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authordebashisdeb2014-06-20 15:42:42 +0530
committerdebashisdeb2014-06-20 15:42:42 +0530
commit83c1bfceb1b681b4bb7253b47491be2d8b2014a1 (patch)
treef54eab21dd3d725d64a495fcd47c00d37abed004 /Engineering_Heat_Transfer
parenta78126bbe4443e9526a64df9d8245c4af8843044 (diff)
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removing problem statements
Diffstat (limited to 'Engineering_Heat_Transfer')
-rw-r--r--Engineering_Heat_Transfer/CHAPTER1.ipynb41
-rw-r--r--Engineering_Heat_Transfer/CHAPTER10.ipynb19
-rw-r--r--Engineering_Heat_Transfer/CHAPTER11.ipynb33
-rw-r--r--Engineering_Heat_Transfer/CHAPTER12.ipynb39
-rw-r--r--Engineering_Heat_Transfer/CHAPTER2.ipynb63
-rw-r--r--Engineering_Heat_Transfer/CHAPTER4.ipynb60
-rw-r--r--Engineering_Heat_Transfer/CHAPTER5.ipynb11
-rw-r--r--Engineering_Heat_Transfer/CHAPTER6.ipynb59
-rw-r--r--Engineering_Heat_Transfer/CHAPTER7.ipynb63
-rw-r--r--Engineering_Heat_Transfer/CHAPTER8.ipynb46
-rw-r--r--Engineering_Heat_Transfer/CHAPTER9.ipynb89
-rw-r--r--Engineering_Heat_Transfer/CHAPTER_3.ipynb38
12 files changed, 0 insertions, 561 deletions
diff --git a/Engineering_Heat_Transfer/CHAPTER1.ipynb b/Engineering_Heat_Transfer/CHAPTER1.ipynb
index d020ad30..9b33caee 100644
--- a/Engineering_Heat_Transfer/CHAPTER1.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER1.ipynb
@@ -27,23 +27,16 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of surface temperature on one side of a firewall\n",
"\n",
- "#Given\n",
"k=9.4 # thermal conductivity in [BTU/hr.ft. \u02daRankine]\n",
"q=6.3 # heat flux in [BTU/s. sq.ft]\n",
"T1=350 # the outside surface temperature of one aide of the wall [\u02daF]\n",
"\n",
- "#Calculation\n",
- "# converting heat flux into BTU/hr sq.ft\n",
"Q=6.3*3600 # [BTU/hr.sq.ft]\n",
"dx=0.5 # thickness in [inch]\n",
- "#converting distance into ft\n",
"Dx=0.5/12.0 # thickness in [ft]\n",
- "# solving for temeprature T2\n",
"T2=T1-(Q*Dx/k) # [\u02daF]\n",
"\n",
- "#Result\n",
"print\"The required temperature on the other side of the firewall is \",round(T2,1),\"F\"\n"
],
"language": "python",
@@ -71,19 +64,15 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of thermal conductivity of aluminium\n",
"\n",
- "#Given\n",
"k_ss=14.4 # thermal conductivity of stainless steel in [W/m.K]\n",
"dt_ss=40 # [K]\n",
"dt_al=8.65 # [K]\n",
"dz_ss=1 # [cm]\n",
"dz_al=3 # [cm]\n",
"\n",
- "#calculation\n",
"k_al=k_ss*dt_ss*dz_al/(dt_al*dz_ss);# thermal conductivity of Al in [W/m.K]\n",
"\n",
- "#result\n",
"print\"The thermal conductivity of aluminium is\",round(k_al,0),\"W/m.K\"\n"
],
"language": "python",
@@ -111,19 +100,15 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of heat transferred by convection\n",
"\n",
- "#Given\n",
"h_c=3 # convective coefficient in [BTU/hr.ft**2\n",
"A=30*18 # Cross sectional area in ft**2\n",
"T_w=140 # Roof surface temperature in degree Fahrenheit\n",
"T_inf=85 # Ambient temperature in degree Fahrenheit\n",
"\n",
- "#Calculation\n",
"dT= (T_w-T_inf)\n",
"Q_c=h_c*A*dT # Convective heat transfer in BTU/hr\n",
"\n",
- "#Result\n",
"print\"The heat transferred by convection is\",round(Q_c,2),\"BTU/hr\"\n"
],
"language": "python",
@@ -151,9 +136,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determining average film conductance\n",
"\n",
- "#Given\n",
"D=0.0243 # diameter in meter\n",
"L=0.2 # length in meter\n",
"A=3.14*D*L # cross-sectional area in sq.m\n",
@@ -164,12 +147,9 @@
"Q=500.0 # volumetric flow rate in cc/s\n",
"density=1000 # density of water in kg/cu.m\n",
"\n",
- "#calculation\n",
"m=Q*density/10**6 # mass flowa rate in kg/s\n",
- "# using definition of specific heat and Newton's law of cooling\n",
"hc=m*cp*(T_b2-T_in)/(A*(T_w-T_in))\n",
"\n",
- "#result\n",
"print\"The average film conductance is \",round(hc,0),\"W/sq.m. K\"\n"
],
"language": "python",
@@ -197,9 +177,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of heat loss rate by radiation\n",
"\n",
- "#Given\n",
"W=14 # width in ft\n",
"L=30.0 # length in ft\n",
"A=W*L # area in ft**2\n",
@@ -207,12 +185,10 @@
"T1=120+460 # driveway surface temperature in degree Rankine\n",
"T2=0 # space temperature assumed to be 0 degree Rankine\n",
"\n",
- "#Calculation\n",
"sigma=0.1714*10**(-8) # value of Stefan-Boltzmann's constant in BTU/(hr.ft**2.(degree Rankine)**4)\n",
"e=0.9 # surface emissivity\n",
"q=sigma*A*e*F_12*((T1)**4-(T2)**4);\n",
"\n",
- "#result\n",
"print\"The heat loss rate by radiation is \",round(q,0),\"BTU/hr\"\n"
],
"language": "python",
@@ -240,18 +216,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of radiation thermal conductance\n",
"\n",
- "#Given\n",
"A=420.0 # area in sq.ft\n",
"T1=580.0 # driveway surface temperature in degree Rankine\n",
"T2=0 # surface temperature assumed to be 0 degree Rankine\n",
"Qr=73320 # heat loss rate in BTU/hr\n",
"\n",
- "#calculation\n",
"hr=Qr/(A*(T1-T2)) # radiation thermal conductance in BTU/(hr.ft**2.(degree Rankine)\n",
"\n",
- "#result\n",
"print\"the radiation thermal conductance is \",round(hr,2),\"BTU/(hr. sq.ft R)\"\n"
],
"language": "python",
@@ -289,18 +261,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Identification of all resistances and their values\n",
- "# Estimation of heat transfer per unit area\n",
- "# Determination of the inside and outside wall temperatures\n",
"\n",
- "#Given\n",
"A=1.0 # assuming A=1 m**2 for convenience\n",
"hc1_avg=15.0 # taking average of extreme values for hc [W/m**2.K]\n",
"k=(0.38+0.52)/2.0 # thermal conductivity of common brick in W/M.k\n",
"L=0.1 #10 cm converted into m\n",
"Rk=(L/(k*A)) # resistance of construction material, assume common brick\n",
"\n",
- "#calcultion\n",
"T_inf1=1000.0 # temperature of exhaust gases in K\n",
"T_inf2=283.0 # temperature of ambient air in K\n",
"Rcl=1/(hc1_avg*A) # resistance on left side of wall [K/W]\n",
@@ -309,7 +276,6 @@
"T_in=T_inf1-Rcl*q #inlet temprature \n",
"T_out=T_inf2+Rc2*q\n",
"\n",
- "#result\n",
"print\"(b)\"\n",
"print\"The resistance on left side of wall is \",round(Rcl,2),\"K/W\"\n",
"print\"The resistance of construction material of wall is\",round(Rk,2),\"K/W\"\n",
@@ -319,7 +285,6 @@
"print\"The inside wall temperature is \",round(T_in,0),\"K\"\n",
"print\"The outside wall temperature is\",round(T_out,1),\"K\"\n",
"\n",
- "#Plot\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -408,9 +373,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of surface temperature\n",
"\n",
- "#Given\n",
"k=0.604 # [BTU/(hr.ft.degree Rankine)]\n",
"hc=3.0 # average value for natural convection in BTU/(hr.ft**2.degree Rankine)\n",
"ew=0.93 \n",
@@ -421,12 +384,9 @@
"T_inf=20+460 # temperature of ambient air in degree Rankine\n",
"T_r=0 # assuming space temperature to be 0 degree Rankine\n",
"\n",
- "#Calculation\n",
- "# LHS of the form a*Tw+b*Tw**4=c\n",
"a=((k/L)+hc) #Coefficient of Tw in the equation\n",
"b=(sigma*ew*f_wr) #Coefficient of Tw**4 in the equation\n",
"c=(k*T1/L)+(hc*T_inf)+(sigma*f_wr*ew*T_r**4) #right hans side of the equation\n",
- "#Soving by try and error\n",
"Tw1=470 #assumed first value of temprature\n",
"LHS1=a*Tw1+b*Tw1**4\n",
"Tw2=480 #assumed 2nd value of temprature\n",
@@ -438,7 +398,6 @@
"Tw5=484.5 #assumed fifth value of temprature\n",
"LHS5=a*Tw5+b*Tw5**4\n",
"\n",
- "#result\n",
"print\"RHS\",round(c,1)\n",
"print\"LHS at surface Temprature 1=\",round(LHS1,1)\n",
"print\"LHS at surface Temprature 2=\",round(LHS2,1)\n",
diff --git a/Engineering_Heat_Transfer/CHAPTER10.ipynb b/Engineering_Heat_Transfer/CHAPTER10.ipynb
index f7b3f4b6..b9bac11c 100644
--- a/Engineering_Heat_Transfer/CHAPTER10.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER10.ipynb
@@ -37,10 +37,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Calculation of the heat-transfer rate and the amount of steam condensed\n",
"\n",
- "#Given\n",
- "# properties of engine oil at (328 + 325)/2 = 326.5 degree F = 320\u00b0F from appendix table C11\n",
"rou_f=0.909*62.4 # density in lbm/ft**3 \n",
"cp=1.037 # specific heat BTU/(lbm-degree Rankine) \n",
"v_f=0.204e-5 # viscosity in ft**2/s \n",
@@ -57,8 +54,6 @@
"W=3.0 # width in ft\n",
"z=0.204*10**-5 # distance from entry of plate in ft\n",
"\n",
- "#Calculation\n",
- "# film thickness is given as follows\n",
"y=((4*kf*v_f*(Tg-Tw)/3600.0)/(rou_f*g*hfg*(1-(rou_v/rou_f))))**(1/4.0) #let y=delta/z**(1/4)\n",
"hz=1665 #From Table 10.1\n",
"hL=(4/3.0)*hz # at plate end\n",
@@ -66,11 +61,9 @@
"q=mf*hfg\n",
"Re=(4*mf/3600)/(W*rou_f*v_f)\n",
"\n",
- "#Result\n",
"print\"The amount of steam condensed is \",round(mf,1),\"lbm/h\"\n",
"print\"The heat transfer rate is \",round(q,0),\"BTU/hr\"\n",
"\n",
- "#plot\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -121,10 +114,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of both the heat that the cooling fluid must remove and the condensation rate. \n",
"\n",
- "#Given\n",
- "# properties of water at (100 + 60)/2 = 80\u00b0C from appendix table C11\n",
"rou_f=974.0 # density in kg/m**3 \n",
"cp_1=4196.0 # specific heat in J/(kg*K) \n",
"v_1=0.364e-6 # viscosity in m**2/s \n",
@@ -139,8 +129,6 @@
"Tw=60\n",
"L=1\n",
"\n",
- "#Calculation\n",
- "# specifications of 1 nominal schedule 40 pipe from appendix F1\n",
"OD=0.03340\n",
"hD=0.782*((g*rou_f*(1-(rou_v/rou_f))*(kf**3)*hfg)/(v_1*OD*(Tg-Tw)))**(1/4.0)\n",
"hD=10720 #According to the book\n",
@@ -148,7 +136,6 @@
"q=hD*math.pi*OD*L*(Tg-Tw)\n",
"mf=q/hfg\n",
"\n",
- "#Result\n",
"print\"The heat flow rate is \",round(q,0),\"W\"\n",
"print\"The rate at which steam condenses is \",round(mf*3600,0),\"kg/hr\"\n"
],
@@ -178,10 +165,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Calculation of (a) the power input to the water for boiling to occur, (b) the evaporation rate of water, and (c) the critical heat flux.\n",
"\n",
- "#Given\n",
