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-rw-r--r--Engineering_Heat_Transfer/CHAPTER2.ipynb63
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diff --git a/Engineering_Heat_Transfer/CHAPTER2.ipynb b/Engineering_Heat_Transfer/CHAPTER2.ipynb
index 556700c2..6e5857d6 100644
--- a/Engineering_Heat_Transfer/CHAPTER2.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER2.ipynb
@@ -27,13 +27,10 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#determination of the heat flow through a composite wall\n",
"\n",
- "#Given\n",
"T3=-10.00\t\t# temperature of inside wall in degree Fahrenheit\n",
"T0=70.0 \t \t# temperature of outside wall in degree Fahrenheit\n",
"dT=T0-T3 # overall temperature difference\n",
- "# values of thermal conductivity in BTU/(hr.ft.degree Rankine) from appendix table B3\n",
"k1=0.38 # brick masonry\n",
"k2=0.02 # glass fibre\n",
"k3=0.063 # plywood\n",
@@ -42,13 +39,11 @@
"dx3=0.5/12.0 # thickness of plywood layer in ft\n",
"A=1.0 # cross sectional area taken as 1 ft**2\n",
"\n",
- "#Calculation\n",
"R1=dx1/(k1*A) # resistance of brick layer in (hr.degree Rankine)/BTU\n",
"R2=dx2/(k2*A) # resistance of glass fibre layer in (hr.degree Rankine)/BTU\n",
"R3=dx3/(k3*A) # resistance of plywood layer in (hr.degree Rankine)/BTU\n",
"qx=(T0-T3)/(R1+R2+R3) \n",
"\n",
- "#Result\n",
"print\"Resistance of brick layer is \",round(R1,3),\"(hr.degree Rankine)/BTU\"\n",
"print\"Resistance of glass fibre layer is \",round(R2,1),\"(hr.degree Rankine)/BTU\"\n",
"print\"Resistance of plywood layer is \",round(R3,3),\"(hr.degree Rankine)/BTU\"\n",
@@ -82,15 +77,11 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of heat transfer through composite wall for materials in parallel\n",
"\n",
- "#Given\n",
- "# values of thermal conductivities in W/(m.K) from appendix table B3\n",
"k1=0.45 # thermal conductivity of brick\n",
"k2a=0.15 # thermal conductivity of pine\n",
"k3=0.814 # thermal conductivity of plaster board\n",
"k2b=0.025 # thermal conductivity of air from appendix table D1\n",
- "# Areas needed fpor evaluating heat transfer in sq.m\n",
"A1=0.41*3 # cross sectional area of brick layer \n",
"A2a=0.038*3 # cross sectional area of wall stud\n",
"A2b=(41-3.8)*0.01*3 # cross sectional area of air layer\n",
@@ -99,7 +90,6 @@
"dx2=0.089 # thickness of wall stud and air layer in m\n",
"dx3=0.013 # thickness of plastic layer in m\n",
"\n",
- "#Calculation\n",
"R1=dx1/(k1*A1) # Resistance of brick layer in K/W\n",
"R2=dx2/(k2a*A2a+k2b*A2b) # Resistance of wall stud and air layer in K/W\n",
"R3=dx3/(k3*A3) # Resistance of plastic layer in K/W\n",
@@ -107,7 +97,6 @@
"T0=0 # temperature of outside wall in degree celsius\n",
"qx=(T1-T0)/(R1+R2+R3) # heat transfer through the composite wall in W\n",
"\n",
- "#result\n",
"print\"Resistance of brick layer is \",round(R1,3),\"k/W\"\n",
"print\"Resistance of wall stud and air layer is \",round(R2,2),\"k/W\"\n",
"print\"Resistance of plastic layer is \",round(R3,3),\"k/W\"\n",
