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diff --git a/Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb new file mode 100644 index 00000000..ba532e83 --- /dev/null +++ b/Heat_Transfer_Principles_And_Applications_by_Dutta/ch3.ipynb @@ -0,0 +1,873 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Heat transfer coefficient" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 Page No : 53" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time required for melting the ice is 4274 s\n" + ] + } + ], + "source": [ + "# Variables\n", + "di = 0.06 \t\t\t#m,initial diameter of iceball\n", + "T1 = 30. \t\t\t#C, room temp.\n", + "T2 = 0. \t\t\t#ice ball temp.\n", + "h = 11.4 \t\t\t#W/m**2 C, heat transfer coefficient\n", + "x = 40. \t\t\t#% for reduction\n", + "rho = 929. \t\t\t#kg/m**3, density of ice\n", + "Lv = 3.35*10**5 \t\t\t#j/kg, latent heat of fusion\n", + "\n", + "# Calculations\n", + "# m = 4/3*math.pi*r**3 \t\t\t#kg,mass of ice ball\n", + "#rate of melting = -dm/dt\n", + "#rate of heat adsorption = -4*math.pi*r**2*rho*dr/dt*lamda\n", + "#at initial time t = 0\n", + "C1 = di/2 \t\t\t#consmath.tant of integration\n", + "#if the volume of the ball is reduced by 40% of the original volume \n", + "r = ((1-x/100)*(di/2)**3)**(1./3)\n", + "#time required for melting umath.sing eq. 1\n", + "t = (di/2-r)/(h*(T1-T2)/(rho*Lv))\n", + "\n", + "# Results\n", + "print \"The time required for melting the ice is %.0f s\"%(t)\n", + "\n", + "# note : rounding off error." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 Page No : 54" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time required for the heating coil is 4.9 s\n" + ] + } + ], + "source": [ + "import math\n", + "from scipy.integrate import quad \n", + "#calculate the time required for the heating coil.\n", + "\n", + "# Variables\n", + "P = 1.*10**3 \t\t\t#W, electrical heating capacity\n", + "V = 220. \t\t\t#V, applied voltage\n", + "d = 0.574*10**-3 \t\t\t#m, diameter of wire\n", + "R = 4.167 \t\t\t#ohm, electrical resistance\n", + "Tr = 21. \t\t\t#C, room temp.\n", + "h = 100. \t\t\t#W/m**2 C, heat transfer coefficient\n", + "rho = 8920. \t\t\t#kg/m**3, density of wire\n", + "cp = 384. \t\t\t#j/kg C, specific heat of wire\n", + "percent = 63. \t\t\t#%, percent of the steady state\n", + "\n", + "#Calculation\n", + "R_ = V**2/P \t\t\t#ohm, total electrical resistance\n", + "l = R_/R \t\t\t#m, length of wire\n", + "A = math.pi*d*l \t\t\t#m**2, area of wire\n", + "Tf = P/(h*A)+Tr \t\t\t#final temp.\n", + "dtf = Tf-Tr \t\t\t#C. steady state temp. rise\n", + "#temp. of wire after 63% rise\n", + "T = Tr+(percent/100)*dtf \n", + "#rate of heat accumulation on the wire\n", + "#d/dt(m*cp*T) (1)\n", + "#rate of heat loss\n", + "#h*A*(T-Tr).........................(2)\n", + "#heat balance eq. (1) = (2)\n", + "m = math.pi*d**2*l*rho/4 \t\t\t#kg. mass of wire\n", + "#integrating heat balance eq.