- "# properties of water at 100\u00b0C = 373 K from appendix table 10.3\n",
"rou_f=958 # density in kg/m**3\n",
"cp_f= 4217 # specific heat in J/(kg*K) \n",
"v_f= 2.91e-7 # viscosity in m**2/s \n",
@@ -195,8 +179,6 @@
"g=9.81\n",
"gc=1\n",
"\n",
- "#calculation\n",
- "# nucleate boiling regime\n",
"Cw=0.0132 # formechanically polished stainless steel from table 10.2\n",
"q_A=(rou_f*v_f*hfg)*((g*rou_f*(1-(rou_g/rou_f)))/(sigma*gc))**(0.5)*((cp_f*(Tw-Tg))/(Cw*hfg*Pr_f**1.7))**3\n",
"A=math.pi*D**2/4.0\n",
@@ -204,7 +186,6 @@
"mf=q/hfg # water evaporation rate\n",
"q_cr=0.18*hfg*(sigma*g*gc*rou_f*rou_g**2)**(0.25)\n",
"\n",
- "#Result\n",
"print\"(a)The power delivered to the water is kW\",round(q/1000,2),\"KW\"\n",
"print\"(b)The water evaporation rate is \",round(mf*3600,2),\"kg/h\"\n",
"print\"(c)The critical heat flux is \",round(q_cr,0),\"W/sq.m\"\n"
diff --git a/Engineering_Heat_Transfer/CHAPTER11.ipynb b/Engineering_Heat_Transfer/CHAPTER11.ipynb
index af58990e..30fa585c 100644
--- a/Engineering_Heat_Transfer/CHAPTER11.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER11.ipynb
@@ -27,15 +27,8 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Calculation of the value of the solid angle subtended by surfaces dA2 and dA3 with respect to dA1 (b) the intensity of emission from dA, in the direction of the other areas (c) the rate at which radiation emitted by dA, is intercepted by the other areas\n",
"\n",
- "#Given\n",
- "# solid angle is calculate using the equation dw=dA*cos(Beta)/r**2\n",
- "# Beta is the angle between the surface normal of a receiver surface and the line connecting the two surfaces\n",
- "# For area A2\n",
- "# dimensions are 1X1 in, so\n",
"\n",
- "#Calculation\n",
"import math\n",
"dA2=(1*1)/144.0\n",
"Beta1=40*math.pi/180.0\n",
@@ -54,7 +47,6 @@
"dq1_2=I_theta2*dA1*math.cos(theta2)*dw2_1 #In book calculation mistake\n",
"dq1_3=I_theta3*dA1*math.cos(theta2)*dw3_1\n",
"\n",
- "#Result\n",
"print\"(a)The solid angle subtended by area dA2 with respect to dA1 is \",round(dw2_1,4),\"sr\"\n",
"print\" The solid angle subtended by area dA3 with respect to dA1 is \",round(dw3_1,4),\"sr\"\n",
"print\"(b) The intensity of radiation emitted from dA1 in the direction of dA2 is \",round(I_theta2,0),\"BTU/(hr.sq.ft.sr)\"\n",
@@ -92,26 +84,19 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Calculation of the value of the solid angle subtended by surfaces dA2 with respect to dA1 (b) the rate at which radiation emitted by dA1 is intercepted by dA2 (c) the irradiation associated with dA2\n",
"\n",
- "#Given\n",
- "# solid angle is calculate using the equation dw=dA*cos(Beta)/r**2\n",
- "# The angle Beta is 0 because the surface normal of dA2 is directed at dA1\n",
"dA2=0.02*0.02\n",
"Beta=0\n",
"r=1\n",
"\n",
- "#Calculation\n",
"import math\n",
"dw2_1=dA2*math.cos(Beta)/r**2\n",
"dA1=dA2\n",
"theta=math.pi*30/180.0\n",
"I_theta=1000# The intensity of radiation leaving dA1 in any direction is 1 000 W/(m**2.sr\n",
"dq1_2=I_theta*dA1*cos(theta)*dw2_1\n",
- "# The irradiation associated with dA2 can be found by dividing the incident radiation by the receiver area\n",
"dQ1_2=dq1_2/dA2\n",
"\n",
- "#Result\n",
"print\"(a)The solid angle subtended by area dA2 with respect to dA1 is \",round(dw2_1,4),\"sr\"\n",
"print\"(b)The rate at which radiation emitted by dA1 is intercepted by dA2 is\",round(dq1_2,5),\"W\"\n",
"print\"(c)The irradiation associated with dA2 is \",round(dQ1_2,3),\"W/sq.m\"\n"
@@ -143,18 +128,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# (a) Calculation of the emissivity of the hole.(b) the rate of radiant emission from the hole\n",
"\n",
- "#Given\n",
"D=2.5/12.0 # diameter in ft\n",
"L=4.5/12.0 # length in ft\n",
"\n",
- "#Calculation\n",
"import math\n",
"A=(2*math.pi*D**2/4)+(math.pi*D*L)\n",
"A_hole=math.pi*(1/(8.0*12.0))**2/4.0\n",
"f=A_hole/A # fraction of area removed\n",
- "# for rolled and polished aluminum, that emissivity = 0.039 from appendix table E1\n",
"emissivity=0.039\n",
"emissivity_hole=emissivity/(emissivity+(1-emissivity)*f)\n",
"\n",
@@ -163,7 +144,6 @@
"qe=emissivity_hole*sigma*T**4\n",
"Qe=A_hole*qe\n",
"\n",
- "#Result\n",
"print\"(a)The emissivity of the hole is %.4f\",round(emissivity_hole,4)\n",
"print\"(b)The heat lost by the hole is \",round(Qe,4),\"BTU/hr\"\n"
],
@@ -193,21 +173,17 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the percentage of total emitted energy that lies in the visible range. \n",
"\n",
- "#Given\n",
"T=2800 #Temprature\n",
"lambda1=4e-7 #Wavelength\n",
"lambda2=7e-7\n",
"hT=lambda1*T\n",
"\n",
- "#Calculation\n",
"lambdaT=lambda2*T\n",
"I1=0.0051 #Fraction of Total Radiation Emitted for lower Wavelength-Temperature Product from Table 11.1\n",
"I2=0.065 #Fraction of Total Radiation Emitted for upper Wavelength-Temperature Product from Table 11.1\n",
"dI=I2-I1\n",
"\n",
- "#Result\n",
"print\"The percentage of total emitted energy that lies in the visible range is\",round(dI*100,0),\"percant\"\n"
],
"language": "python",
@@ -235,17 +211,12 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Estimation of the surface temperature of the sun and the emitted heat flux\n",
"\n",
- "#Given\n",
"lambda_max=0.5e-6 # maximum wavelength in m\n",
- "# From Wien\u2019s Displacement Law we can write lambda_max*T=2.898e-3 m.K\n",
"T=2.898e-3/lambda_max\n",
- "# The heat flux is given by the Stefan-Boltzmann Equation as q=sigma*T**4\n",
"sigma=5.675e-8 # value of Stefan-Boltzmann constant in W/(m**2.K**4)\n",
"q=sigma*T**4\n",
"\n",
- "#result\n",
"print\"The Surface Temperature of the Sun is \",round(T,2),\"K\"\n",
"print\"The heat flux emitted is \",round(q,0),\"W/sq.m\"\n"
],
@@ -275,9 +246,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# (a) Calculation of the rate at which the sun\u2019s radiant energy is transmitted through the glass windshield. The interior of the car is considered to be a black body that radiates at 100\u00b0F. (b) Calculation of the rate at which radiant energy from the car interior is transmitted through the glass windshield. \n",
"\n",
- "#Given\n",
"lambda1=300e-9 # lower limit of wavelength\n",
"lambda2=380e-9 # upper limit of wavelength\n",
"T=5800\n",
@@ -291,7 +260,6 @@
"q_in=t*q # energy transmitted from the sun through the glass\n",
"\n",
"\n",
- "#Calculation\n",
"Tb=311 # temperature of black body source in K\n",
"lambda1_Tb=lambda1*Tb\n",
"lambda2_Tb=lambda2*Tb\n",
@@ -299,7 +267,6 @@
"t_b=dI_b*0.68 # transmissivity\n",
"q_out=t_b*q\n",
"\n",
- "#Result\n",
"print\"(a)The energy transmitted from the sun through the glass is \",round(q_in,1),\"W/sq.m\"\n",
"print\"(b)the rate at which radiant energy from the car interior is transmitted through the glass windshield is\",q_out,\"W/sq.m\"\n"
],
diff --git a/Engineering_Heat_Transfer/CHAPTER12.ipynb b/Engineering_Heat_Transfer/CHAPTER12.ipynb
index 1d32142c..dacb5289 100644
--- a/Engineering_Heat_Transfer/CHAPTER12.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER12.ipynb
@@ -27,27 +27,20 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat transferred by radiation from dA1 to A. \n",
"\n",
- "#Given\n",
- "# The view factor Fd1_2 can be calculated as Fd1_2=Fd1_3+Fd1_4+Fd1_5\n",
- "# For Fd1_3\n",
"a_13=100.0\n",
"b_13=250.0\n",
"c_13=100.0\n",
"X_13=a_13/c_13\n",
"Y_13=b_13/c_13\n",
"Fd1_3=0.17 # value for Fd1_3 corresponding to above calculated values of a/c and b/c\n",
- "# For Fd1_4\n",
"a_14=300\n",
"b_14=50\n",
"c_14=100\n",
"\n",
- "#Calculation\n",
"X_14=a_14/c_14\n",
"Y_14=b_14/c_14\n",
"Fd1_4=0.11 #value for Fd1_4 corresponding to above calculated values of a/c and b/c\n",
- "# For Fd1_5\n",
"a_15=100\n",
"b_15=50\n",
"c_15=100\n",
@@ -60,7 +53,6 @@
"T2=560\n",
"q12_A1=sigma*Fd1_2*(T1**4-T2**4)\n",
"\n",
- "#result\n",
"print\"The net heat transferred is \",round(q12_A1,1),\"BTU/(hr.sq.ft)\"\n"
],
"language": "python",
@@ -88,9 +80,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat transferred to the conveyed parts for the conditions given\n",
"\n",
- "#Given\n",
"import math\n",
"L1=1\n",
"angle=math.pi*45/180.0\n",
@@ -99,12 +89,10 @@
"T1=303\n",
"T2=473\n",
"\n",
- "#Calculation\n",
"sigma=5.67e-8 # Stefan-Boltzmann constant\n",
"q21_A2=sigma*(T2**4-T1**4)*((L1/L2)+1-(L3/L2))/2.0\n",
"q31_A3=sigma*(T2**4-T1**4)*((L1/L2)-1+(L3/L2))/2.0\n",
"\n",
- "#result\n",
"print\"The heat transferred from A3 to A1 is \",round(q31_A3,0),\" W/sq.m\"\n"
],
"language": "python",
@@ -132,17 +120,12 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat exchanged between the two plates\n",
"\n",
- "#Given\n",
- "# The view factor can be found with the crossed-string method\n",
- "# from figure 12.13(b)\n",
"ac=1\n",
"bd=1\n",
"ad=(9+1)**0.5\n",
"bc=ad\n",
"\n",
- "#calculation\n",
"crossed_strings=ad+bc\n",
"uncrossed_strings=ac+bd\n",
"L1_F12=(1/2.0)*(crossed_strings-uncrossed_strings)\n",
@@ -153,7 +136,6 @@
"T2=460\n",
"q12_A1=sigma*(T1**4-T2**4)*F12\n",
"\n",
- "#Result\n",
"print\"The heat transfer rate is \",round(q12_A1,0),\"W/sq m\"\n"
],
"language": "python",
@@ -181,14 +163,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat that must be supplied to each of the isothermal surfaces, and also the temperature of the insulated surface. \n",
"\n",
- "#Given\n",
- "# we can apply the equations as follows\n",
- "# q1=sigma*A1*[(T1**4-T2**4)F12+(T1**4-T3**4)F13]..... (1)\n",
- "# q2=sigma*A2*[(T2**4-T1**4)F21+(T2**4-T3**4)F23]..... (2)\n",
- "# q3=sigma*A3*[(T3**4-T1**4)F31+(T3**4-T2**4)F32]..... (3)\n",
- "# given data:\n",
"T1=1000.0\n",
"T3=500.0\n",
"q2=0\n",
@@ -199,13 +174,11 @@
"F31=1/2.0\n",
"F32=1/2.0\n",
"\n",
- "#Calculation\n",
"T2=((T1**4+T3**4)/2.0)**(1/4.0) # using equation (2)\n",
"sigma=0.1714e-8 # Stefan-Boltzmann constant\n",
"q1_A1=sigma*((T1**4-T2**4)*F12+(T1**4-T3**4)*F13) # using equation (1)\n",