@@ -141,9 +130,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of heat transfer rate and overall heat transfer coefficient\n",
"\n",
- "#Given\n",
"k1=24.8 # thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2 \n",
"k2=0.02 # thermal conductivity of styrofoam steel in BTU/(hr.ft.degree Rankine)\n",
"k3=0.09 # thermal conductivity of fibreglass in BTU/(hr.ft.degree Rankine)\n",
@@ -154,9 +141,7 @@
"dx2=0.75/12.0 # thickness of styrofoam in ft\n",
"dx3=0.25/12.0 # thickness of fiberglass in ft\n",
"\n",
- "#Calculation \n",
"\n",
- "# Resistances in (degree Fahrenheit.hr)/BTU\n",
"Rc1=1/(hc1*A) # Resistance from air to sheet metal\n",
"Rk1=dx1/(k1*A) # Resistance of steel layer\n",
"Rk2=dx2/(k2*A) # Resistance of styrofoam layer\n",
@@ -167,7 +152,6 @@
"T_inf2=32 # temperature of mixture of ice and water in degree F\n",
"q=U*A*(T_inf1-T_inf2)\n",
"\n",
- "#result\n",
"print\"Resistance from air to sheet metal: \",round(Rc1,3),\"degree F.hr/BTU\"\n",
"print\"Resistance of steel layer is \",round(Rk1,4),\"degree F.hr/BTU\"\n",
"print\"Resistance of styrofoam layer is \",round(Rk2,3),\"degree F.hr/BTU\"\n",
@@ -208,21 +192,16 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of the heat transfer through the pipe wall per unit length of pipe.\n",
"\n",
- "#given\n",
"k=14.4 # thermal conductivity of 304 stainless steel in W/(m.K) from appendix table B2\n",
- "# dimensions of steel pipes in cm from appendix table F1\n",
"D2=32.39 #Diameter (cm)\n",
"D1=29.53\n",
"T1=40 #Temprature\n",
"T2=38\n",
"\n",
- "#Calculation\n",
"import math\n",
"Qr_per_length=(2*3.14*k)*(T1-T2)/math.log(D2/D1)#format(6)\n",
"\n",
- "#Result\n",
"print\"The heat transfer through the pipe wall per unit length of pipe is \",round(Qr_per_length/1000,2),\"kw/m\"\n"
],
"language": "python",
@@ -250,12 +229,9 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of the heat gain per unit length\n",
"\n",
- "#Given\n",
"k1=231 # thermal conductivity of copper in BTU/(hr.ft.degree Rankine)from appendix table B1 \n",
"k2=0.02 # thermal conductivity of insuLtion in BTU/(hr.ft.degree Rankine)\n",
- "# Specifications of 1 standard type M copper tubing from appendix table F2 are as follows\n",
"D2=1.125/12 # outer diameter in ft\n",
"D1=0.08792 # inner diameter in ft\n",
"R2=D2/2 # outer radius\n",
@@ -268,7 +244,6 @@
"T3=70 # temperature of surface temperature of insulation degree fahrenheit\n",
"q_per_L=(T1-T3)/(LRk1+LRk2) # heat transferred per unit length in BTU/(hr.ft)\n",
"\n",
- "#Result\n",
"print\"The heat transferred per unit length is \",round(q_per_L,2),\" BTU/(hr.ft\"\n"
],
"language": "python",
@@ -296,12 +271,9 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the overall heat transfer coefficient\n",
"\n",
- "#Given data\n",
"k12=24.8 # thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2 \n",
"k23=.023 # thermal conductivity of glass wool insulation in BTU/(hr.ft.degree Rankine)from appendix table B3 \n",