\n", + "\n", + "def f6(T): \n", + " return 1/((P/(m*cp))-((h*A)/(m*cp))*(T-Tr))\n", + "\n", + "t = quad(f6,21,322)[0]\n", + "\n", + "# Results\n", + "print \"The time required for the heating coil is %.1f s\"%(t)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3 Page No : 56" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " the heat transfer coefficient is 11.63 W/m**2 C \n", + "So there is no heat flow at other surface of the wall \n", + "average volumetric rate of heat generation is 6396 W/m**3\n" + ] + } + ], + "source": [ + "# Variables\n", + "t = 0.2 \t\t\t#m, thickness of wall\n", + "k = 1.163 \t\t\t#W/m C, thermal conductivity of material\n", + "Ta = 30 \t\t\t#C, ambient temp\n", + "\n", + "# Calculations and Results\n", + "#(a) at x = 0.2 let T = T1 at x = x1\n", + "x1 = 0.2\n", + "T1 = 250-2750*x1**2\n", + "#let D = dT/dx\n", + "D = -5500*0.2 \t\t\t#C/m, at x = 0.2\n", + "h = -k*D/(T1-Ta)\n", + "print \" the heat transfer coefficient is %.2f W/m**2 C \"%(h)\n", + "\n", + "#(b)at other surface of wall, x = 0 = x2 (say)\n", + "x2 = 0\n", + "a = -5500*0\n", + "print \"So there is no heat flow at other surface of the wall \"\n", + "\n", + "#(c)\n", + "A = 1 \t\t\t#m**2, area\n", + "Vw = A*x1 \t\t\t#m**3, volume of wall\n", + "HL = h*(T1-Ta) \t\t\t#W, heat loss from unit area\n", + "Vav = HL/x1\n", + "print \"average volumetric rate of heat generation is %.0f W/m**3\"%(Vav)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 Page No : 61" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The rate of heat loss is 150.9 W\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve \n", + "import math\n", + "# Variables\n", + "id_ = 97.*10**-3 \t\t\t#m,internal diameter of steam pipe\n", + "od = 114.*10**-3 \t\t\t#m,outer diameter of steam pipe\n", + "pr = 30. \t\t\t#bar, absolute pressure os saturated steam\n", + "Ti = 234. \t\t\t#C, temp. at 30 bar absolute pressure\n", + "Ts = 55. \t\t\t#C, skin temp.\n", + "To = 30. \t\t\t#C, ambient temp.\n", + "kc = 0.1 \t\t\t#W/m C, thermal conductivity of wool\n", + "kw = 43. \t\t\t#W/m C, thermal conductivity of pipe\n", + "h = 8. \t\t\t#W/m**2 C, external air film coefficient \n", + "L = 1. \t\t\t#m, assume length\n", + "\n", + "#Calculation\n", + "ri = id_/2 \t\t\t#m, \n", + "r1 = (114.*10**-3)/2 \t\t\t#m,outer radius of steam pipe\n", + "\n", + "#thermal resistance of insulation\n", + "#Ri = math.log(ro/r1)/(2*math.pi*L*kc)\n", + "#Thermal resistance of pipe wall\n", + "Rp = math.log(r1/ri)/(2*math.pi*L*kw)\n", + "#RT = Ri+Rp\n", + "DF = Ti-Ts \t\t\t#C, driving force\n", + "#At steady state the rate of heat flow through the insulation\n", + "# and the outer air film are equ\n", + "\n", + "#by trial and error method :\n", + "def f(ro): \n", + " return (Ti-Ts)/(math.log(ro/r1)/kc+math.log(r1/ri)/kw)-(h*ro*(Ts-To))\n", + "ro = fsolve(f,0.1)\n", + "th = ro-r1 \t\t\t#m, required thickness of insulation\n", + "Q = 2*math.pi*ro*h*L*(Ts-To)\n", + "\n", + "# Results\n", + "print \"The rate of heat loss is %.1f W\"%(Q)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 Page No : 62" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "effective thickness of air is 0.75 mm\n", + "effective thickness of liquid films is 2.6 mm.