"q3_A3=sigma*((T3**4-T1**4)*F31+(T3**4-T2**4)*F32) # using equation (3)\n",
"\n",
- "#result\n",
"print\"The temperature is \",round(T2,1),\"R\"\n",
"print\"The heat flux through area A1 is\",round(q1_A1,0),\"BTU/(hr.sq.ft)\"\n",
"print\"The heat flux through area A3 is\",round(q3_A3,0),\"BTU/(hr.sq.ft)\"\n",
@@ -239,10 +212,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat lost by the oven through its top surface. \n",
"\n",
- "#Given\n",
- "# all energy leaving A1 is intercepted by A2 and vice versa\n",
"F12=1\n",
"F21=1\n",
"F11=0 # the surfaces are flat\n",
@@ -253,11 +223,9 @@
"T2=323\n",
"sigma=5.67e-8 # Stefan-Boltzmann constant\n",
"\n",
- "#Calculation\n",
"q1=(sigma*(T1**4-T2**4))/((1/emissivity1)+(1/emissivity2)-1)\n",
"q2=-q1\n",
"\n",
- "#Result\n",
"print\"The heat lost through bottom surface is \",round(q1,1),\"W/sq m\"\n",
"print\"The heat lost through top surface is \",round(q2,1),\"W/sq m\"\n"
],
@@ -287,13 +255,10 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the net heat exchanged between the dish and the surroundings by radiation at the instant the dish is removed from the oven. Perform the calculations (a) if the dish and surroundings behave like black bodies, and again (b) if the dish has an emissivity of 0.82 and the surroundings have an emissivity of 0.93.\n",
"\n",
- "#Given\n",
"D=1.0 # diameter in ft\n",
"L=6/12.0 # length in ft\n",
"\n",
- "#Calculation\n",
"A=2*math.pi*D**2/4+math.pi*D*L\n",
"F12=1 # the view factor between the dish and the surroundings is unity\n",
"T1=810\n",
@@ -301,15 +266,11 @@
"sigma=0.1714e-8 # Stefan-Boltzmann constant\n",
"q1=sigma*A*(T1**4-T2**4)*F12\n",
"\n",
- "# For gray-surface behavior, we can apply the following Equation\n",
- "# q1/(A1e1)-[F11*(q1/A1)*(1-e1)/e1+F12*(q2/A2)*(1-e2)/e2]=sigma*T1**4-(F11*sigma*T1**4+F12*sigma*T2**4)... equation (1)\n",
"F11=0\n",
"e1=0.82\n",
"e2=0.93\n",
- "# putting q2/A2=0 in equation (1) as A2 tends to infinity\n",
"q1_=A*e1*(sigma*T1**4-F12*sigma*T2**4)\n",
"\n",
- "#Result\n",
"print\"(a)The heat exchanged between the dish and the surroundings is\",round(q1,0),\"BTU/hr\"\n",
"print\"(b)The heat exchanged between the dish and the surroundings for the second case is \",round(q1_,0),\"BTU/hr\"\n"
],
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"
diff --git a/Engineering_Heat_Transfer/CHAPTER4.ipynb b/Engineering_Heat_Transfer/CHAPTER4.ipynb
index 918045d0..a15cf189 100644
--- a/Engineering_Heat_Transfer/CHAPTER4.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER4.ipynb
@@ -27,9 +27,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of response time\n",
"\n",
- "#Given\n",
"k=12.0 # thermal conductivity in BTU/(hr.ft.degree Rankine) \n",
"c=0.1 # specific heat in BTU/(lbm.degree Rankine) \n",
"D=0.025/12.0 # diameter in ft\n",
@@ -40,15 +38,11 @@
"As=3.14*D**2 # surface area in sq.ft\n",
"Vs=3.14*D**(0.5) # volume in cu.ft\n",
"\n",
- "#calculation\n",
- "#As/Vs=6/D\n",
"import math\n",
"reciprocal_timeconstant=(hc*6)/(density*D*c)\n",
- "# selecting T=139 F as T=140 gives an infinite time through the equation (T-T_inf)/(T_i-T_inf)=exp(-hc*As/density*Vs*c)t\n",
"T=139\n",
"t=math.log((T-T_inf)/(T_i-T_inf))/(-reciprocal_timeconstant)\n",
"\n",
- "#result\n",
"print\"The response time of the junction is %.1f s\",round(t*3600,2),\"s\"\n"
],
"language": "python",
@@ -76,10 +70,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of temperature of metal and cumulative heat rate\n",
- "# properties of aluminium from appendix table B1\n",
"\n",
- "#Given\n",
"k=236.0 # thermal conductivity in W/(m.K)\n",
"Cp=896.0 # specific heat in J/(kg.K)\n",
"sp_gr=2.702 # specific gravity\n",
@@ -88,13 +79,11 @@
"L=0.60 # length in m\n",
"hc=550.0 # unit surface conductance between the metal and the bath in W/(K.sq.m)\n",
"\n",
- "#calculation\n",
"import math\n",
"Vs=(math.pi*D**2*L)/4.0 # Volume in cu.m\n",
"As=(2*math.pi*D**2/4.0)+(math.pi*D*L) # surface area in sq.m\n",
"import math\n",
"Bi=(hc*Vs)/(k*As) # Biot Number\n",
- "# Biot number is less than 1 hence lump capacitance equations apply\n",
"T_i=50.0 # initial temperature in degree celsius\n",
"T_inf=2.0 # temperature of ice water bath in degree celsius\n",
"t=60.0 # time=1 minute=60 s\n",
@@ -102,7 +91,6 @@
"T=T_inf+(T_i-T_inf)*math.exp(-(hc*As_*t)/(density*Vs*Cp))\n",
"Q=density*Vs*Cp*(T_inf-T_i)*(1-math.exp(-(hc*As_*t)/(density*Vs*Cp)))\n",
"\n",
- "#result\n",
"print\"(a)The temperature of aluminium is\",round(T,1),\"C\"\n",
"print\"(b)The cumulative heat transferred is \",round(-Q/1000,1),\"KJ\"\n"
],
@@ -132,9 +120,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Determine the time required and temprature profile\n",
"\n",
- "#Given\n",
"hc=30\n",
"L=0.24\n",
"k=1.25 #Conductivity\n",
@@ -142,7 +128,6 @@
"rou=550\n",
"Fo=0.4 #Fourier no\n",
"\n",
- "#Calculation\n",
"Bi=hc*L/k\n",
"alpha=k/(rou*c)\n",
"Tc=150\n",
@@ -150,7 +135,6 @@
"T_i=50\n",
"t=(L**2*Fo)/(alpha)\n",
"TC1=0.82 #Centreline temprature\n",
- "#from table at x/l=0.4\n",
"T=0.71*(T_i-T_inf)*TC1\n",
"x=0.4*L\n",
"Ti=149\n",
@@ -182,7 +166,6 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Calculate the time required\n",
"hc=6 #Surface Conductance\n",
"D=0.105 #Orange Diameter\n",
"k=0.431 #Thermal conductivity \n",
@@ -190,11 +173,9 @@
"rou=998 #Density\n",
"Fo=1.05 #Fourier no.\n",
"\n",
- "#Calculation\n",
"import math\n",
"Vs=math.pi*D**3/6\n",
"As=math.pi*D**2\n",
- "# calculating Biot Number for lumped capacitance approach\n",
"Bi=hc*Vs/(k*As)\n",
"Bi_=hc*(D/2)/(k)\n",
"alpha=k/(rou*c)\n",
@@ -205,7 +186,6 @@
"a=Bi_**2*Fo\n",
"Q=0.7*rou*c*(math.pi/6.0*(Fo**3))*(T_i-T_inf)\n",
"\n",
- "#Result\n",
"print\"The time required is \",round(t/3600,2),\"hr\"\n",
"print\"The heat transfered is\",round(Q/1000,2),\"kj\""
],
@@ -235,7 +215,6 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Estimate the depth of freeze line\n",
" \n",
"D=0.105 #diameter\n",
"k=0.3 #Thermal conductivity \n",
@@ -245,8 +224,6 @@
"alpha=k/(sp_gr*rou_water*c)\n",
"t=3*30*24\n",
"\n",
- "#Calculation\n",
- "# Bi_math.sqrt(Fo) is infinite\n",
"T_inf=10\n",
"Ts=10\n",
"T=32\n",
@@ -255,7 +232,6 @@
"variable_fig4_12=0.38 #The value of x/(2*(alpha*t)**0.5) from figure 4.12\n",
"x=2*math.sqrt(alpha*t)*variable_fig4_12\n",
"\n",
- "#result\n",
"print\"The depth of the freeze line in soil is ft\",round(x,2),\"ft\"\n"
],
"language": "python",
@@ -283,18 +259,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# properties of aluminium from appendix table B1\n",
"\n",
- "#Given\n",
"k_al=236\n",
"p_al=2.7*1000\n",
"c_al=896\n",
- "# properties of oak from appendix table B3\n",
"k_oak=0.19\n",
"p_oak=0.705*1000\n",
"c_oak=2390\n",
"\n",
- "#Calculation\n",
"import math\n",
"math.sqrt_kpc_al=math.sqrt(k_al*p_al*c_al)\n",
"kpc_R=4\n",
@@ -304,7 +276,6 @@
"math.sqrt_kpc_oak=math.sqrt(k_oak*p_oak*c_oak)\n",
"T_oak=(T_Li*(math.sqrt_kpc_oak)+T_Ri*math.sqrt(kpc_R))/(math.sqrt_kpc_oak+math.sqrt(kpc_R))\n",
"\n",
- "#Result\n",
"print\"The temperature of aluminium is felt as \",round(T_al,2),\"C\"\n",
"print\"The temperature of oak is felt as %.1f degree celsius\",round(T_oak,1),\"C\"\n",
"print\"So oak will feel warmer to the touch than will the aluminium\"\n"
@@ -336,10 +307,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# properties of water at 68 degree fahrenheit from appendix table C11\n",
"\n",
- "#Given\n",
- "# The given properties are\n",
"rou=62.46\n",
"cp=0.9988\n",
"k=0.345\n",
@@ -347,7 +315,6 @@
"D=2.5/12.0\n",
"L=4.75/12.0\n",
"\n",
- "#Calculation\n",
"Vs=math.pi*D**2*L/4\n",
"As=(math.pi*D*L)+(math.pi*D**2)/2\n",
"Lc=Vs/As\n",
@@ -355,19 +322,16 @@
"Bi=hc*Lc/k\n",
"t=4\n",
"\n",
- "# for the cylinder solution\n",
"Fo_cylinder=alpha*t/(D/2)**2\n",
"Bi_cylinder=hc*(D/2)/k\n",
"reciprocal_Bi_cylinder=1/Bi_cylinder\n",
"dim_T_cylinder=0.175 #The value of dimensionless temperature of cylinder from figure 4.7a at corresponding values of Fo and 1/Bi\n",
"\n",
- "# for the infinite plate solution\n",
"Fo_plate=alpha*t/(L/2)**2\n",
"Bi_plate=hc*L/(2*k)\n",
"reciprocal_Bi_plate=1/Bi_plate\n",
"dim_T_plate=0.55 #The value of dimensionless temperature of infinite plate from figure 4.7a at corresponding values of Fo and 1/Bi\n",
"\n",
- "# For short cylinder problem\n",
"dim_T_shortcylinder=dim_T_cylinder*dim_T_plate\n",
"T_inf=30\n",
"T_i=72\n",
@@ -377,7 +341,6 @@
"dim_Tw_shortcylinder=dim_Tw_cylinder*dim_Tw_plate\n",
"Tw=dim_Tw_shortcylinder*(Tc-T_inf)+T_inf\n",
"\n",
- "#Result\n",
"print\"The temperature at centre of can is %.1f degree celsius\",round(Tc,0),\"F\"\n",
"print\"The bear temperature near the metal of the can is\",round(Tw,0),\"F\"\n",
"\n"
@@ -408,15 +371,11 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Determine the time to reach centre temp. 50 C\n",
"\n",
- "#Given Data\n",
- "#The given properties are\n",
"rou=7817 #Density\n",
"c=461 #Specific heat \n",
"k=14.4 #Thermal conductivity \n",
"alpha=.387e-5\n",
- "#The dimension are\n",
"L1=0.03\n",
"L2=0.03\n",
"L3=0.04\n",
@@ -424,19 +383,14 @@
"T_i=95 #Internal temprature \n",
"T_inf=17 #Temprature at infinity\n",
"\n",
- "#Calculation\n",
- "# for infinite plate\n",
"L=L1/2\n",
"hc=50\n",
"reciprocal_Bi_plate=k/(hc*L)\n",
"Tinf=0.085 #Temprature distribution for infinite plate\n",
"Tsi=0.225 #Temprature distribution for semi infinite plate\n",
- "#Temprature at a depth of 4 cm\n",
"T=(Tinf**2)*(1-Tsi)*(T_i-T_inf)+T_inf\n",
- "#From the table\n",
"t=350\n",
"\n",
- "#Result\n",
"print\"At a time 3000s The temprature is \",round(T,1),\"C\"\n",
"print\"From the table The time requires to reach tempratue 50C is \",t,\"s\""
],
@@ -476,19 +430,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Determine the time that will pass before the heat added\n",
"\n",
- "#Given\n",
"rou=0.5*1000\n",
"cp=837\n",
"k=0.128\n",
"alpha=0.049e-5\n",
"Ti=20 #Initial temprature\n",
- "# let Fo=0.5 and dx=0.05\n",