- "# Specifications of 6 nominal, schedule 40 pipe (no schedule was specified, so the standard is assumed) from appendix table F1 are as follows\n",
"D2=6.625/12.0 # outer diameter in ft\n",
"D1=0.5054 # inner diameter in ft\n",
"t=2/12.0 # wall thickness of insulation in ft\n",
@@ -309,10 +281,8 @@
"hc1=12 # convection coefficient between the air and the pipe wall in BTU/(hr. sq.ft.degree Rankine).\n",
"hc2=1.5 # convection coefficient between the glass wool and the ambient air in BTU/(hr. sq.ft.degree Rankine).\n",
"\n",
- "#calculation\n",
"U=1/((1/hc1)+(D1*log(D2/D1)/k12)+(D1*log(D3/D2)/k23)+(D1/(hc2*D3)))\n",
"\n",
- "#Result\n",
"print\"Overall heat transfer coefficient is \",round(U,4),\" BTU/(hr.sq.ft. Fahrenheit)\"\n"
],
"language": "python",
@@ -340,9 +310,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the thermal contact resistance\n",
"\n",
- "#Given\n",
"k=14.4 # thermal conductivity of 304 stainless steel in W/(m.K)from appendix table B2 \n",
"T1=543.0 # temperature in K at point 1\n",
"T2=460.0 # temperature in K at point 2\n",
@@ -360,14 +328,12 @@
"T7=349 # temperature in K at point 7\n",
"T8=337 # temperature in K at point 8\n",
"\n",
- "#Calculation\n",
"qz_per_A=k*dT/dz12 # heat flow calculated in W/m**2 calculated using Fourier's law\n",
"T_ial=T5-(dz5i*(T5-T6)/dz56) # temperature of aluminium interface in K\n",
"T_img=dzi8*(T7-T8)/dz78+T8 # temperature of magnesium interface in K\n",
"T_img_=355.8 #Approx value in the book\n",
"Rtc=(T_ial-T_img_)/(qz_per_A)\n",
"\n",
- "#Result\n",
"\n",
"print\"The required thermal contact resistance is\",round(Rtc,7),\"K sq.m/W\"\n"
],
@@ -406,36 +372,26 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of temperature profile, heat transferred, efficiency, effectiveness.\n",
"\n",
- "#Given\n",
"import math\n",
"k=24.8 # thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2\n",
"D=(5.0/16.0)/12.0 # diameter of the rod in ft\n",
"P=(math.pi*D) # Circumference of the rod in ft\n",
- "#print\"The perimeter is %.4f ft\",P)\n",
"A=(math.pi/4)*D**2 # Cross sectional area of the rod in sq.ft\n",
- "#print\"The Cross sectional area is %.6f sq.ft\",A)\n",
"hc=1.0 # assuming the convective heat transfer coefficient as 1 BTU/(hr. sq.ft. degree Rankine)\n",
"\n",
- "#Calculation\n",
"m=math.sqrt(hc*P/(k*A))\n",
- "#print\"The value of parameter m is: /ft\",m)\n",
"L=(9/2.0)/12.0 # length of rod in ft\n",
- "# using the equation (T-T_inf)/(T_w-T_inf)=(cosh[m(L-z)])/(cosh(mL)) for temperature profile\n",
"T_inf=70.0\n",
"T_w=200.0\n",
"dT=T_w-T_inf\n",
"const=dT/math.cosh(m*L)\n",
"\n",
- "#(b) the heat transferred can be calculated using the equation qz=k*A*m*(T_w-T_inf)*tanh(m*L)\n",
"qz=k*A*m*dT*tanh(m*L)\n",
"\n",
- "#(c)\n",
"mL=m*L\n",
"efficiency=0.78 # from fig. 2.30\n",
"\n",
- "#(d)\n",
"effectiveness=math.sqrt(k*P/(hc*A))*tanh(mL)\n",
"\n",
"\n",
@@ -444,7 +400,6 @@
"print\"(d)The effectiveness is found to be\",round(effectiveness,1)\n",
"\n",
"\n",
- "#Plot\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -534,9 +489,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of heat transferred\n",
"\n",
- "#Given\n",
"k=136.0 # thermal conductivity of aluminium in BTU/(hr.ft.degree Rankine)from appendix table B1\n",
"L=9/(8*12.0) #length in ft\n",
"W=9/(4*12.0) #width in ft\n",
@@ -545,7 +498,6 @@
"T_w=1000.0 # the root temperature in degree fahrenheit\n",
"T_inf=90.0 # the ambient temperature in degree fahrenheit\n",
"\n",
- "#Calculation\n",
"import math\n",
"m=math.sqrt(hc/(k*delta))\n",
"P=2*W\n",
@@ -555,7 +507,6 @@
"Lc=L+delta\n",
"qz3=k*A*m*(T_w-T_inf)*math.tanh(m*L*(1+delta/Lc))\n",
"\n",
- "#Result\n",
"print\"(a)The heat transferred is \",round(qz1,2),\"BTU/hr\"\n",
"print\"(b)The heat transferred is \",round(qz2,2),\"BTU/hr In the book the answer is incorrect\"\n",
"print\"(c)The heat transferred is \",round(qz3,2),\" BTU/hr\"\n"
@@ -587,20 +538,15 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of optimum fin length and heat transferred by fin\n",
"\n",
- "#Given\n",
"k=8.32 # thermal conductivity BTU/(hr.ft.degree Rankine)\n",
"hc=400.0 # the convective heat transfer coefficient given in BTU/(hr.ft**2. degree Rankine)\n",
"\n",
- "#Calculation\n",
"import math\n",
"delta_opt=0.55/(12*2)\n",
- "# determination of dimension of one fin using the equation delta_opt=0.583*hc*Lc**2/k\n",
"Lc=math.sqrt(delta_opt*k/(0.583*hc))\n",
"\n",
"A=Lc*delta_opt\n",
- "# determination of parameter for finding out efficiency from graph\n",
"parameter=Lc**1.5*math.sqrt(hc/(k*A))\n",
"efficiency=0.6\n",
"W=1/(2.0*12.0) # width in ft\n",
@@ -610,7 +556,6 @@
"delta=W/2.0 \n",
"q_ac=efficiency*hc*2*W*math.sqrt(L**2+delta**2)*(T_w-T_inf)\n",
"\n",
- "#Result\n",
"print\"(a)The optimum length is \",round(Lc*12,2),\"inch\"\n",
"print\"(b)The actual heat transferred is \",round(q_ac,2),\"BTU/hr. NOTE: In the book answer is incorrect\"\n"
],
@@ -640,10 +585,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of heat transferred and fin effectiveness\n",
"\n",
- "#Given\n",
- "#parameters of the problem are\n",
"N=9 # number of fins\n",
"delta=0.003/2.0 \n",
"L=0.025\n",
@@ -656,9 +598,7 @@
"hc=15 \n",
"k=52 # thermal conductivity of cast iron in W/(m.K)from appendix table B2\n",
"\n",
- "#calculation\n",
"import math\n",
- "#(a)\n",
"Ap=2*delta*Lc\n",
"As=2*math.pi*(R2c**2-R1**2)\n",
"radius_ratio=R2c/R1 # for finding efficiency from figure 2.38\n",
@@ -670,15 +610,12 @@
"qw=hc*Asw*(T_w-T_inf)\n",
"q=qf+qw\n",
"\n",
- "#(b)H=N*(Sp+2*delta) # height of cylinder\n",
"H=N*(Sp+2*delta)\n",
"Aso=2*math.pi*R1*H # surface area without fins\n",
"qo=hc*Aso*(T_w-T_inf)\n",
"\n",
- "#(c)\n",
"effectiveness=q/qo # effectiveness defined as ratio of heat transferred with fins to heat transferred without fins\n",
"\n",
- "#Result\n",
"print\"(a)The total heat transferred from the cylinder is \",round(q,0),\"W\"\n",
"print\"(b)The Heat transferred without fins is W\",round(qo,0),\"W\"\n",
"print\"(c)The fin effectiveness is \",round(effectiveness,2)\n"
generated by cgit (git 2.34.1) at 2025-06-06 21:40:29 +0000