\n", + "the overall heat transfer coefficient based on i.d of pipe is 2.707 W/m**2 C\n", + "the overall heat transfer coefficient based on od of pipe is 1.025 W/m**2 C\n", + "the percentage of total resistance offered by air film. is 10.25 percent\n", + "Rate of heat loss is 21.2 W\n", + "insulation skin temp.is 32.8 C\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "w1 = 8. \t\t\t#%, solubility of alcohol\n", + "w2 = 92. \t\t\t#%, solubility of water\n", + "k1 = 0.155 \t\t\t#W/m C, thermal conductivity of alcohol\n", + "k2 = 0.67 \t\t\t#W/m C thermal conductivity of water\n", + "ka = 0.0263 \t\t\t#W/m C thermal conductivity of air\n", + "kw = 45. \t\t\t#W/m Cthermal conductivity of pipe wall\n", + "ki = 0.068 \t\t\t#W/m C , thermal cond. of glass\n", + "id_ = 53.*10**-3 \t\t\t#m, internal diameter of pipe\n", + "od = 60.*10**-3 \t\t\t#m, outer diameter of pipe\n", + "t = 0.04 \t\t\t#m, thickness of insulation\n", + "hi = 800. \t\t\t#W/m**2 C, liquid film coefficient\n", + "ho = 10. \t\t\t#W/m**2 C, air film coefficient\n", + "L = 1. \t\t\t#m, length of pipe\n", + "T1 = 75. \t\t\t#C, initial temp.\n", + "T2 = 28. \t\t\t#C, ambient air temp.\n", + "\n", + "# Calculations and Results\n", + "#(a)\n", + "km = (w1/100)*k1+(w2/100)*k2-0.72*(w1/100)*(w2/100)*(-(k1-k2))\n", + "deli = km/hi \t\t\t#m, effective thickness of liquid film\n", + "delo = ka/ho \t\t\t#m, effective thickness of air film\n", + "print \"effective thickness of air is %.2f mm\"%(deli*10**3)\n", + "print \"effective thickness of liquid films is %.1f mm.\"%(delo*10**3)\n", + "\n", + "#(b)\n", + "Ai = 2*math.pi*id_/2*L \t\t\t#m**2, inside area\n", + "ri = id_/2 \t\t\t#m,inside radius of pipe\n", + "r_ = od/2 \t\t\t#m, outside radius of pipe\n", + "ro = r_+t \t \t\t#m, outer radius of insulation\n", + "Ao = 2*math.pi*ro*L \t\t \t#m**2, outer area\n", + "#from eq. 3.11, overall heat transfer coefficient\n", + "Ui = 1/(1/hi+(Ai*math.log(r_/ri))/(2*math.pi*L*kw)+(Ai*math.log(ro/r_))/(2*math.pi*L*ki)+Ai/(Ao*ho))\n", + "print \"the overall heat transfer coefficient based on i.d of pipe is %.3f W/m**2 C\"%(Ui)\n", + "\n", + "#(c)\n", + "#frim eq. 3.14\n", + "Uo = Ui*Ai/Ao \n", + "print \"the overall heat transfer coefficient based on od of pipe is %.3f W/m**2 C\"%(Uo)\n", + "\n", + "#(d)\n", + "R = 1/(Ui*Ai) \t\t\t#C/W, total heat transfer resistance\n", + "Rair = 1/(Ao*ho) \t\t\t#C/W, heat transfer resistance of air film\n", + "p = Rair/R\n", + "print \"the percentage of total resistance offered by air film. is %.2f percent\"%(p*100)\n", + "\n", + "#(e)\n", + "Q = Ui*Ai*(T1-T2)\n", + "print \"Rate of heat loss is %.1f W\"%(Q)\n", + "\n", + "#(f)\n", + "Ts = Uo*Ao*(T1-T2)/(ho*Ao)+T2\n", + "print \"insulation skin temp.is %.1f C\"%(Ts)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 Page No : 64" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Inlet liquid temp. should be 82 C \n", + " the insulation skin temp. at the flat top surface is 35 C \n", + "similarly the insulation skin temp at cylindrical surface is 38 C\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "id_ = 1.5 \t\t\t#m, internal diameter of math.tank\n", + "h = 2.5 \t\t\t#m, height of math.tank\n", + "t1 = 0.006 \t\t\t#m, thickness of wall\n", + "t2 = 0.04 \t\t\t#m, thickness of insulation\n", + "Ta = 25. \t\t\t#C, ambient temp.\n", + "T1 = 80. \t\t\t#C, outlet temp. of liquid\n", + "cp = 2000. \t\t\t#j/kg C, specific heat of liquid\n", + "FR = 700./3600 \t\t\t#KG/s, Liquid flow rate\n", + "\n", + "# Calculations and Results\n", + "ri = id_/2+t1 \t\t\t#m, inner radius of insulation\n", + "ro = ri+t2 \t\t\t#m, outer radius of insulation\n", + "ki = 0.05 \t\t\t#W/m C, thermal conductivity of insulation\n", + "hc = 4 \t\t\t#W/m**2 C, heat transfer coefficient at cylindrical surface\n", + "ht = 5.5 \t\t\t#W/m**2 C, heat transfer coefficient at flat surface\n", + "l = h+t1+t2 \t\t\t#m, height of the top of insulation\n", + "#fromm eq. 3.10\n", + "#heat transfer resistance of cylindrical wall\n", + "Rc = math.log(ro/ri)/(2*math.pi*l*ki)+1/(2*math.pi*ro*l*hc)\n", + "#heat transfer resistance of flat insulated top surface\n", + "Ri = (1/(math.pi*ro**2))*((ro-ri)/ki+1/ht)\n", + "tdf = T1-Ta \t\t\t#C, temp. driving force\n", + "Q = tdf/Rc + tdf/Ri \t\t\t#W, total rate of heat loss\n", + "Tt = Q/(FR*cp)+T1 \t\t\t#C, inlet temp. of liquid\n", + "print \"Inlet liquid temp. should be %.0f C \"%(Tt)\n", + "Q1 = tdf/Ri \t\t\t#W, rate of heat loss from flat surface\n", + "T1 = Q1/(math.pi*ro**2*ht)+Ta \n", + "print \" the insulation skin temp. at the flat top surface is %.0f C \"%(T1)\n", + "#similarly\n", + "T2 = 38\n", + "print \"similarly the insulation skin temp at cylindrical surface is %.0f C\"%(T2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7 Page No : 66" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the heat imput to the boiling.is 191.2 W\n" + ] + } + ], + "source": [ + "import math\n", + "# Variables\n", + "id_ = 2.5*10**-2 \t\t\t#m, internal diameter of glass tube\n", + "t = 0.3*10**-2 \t\t\t#m, thickness of wall\n", + "l = 2.5 \t\t\t#m, length of nichrome wire\n", + "L = 0.12 \t\t\t#m, length of steel covered with heating coil\n", + "Re = 16.7 \t\t\t#ohm, electrical resistance\n", + "ti = 2.5*10**-2 \t\t\t#m, thickness of layer of insulation\n", + "kg = 1.4 \t\t\t#W/m C, thermal conductivity of glass\n", + "ki = 0.041 \t\t\t#W/m C, thermal conductivity of insulation\n", + "T1 = 91. \t\t\t#C, boiling temp. of liquid\n", + "T2 = 27. \t\t\t#C, ambient temp.\n", + "ho = 5.8 \t\t\t#W/m **2 C outside air film coefficient\n", + "V = 90. \t\t\t#V, voltage\n", + "\n", + "#Calculation\n", + "Rc = Re*l \t\t\t#ohm, resistance of heating coil\n", + "Q = V**2/Rc \t\t\t#W, rate of heat generation\n", + "ri = id_/2 \t\t\t#m, inner radius of glass tube\n", + "r_ = ri+t \t\t\t#m, outer radius of glass tube\n", + "ro = r_+ti \t\t\t#m,outer radius of insulation\n", + "#heat transfer resistance of glass wall\n", + "Rg = math.log(r_/ri)/(2*math.pi*L*kg)\n", + "#combined resistance of insulation and outer air film\n", + "Rt = math.log(ro/r_)/(2*math.pi*L*ki)+1/(2*math.pi*ro*L*ho)\n", + "#Rate of heat input to the boiling liquid in steel = Q1 = (Ts-T1)/Rg\n", + "#Rate of heat loss through insulation ,Q2 = (Ts-To)/(Rt)\n", + "#Q1+Q2 = Q\n", + "Ts = (Q+ T1/Rg +T2/Rt)/(1/Rg +1/Rt)\n", + "Q1 = (Ts-T1)/Rg\n", + "Q2 = Q-Q1\n", + "\n", + "# Results\n", + "print \"the heat imput to the boiling.is %.1f W\"%(Q1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.8 Page No : 68" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum allowable current is 54.04 A\n", + "remp. at the centre of wire is 90.005 C\n", + "The temprature at the outer surface of insulation is 80.3 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "ri = 1.3*10**-3 \t\t\t#m, radius of 10 gauge wire\n", + "t = 1.3*10**-3 \t\t\t#m, thickness of rubber insulation\n", + "Ti = 90. \t\t\t#C, temp. 0f insulation\n", + "To = 30. \t\t\t#C, ambient temp.\n", + "h = 15. \t\t\t#W/m**2 C, air film coefficient\n", + "km = 380. \t\t\t#W/m C, thermal cond. of copper\n", + "kc = 0.14 \t\t\t#W/m C, thermal cond. of rubber(insulation)\n", + "Rc = 0.422/100 \t\t\t#ohm/m, eletrical resistance of copper wire\n", + "\n", + "# Calculations and Results\n", + "Tcmax = 90. \t\t\t#X, the maximum temp. in insulation\n", + "ro = ri+t \t\t\t#m, outside radius of 10 gauge wire\n", + "Sv = ((Tcmax-To)*(2*kc/ri**2))/(math.log(ro/ri)+kc/(h*ro))\n", + "I = (math.pi*ri**2*Sv/Rc)**0.5 \t\t\t#A, Current strength\n", + "print \"maximum allowable current is %.2f A\"%(I)\n", + "\n", + "#(b) at r = 0\n", + "Tm = To+(ri**2*Sv/2)*(1/km+(math.log(ro/ri))/kc+1/(h*ro))\n", + "print \"remp. at the centre of wire is %.3f C\"%(Tm)\n", + "\n", + "#at r = ro\n", + "Tc = 30+(ri**2*Sv/(2*kc))*(kc/(h*ro))\n", + "print \"The temprature at the outer surface of insulation is %.1f C\"%(Tc)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9 Page No : 72" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "At x = 0.1 the temp. at the surface of slab A is 430 C\n", + "At x = 0.35 the temp. at the surface of slab A is 318 C\n", + " the maximum Temp. in A occurs at 0.2045 m\n", + " the maximum Temp. in A is 550.2 TAmax \n", + "temp. gradient at interface 2 of the slabs A is 2300 C/W\n", + "temp. gradient at interface 3 of the slabs A is -3200 C/W\n", + "temp. gradient at interface 2 of the slabs B is 3450 C/W\n", + "temp. gradient at interface 1 of the slabs B is 3450 C/W\n", + "temp. gradient at interface 3 of the slabs C is -1600 C/W\n", + "temp. gradient at interface 4 of the slabs C is -1600 C/W\n", + "The heat transfer coefficient at one surface of solid fluid interface is 766.7 W/m**2 C\n", + "The heat transfer coefficient at other surface of solid fluid interface is 1129 W/m**2 C\n" + ] + } + ], + "source": [ + "# Variables\n", + "tA = 0.25 \t\t\t#m, thickness of slab A\n", + "tB = 0.1 \t\t\t#m, thickness of slab B\n", + "tC = 0.15 \t\t\t#m, thickness of slab C\n", + "kA = 15. \t\t\t#W/m C, thermal comductivity of slab A\n", + "kB = 10. \t\t\t#W/m C, thermal comductivity of slab B\n", + "kC = 30. \t\t\t#W/m C, thermal comductivity of slab C\n", + "#Temprature distribution in slab A\n", + "T1 = 40. \t\t\t#C, fluid temp.\n", + "T2 = 35. \t\t\t#C, medium temp.\n", + "\n", + "# Calculations and Results\n", + "#(a)\n", + "x1 = tB \n", + "TA1 = 90.+4500*x1-11000*x1**2\n", + "#similarly at the right surface\n", + "x2 = tA+tB\n", + "TA2 = 90.+4500*x2-11000*x2**2\n", + "#let dTA/dx = D\n", + "D = 0 \t\t\t#for maximum temp.\n", + "x3 = 4500./22000\n", + "TAmax = 90.+4500*x3-11000*x3**2\n", + "print \"At x = 0.1 the temp. at the surface of slab A is %.0f C\"%(TA1)\n", + "print \"At x = 0.35 the temp. at the surface of slab A is %.0f C\"%(TA2)\n", + "print \" the maximum Temp. in A occurs at %.4f m\"%(x3)\n", + "print \" the maximum Temp. in A is %.1f TAmax \"%(TAmax)\n", + "\n", + "#(b)\n", + "#At the interface 2\n", + "D1 = 4500-2.*11000*x1 \t\t\t#C/W, D1 = dTA/dx, at x = 0.1\n", + "#At the interface 3\n", + "D2 = 4500-2.*11000*x2 \t\t\t#D12 = dTA/dx, at x = 0.35\n", + "#Temprature gradient in slab B and C\n", + "#by umath.sing the continuity of heat flux at interface (2)\n", + "D3 = -kA*D1/(-kB) \t\t\t#D3 = dTB/dx, at x = 0.1\n", + "#at interface (1)\n", + "D4 = D3 \t\t\t#D4 = dTB/dx at x = 0\n", + "#similarly \n", + "D5 = -1600. \t\t\t#C/W, dTB/dx, x = 0.35\n", + "D6 = D5 \t\t\t#at interface 4\n", + "print \"temp. gradient at interface 2 of the slabs A is %.0f C/W\"%(D1)\n", + "print \"temp. gradient at interface 3 of the slabs A is %.0f C/W\"%(D2)\n", + "print \"temp. gradient at interface 2 of the slabs B is %.0f C/W\"%(D3)\n", + "print \"temp. gradient at interface 1 of the slabs B is %.0f C/W\"%(D4)\n", + "print \"temp. gradient at interface 3 of the slabs C is %.0f C/W\"%(D5)\n", + "print \"temp. gradient at interface 4 of the slabs C is %.0f C/W\"%(D6)\n", + "\n", + "#(c)\n", + "#from D3 = 3450 and TB = beeta1*x+beeta2\n", + "beeta1 = 3450.\n", + "beeta2 = 85.\n", + "x = 0.\n", + "TB = beeta1*x+beeta2\n", + "#similary\n", + "TC = 877.5-1600*x\n", + "h1 = -kB*D4/(T1-TB)\n", + "#similarly\n", + "h2 = 1129.\n", + "print \"The heat transfer coefficient at one surface of solid fluid interface is %.1f W/m**2 C\"%(h1)\n", + "print \"The heat transfer coefficient at other surface of solid fluid interface is %.0f W/m**2 C\"%(h2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.10 Page No : 79" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the percentage increase in the rate of heat transfer is 103.6 percent \n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "id_ = 78.*10**-3 \t\t\t#m, actual internal dia of pipe\n", + "tw = 5.5*10**-3 \t\t\t#m, wall thickness\n", + "nl = 8. \t\t\t#no. of longitudinal fins\n", + "tf = 1.5*10**-3 \t\t\t#m, thickness of fin\n", + "w = 30.*10**-3 \t\t\t#m,breadth of fin\n", + "kf = 45. \t\t\t#W/m C, thermal conductivity of fin \n", + "Tw = 150. \t\t\t#C, wall temp.\n", + "To = 28. \t\t\t#C, ambient temp.\n", + "h = 75. \t\t\t#W/m**2C, surface heat transfer coefficient\n", + "\n", + "#Calculation\n", + "#from eq. 3.27\n", + "e = math.sqrt(2*h/(kf*tf)) \n", + "n = (1./(e*w))*math.tanh(e*w) \t\t\t#efficiency of fin\n", + "L = 1. \t\t\t#m, length of fin\n", + "Af = 2.