"dt=0.5*(0.05)**2/alpha\n",
"\n",
- "#Calculation\n",
- "#Temprature at 1<m<6 and p>0 till T6>20\n",
"p=0\n",
"T0=200\n",
"m=1\n",
@@ -504,7 +453,6 @@
"m=6\n",
"T61=(Ti+Ti)/2.0\n",
"\n",
- "#For next time interval\n",
"p=1\n",
"m=1\n",
"T12=(Ti+T0)/2.0\n",
@@ -518,11 +466,9 @@
"T52=(Ti+T41)/2.0\n",
"m=6\n",
"T62=(Ti+T51)/2.0\n",
- "#Repeating it\n",
"t=4.97\n",
"print\"The time that will pass before the heat added\",t,\"hr\"\n",
"\n",
- "#Plot\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -567,24 +513,18 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of time required to cool to a certain temperature\n",
"\n",
- "#Given\n",
- "#The properties are\n",
"rou=7.817*62.4 #density\n",
"c=0.110\n",
"k=8.32\n",
"alpha=0.417e-4\n",
"dx=1/12.0\n",
- "# taking Fo=1\n",
"Fo=1\n",
"\n",
- "#Calculation\n",
"dt=Fo*dx**2/alpha\n",
"n=8 #Enter the number of time intervals from Saulev plot\n",
"time=n*dt\n",
"\n",
- "#result\n",
"print\"The required time is hr\",round(time/3600,2),\"hr\"\n"
],
"language": "python",
diff --git a/Engineering_Heat_Transfer/CHAPTER5.ipynb b/Engineering_Heat_Transfer/CHAPTER5.ipynb
index 9ab19382..eed3e1b0 100644
--- a/Engineering_Heat_Transfer/CHAPTER5.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER5.ipynb
@@ -27,21 +27,16 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#How much heat must be added for the two cases.\n",
"\n",
- "#Given\n",
- "# properties of CO at 300K from appendix table D2\n",
"Cp=871\n",
"Gamma=1.3\n",
"\n",
- "#calculation\n",
"Cv=Cp/Gamma\n",
"dT=20\n",
"m=5\n",
"Qp=m*Cp*dT\n",
"Qv=m*Cv*dT\n",
"\n",
- "#Result\n",
"print\" The heat required at constant pressure is \",Qp/1000,\"kj\"\n",
"print\"The heat required at constant volume is \",Qv/1000,\"kj\"\n"
],
@@ -71,27 +66,21 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#verify the valve for Voluemetric thermal coefficient\n",
"\n",
- "#Given\n",
- "# properties of Freon-12 from appendix table C3\n",
"T1_Fr=-50\n",
"T2_Fr=-40\n",
"rou1_Fr=1.546*1000\n",
"rou2_Fr=1.518*1000\n",
"\n",
- "#Calculation\n",
"beta_Fr=-(rou1_Fr-rou2_Fr)/(rou1_Fr*(T1_Fr-T2_Fr))\n",
"beta_acc_Fr=2.63e-3 # the accurate value of volumetric thermal expansion coefficient for Freon-12\n",
"error_Fr=(beta_acc_Fr-beta_Fr)*100/beta_acc_Fr\n",
- "# properties of helium from appendix table D3\n",
"T1_He=366\n",
"T2_He=477\n",
"rou1_He=0.13280\n",
"rou2_He=0.10204\n",
"beta_He=-(rou1_He-rou2_He)/(rou1_He*(T1_He-T2_He))\n",
"\n",
- "#REsult\n",
"print\"The volumetric thermal expansion coefficient calculated for Freon-12 is \",round(beta_Fr,6),\"1/K\"\n",
"print\"The error introduced in the case of Freon-12 is percent\",round(error_Fr,0),\"percent\"\n",
"print\"The volumetric thermal expansion coefficient calculated for Freon-12 is \",round(beta_He,6),\"1/K\"\n"
diff --git a/Engineering_Heat_Transfer/CHAPTER6.ipynb b/Engineering_Heat_Transfer/CHAPTER6.ipynb
index f8b8ec9f..318545d4 100644
--- a/Engineering_Heat_Transfer/CHAPTER6.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER6.ipynb
@@ -27,38 +27,29 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the fluid outlet tetnperature and the tube-wall temperature at the outlet.\n",
"\n",
- "#Given\n",
- "#properties of ethylene glycol at 20 degree celsius from appendix table C5\n",
"Cp_20=2382\n",
"rou_20=1.116*1000\n",
"v_20=19.18e-6\n",
"kf_20=0.249\n",
"a_20=0.939e-7\n",
"Pr_20=204.0\n",
- "# specifications of 1/2 standard type M seamless copper water tubing from appendix table F2\n",
"OD=1.588/100.0\n",
"ID=1.446/100.0\n",
"A=1.642e-4\n",
"Q=3.25e-6\n",
"\n",
- "#Calculation\n",
"V=Q/A\n",
- "# calculation of Reynold's Number to check flow regime\n",
"Re=V*ID/v_20\n",
- "# since Re>he 2100, the flow regime is laminar and the hydrodynamic length can be calculated as\n",
"Z_h=0.05*ID*Re\n",
"Tbi=20 # bulk-fluid inlet temperature in degree celsius\n",
"qw=2200 # incident heat flux in W/m**2\n",
"L=3 # Length of copper tube in m\n",
"R=ID/2 # inner radius in m\n",
"Tbo=Tbi+(2*qw*a_20*L)/(V*kf_20*R)\n",
- "# This result is based on fluid properties evaluated at 20\u00b0C. taken as a first approximation\n",
"Z_t=0.05*ID*Re*Pr_20\n",
"Two=Tbo+(11*qw*ID)/(48*kf_20) # The wall temperature at outlet in degree celsius\n",
"\n",
- "#Result\n",
"print\"The bulk-fluid outlet temperature is degree celsius\",round(Tbo,0),\"C\"\n",
"print\"The wall temperature at outlet is degree celsius\",round(Two,0),\"C\"\n"
],
@@ -88,18 +79,14 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of average convection coefficient\n",
"\n",
- "#Given\n",
"T_avg=(140+70)/2.0\n",
- "# properties of water at average bulk temperature from appendix table C11\n",
"rou=0.994*62.4\n",
"kf=0.363\n",
"cp=0.9980\n",
"a=5.86e-3\n",
"v=0.708e-5\n",
"Pr=4.34\n",
- "# specifications of 1 standard type M copper tube from appendix table F2\n",
"OD=1.125/12.0 # outer diameter in ft\n",
"ID=0.8792 # inner diameter in ft\n",
"A=0.006071 # cross sectional area in sq.ft\n",
@@ -112,7 +99,6 @@
"Tbi=70.0\n",
"hL=-(rou*V*ID*cp*math.log((Tw-Tbo)/(Tw-Tbi)))/(4*L)\n",
"\n",
- "#result\n",
"print\"The average convective coefficient is \",round(hL/10,1),\"BTU/(hr. sq.ft.degree Rankine\"\n"
],
"language": "python",
@@ -151,15 +137,11 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the variation of wall temperature with length up to the point where the flow becomes fully developed.\n",
"\n",
- "#Given\n",
- "# properties of milk \n",
"kf=0.6 # thermal conductivity in W/(m-K)\n",
"cp=3.85*1000 # specific heat in J/(kg*K)\n",
"rou=1030 # density in kg/m**3\n",
"mu=2.12e3 # viscosity in N s/m**2\n",
- "# specifications of 1/2 standard type K tubing from appendix table F2\n",
"OD=1.588/100 # outer diameter in m\n",
"ID=1.340/100 # inner diameter in m\n",
"A=1.410e-4 # cross sectional area in m**2\n",
@@ -167,10 +149,7 @@
"V=0.1\n",
"mu=2.12e-3\n",
"\n",
- "#Calculation\n",
- "# determination of flow regime\n",
"Re=rou*V*ID/(mu)\n",
- "# The flow being laminar, the hydrodynamic entry length is calculated as follows\n",
"ze=0.05*ID*Re\n",
"Tbo=71.7 # final temperature in degree celsius\n",
"Tbi=20 # initial temperature in degree celsius\n",
@@ -180,11 +159,9 @@
"Pr=(cp*mu)/kf # Prandtl Number\n",
"zf=0.05*ID*Re*Pr\n",
"\n",
- "#result\n",
"print\"The heat flux is \",round(qw,0),\"W/sq.m\"\n",
"print\"The power required is \",round(q,0),\"W\"\n",
"print\"The length required for flow to be thermally developed is\",round(zf,1),\"m\"\n",
- "#plot\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -257,56 +234,40 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Graph the variation between fluid and wall bulb tempratures.\n",
- "#and how long it will take for the water volume to solidify.\n",
"\n",
- "#Given\n",
- "# The average bulk temperature of the Freon-12 is [-4O +(-4)]/2 = -22\u00b0F\n",
- "# properties of Freon-12 at average bulk temperature\n",
"kf=0.04 # thermal conductivity in BTU/(hr.ft.\u00b0R) \n",
"cp=0.2139 # specific heat in BTU/(lbm-\u00b0R)\n",
"rou= 1.489*(62.4) # density in lbm/cu.ft\n",
"v=0.272e-5 # viscosity in sq.ft/s\n",
"a=2.04e-3 # diffusivity in sq.ft/hr\n",
"Pr=4.8 # Prandtl Number\n",
- "# specifications of 3/8 standard type K copper tubing from appendix table F2\n",
"OD=0.5/12.0 # outer diameter in ft\n",
"ID=0.03350 # inner diameter in ft\n",
"A=0.0008814 # cross sectional area in sq.ft\n",
- "# Laminar conditions are asssumed\n",
"z=5.0\n",
"Tw=32.0\n",
"Tbo=-4.0\n",
"Tbi=-40.0\n",
"L=5.0\n",
"\n",
- "#Calculation\n",
"x=2*a*L/((kf*ID/2.0)*(math.log((Tw-Tbo)/(Tw-Tbi)))) #x=V/hl\n",
- "#1/Gz=9.09/V\n",
- "#from the table unless convergence is achieved\n",
"V=336.0 #ft/h\n",
"V_final=V/3600.0 #ft/s\n",
"hl_=V_final/(x) #\n",
"\n",
- "# checking the laminar-flow assumption by calculating the Reynolds number\n",
"Re=(V_final/3600.0)*ID/v\n",
- "# The flow is laminar\n",
"m_Fr=rou*A*V_final\n",
"As=math.pi*ID*L\n",
"q=hl_*As*((Tw-Tbo)-(Tw-Tbi))/(log((Tw-Tbo)/(Tw-Tbi)))\n",
"q_check=m_Fr*cp*(Tbo-Tbi)\n",
"rou_water=1.002*62.4 # density of water in lbm/ft**3 from appendix table C11\n",
"m_water=rou_water*L*(2/12.0)*(3/12.0)\n",
- "# to remove 144 BTU/lbm of water, the time required is caalculated as below\n",
"t=144*m_water/(-q*3600)\n",
"\n",
- "#result\n",
"print\"The mass flow rate of Freon-12 is \",round(m_Fr*3600,2),\"lbm/hr\"\n",
"print\"The required time is \",round(t,0),\"hr\"\n",
"\n",
"\n",
- "#plot\n",
- "#Constant wall temprature with length\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -336,8 +297,6 @@
"show(a2)\n",
"show(a3)\n",
"\n",
- "#2nd Plot\n",
- "#Nusslet number with dimensionless length\n",
"x1=[0.001,0.01,0.1,1]\n",
"Nu1=[31,11,5.5,5.2]\n",
"Nu2=[25,10,5.3,5.1]\n",
@@ -412,18 +371,12 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination for the power required for heating and the wall temperature at the outlet. \n",
"\n",
- "#Given\n",
- "# The liquid properties are evaluated at the mean temperature of (80 + 20)/2 = 50\u00b0C.\n",
- "# specifications of 1 standard type K copper water tubing from appendix table F2\n",
"OD = 2.858/100.0 # outer diameter in m \n",
"ID = 2.528/100.0 # inner diameter in m \n",
"A = 5.019e-4 # cross sectional area in sq.m\n",
- "# 1 oz = 2.957e-5 m**3\n",
"Q=80*2.957e-5/120 # The volume flow rate of water (at 20\u00b0C) in cu.m/s\n",
"p_20= 1.000*1000 # density of water at 20\u00b0C in kg/cu.m\n",
- "# properties of water at 50\u00b0C from appendix table C11\n",
"p_50= 0.990*(1000) # density in kg/m3 \n",
"cp= 4181 # specific heat in J/(kg*K) \n",
"v = 0.586e-6 # viscosity in sq.m/s \n",
@@ -431,7 +384,6 @@
"a = 1.533e-7 # diffusivity in sq.m/s \n",
"Pr = 3.68 # Prandtl number\n",
"\n",
- "#CALCULATION\n",
"import math\n",
"mass_flow=p_20*Q # mass flow rate through the tube in kg/s\n",
"L=3 # length of tube in m\n",
@@ -444,7 +396,6 @@
"print\"The power required is\",round(q,0),\"W\"\n",
"V=mass_flow/(p_50*A) # average velocity at 50 \u00b0C\n",
"Re=(V*ID)/v # Reynold's Number\n",
- "# The flow is laminar so we can use Figure 6.12 to obtain the information needed on Nusselt number and to find hz\n",
"inv_Gz=L/(Re*ID*Pr) # The inverse Graetz number at tube end, based on 50\u00b0C conditions\n",
"Nu=6.9 #value of corresponding Nusselts Number from figure 6.12\n",
"hz=(Nu*kf)/ID\n",
@@ -478,12 +429,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#Determine heat gained by the duct and wall temprature\n",
"\n",
- "#Given\n",
- "# determibation of heat gained\n",
- "# air properties to be calculated at T=(72+45)/2=58.5 degree Fahrenheit\n",
- "# properties at T=58.5 degree fahrenheit from appendix table D1\n",
"p = 0.077 # density in lbm/ft^3 \n",
"cp = 0.240 # specific heat in BTU/(lbm.degree Rankine) \n",
"v = 15.28e-5 # viscosity in ft^2/s \n",
@@ -495,16 +441,12 @@
"Tbo=72 # outlet temperature in degree Fahrenheit\n",
"Tbi=45 # inlet temperature in degree Fahrenheit\n",
"A=math.pi*(D**2)/4 # cross sectional area of duct in ft^2\n",
- "# density at outlet temperature in lbm/ft^3 \n",
"rou_o=.0748\n",
"V=10 # average velocity in ft/s\n",
"mass_flow=rou_o*A*V\n",
"\n",
- "#Calculation\n",
- "# average velocity evaluated by using the average bulk temperature\n",
"V_avg=mass_flow/(p*A)\n",
"Re=(V_avg*D)/v\n",
- "# the flow is in turbulent regime\n",
"q=mass_flow*cp*(Tbo-Tbi)\n",
"hc=1 # convection coefficient between the outside duct wall \n",
"T_inf=105 # The temperature of attic air surrounding the duct in degree Fahrenheit\n",
@@ -512,7 +454,6 @@
"qw=(T_inf-Tbo)/((1/hc)+(1/hz)) \n",
"Two=qw*(1/hz)+Tbo # The wall temperature at exit in degree Fahrenheit\n",
"\n",
- "#result\n",
"print\"The heat gained by air is\",round(q,3),\"BTU\"\n",
"print\"The wall temperature at exit is \",round(Two,1),\"F\"\n"
],
diff --git a/Engineering_Heat_Transfer/CHAPTER7.ipynb b/Engineering_Heat_Transfer/CHAPTER7.ipynb
index 3c0bf86c..b246db44 100644
--- a/Engineering_Heat_Transfer/CHAPTER7.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER7.ipynb
@@ -37,18 +37,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of boundary layer growth with length\n",
"\n",
- "#Given\n",
- "# properties of air at 27 degree celsius from appendix table D.1\n",
"rou=1.177 # density in kg/cu.m\n",
"v=15.68e-6 # viscosity in sq.m/s\n",
"L=0.5 # length in m\n",
"V_inf=1; # air velocity in m/s\n",
"Re= (V_inf*L)/v # Reynolds Number\n",
"\n",
- "#(a)plot\n",
- "#delta=1.98*10**-2*x**(1/2)\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -61,7 +56,6 @@
"plt.ylim((0,0.015))\n",
"a1=plot(x,t)\n",
"\n",
- "#(b)plot\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -74,14 +68,12 @@
"plt.ylim((0,0.010))\n",
"a1=plot(x,t)\n",
"\n",
- "# (c)calculation of absolute viscosity\n",
"gc=1\n",
"mu=rou*v/gc\n",
"b=1 # width in m\n",
"Df=0.664*V_inf*mu*b*(Re)**0.5\n",
"\n",
"\n",
- "#result\n",
"print\"(a)plot between boundary layer growth with distance\"\n",
"print\"(b)velocity distribution with distance\"\n",
"print\"(c)The skin-drag including both sides of plate is \",round(2*Df,4),\"n\"\n"
@@ -131,10 +123,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of temperature profile\n",
"\n",
- "#Given\n",
- "# properties of water at (40 + 100)/2 = 70\u00b0F = 68\u00b0F from appendix table C11\n",
"rou= 62.4 # density in Ibm/ft^3 \n",
"cp=0.9988 # specific heat BTU/(lbm-degree Rankine) \n",
"v= 1.083e-5 # viscosity in sq.ft/s \n",
@@ -143,10 +132,8 @@
"Pr = 7.02 # Prandtl Number\n",
"V=1.2 # velocity in ft/s\n",
"\n",
- "#Calculation\n",
"x1=1\n",
"Re1=(V*x1)/v # Reynolds Number at x=1 ft\n",
- "# since Reynolds Number is less than 5*10^5, the flow is laminar\n",
"hL1=0.664*Pr**(1/3.0)*Re1**0.5*kf/x1\n",
"Tw=100 # temperature of metal plate in degree fahrenheit\n",
"T_inf=40 # temperature of water in degree fahrenheit\n",
@@ -156,7 +143,6 @@
"print\"The heat transferred to water over the plate is\",round(q1,0),\"BTU/hr\"\n",
"x2=2\n",
"Re2=(V*x2)/v # Reynolds Number at x=1 ft\n",
- "# since Reynolds Number is less than 5*10^5, the flow is laminar\n",
"hL2=0.664*Pr**(1/3.0)*Re2**0.5*kf/x2\n",
"Tw=100 # temperature of metal plate in degree fahrenheit\n",
"T_inf=40 # temperature of water in degree fahrenheit\n",
@@ -164,7 +150,6 @@
"q2=hL2*A2*(Tw-T_inf)\n",
"print\"The heat transferred to water over the plate is \",round(q2,0),\"BTU/hr\"\n",
"\n",
- "#PLot for temprature distribution\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -177,7 +162,6 @@
"plt.ylim((0,0.004))\n",
"a1=plot(T,y)\n",
"\n",
- "#PLot for influence of prandtl no on boundary layer thickness\n",
"import matplotlib.pyplot as plt\n",
"fig = plt.figure()\n",
"ax = fig.add_subplot(111)\n",
@@ -232,10 +216,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the average convection coefficient and the total drag force exerted on the plate.\n",
"\n",
- "#Given\n",
- "# properties of air at (300 + 400)/2 = 350 K from appendix table D1\n",
"rou=0.998\t # density in kg/cu.m\n",
"cp=1009 \t\t# specific heat in J/(kg*K) \n",
"v=20.76e-6 \t# viscosity in sq.m/s \n",
@@ -247,15 +228,12 @@
"b=0.5 \t\t# width in m\n",
"Re=V*L/v \t# Reynolds number at plate end\n",
"\n",
- "#calculation\n",
- "# Nu=h*L/k= 0.664 Re**(1/2)Pr**(1/3)\n",
"h=k*0.664*Re**(0.5)*Pr**(1/3.0)/L\t # The average convection coefficient in W/(sq.m.K)\n",
"Df=0.664*V*rou*v*b*(Re)**0.5 \t # drag force in N\n",
"hx=(1/2.0)*h # local convective coefficient\n",
"delta=5*L/(Re)**0.5 # The boundary-layer thickness at plate end\n",
"delta_t=delta/(Pr)**(1/3.0)\n",
"\n",
- "#Result\n",
"print\"The local convective coefficient is \",round(hx,2),\"W/(sq.m.K)\"\n",
"print\"The boundary-layer thickness at plate end is \",round(delta*100,2),\"cm\"\n",
"print\"The thermal-boundary-layer thickness is \",round(delta_t*100,2),\"cm\""
@@ -287,10 +265,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the maximum heater-surface temperature for given conditions\n",
"\n",
- "#Given\n",
- "# fluid properties at (300 degree R + 800 degree R)/2 = 550 degree R=540degree R from Appendix Table D.6\n",
"rou= 0.0812 # density in Ibm/ft^3 \n",
"cp=0.2918 # specific heat BTU/(lbm-degree Rankine) \n",
"v= 17.07e-5 # viscosity in ft^2/s \n",
@@ -298,16 +273,13 @@
"a = 0.8862 # diffusivity in ft^2/hr \n",
"Pr = 0.709 # Prandtl Number\n",
"\n",
- "#calculation\n",
"qw=10/(1.5*10.125)*(1/.2918)*144 # The wall flux \n",
"V_inf=20 # velocity in ft/s\n",
"L=1.5/12 # length in ft\n",
"Re_L=V_inf*10*L/v # Reynolds number at plate end\n",
- "# So the flow is laminar and we can find the wall temperature at plate end as follows\n",
"T_inf=300 # free stream temperature in degree Rankine\n",
"Tw=T_inf+(qw*L*10/(kf*0.453*(Re_L)**0.5*(Pr)**(1/3.0)))\n",
"\n",
- "#Result\n",
"print\"The maximum heater surface temperature is \",round(Tw,0),\"R\"\n"
],
"language": "python",
@@ -335,10 +307,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# validation of the equation st.(Pr)**(2/3)=Cd/2 where St: Stanton Number Pr:Prandtl Number Cd: Drag Coefficient\n",
"\n",
- "#Given\n",
- "# values of parameters from example 7.4\n",
"rou= 0.0812 # density in Ibm/ft**3 \n",
"cp=0.2918 # specific heat BTU/(lbm-degree Rankine) \n",
"v= 17.07e-5 # viscosity in ft**2/s \n",
@@ -350,14 +319,12 @@
"qw=324.0 # The wall flux in BTU/(hr.ft**2)\n",
"V_inf=20 # velocity in ft/s\n",
"\n",
- "#Calculation\n",
"hx=qw/(Tw-T_inf) # The convection coefficient\n",
"LHS=(hx/3600.0)*(Pr)**(2/3.0)/(rou*cp*V_inf)\n",
"Re_L=1.46e+005 # Reynolds number at plate end\n",
"RHS=0.332*(Re_L)**(-0.5)\n",
"err=(LHS-RHS)*100/LHS\n",
"\n",
- "#Result\n",
"print\"The convection coefficient is BTU/(hr.sq.ft.degree R)\",hx\n",
"print\"The error is \",round(err,0),\"percent\"\n",
"print\"Since the error is only 3 percent, the agreement is quite good\"\n"
@@ -389,10 +356,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Estimation of the drag due to skin friction\n",
"\n",
- "#Given\n",
- "# properties of water at 68\u00b0F from Appendix Table C.11\n",
"rou= 62.4 # density in Ibm/cu.ft\n",
"v= 1.083e-5 # viscosity in sq.ft/s \n",
"V_inf=5*.5144/.3048 # barge velocity in ft/s using conversion factors from appendix table A1\n",
@@ -402,10 +366,8 @@
"gc=32.2\n",
"b=12 # width in ft\n",
"\n",
- "#Calculation\n",
"Df=(Cd*rou*V_inf**2*b*L)/(2*gc)\n",
"\n",
- "#Result\n",
"print\"The drag force is \",round(Df,0),\"lbf\"\n"
],
"language": "python",
@@ -433,10 +395,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of wattage requirement\n",
"\n",
- "#Given\n",
- "# properties of carbon dioxide at a film temperature of (400+600)/2 = 500 K from appendix table D2\n",
"rou= 1.0732 # density in kg/m**3 \n",
"cp= 1013 # specific heat in J/(kg*K) \n",
"v= 21.67e-6 # viscosity in m**2/s \n",
@@ -452,8 +411,6 @@
"Tw=600 # temperature of heater surface in K\n",
"T_inf=400 # temperature of carbon dioxide in K\n",
"\n",
- "#calculation\n",
- "#Nux=0.664*Rex**(1/2)*Pr**(1/3)\n",
"y=0.664*Pr**(1/3.0)*k*(vel/v)**(1/2.0) #y=h1*x1**(1/2)\n",
"x1=0.04 # m\n",
"h1=y/x1**(0.5)\n",
@@ -474,24 +431,19 @@
"Q2=h4*x4*(b)*(Tw-T_inf) #Q=q1+q2+q3+q4\n",
"q4=Q2-Q1\n",
"\n",
- "#at the end of 5th heater\n",
"Re5=vel*5*x1/(v)\n",
"\n",
- "#Nux=10.0359*(Rex**(0.8)-830)*Pr**(1/3)\n",
"h5=0.0359*(Re5**(0.8)-830)*Pr**(1/3.0)*(k/(11.9*x1))\n",
"x5=0.2\n",
"Q3=h5*x5*(b)*(Tw-T_inf) #Q3=q1+q2+q3+q4+q5\n",
"q5=Q3-Q2\n",
"\n",
- "#at the end of 6th heater\n",
"Re6=vel*6*x1/(v)\n",
- "#Nux=10.0359*(Rex**(0.8)-830)*Pr**(1/3)\n",
"h6=0.0359*(Re5**(0.8)-830)*Pr**(1/3.0)*(k/(10.3*x1))\n",
"x6=0.24\n",
"Q4=h6*x6*(b)*(Tw-T_inf) #Q4=q1+q2+q3+q4+q5+q6\n",
"q6=round(Q4,0)-round(Q3,0)\n",
"\n",
- "#Result\n",
"print\"The wattage required for 1st heater is \",round(q1,0),\"W\"\n",
"print\"The wattage required for 2nd heater is \",round(q2,0),\"W\"\n",
