\n", + "print \"\\n Rate of heat dissipation is \",Q ,\" W\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.11:pg-765" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.11\n", + "\n", + "\n", + " Time required for heating operation is 27.6219838873 s\n" + ] + } + ], + "source": [ + "d1 = 2.0 # Diameter of steel rod in cm\n", + "d2 = 16.0 # Diameter of cylindrical furnace in cm\n", + "e1 = 0.6 # emissivity of inner surface\n", + "e2 = 0.85 # emissivity of rod surface\n", + "T = 1093.0 # Inner surface temperature of furncae in degree celcius\n", + "Tr1 = 427.0 # Initial temperature of rod in degree celcius\n", + "Tr2 = 538.0 # Initial temperature of rod in degree celcius\n", + "sigma = 5.67e-8 # Constant\n", + "rho = 7845.0 # density in kg/ m**3\n", + "c = 0.67 # Specific heat capacity in kJ/kgK\n", + "print \"\\n Example 18.11\\n\"\n", + "A_ratio = d1/d2 # Surface area ratio of cylindrical bodies\n", + "F12 = (1/((1/e1)+(A_ratio*(1/e2 -1))))\n", + "A1 = math.pi*d1*1e-2*1 # Surface area of rod\n", + "T1 = Tr1+273\n", + "T2 = T +273\n", + "T3 = Tr2 +273\n", + "Qi = sigma*A1*F12*(T1**4-T2**4)\n", + "Qe = sigma*A1*F12*(T3**4-T2**4)\n", + "\n", + "Q_avg = abs((Qi+Qe)/2)\n", + "tau = rho*c*(1e-4)*math.pi*(Tr2-Tr1)/(Q_avg*(1e-3))\n", + "\n", + "# Time required for heating operation \n", + "print \"\\n Time required for heating operation is \",tau ,\" s\"\n", + "\n", + "#The answers vary due to round off error\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex18.12:pg-765" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 18.12\n", + "\n", + "\n", + " Net heat transfer between two cylinders is 7297.2729358 W/m length\n", + "\n", + " Example 18.12\n", + "\n", + "\n", + " Net heat transfer between two cylinders is 7297.2729358 W/m length\n" + ] + } + ], + "source": [ + "\n", + "d1 = 10.0 # Diameter of inner cylinder in cm\n", + "d2 = 20.0 # Diameter of outer cylinder in cm\n", + "e1 = 0.65 # emissivity of inner surface\n", + "e2 = 0.4 # emissivity of outer surface\n", + "T1 = 1000.0 # Inner surface temperature in K\n", + "T2 = 500.0 # outer suface temperature in K\n", + "sigma = 5.67e-8 # Constant\n", + "print \"\\n Example 18.12\\n\"\n", + "A1 = math.pi*d1*1e-2\n", + "A2 = math.pi*d2*1e-2\n", + "R =(((1-e1)/(e1*A1))+((1-e2)/(e2*A2))+(1/(A1*1)))\n", + "Eb1 = sigma*T1**4\n", + "Eb2 = sigma*T2**4\n", + "Q = (Eb1-Eb2)/R # Net heat transfer between two cylinders\n", + "print \"\\n Net heat transfer between two cylinders is \",Q ,\" W/m length\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n", + "d1 = 10.0 # Diameter of inner cylinder in cm\n", + "d2 = 20.0 # Diameter of outer cylinder in cm\n", + "e1 = 0.65 # emissivity of inner surface\n", + "e2 = 0.4 # emissivity of outer surface\n", + "T1 = 1000.0 # Inner surface temperature in K\n", + "T2 = 500.0 # outer surface temperature in K\n", + "sigma = 5.67e-8 # Constant\n", + "print \"\\n Example 18.12\\n\"\n", + "A1 = math.pi*d1*1e-2\n", + "A2 = math.pi*d2*1e-2\n", + "R =(((1-e1)/(e1*A1))+((1-e2)/(e2*A2))+(1/(A1*1)))\n", + "Eb1 = sigma*T1**4\n", + "Eb2 = sigma*T2**4\n", + "Q = (Eb1-Eb2)/R # Net heat transfer between two cylinders\n", + "print \"\\n Net heat transfer between two cylinders is \",Q ,\" W/m length\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + } + ], + "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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |