{
"metadata": {
"name": "",
"signature": "sha256:e60aa082c3bd1f8f9672b5ca74c6357f1b26803a7fb296a1d0a644ab9efb4ea1"
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"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 19: Insulation and Refractory"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.1, Page number: 413"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"H = 2.5 #Height of wall (m)\n",
"W = 4 #Width of wall (m)\n",
"h = 11 #Convective heat transfer coefficient (W/m^2.K)\n",
"T1 = 24 #Outside surface temperature (\u00b0C)\n",
"T3 = -15 #Outside air temperature (\u00b0C)\n",
"L = 7.62/10**3 #Insulation thickness (m)\n",
"k = 0.04 #Thermal conductivity of wool (W/m.K)\n",
"\n",
"#Calculation:\n",
"A = H*W #Heat transfer area (m^2)\n",
"Q = h*A*(T1-T3) #Heat transfer rate (W)\n",
"Ri = L/(k*A) #Insuation resistance (K/W)\n",
"Rc = 1/(h*A) #Convective resitance (K/W)\n",
"R = Ri+Rc #Total resistance (K/W)\n",
"Qt = (T1-T3)/R #Revised heat transfer rate (Btu/h)\n",
" \n",
"#Result:\n",
"print \"1. The heat transfer rate without insulation is :\",round(Q),\" W .\"\n",
"print \"Or, the heat transfer rate without insulation is :\",round(Q*3.412),\" Btu/h .\"\n",
"print \"2. The revised heat transfer rate with insulation is :\",round(Qt),\" W .\"\n",
"print \"Or, the revised heat transfer rate with insulation is :\",round(Qt*3.412),\" Btu/h .\"\n",
"print \"There is a calculation mistake in book.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1. The heat transfer rate without insulation is : 4290.0 W .\n",
"Or, the heat transfer rate without insulation is : 14637.0 Btu/h .\n",
"2. The revised heat transfer rate with insulation is : 1386.0 W .\n",
"Or, the revised heat transfer rate with insulation is : 4729.0 Btu/h .\n",
"There is a calculation mistake in book.\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.2, Page number: 414"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#Variable declaration:\n",
"#From example 19.1:\n",
"T1 = 24 #Outside surface temperature (\u00b0C)\n",
"Ri = 0.0191 #Insulation resistance (K/W)\n",
"Q = 1383 #Revised heat transfer rate (Btu/h)\n",
"\n",
"#Calculation:\n",
"T2 = T1-Q*Ri #Temperature at outer surface of insulation (\u00b0C)\n",
"\n",
"#Result:\n",
"print \"The temperature at the outer surface of the insulation is :\",round(T2,1),\" \u00b0C .\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature at the outer surface of the insulation is : -2.4 \u00b0C .\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.3, Page number: 415"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#Variable declaration:\n",
"#From example 19.1:\n",
"h = 11 #Convective heat transfer coefficient (W/m^2.K)\n",
"L = 7.62/10**3 #Insulation thickness (m)\n",
"k = 0.04 #Thermal conductivity of wool (W/m.K)\n",
"\n",
"#Calculation:\n",
"Bi = h*L/k #Biot number\n",
"\n",
"#Result:\n",
"print \"The Biot nmuber is :\",round(Bi,1),\" .\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Biot nmuber is : 2.1 .\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.4, Page number: 415"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"k = 0.022 #Thermal conductivity of glass wool (Btu/h.ft.\u00b0F)\n",
"T1 = 400 #Inside wall temperature (\u00b0F)\n",
"T2 = 25 #Outside wall temperature (\u00b0C)\n",
"L = 3/12 #Length of insulation cover (ft)\n",
"\n",
"#Calculation:\n",
"T_2 = T2*(9/5)+32 #Outside wall temperature in fahrenheit scale (\u00b0F)\n",
"QbyA = k*(T1-T_2)/L #Heat flux across the wall (Btu/h.ft^2)\n",
"\n",
"#Result:\n",
"print \"The heat flux across the wall is :\",round(QbyA,1),\" Btu/h.ft^2 .\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The heat flux across the wall is : 28.4 Btu/h.ft^2 .\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.5, Page number: 415"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"w = 8 #Width of wall (m)\n",
"H = 3 #Height of wall (m)\n",
"h = 21 #Convective heat transfer coefficient between the air and the surface (W/m^2.K)\n",
"T1 = -18 #Outside surace of wall temperature (\u00b0C)\n",
"T3 = 26 #Surrounding air temperature (\u00b0C)\n",
"l1 = 80/100 #Reduction in cooling load\n",
"k = 0.0433 #Thermal conductivity of cork board insulation (W/m.K)\n",
"T = 12000 #Units Btu/h in 1 ton of refrigeration\n",
"\n",
"#Calculation:\n",
"A = w*H #Heat transfer area (m^2) (part 1)\n",
"Q1 = h*A*(T1-T3) #Rate of heat flow in the absence of insulation (W)\n",
"Q2 = Q1*3.4123/T #Rate of heat flow in the absence of insulation (ton of refrigeration)\n",
"l2 = 1-l1 #Reduced cooling load (part 2)\n",
"Q3 = l2*Q1 #Heat rate with insulation (W)\n",
"Rt = (T1-T3)/Q3 #Total thermal resistance (\u00b0C/W)\n",
"R2 = 1/(h*A) #Convection thermal resistance (\u00b0C/W)\n",
"R1 = Rt-R2 #Insulation conduction resistance (\u00b0C/W)\n",
"L = R1*k*A #Required insulation thickness (m)\n",
"\n",
"#Result:\n",
"print \"1. The rate of heat flow through the rectangular wall without insulation is :\",round(Q1/10**3,2),\" kW .\"\n",
"print \"Or, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is :\",round(Q2,1),\" ton of refrigeration .\"\n",
"if (Q1<0):\n",
" print \" The negative sign indicates heat flow from the surrounding air into the cold room.\"\n",
"else :\n",
" print \" The positive sign indicates heat flow from the surrounding air into the cold room.\"\n",
"print \"2. The required thickness of the insulation board is :\",round(L*10**3,2),\" mm .\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1. The rate of heat flow through the rectangular wall without insulation is : -22.18 kW .\n",
"Or, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is : -6.3 ton of refrigeration .\n",
" The negative sign indicates heat flow from the surrounding air into the cold room.\n",
"2. The required thickness of the insulation board is : 8.25 mm .\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.6, Page number: 417"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#Variable declaration:\n",
"#From example 19.5:\n",
"Q = -4435.2 #Heat rate with insulation (W)\n",
"R2 = 0.00198 #Convection thermal resistance (\u00b0C/W)\n",
"T3 = 26 #Surrounding air temperature (\u00b0C)\n",
"h = 21 #Convective heat transfer coefficient between the air and the surface (W/m^2.K)\n",
"k = 0.0433 #Thermal conductivity of cork board insulation (W/m.K)\n",
"L = 0.00825 #Required insulation thickness (m)\n",
"\n",
"#Calculation:\n",
"T2 = T3+Q*R2 #Interface temperature (\u00b0C) (part 1)\n",
"Bi = h*L/k #Biot number (part 2)\n",
"\n",
"#Result:\n",
"print \"1. The interface temperature is :\",round(T2,2),\" \u00b0C .\"\n",
"print \"2. The Biot number is :\",round(Bi),\" .\"\n",
"print \"3. Theoretical part.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1. The interface temperature is : 17.22 \u00b0C .\n",
"2. The Biot number is : 4.0 .\n",
"3. Theoretical part.\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.7, Page number: 417"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"from math import pi,log\n",
"\n",
"#Variable declaration:\n",
"D2 = 0.5/10**3 #External diameter of needle (m)\n",
"h3 = 12 #Heat transfer coefficient (W/m^2.K)\n",
"L = 1 #Insulation thickness (m)\n",
"T1 = 95 #Reactant temperature (\u00b0C)\n",
"T3 = 20 #Ambient air temperature (\u00b0C)\n",
"k1 = 16 #Thermal conductivity of needle (W/m.K)\n",
"k3 = 0.0242 #Thermal conductivity of air (W/m.K)\n",
"D3 = 2/10**3 #Diameter of rubber tube (m)\n",
"\n",
"#Calculation:\n",
"r2 = D2/2 #External radius of needle (m)\n",
"r3 = D3/2 #Radius of rubber tube (m)\n",
"Rt1 = 1/(h3*(2*pi*r2*L)) #Thermal resistance (\u00b0C/W)\n",
"Q1 = (T1-T3)/Rt1 #Rate of heat flow in the absence of insulation (W)\n",
"Bi = h3*D2/k1 #Biot number \n",
"Nu = h3*D2/k3 #Nusselt number\n",
"R2 = log(r3/r2) #Thermal resistance of needle (\u00b0C/W)\n",
"R3 = 1/(h3*(2*pi*r3*L)) #Thermal resistance of rubber tube (\u00b0C/W)\n",
"Rt2 = R2+R3 #Total thermal resistance (\u00b0C/W)\n",
"Q2 = (T1-T3)/Rt2 #Rate of heat loss (W)\n",
"\n",
"#Result:\n",
"print \"1. The rate of the heat loss from the hypodermic needle with the rubber insulation is :\",round(Q1,2),\" W .\"\n",
"print \" The rate of the heat loss from the hypodermic needle without the rubber insulation is :\",round(Q2,2),\" W .\"\n",
"print \"2. The Biot number is :\",round(Bi,6),\" .\"\n",
"print \" The nusselt number is :\",round(Nu,3),\" .\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1. The rate of the heat loss from the hypodermic needle with the rubber insulation is : 1.41 W .\n",
" The rate of the heat loss from the hypodermic needle without the rubber insulation is : 5.12 W .\n",
"2. The Biot number is : 0.000375 .\n",
" The nusselt number is : 0.248 .\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.9, Page number: 420"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from __future__ import division\n",
"from math import log, pi\n",
"\n",
"#Variable declaration:\n",
"h = 140 #Convention heat transfer coefficient (W/m^2.K)\n",
"D1 = 10/10**3 #Rod diameter (m)\n",
"L = 2.5 #Rod length (m)\n",
"T1 = 200 #Surface temperature of rod (\u00b0C)\n",
"T2 = 25 #Fluid temperature (\u00b0C)\n",
"k = 1.4 #Thermal conductivity of bakellite (W/m.K)\n",
"l = 55/10**3 #Insulation thickness (m)\n",
"\n",
"#Calculation:\n",
"Q1 = h*pi*D1*L*(T1-T2) #Rate of heat transfer for the bare rod (W) (part 1)\n",
"Bi = 2 #Critical Biot number (part 2)\n",
"D2 = Bi*k/h #Critical diameter associated with the bakelite coating (m)\n",
"r2 = D2/2 #Critical radius associated with the bakelite coating (m)\n",
"r1 = D1/2 #Rod radius (m)\n",
"R1 = log(r2/r1)/(2*pi*k*L) #Insulation conduction resistance (\u00b0C/W)\n",
"R2 = 1/(h*(2*pi*r2*L)) #Convection thermal resistance (\u00b0C/W)\n",
"Rt1 = R1+R2 #Total thermal resistance (\u00b0C/W)\n",
"Qc = (T1-T2)/Rt1 #Heat transfer rate at the critical radius (W)\n",
"r3 = r1+l #New radius associated with the bakelite coating after insulation (m) (part 3)\n",
"R3 = log(r3/r1)/(2*pi*k*L) #Insulation conduction bakelite resistance (\u00b0C/W)\n",
"R4 = 1/(h*(2*pi*r3*L)) #Convection bakelite thermal resistance (\u00b0C/W)\n",
"Rt2 = R3+R4 #Total bakelite thermal resistance (\u00b0C/W)\n",
"Q2 = (T1-T2)/Rt2 #Heat transfer rate at the bakelite critical radius (W)\n",
"Re = ((Q1-Q2)/Q1)*100 #Percent reduction in heat transfer rate relative to the case of a bare rod (%)\n",
"\n",
"#Result:\n",
"print \"1. The rate of heat transfer for the bare rod is :\",round(Q1),\" W .\"\n",
"print \"2. The critical radius associated with the bakelite coating is :\",round(r2*10**3),\" mm.\"\n",
"print \" & the heat transfer rate at the critical radius is :\",round(Qc),\" W .\"\n",
"print \"3. The fractional reduction in heat transfer rate relative to the case of a bare rod is :\",round(Re,1),\" % .\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1. The rate of heat transfer for the bare rod is : 1924.0 W .\n",
"2. The critical radius associated with the bakelite coating is : 10.0 mm.\n",
" & the heat transfer rate at the critical radius is : 2273.0 W .\n",
"3. The fractional reduction in heat transfer rate relative to the case of a bare rod is : 24.6 % .\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.10, Page number: 421"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from __future__ import division\n",
"from math import pi, log\n",
"\n",
"#Variable declaration:\n",
"r1 = 1.1/100 #Inside radius of pipe (m)\n",
"r2 = 1.3/100 #Outside radius of pipe (m)\n",
"r3 = 3.8/100 #Outside radius of asbestos insulation (m)\n",
"L = 1 #Length of tube (m)\n",
"h1 = 190 #Heat transfer coefficient from ethylene glycol to the stainless steel pipe (W/m^2.K)\n",
"k2 = 19 #Thermal conductivity of pipe (W/m.K)\n",
"h2 = 14 #Outside heat transfer coefficient from the air to the surface of the insulation (W/m^2.K)\n",
"k3 = 0.2 #Thermal conductivity of asbestos (W/m.K)\n",
"T1 = 124 #Hot ethylene glycol temperature (\u00b0C)\n",
"T5 = 2 #Surrounding air temperature (\u00b0C)\n",
"k4 = 0.0242 #Thermal conductivity of air (W/m.K)\n",
"\n",
"#Calculation:\n",
"A1 = 2*pi*r1*L #Inside surface area of pipe (m^2) (part1)\n",
"A2 = 2*pi*r2*L #Outside surface area of pipe (m^2)\n",
"A3 = 2*pi*r3*L #Outside surface area of asbestos insulation (m^2)\n",
"R1 = 1/(h1*A1) #Inside convection resistance (\u00b0C/W)\n",
"R2 = log(r2/r1)/(2*pi*k2*L) #Conduction resistance through the tube (\u00b0C/W)\n",
"R3 = 1/(h2*A2) #Outside convection resistance (\u00b0C/W)\n",
"Rt1 = R1+R2+R3 #Total resistance without insulation (\u00b0C/W)\n",
"Q1 = (T1 - T5)/Rt1 #Heat transfer rate without insulation (W)\n",
"R4 = log(r3/r2)/(2*pi*k3*L) #Conduction resistance associated with the insulation (\u00b0C/W) (part 2)\n",
"R5 = 1/(h2*A3) #Outside convection resistance (\u00b0C/W)\n",
"Rt2 = R1+R2+R4+R5 #Total rsistance with the insulation (\u00b0C/W)\n",
"Q2 = (T1-T5)/Rt2 #Heat transfer rate with the insulation (W)\n",
"U1 = 1/(Rt2*A1) #Overall heat transfer coefficient based on the inside area (W/m^2.K) (part 3)\n",
"U3 = 1/(Rt2*A3) #Overall heat transfer coefficient based on the outside area (W/m^2.K) (part 4)\n",
"T3 = T1-(R1+R2)*Q2 #Temperature at the steel\u2013insulation interface (\u00b0C) (part 5)\n",
"Bi1 = h2*(2*r3)/k3 #Outside Biot number (part 6)\n",
"Bi2 = h1*(2*r1)/k2 #Inside Biot number\n",
"Nu = h1*(2*r1)/k4 #Nusselt number of the air\n",
"rlm = (r3-r2)/log(r3/r2) #Log mean radius of the insulation (m) (part 7)\n",
"\n",
"#Result:\n",
"print \"1. The rate of heat transfer without insulation is :\",round(Q1,1),\" W.\"\n",
"print \"2. The rate of heat transfer with insulation is :\",round(Q2,1),\" W.\"\n",
"print \"3. The overall heat transfer coefficient based on the inside area of the tube is :\",round(U1,2),\" W/m^2.K .\"\n",
"print \"4. The overall heat transfer coefficient based on the outside area of the insulation is :\",round(U3,1),\" W/m^2.K .\"\n",
"print \"5. The temperature, T3, at the steel\u2013insulation interface is :\",round(T3,1),\" \u00b0C.\"\n",
"print \"6. The inside Biot numbers is :\",round(Bi2,2),\" .\"\n",
"print \" The outside Biot numbers is :\",round(Bi1,2),\" .\"\n",
"print \" The Nusselt number is :\",round(Nu,1),\" .\"\n",
"print \"7. The log mean radius of insulation is :\",round(rlm*100,2),\" cm.\"\n",
"print \"There is a printing mistake in book for unit in part 7.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1. The rate of heat transfer without insulation is : 128.1 W.\n",
"2. The rate of heat transfer with insulation is : 99.2 W.\n",
"3. The overall heat transfer coefficient based on the inside area of the tube is : 11.76 W/m^2.K .\n",
"4. The overall heat transfer coefficient based on the outside area of the insulation is : 3.4 W/m^2.K .\n",
"5. The temperature, T3, at the steel\u2013insulation interface is : 116.3 \u00b0C.\n",
"6. The inside Biot numbers is : 0.22 .\n",
" The outside Biot numbers is : 5.32 .\n",
" The Nusselt number is : 172.7 .\n",
"7. The log mean radius of insulation is : 2.33 cm.\n",
"There is a printing mistake in book for unit in part 7.\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.11, Page number: 424"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from __future__ import division\n",
"from math import pi\n",
"\n",
"#Variable declaration:\n",
"h1 = 800 #Heat transfer coefficient for steam condensing inside coil (Btu/h.ft^2.\u00b0F)\n",
"h2 = 40 #Heat transfer coefficient for oil outside coil (Btu/h.ft^2.\u00b0F)\n",
"h3 = 40 #Heat transfer coefficient for oil inside tank wal (Btu/h.ft^2.\u00b0F)\n",
"h4 = 2 #Heat transfer coefficient for outer tank wall to ambient air (Btu/h.ft^2.\u00b0F)\n",
"k1 = 0.039 #Thermal conductivity of insulation layer (Btu/h.ft.\u00b0F)\n",
"l1 = 2/12 #Thickness of insulation layer (ft)\n",
"D = 10 #Diameter of tank (ft)\n",
"H = 30 #Height of tank (ft)\n",
"k2 = 224 #Thermal conductivity of copper tube (Btu/h.ft.\u00b0F)\n",
"l2 = (3/4)/12 #Thickness of insulation layer (ft)\n",
"T1 = 120 #Temperature of tank (\u00b0F)\n",
"T2 = 5 #Outdoor temperature (\u00b0F)\n",
"\n",
"#Calculation:\n",
"Uo1 = 1/(1/h3+(l1/k1)+1/h4) #Overall heat transfer coefficient for tank (Btu/h.ft^2.\u00b0F)\n",
"At = pi*(D+2*l1)*H #Surface area of tank (ft^2)\n",
"Q = Uo1*At*(T1-T2) #Heat transfer rate lost from the tank (Btu/h)\n",
"#From table 6.3:\n",
"l2 = 0.049/12 #Thickness of coil (ft)\n",
"A = 0.1963 #Area of 18 guage, 3/4-inch copper tube (ft^2/ft)\n",
"Uo2 = 1/(1/h2+(l2/k2)+1/h1) #Overall heat transfer coefficient for coil (Btu/h.ft^2.\u00b0F)\n",
"#From steam tables:\n",
"Tst = 240 #Temperature for 10 psia (24.7 psia) steam (\u00b0F)\n",
"Ac = Q/(Uo2*(Tst-T1)) #Area of tube (ft^2)\n",
"L = Ac/A #Lengt of tube (ft)\n",
"\n",
"#Result:\n",
"print \"The length ofcopper tubing required is :\",round(L,1),\" ft .\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The length ofcopper tubing required is : 26.0 ft .\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.12, Page number: 426"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"from __future__ import division\n",
"from math import pi, log\n",
"from numpy import array,log as LOG\n",
"\n",
"#Variable declaration:\n",
"#For 1-inch pipe schedule 40:\n",
"Di = 1.049/12 #Inside diameter (ft)\n",
"Do = 1.315/12 #Outside diameter (ft)\n",
"L = 8000 #Length of pipe (ft)\n",
"hi = 2000 #Heat transfer coefficient inside of the pipe (Btu/h.ft^2.\u00b0F)\n",
"ho = 100 #Outside heat transfer coefficient (Btu/h.ft.\u00b0F)\n",
"kl = 0.01 #Thermal conductivity of insulation (Btu/h.ft.\u00b0F)\n",
"T1 = 240 #Steam temperature (\u00b0F)\n",
"T2 = 20 #Air temperature (\u00b0F)\n",
"k = 24.8 #Thermal conductivity for steel (Btu/h.ft.\u00b0F)\n",
"Dxl = array([3/8,1/2,3/4,1])/12 #thickness(ft)\n",
"amt = array([1.51,3.54,5.54,8.36])/6 #Cost per feet($) \n",
"\n",
"#Calculation:\n",
"D_ = (Do-Di)/log(Do/Di) #Log-mean diameter of the pipe (ft)\n",
"Dl = Do+2*(Dxl) #Insulation thickness (ft)\n",
"D_l = (Dl-Do)/LOG(Dl/Do) #Log mean diameter of pipe (ft)\n",
"Dxw = (Do-Di)/2 #Pipe thickness (ft)\n",
"Rw = Dxw/(k*pi*D_*L) #Wall resistance ((Btu/h.\u00b0F)^-1)\n",
"Ri = 1/(hi*pi*Di*L) #Inside steam convection resistance ((Btu/h.\u00b0F)^-1)\n",
"Rl = Dxl/(kl*pi*D_l*L) #Insulation resistance ((Btu/h.\u00b0F)^-1)\n",
"Ro = 1/(ho*pi*Dl*L) #Outside air convection resistance ((Btu/h.\u00b0F)^-1)\n",
"R = Ri+Rw+Rl+Ro #Total resistance ((Btu/h.\u00b0F)^-1)\n",
"Uo = 1/(R*pi*Dl*L) #Overall outside heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n",
"Ui = 1/(R*pi*Di*L) #Overall inside heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n",
"dT = T1-T2\n",
"Ai = pi*Di*L #Inside area (ft^2)\n",
"Q = Ui*Ai*dT #Energy loss (Btu/h)\n",
"def energyPerDollar(Q1,Q2,amt1,amt2):\n",
" return round((Q1-Q2)/(8000*(amt2-amt1)),1)\n",
"\n",
"#Results:\n",
"print \"Energy saved per dollar ingoing from 3/8 to 1/2 inch is :\",energyPerDollar(Q[0],Q[1],amt[0],amt[1]),' Btu/h.$'\n",
"print \"Energy saved per dollar ingoing from 1/2 to 3/4 inch is :\",energyPerDollar(Q[1],Q[2],amt[1],amt[2]),' Btu/h.$'\n",
"print \"Energy saved per dollar ingoing from 3/4 to 1 inch is :\",energyPerDollar(Q[2],Q[3],amt[2],amt[3]),' Btu/h.$'"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Energy saved per dollar ingoing from 3/8 to 1/2 inch is : 18.2 Btu/h.$\n",
"Energy saved per dollar ingoing from 1/2 to 3/4 inch is : 18.8 Btu/h.$\n",
"Energy saved per dollar ingoing from 3/4 to 1 inch is : 6.8 Btu/h.$\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"ILLUSTRATIVE EXAMPLE 19.16, Page number: 434"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"ki = 0.44 #Thermal conductivity of insulation (Btu/h.ft.\u00b0F)\n",
"ho = 1.32 #Air flow coefficient (Btu/h.ft^2.\u00b0F)\n",
"OD = 2 #Outside diameter of pipe (in)\n",
"\n",
"#Calculation:\n",
"rc = (ki/ho)*12 #Outer critical radius of insulation (in)\n",
"ro = OD/2 #Outside radius of pipe (in)\n",
"L = rc-ro #Critical insulation thickness (in)\n",
"\n",
"#Result:\n",
"print \"The outer critical radius of insulation is :\",round(rc),\" in .\"\n",
"if rorc, the heat loss will decrease as insulation is added.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The outer critical radius of insulation is : 4.0 in .\n",
"Since, ro