"print\"The wattage required for 3rd heater is \",round(q3,1),\"W\"\n",
@@ -529,10 +481,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Estimation of force exerted on the pole\n",
"\n",
- "#Given\n",
- "# properties of air at given conditions from appendix table D1\n",
"rou= 0.0735 # density in Ibm/ft**3 \n",
"v= 16.88e-5 # viscosity in ft**2/s \n",
"V=20*5280/3600.0 # flow velocity in ft/s\n",
@@ -543,7 +492,6 @@
"Cd_cylinder=1.1 # value of Cd for smooth cylinder from figure 7.22\n",
"A_cylinder=D*L # frontal area of pole\n",
"\n",
- "#Calculation\n",
"Df_cylinder=Cd_cylinder*(0.5)*rou*V**2*A_cylinder/gc\n",
"D_square=2/12.0 # length of square part of pole\n",
"L_square=4\n",
@@ -579,10 +527,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of required current\n",
"\n",
- "#Given\n",
- "# properties of air at film temperature (300 + 500)/2 = 400 K from appendix table D1\n",
"rou= 0.883 # density in kg/cu.m\n",
"cp= 1014 # specific heat in J/(kg*K) \n",
"v= 25.90e-6 # viscosity in sq.m/s \n",
@@ -596,7 +541,6 @@
"C=0.911 #value of C for cylinder from table 7.4\n",
"m=0.385 #value of m for cylinder from table 7.4\n",
"\n",
- "#calculation\n",
"hc=kf*C*(Re_D)**m*(Pr)**(1/3)/D # the convection coefficient in W/(m**2.K)\n",
"Tw=500 # air stream temperature in K\n",
"T_inf=300 # wire surface temperature in K\n",
@@ -606,7 +550,6 @@
"Resistance=resistivity*(L/(math.pi*D**2)) # resistance in ohm\n",
"i=(qw*100/Resistance)**0.5 # current in ampere\n",
"\n",
- "#Result\n",
"print\"The current is %.1f A\",round(i,1),\"A\"\n"
],
"language": "python",
@@ -634,17 +577,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Calculation of the pressure drop for the air passing over the tubes and the heat transferred to the air.\n",
"\n",
- "#Given\n",
- "# properties of air at 70 + 460 = 530 degree R = 540 degree R from appendix table D1\n",
"rou= 0.0735 # density in Ibm/cu.ft \n",
"cp=0.240 # specific heat BTU/(lbm-degree Rankine) \n",
"v= 16.88e-5 # viscosity in sq.ft/s \n",
"kf = 0.01516 # thermal conductivity in BTU/(hr.ft.degree Rankine) \n",
"a = 0.859 # diffusivity in sq.ft/hr \n",
"Pr = 0.708 # Prandtl Number\n",
- "# specifications of 3/4 standard type K copper tubing from appendix table F2\n",
"OD=0.875/12 # outer diameter in ft\n",
"ID=0.06208 # inner diameter in ft\n",
"A=0.003027 # cross sectional area in sq.ft\n",
@@ -653,7 +592,6 @@
"sT=1.3/12\n",
"V_inf=12 # velocity of air in ft/s\n",
"\n",
- "#calculation\n",
"V1=(sT*V_inf)/(sT-OD) # velocity at area A1 in ft/s\n",
"sD=((sL)**2+(sT/2)**2)**0.5 # diagonal pitch in inch\n",
"V2=(sT*V_inf)/(2*(sD-OD))\n",
@@ -676,7 +614,6 @@
"T_inf=70 # air temperature in degree F\n",
"q=hc*As*(Tw-T_inf) # heat transferred\n",
"\n",
- "#result\n",
"print\"The pressure drop is \",round(dP/147,3),\"psi\"\n",
"print\"The heat transferred is\",round(q,1),\" BTU/hr\"\n"
],
diff --git a/Engineering_Heat_Transfer/CHAPTER8.ipynb b/Engineering_Heat_Transfer/CHAPTER8.ipynb
index a70f7944..759f39bd 100644
--- a/Engineering_Heat_Transfer/CHAPTER8.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER8.ipynb
@@ -27,10 +27,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat transferred to the wall.\n",
"\n",
- "#Given\n",
- "# air properties at (400+120)/2 =260 degree F = 720 degree R from Appendix Table D1\n",
"rou= 0.0551 # density in Ibm/cu.ft \n",
"cp=0.2420 # specific heat BTU/(lbm-degree Rankine) \n",
"v= 27.88e-5 # viscosity in sq.ft/s \n",
@@ -41,8 +38,6 @@
"Tw=400.0+460.0 # inside wall temperature in degree R\n",
"Beta=1/T_inf\n",
"\n",
- "#IN THE BOOK THERE IS CALCULATION MISTAKE IN Beta\n",
- "#Accoding to book Beta=0.00116 /R\n",
"Beta_=0.00116\n",
"gc=32.2\n",
"L=1.0 # length of wall in ft\n",
@@ -53,7 +48,6 @@
"A=L*W # cross sectional area in sq.ft\n",
"qw=hL*A*(Tw-T_inf)\n",
"\n",
- "#Result\n",
"print\"The heat transferred is\",round(qw,0),\"BTU/hr\"\n"
],
"language": "python",
@@ -81,10 +75,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of heat lost through the glass per unit area\n",
"\n",
- "#Given\n",
- "# properties of air at 0 + 273 K=273 K from appendix table D1\n",
"rou1=1.295 # density in kg/cu.m\n",
"cp1=1005.5 # specific heat in J/(kg*K) \n",
"v1=12.59e-6 # viscosity in sq.m/s \n",
@@ -94,7 +85,6 @@
"T_inf1=0 # inside and outside temperature in K\n",
"Beta1=1/(T_inf1+273.0) # volumetric thermal expansion coefficient at 295 K and 273 K\n",
"\n",
- "# properties of air at 22 + 273 = 295 K = 300 K(approx) \n",
"rou2=1.177 # density in kg/cu.m\n",
"cp2=1005 # specific heat in J/(kg*K) \n",
"v2=15.68e-6 # viscosity in sq.m/s \n",
@@ -108,7 +98,6 @@
"t=0.005 # thickness of glass\n",
"L=0.60 # window length in m\n",
"k=0.81 # thermal conductivity of glass from appendix table B3\n",
- "# for first guess\n",
"Tw1=18\n",
"Tw2=4\n",
"Ra1=(g*Beta1*(Tw2-T_inf1)*L**3)/(v1*a1)\n",
@@ -119,18 +108,15 @@
"Tw2_=T_inf2-(q1/hL2)\n",
"Tw1_=q1/hL1+T_inf1\n",
"\n",
- "#Using these temprature as second estimates\n",
"Ra1_=3.7*10**8\n",
"hL1_=2.92\n",
"Ra2_=2.31*10**8\n",
"hL2_=2.80\n",
"q2=(T_inf2-T_inf1)/((1/hL2_)+(t/k)+(1/hL1_))\n",
"\n",
- "#The wall temprature are\n",
"Tw2final=q2-T_inf2\n",
"Tw1final=10.7\n",
"\n",
- "#result\n",
"print\"The heat loss is \",round(q2,1),\" W/sq.m\"\n",
"\n",
"\n"
@@ -160,9 +146,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of heat loss through the side.\n",
"\n",
- "#Given\n",
"rou= 0.0735 # density in Ibm/cu.ft \n",
"cp=0.240 # specific heat BTU/(lbm-degree Rankine) \n",
"v= 16.88e-5 # viscosity in sq.ft/s \n",
@@ -179,7 +163,6 @@
"hc=(kf/L)*(0.825+((0.387*(Ra)**(1/6.0))/(1+(0.492/Pr)**(9/16.0))**(8/27.0)))**2\n",
"q=hc*L*W*(Tw-T_inf)\n",
"\n",
- "#Result\n",
"print\"The heat gained is %d BTU/hr\",round(q,0),\"BTU/hr\"\n"
],
"language": "python",
@@ -207,8 +190,6 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the variation of average convection coefficient with distance\n",
- "# properties of air at (65 + 20)/2 = 42.5 degree C =315 K. from appendix table D1\n",
"rou= 1123 # density in kg/m^3 \n",
"cp= 1006.7 # specific heat in J/(kg*K) \n",
"v= 17.204e-6 # vismath.cosity in m^2/s \n",
@@ -222,9 +203,7 @@
"Tw=65 # roof surface temperature in degree C\n",
"Beta=1/(T_inf+273.0) # volumetric thermal math.expansion coefficient in per K\n",
"\n",
- "#Calculation\n",
"import math\n",
- "# determination of Laminar-turbulent transition length by Vliet equation Ra=3x10^5xmath.exp(0.1368math.cos(90-theta))\n",
"x=((3e5*math.exp(0.1368*math.cos(90-theta))*v*a)/(g*math.cos(theta)*Beta*(Tw-T_inf)))**(1/3.0)\n",
"x=0.051\n",
"print\"The Laminar-turbulent transition length by Vliet equation is \",round(x,3),\"m\\n\\n\"\n",
@@ -290,10 +269,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determine if heat is lost lose more heat through its upper surface or one of its vertical sides\n",
"\n",
- "#Given\n",
- "# properties of air at (100 + 60)/2 = 80\u00b0F = 540 degree R from appendix table D1\n",
"rou= 0.0735 # density in lbm/cu.ft\n",
"cp=0.240 # specific heat BTU/(lbm-degree Rankine) \n",
"v= 16.88e-5 # viscosity in sq.ft/s \n",
@@ -306,18 +282,15 @@
"L=2.0 # length in ft\n",
"W=2.0 # width in ft\n",
"\n",
- "#Calculation\n",
"Beta=1/(T_inf+460.0) # volumetric thermal expansion coefficient in per degree Rankine\n",
"Ra=(g*Beta*(Tw-T_inf)*L**3)/(v*a/3600.0)\n",
"hc=(kf/L)*(0.68+(0.670*Ra**(0.25))/(1+(0.492/Pr)**(9/16.0))**(4/9.0))\n",
"q1side=hc*L*W*(Tw-T_inf)\n",
- "# For the top, we have a heated surface facing upward, The characteristic length is determined as follows\n",
"Lc=0.5\n",
"Ra_L=(g*Beta*(Tw-T_inf)*Lc**3)/(v*a/3600.0) # Rayleigh number based on characteristic length\n",
"hc_L=(kf/Lc)*0.54*(Ra_L)**(1/4.0)\n",
"qtop=hc_L*L*W*(Tw-T_inf)\n",
"\n",
- "#Result\n",
"print\"The heat transferred from one side is \",round(q1side,1),\"BTU/hr\"\n",
"print\"The heat transferred from top is \",round(qtop,0),\"BTU/hr\"\n",
"\n",
@@ -355,10 +328,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of heat lost from the insulation by convection\n",
"\n",
- "#Given\n",
- "# properties of air at (50 + 5)/2 = 27.5 degree C = 300 K from appendix table D1\n",
"rou= 1.177 # density in kg/cu.m\n",
"cp= 1005.7 # specific heat in J/(kg*K) \n",
"v= 15.68e-6 # viscosity in sq.m/s \n",
@@ -371,20 +341,16 @@
"T_inf=5.0 # ambient air temperature in degree C\n",
"Tw=50.0 # outside surface temperature in degree C\n",
"\n",
- "#Calculation\n",
"import math\n",
"Beta=1/(T_inf+273.0) # volumetric thermal expansion coefficient in per K\n",
"Ra=(g*Beta*(Tw-T_inf)*D**3)/(v*a)\n",
- "# for horizontal pipe, the convective coefficient is determined as follows\n",
"hc_h=(kf/D)*(0.60+(0.387*Ra**(1/6.0))/(1+(0.559/Pr)**(9/16.0))**(8/27.0))**2\n",
"As=math.pi*D*L\n",
"q_hor=hc_h*As*(Tw-T_inf)\n",
- "# for vertical pipe, the convective coefficient is determined as follows\n",
"hc_v=(kf/D)*0.6*(Ra*(D/L))**(1/4.0)\n",
"q_ver=hc_v*As*(Tw-T_inf)\n",
"q=round(q_ver,0)+round(q_hor,0)\n",
"\n",
- "#Result\n",
"print\"The heat transferred from the horizontal length of 4 m is \",round(q_hor,0),\"W\"\n",
"print\"The heat transferred from the vertical length of 4 m is \",round(q_ver,0),\"W\"\n",
"print\"nThe total heat lost from the pipe is \",round(q,2),\"W\"\n"
@@ -416,10 +382,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determinion of the convection coefficient about the ice cube\n",
"\n",
- "#Given\n",
- "# properties of air at (0 + 70)/2 = 35\u00b0F == 495 degree R from appendix table D1\n",
"rou= 0.0809 # density in lbm/cu.ft \n",
"cp=0.240 # specific heat BTU/(lbm-degree Rankine) \n",
"v= 13.54e-5 # viscosity in sq.ft/s \n",
@@ -430,12 +393,10 @@
"T_inf=70.0 # ambient temperature in degree F\n",
"g=32.2\n",
"Beta=1/(T_inf+460.0) # volumetric thermal expansion coefficient in per degree Rankine\n",
- "# The characteristic length is found by using King Equation\n",
"Lc=1/((1/1)+(1/1.2))\n",
"Ra=(g*Beta*abs(Tw-T_inf)*Lc**3)/(v*a/3600.0)\n",
"hc=(kf/Lc)*0.6*(Ra)**(1/4.0)\n",
"\n",
- "#Result\n",
"print\"The value of convection coefficient is \",round(hc,2),\"BTU/(hr.sq.ft.degree R)\"\n"
],
"language": "python",
@@ -463,10 +424,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of the maximum amount of heat that fins can transfer\n",
"\n",
- "#Given\n",
- "# properties of air at (100 + 35)/2 = 67.5 degree C from appendix table D1\n",
"rou= 0.998 # density in kg/cu.m\n",
"cp= 1009.0 # specific heat in J/(kg*K) \n",
"v= 20.76e-6 # viscosity in sq.m/s \n",
@@ -477,7 +435,6 @@
"T_inf=35 # ambient air temperature in degree C\n",
"Tw=100 # surface temperature in degree C\n",
"Beta=1/(T_inf+273.0) # volumetric thermal expansion coefficient in per K\n",
- "# properties of aluminium from appendix table B1\n",
"rou_Al=2702 # density in kg/cu.m\n",
"k_Al=236 # thermal conductivity in W/(m.K)\n",
"cp_Al=896 # specific heat in J/(kg*K) \n",
@@ -485,15 +442,12 @@
"b=46/100.0\n",
"w=24/100.0\n",
"\n",
- "#Calculation\n",
- "# Applying the Bar-Cohen Equations\n",
"zeta=((w*v**2)/(g*Beta*(Tw-T_inf)*Pr))**(1/4.0)\n",
"L=1.54*(k_Al/kf)**(1/2)*zeta\n",
"S=2.89*zeta\n",
"q=(b*w*(Tw-T_inf)*1.3*(k_Al*kf)**(1/2.0))/(6*zeta)\n",
"N=b/(2*S)\n",
"\n",
- "#result\n",
"print\"The heat transfer rate is \",round(q,0),\"W\"\n",
"print\"The number of fins can be atmost\",round(N,0)\n",
"\n"
diff --git a/Engineering_Heat_Transfer/CHAPTER9.ipynb b/Engineering_Heat_Transfer/CHAPTER9.ipynb
index c1ada237..d66004ef 100644
--- a/Engineering_Heat_Transfer/CHAPTER9.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER9.ipynb
@@ -27,24 +27,16 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# determination of counterflow and parallel-flow configurations. \n",
"\n",
- "#Given\n",
- "# temperatures of hot fluid in degree C\n",
"T1=100.0 \n",
"T2=75.0\n",
- "# temperatures of cold fluid in degree C\n",
"t1=5.0\n",
"t2=50.0\n",
"\n",
- "#Calculation\n",
- "# for counterflow\n",
"import math\n",
"LMTD_counter=((T1-t2)-(T2-t1))/(math.log((T1-t2)/(T2-t1)))\n",
- "# for parallel flow\n",
"LMTD_parallel=((T1-t1)-(T2-t2))/(math.log((T1-t1)/(T2-t2)))\n",
"\n",
- "#Result\n",
"print\"The LMTD for counter flow configuration is \",round(LMTD_counter,1),\"C\"\n",
"print\"The LMTD for parallel flow configuration is \",round(LMTD_parallel,2),\"C\""
],
@@ -74,24 +66,16 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the LMTD for both counterflow and parallel-flow configurations. \n",
"\n",
- "#Given\n",
- "# temperatures of hot fluid in degree F\n",
"T1=250\n",
"T2=150\n",
- "# temperatures of cold fluid in degree F\n",
"t1=100\n",
"t2=150\n",
"\n",
- "#calculation\n",
- "# for counterflow\n",
"import math\n",
"LMTD_counter=((T1-t2)-(T2-t1))/(math.log((T1-t2)/(T2-t1)));\n",
- "# for parallel flow\n",
"LMTD_parrelel=0\n",
"\n",
- "#Result\n",
"print\"The LMTD for counter flow configuration is\",round(LMTD_counter,1),\"C\"\n",
"print\"if parallel flow is to give equal outlet temperatures,then the area needed must be infinite which is not feasible economically.\"\n",
"print\"The LMTD for parrelel flow configuration is\",LMTD_parrelel,\"C\"\n"
@@ -123,10 +107,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the outlet temperature of the ethylene glycol for counterflow.\n",
"\n",
- "#Given\n",
- "# properties of air at (195 + 85)/2 = 140\u00b0F. from appendix table CII\n",
"rou_1= 0.985*62.4 # density in lbm/ft**3 \n",
"cp_1=0.9994 # specific heat BTU/(lbm-degree Rankine) \n",
"v_1= 0.514e-5 # viscosity in ft**2/s \n",
@@ -135,7 +116,6 @@
"Pr_1 = 3.02 # Prandtl Number \n",
"m_1=5000 # mass flow rate in lbm/hr\n",
"T_1=195 # temperature in degree F\n",
- "# properties of ethylene glycol at 140 degree F from Appendix Table C.5\n",
"rou_2= 1.087*62.4 # density in lbm/ft**3 \n",
"cp_2=0.612 # specific heat BTU/(lbm-degree Rankine) \n",
"v_2= 5.11e-5 # viscosity in ft**2/s \n",
@@ -144,32 +124,25 @@
"Pr_2 = 51 # Prandtl Number \n",
"m_2=12000 # mass flow rate in lbm/hr\n",
"T_2=85 # temperature in degree F\n",
- "# specifications of seamless copper water tubing (subscripts: a = annulus, p = inner pipe or tube) from appendix table F2\n",
"ID_a=0.1674\n",
"ID_p=0.1076\n",
"OD_p=1.375/12\n",
- "# Flow Areas\n",
"A_p=math.pi*ID_p**2/4\n",
"A_a=math.pi*((ID_a)**2-(OD_p)**2)/4\n",
"\n",
- "# Annulus Equivalent Diameters\n",
"D_h=ID_a-OD_p\n",
"D_e=(ID_a**2-OD_p**2)/(OD_p)\n",
"\n",
- "# Reynolds Numbers \n",
"Re_1=(m_1/3600.0)*(ID_p)/(v_1*rou_1*A_p)\n",
"Re_2=(m_2/3600.0)*(D_e)/(v_2*rou_2*A_a)\n",
"\n",
- "# Nusselt numbers\n",
"Nu_1=0.023*(Re_1)**(4/5.0)*(Pr_1)**0.3\n",
"Nu_2=0.023*(Re_2)**(4/5.0)*(Pr_2)**0.4\n",
"\n",
- "# Convection Coefficients \n",
"h_1i=Nu_1*kf_1/ID_p\n",
"h_1o=h_1i*ID_p/OD_p\n",
"h_2=Nu_2*kf_2/D_e\n",
"\n",
- "# Exchanger Coefficient \n",
"Uo=1/((1/h_1o)+(1/h_2))\n",
"R=(m_2*cp_2)/(m_1*cp_1)\n",
"L=20\n",
@@ -205,11 +178,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of (a) no. of exchangers required, (b) the overall coefficient of (all) the exchanger(s), and (c) the pressure drop for each stream. \n",
"\n",
- "#Given\n",
- "# assuming counterflow arrangement\n",
- "# properties of air at 323 K. from appendix table D1\n",
"rou_1= 1.088 \t\t# density in kg/m**3 \n",
"cp_1= 1007\t\t # specific heat in J/(kg*K) \n",
"v_1= 18.2e-6\t\t # viscosity in m**2/s \n",
@@ -217,10 +186,8 @@
"kf_1= 0.02814 \t\t# thermal conductivity in W/(m.K)\n",
"a_1 = 0.26e-4 \t\t# diffusivity in m**2/s \n",
"m_1=100 \t\t # mass flow rate in kg/hr\n",
- "# temperatures in K\n",
"t1_air=20+273 \n",
"t2_air=80+273\n",
- "# properties of carbon dioxide at [600 + (20 + 273)]/2 = 480 = 500 K. from appendix table D2\n",
"rou_2= 1.0732\t\t # density in kg/m**3 \n",
"cp_2= 1013 \t\t# specific heat in J/(kg*K) \n",
"v_2= 21.67e-6 \t #viscosity in m**2/s \n",
@@ -228,71 +195,53 @@
"kf_2= 0.03352 \t\t# thermal conductivity in W/(m.K)\n",
"a_2 = 0.3084e-4 \t\t# diffusivity in m**2/s \n",
"m_2=90\t\t\t # mass flow rate in kg/hr\n",
- "# temperatures in K\n",
"T1_CO2=600 \n",
- "# specifications of seamless copper tubing from appendix table F2\n",
"ID_a=.098\n",
"ID_p=.07384\n",
"OD_p=.07938\n",
"\n",
- "#calculation\n",
"import math\n",
- "# Flow Areas\n",
"A_p=math.pi*ID_p**(2)/4.0\n",
"A_a=math.pi*((ID_a)**2-(OD_p)**2)/4.0\n",
"\n",
- "# Heat Balance \n",
"q_air=(m_1/3600.0)*(cp_1)*(t2_air-t1_air)\n",
"T2_CO2=T1_CO2-(q_air/(m_2*cp_2/3600.0))\n",
"\n",
- "# Log-Mean Temperature Difference\n",
"LMTD_counter=((T1_CO2-t2_air)-(T2_CO2-t1_air))/(log((T1_CO2-t2_air)/(T2_CO2-t1_air)))\n",
- "# Annulus Equivalent Diameters\n",
"D_h=ID_a-OD_p\n",
"D_e=(ID_a**2-OD_p**2)/(OD_p)\n",
"\n",
- "# Reynolds Numbers \n",
"Re_1=(m_1/3600.0)*(ID_p)/(v_1*rou_1*A_p)\n",
"Re_2=(m_2/3600.0)*(D_e)/(v_2*rou_2*A_a)\n",
"\n",
- "# Nusselt numbers\n",
"Nu_1=0.023*(Re_1)**(0.8)*(Pr_1)**0.3\n",
"Nu_2=0.023*(Re_2)**(0.8)*(Pr_2)**0.4\n",
"\n",
- "# Convection Coefficients \n",
"\n",
"h_1i=Nu_1*kf_1/ID_p\n",
"h_1o=h_1i*ID_p/OD_p\n",
"h_2=Nu_2*kf_2/D_e\n",
"\n",
- "# Fouling Factors in (m**2.K)/W\n",
"Rd_air=0.0004\n",
"Rd_CO2=0.002\n",
"\n",
- "# exchanger coefficients\n",
"Uo=1/((1/h_1o)+(1/h_2))\n",
"Uo=1/((1/Uo)+Rd_air+Rd_CO2)\n",
"\n",
- "# area required\n",
"A=q_air/(Uo*LMTD_counter)\n",
"\n",
- "# surface area of one exchanger is A=math.pi*OD*L, so\n",
"L=(A/(math.pi*OD_p)) # length of each exchanger\n",
"L_available=2 # available exchanger length\n",
"N=L_available/L # no. of exchangers\n",
"\n",
- "#friction factors\n",
"fp=0.0245 #friction factor for air fom figure 6.14 corresponding to Re\n",
"fa=0.033 #friction factor for cCO2fom figure 6.14 corresponding to Re\n",
- "# Velocities\n",
"V_air=(m_1/3600.0)/(rou_1*A_p)\n",
"V_CO2=(m_2/3600.0)/(rou_2*A_a)\n",
"\n",
- "# pressure drops\n",
"dP_p=(fp*L_available*rou_1*V_air**2)/(ID_p*2)\n",
"dP_a=((rou_2*V_CO2**2)/2.0)*((fa*L_available/D_h)+1)\n",
"\n",
- "#Result\n",
"print\"(a)The number of exchangers is \",round(N,0)\n",
"print\"(b)The overall exchanger coefficient is \",round(Uo,1),\" W/(sq.m.K)\"\n",
"print\"(c)The pressure drop for tube side is \",round(dP_p,2),\"Pa\"\n",
@@ -326,10 +275,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the outlet temperature of the distilled water and the pressure drop for each stream. \n",
"\n",
- "#Given\n",
- "# properties of (distilled) water at 104\u00b0F from appendix table CII\n",
"rou_1= 0.994*62.4 # density in lbm/ft**3 \n",
"cp_1=0.998 # specific heat BTU/(lbm-degree Rankine) \n",
"v_1= 0.708e-5 # viscosity in ft**2/s \n",
@@ -338,7 +284,6 @@
"Pr_1 = 4.34 # Prandtl Number \n",
"m_1=170000 # mass flow rate in lbm/hr\n",
"T1=110.0 # temperature in degree F\n",
- "# properties of (raw) water at 68\u00b0F from Appendix Table C11\n",
"rou_2= 62.4 # density in lbm/ft**3 \n",
"cp_2=0.9988 # specific heat BTU/(lbm-degree Rankine) \n",
"v_2= 1.083e-5 # viscosity in ft**2/s \n",
@@ -347,40 +292,32 @@
"Pr_2 = 7.02 # Prandtl Number \n",
"m_2=150000 # mass flow rate in lbm/hr\n",
"t1=65 # temperature in degree F\n",
- "# specifications of 3/4-in-OD, 18-BWG tubes, from table 9.2\n",
"OD=3/(4*12.0)\n",
"ID=0.652/12.0\n",
"OD_p=1.375/12.0\n",
"Nt=224.0 # from table 9.3\n",
"Np=2 # no. of tube passes\n",
- "# Shell dimensions and other miscellaneous data\n",
"Ds=17.25/12.0\n",
"Nb=15.0 # no. of baffles\n",
"B=1\n",
"sT=15/(16*12.0)\n",
"C=sT-OD\n",
"\n",
- "#CALCULATION\n",
"import math\n",
- "# flow areas\n",
"At=(Nt*math.pi*ID**2)/(4*Np)\n",
"As=(Ds*C*B)/sT\n",
"\n",
- "# Shell Equivalent Diameter \n",
"De=4*((sT/2.0)*(0.86*sT)-(math.pi*OD**2/8.0))/(math.pi*OD/2.0)\n",
"\n",
- "# Reynolds Numbers \n",
"Re_s=(m_1/3600.0)*(De)/(v_1*rou_1*As)\n",
"Re_t=(m_2/3600.0)*(ID)/(v_2*rou_2*At)\n",
"\n",
- "# Nusselt numbers\n",
"Nu_t=0.023*(Re_t)**(0.8)*(Pr_2)**0.4\n",
"Nu_s=0.36*(Re_s)**(0.55)*(Pr_1)**(1/3.0)\n",
"h_ti=Nu_t*kf_2/ID\n",
"h_to=h_ti*ID/OD\n",
"h_s=Nu_s*kf_1/De\n",
"\n",
- "# Exchanger Coefficient \n",
"Uo=1/((1/h_to)+(1/h_s))\n",
"R=(m_2*cp_2)/(m_1*cp_1)\n",
"L=16\n",
@@ -389,20 +326,16 @@
"S=0.58 #value of S from fig. 9.13 Ten Broeck graph corresponding to the value of (UoAo)/(McCpc)\n",
"t2=S*(T1-t1)+t1\n",
"T2=T1-R*(t2-t1)\n",
- "#friction factors\n",
"ft=0.029 #friction factor for raw water fom figure 6.14 corresponding to Reynolds Number calculated above\n",
"fs=0.281 #friction factor for distilled water fom figure 6.14 corresponding to Reynolds Number calculated above\n",
"\n",
- "# Velocities\n",
"V_t=(m_2/3600.0)/(rou_2*At)\n",
"V_s=(m_1/3600.0)/(rou_1*As)\n",
"\n",
- "# pressure drops\n",
"gc=32.2\n",
"dP_t=(rou_2*V_t**2)*((ft*L*Np/ID)+4*Np)/(2*gc)\n",
"dP_s=((rou_1*V_s**2)*(fs*Ds*(Nb+1)))/(2*gc*De)\n",
"\n",
- "#Result\n",
"print\"Outlet Temperatures of raw water is \",round(t2,1),\"F\"\n",
"print\"Outlet Temperatures of distilled water is \",round(T2,1),\"F\"\n",
"print\"\\nThe pressure drop for tube side is\",round(dP_t/147,1),\"psi\"\n",
@@ -424,26 +357,18 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Using the effectiveness-NTU method to calculate the outlet temperatures of the fluids\n",
"\n",
- "#Given\n",
- "# Data from Example 9.5\n",
- "# properties of (distilled) water at 104\u00b0F \n",
"m_1=170000 \t\t# mass flow rate in lbm/hr\n",
"T1=110 \t\t\t# temperature in degree F\n",
"cp_1=0.998\t\t # specific heat BTU/(lbm-degree Rankine) \n",
- "# properties of (raw) water at 68\u00b0F \n",
"m_2=150000 \t\t# mass flow rate in lbm/hr\n",
"t1=65 \t\t\t# temperature in degree F\n",
"cp_2=0.9988\t # specific heat BTU/(lbm-degree Rankine) \n",
"Uo=350 \t\t\t# exchanger coefficient\n",
"Ao=703.7\n",
- "# The effectiveness-NTU approach is used when the overall heat transfer coefficient is known\n",
- "# determining the capacitances\n",
"mcp_raw=m_2*cp_2\n",
"mcp_distilled=m_1*cp_1\n",
"\n",
- "# determination of parameters for determining effectiveness\n",
"mcp_min_max=mcp_raw/mcp_distilled\n",
"UA_mcpmin=(Uo*Ao)/(mcp_raw)\n",
"effectiveness=0.58 \t\t#value of effectiveness from figure 9.15 corresponding to the above calculated values of capacitance ratio and (UoAo/mcp_min)\n",
@@ -452,7 +377,6 @@
"t2=(q/mcp_raw)+t1\n",
"T2=T1-(q/mcp_distilled)\n",
"\n",
- "#Result\n",
"print\"The Outlet temperature is Raw Water is\",round(t2,1),\"F\"\n",
"print\"The Outlet temperature is disilled Water is\",round(T2,1),\"F\"\n"
],
@@ -482,8 +406,6 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# (a) Determine the UA product for the exchanger. (b) Calculate the exit temperatures for the exchanger, assuming that only the inlet temperatures are known\n",
- "# properties of engine oil at (190 + 158)/2 = 174\u00b0F = 176 degree F from appendix table C4\n",
"rou_1= 0.852*62.4\t\t # density in lbm/ft**3 \n",
"cp_1=0.509\t\t # specific heat BTU/(lbm-degree Rankine) \n",
"v_1=0.404e-3 \t\t# viscosity in ft**2/s \n",
@@ -491,10 +413,8 @@
"a_1=2.98e-3 \t# diffusivity in ft**2/hr \n",
"Pr_1=490.0 \t\t# Prandtl Number \n",
"m_1=39.8 \t\t # mass flow rate in lbm/min\n",
- "# temperatures in degree F\n",
"T1=190.0\n",
"T2=158.0\n",
- "# properties of air at (126 + 166)/2 = 146\u00b0F = 606 degree R from appendix table D1\n",
"rou_2= 0.0653\t\t # density in lbm/ft**3 \n",
"cp_2=0.241\t\t # specific heat BTU/(lbm-degree Rankine) \n",
"v_2= 20.98e-5 \t\t# viscosity in ft**2/s \n",
@@ -502,38 +422,29 @@
"a_2 = 1.066 \t\t# diffusivity in ft**2/hr \n",
"Pr_2 = 0.706 \t\t# Prandtl Number \n",
"m_2=67.0 \t\t\t# mass flow rate in lbm/min\n",
- "# temperatures in degree F\n",
"t1=126.0\n",
"t2=166.0\n",
- "# Heat Balance\n",
"q_air=m_2*cp_2*60*(t2-t1)\n",
"q_oil=m_1*cp_1*60*(T1-T2)\n",
"\n",
- "# for counterflow\n",
"import math\n",
"LMTD=((T1-t2)-(T2-t1))/(math.log((T1-t2)/(T2-t1)))\n",
- "# Frontal Areas for Each Fluid Stream\n",
"Area_air=(9.82*8)/144.0\n",
"Area_oil=(3.25*9.82)/144.0\n",
"\n",
- "# Correction Factors (parameters calculated first)\n",
"S=(t2-t1)/(T1-t1)\n",
"R=(T1-T2)/(t2-t1)\n",
"F=0.87 #value of correction factor from figure 9.21a corresponding to above calculated values of S and R\n",
- "# Overall Coefficient (q = U*A*F*LMTD)\n",
"UA=q_air/(F*LMTD)\n",
- "# determining the capacitances\n",
"mcp_air=m_2*cp_2*60\n",
"mcp_oil=m_1*cp_1*60\n",
"\n",
- "# determination of parameters for determining effectiveness\n",
"mcp_min_max=mcp_air/mcp_oil\n",
"NTU=(UA/mcp_air)\n",
"effectiveness=0.62 \t\t#effectiveness from fig 9.21b corresponding to the values of capacitance ratio \n",
"t2_c=(T1-t1)*effectiveness+t1\n",
"T2_c=T1-(mcp_air)*(t2_c-t1)/(mcp_oil)\n",
"\n",
- "#Result\n",
"print\"The Overall Coefficient is \",round(UA,0),\" BTU/(hr. degree R)\"\n",
"print\"Calculated outlet temprature are:\"\n",
"print\"Outlet temprature for air\",round(t2_c,1),\"F\"\n",
diff --git a/Engineering_Heat_Transfer/CHAPTER_3.ipynb b/Engineering_Heat_Transfer/CHAPTER_3.ipynb
index 93f29ad1..98a7ff3d 100644
--- a/Engineering_Heat_Transfer/CHAPTER_3.ipynb
+++ b/Engineering_Heat_Transfer/CHAPTER_3.ipynb
@@ -27,12 +27,8 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat-flow rate from one tube \n",
- "# specifications of 1 standard type K from table F2\n",
"\n",
- "#iven\n",
"OD=0.02858 # outer diameter in m\n",
- "# from figure 3.11\n",
"M=8.0 # total number of heat-flow lanes\n",
"N=6.0 # number of squares per lane\n",
"S_L=M/N # conduction shape factor\n",
@@ -40,11 +36,9 @@
"T1=85 # temperature of tube surface\n",
"T2=0 # temperature of ground beneath the slab\n",
"\n",
- "#Calculation\n",
"q_half=k*S_L*(T1-T2)\n",
"q=2*q_half\n",
"\n",
- "#Result\n",
"print\"The total heat flow per tube is\",round(q,0),\" W/m\"\n"
],
"language": "python",
@@ -72,11 +66,7 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat transferred from the buried pipe per unit length\n",
- "# shape factor number 8 is selected from table 3.1\n",
- "# specifications of 10 nominal, schedule 80 pipe from table F1\n",
"\n",
- "#Given\n",
"import math\n",
"OD=10.74/12 # diameter in ft\n",
"R=OD/2\n",
@@ -113,34 +103,25 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the heat lost through the walls, using the shape-factor method. \n",
- "#(b) Repeat the calculations but neglect the effects of the corners that is, . \n",
"\n",
- "#Given\n",
"k = 1.07 # thermal conductivity of silica brick from appendix table B3 in W/(m.K)\n",
"S1_A=0.138*0.138/0.006\n",
"nA=2\n",
"\n",
- "# Calculation of total shape factor\n",
- "# From figure 3.12, for component A\n",
"St_A=nA*S1_A # Total shape factor of component A\n",
"\n",
- "# For component B\n",
"S1_B=0.138*0.188/0.006\n",
"nB=4\n",
"St_B=nB*S1_B # Total shape factor of component B\n",
"\n",
- "# For component C\n",
"S3_C=0.15*0.006\n",
"nC=8\n",
"St_C=nC*S3_C # Total shape factor of component C\n",
"\n",
- "# For component D\n",
"S2_D=0.54*0.188\n",
"nD=4\n",
"St_D=nD*S2_D # Total shape factor of component D\n",
"\n",
- "# For component E\n",
"S2_E=0.138*0.54\n",
"nE=8\n",
"St_E=nE*S2_E # Total shape factor of component E\n",
@@ -153,9 +134,7 @@
"q_1=k*S*(T1-T2)\n",
"Error=(q-q_1)/q\n",
"\n",
- "#result\n",
"print\"(a)The heat transferred through the walls of the furnace is \",round(q/1000,1),\"kw\"\n",
- "# Neglecting the effects of the edges and corners, the shape factor for all walls is found as \n",
"print\"(b The heat transferred is\",q_1/1000,1,\"kw\"\n",
"print\" The error introduced by neglecting heat flow through the edges and corners is \",round(Error*100,1),\" percent\"\n"
],
@@ -186,26 +165,17 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "# Determination of the conduction shape factor for the underground portion of the configuration\n",
- "# specifications of 4 nominal, schedule 40 pipe from table F1\n",
"\n",
- "#Given\n",
"OD=4.5/12.0 # diameter in ft\n",
"R=OD/2.0\n",
"\n",
- "# For pipe A\n",
- "#calculation\n",
"import math\n",
"L_A=4.5 # length in ft\n",
- "# shape factor number 9 is selected from table 3.1\n",
"S_A=(2*math.pi*L_A)/(math.log(2*(L_A)/R))\n",
- "# For pipe B\n",
"L_B=18 # length in ft\n",
- "# shape factor number 9 is selected from table 3.1\n",
"S_B=(2*math.pi*L_B)/(math.acosh(L_A/R))\n",
"S=2*S_A+S_B\n",
"\n",
- "#Result\n",
"print\"The total conduction shape factor for the system is\",round(S,1)\n"
],
"language": "python",
@@ -243,18 +213,13 @@
"cell_type": "code",
"collapsed": false,
"input": [
- "#(a)plot graph the temperature distribution existing within the rod. \n",
- "#(b) Use the numerical formulation of this section to obtain the temperature distribution.\n",
- "#(c) Compare the two models to determine how well the numerical results approximate the exact results\n",
"\n",
- "#Given\n",
"h=1.1 # convective coefficient in BTU/(hr.ft^2. degree R)\n",
"Tw=200.0\n",
"T_inf=68.0 # ambient temperature\n",
"k=0.47 # thermal conductivity in BTU/(hr.ft.degree R) from table B3\n",
"D=0.25/12 # diameter in ft\n",
"\n",
- "#Calculation\n",
"A=math.pi*D**(2)/4.0 # cross sectional area in ft^2\n",
"P=math.pi*D # perimeter in ft\n",
"L=6/12.0 # length in ft\n",
@@ -264,19 +229,16 @@
"de=dz/L\n",
"K=2+(mL*de)**2\n",
"\n",
- "#Tempature can be calculated as\n",
"T4=T_inf+(Tw-T_inf)*(2/(K**4-4*K**2+2))\n",
"T3=T_inf+(Tw-T_inf)*(K/(K**4-4*K**2+2))\n",
"T2=T_inf+(Tw-T_inf)*((K**2-1)/(K**4-4*K**2+2))\n",
"T1=T_inf+(Tw-T_inf)*((K**3-3*K)/(K**4-4*K**2+2))\n",
"\n",
- "#result\n",
"print\"The temprature distribution is T4=\",round(T4,2),\"F\"\n",
"print\"The temprature distribution is T3=\",round(T3,2),\"F\"\n",
"print\"The temprature distribution is T2=\",round(T2,2),\"F\"\n",
"print\"The temprature distribution is T1=\",round(T1,2),\"F\"\n",
"\n",
- "#(b)plot\n",
"import matplotlib.pyplot as plt\n",
"import numpy as np\n",
"T=[200.0,77.33,68.66,68.05,68]\n",
generated by cgit (git 2.34.1) at 2025-06-02 22:02:40 +0000