*L*w \t\t\t#m**2, area of math.single fin\n", + "Atf = nl*Af \t\t\t#m**2 total area of fin\n", + "Qmax = h*Atf*(Tw-To) \t\t\t#W, maximum rate of heat transfer\n", + "Qa = n*Qmax \t\t\t#W, actual rate of heat transfer\n", + "Afw = L*tf \t\t\t#m**2, area of contact of fin with pipe wall\n", + "Atfw = Afw*nl \t\t\t#m**2 , area of contact of all fin with pipe wall\n", + "ro = id_/2+tw \t\t\t#m, outer pipe radius\n", + "A = 2*math.pi*L*ro \t\t\t#m**2 area per meter\n", + "Afree = A-Atfw \t\t\t#m**2, free outside area of finned pipe\n", + "#Rate of heat transfer from free area of pipe wall\n", + "Q1 = h*Afree*(Tw-To) \t\t\t#W, \n", + "#total rate of hewat gtransfer from finned pipe\n", + "Qtotal = Qa+Q1 \t\t\t#W\n", + "#Rate of heat transfer fromm unfinned pipe\n", + "Q2 = h*A*(Tw-To)\n", + "per = (Qtotal-Q2)/Q2\n", + "\n", + "# Results\n", + "print \"the percentage increase in the rate of heat transfer is %.1f percent \"%(per*100)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.11 Page No : 80" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer in the absence of contact resistance is 11.585 KW\n", + "The actual rate of heat loss is 5.18kW is much less than this value. So there is a thermal contact resistance at the interface between the layers \n", + "The contact resistance is 0.001067 C/W \n", + "contact heat transfer coefficient is 298.2 W/m**2 C \n", + "The temprature jump is 5.5 C\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "# Variables\n", + "id_ = 90.*10**-2 \t\t\t#m, internal diameter of steel\n", + "od = 110.*10**-2 \t\t\t#m, outer diameter of steel\n", + "Ti = 180. \t\t\t#C, inside temp. of steel\n", + "To = 170. \t\t\t#C, outside temp. of steel\n", + "k = 37. \t\t\t#W/m C, thermal conductivity of alloy\n", + "Q = 5.18*10**3 \t\t\t#W, Rate of heat loss\n", + "\n", + "# Calculations and Results\n", + "ri = id_/2 \t\t\t#m, inside radius of shell\n", + "ro = od/2 \t\t\t#m, outside radius of shell\n", + "r_ = 0.5 \t\t\t#m, boundary between the layers\n", + "L = 1 \t\t\t#m, length of shell\n", + "#Rate of heat transfer in the absence of contact resistance\n", + "Q1 = 2*math.pi*L*k*(Ti-To)/(math.log(ro/ri)) \n", + "print \"Rate of heat transfer in the absence of contact resistance is %.3f KW\"%(Q1/1000)\n", + "print \"The actual rate of heat loss is 5.18kW is much less than this value\\\n", + ". So there is a thermal contact resistance at the interface between the layers \"\n", + "\n", + "#(b)\n", + "Ri = (math.log(r_/ri)/(2*math.pi*L*k)) \t\t\t#C/W, resistance of inner layer\n", + "Ro = (math.log(ro/r_)/(2*math.pi*L*k)) \t\t\t#C/W, resistance of outer layer\n", + "Rc = ((Ti-To)/(Q))-(Ri+Ro) \t\t\t#C/W, contact resistance\n", + "print \"The contact resistance is %f C/W \"%(Rc)\n", + "Ac = 2*math.pi*L*r_ \t\t\t#m**2, area of contact surface of shell\n", + "hc = 1/(Ac*Rc) \t\t\t #W/m**2 c, contact heat transfer coefficient\n", + "print \"contact heat transfer coefficient is %.1f W/m**2 C \"%(hc)\n", + "\n", + "#(c)\n", + "dt = Q/(hc*Ac)\n", + "print \"The temprature jump is %.1f C\"%(dt)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.12 Page No : 84" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the critical thickness is 35.29 mm\n" + ] + } + ], + "source": [ + "# Variables\n", + "d = 5.2*10**-3 \t\t\t#m, diameter of copper wire\n", + "ri = d/2 \t\t\t#inner radius of insulation\n", + "kc = 0.43 \t\t\t#W/m C, thermal conductivity of PVC\n", + "Tw = 60. \t\t\t#C, temp. 0f wire\n", + "h = 11.35 \t\t\t#W/m**2 C, film coefficient\n", + "To = 21. \t\t\t#C, ambient temp.\n", + "\n", + "#calculation\n", + "Ro = kc/h \t\t\t#m,critical outer radius of insulation\n", + "t = Ro-ri\n", + "\n", + "# Results\n", + "print \"the critical thickness is %.2f mm\"%(t*10**3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.13 Page No : 85" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ro = 3.5 cm \n", + "Radius of bare pipe is larger than outer radius of insulation So critical insulation thickness does not exist \n" + ] + } + ], + "source": [ + "# calculate the critical insulation thickness.\n", + "\n", + "# Variables\n", + "d = 15.*10**-2 \t\t\t#m, length of steam main\n", + "t = 10.*10**-2 \t\t\t#m, thickness of insulation\n", + "ki = 0.035 \t\t\t#W/m C, thermal conductivity of insulation\n", + "h = 10. \t\t\t#W/m**2 C, heat transfer coefficient\n", + "\n", + "#calculation\n", + "#from eq. 3.29\n", + "ro = ki/h\n", + "\n", + "# Results\n", + "print \"ro = %.1f cm \"%(ro*10**3)\n", + "print \"Radius of bare pipe is larger than outer radius of insulation So critical \\\n", + " insulation thickness does not exist \"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.14 Page No : 87" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The optimum insulation thickness is 71 mm\n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve\n", + "import math\n", + "\n", + "# Variables\n", + "Ti = 172. \t\t\t#C, saturation temp.\n", + "To = 20. \t\t\t#C, ambient temp.\n", + "Cs = 700. \t\t\t#per ton, math.cost of steam\n", + "Lv = 487. \t\t\t#kcal/kg, latent heat of steam\n", + "ho = 10.32 \t\t\t#kcal/h m**2 C, outer heat transfer coefficient\n", + "kc = 0.031 \t\t\t#W/m C, thermal conductivity of insulation\n", + "n = 5. \t\t\t#yr, service life of insulation\n", + "i = 0.18 \t\t\t#Re/(yr)(Re), interest rate\n", + "\n", + "#Calculation\n", + "di = 0.168 \t\t\t#m, inner diameter of insulation\n", + "#Cost of insulation\n", + "Ci = 17360.-(1.91*10**4)*di \t\t\t#Rs/m**3\n", + "Ch = Cs/(1000*Lv) \t\t\t#Rs/cal, math.cost of heat energy in steam\n", + "sm = 1./(1+i)+1/(1+i)**2+1/(1+i)**3+1/(1+i)**4+1/(1+i)**n\n", + "#from eq. 3.33\n", + "ri = di/2 \t\t\t#m inner radius of insulation\n", + "L = 1 \t\t\t#m, length of pipe\n", + "#Pt = Ch*sm*2*math.pi*ri*L*( 1/(((ri/kc)*('math.log(ro/ri)'))+ri/(ho*ro)))*7.2*10**3*(Ti-To)+math.pi*(ro**2-ri**2)*L*Ci\n", + "#On differentiating , dpt/dro = -957.7*((1/ro)-(0.003/ro**2))/(math.log(ro)+(0.003/ro)+2.477)**2\n", + "def f(ro): \n", + " return -957.7*((1/ro)-(0.003/ro**2))/(math.log(ro)+(0.003/ro)+2.477)**2+98960*ro\n", + "ro = fsolve(f,0.1)\n", + "t = ro-ri\n", + "\n", + "# Results\n", + "print \"The optimum insulation thickness is %.0f mm\"%(t*1000)\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": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |