{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Conduction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.1,Page no:2.18"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Thickness of insulation\n",
"\n",
"#Variable declaration\n",
"A=1 #Area of heat transfer[sq metre]\n",
"Q=450 #Rate of heat loss/unit area[W/ sq mtre]\n",
"dT=400 #Temperature difference across insulation layer[K]\n",
"k=0.11 #k for asbestos[W/(m.K)]\n",
"\n",
"#Calculation\n",
"#Q=(k* A*dT)/x\n",
"x1=(k*A*dT)/Q\n",
"X1=x1*1000 \n",
"#for fire clay insulation\n",
"k=0.84 #For fire clay insulation[W/(m.K)]\n",
"x=(k*A*dT)/Q \n",
"X=x*1000 \n",
"\n",
"#Result\n",
"print\"Ans.(A).Thickness of asbestos is:\",round(X1),\"mm\"\n",
"print\"Ans.(B)Thickness of fire clay insulation is:\",round(X),\"mm\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ans.(A).Thickness of asbestos is: 98.0 mm\n",
"Ans.(B)Thickness of fire clay insulation is: 747.0 mm\n"
]
}
],
"prompt_number": 125
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.2,Page no:2.18"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss per metre\n",
"\n",
"#Variable declaration\n",
"L=1.0 # Length of pipe[m]\n",
"r1=(50.0/2.0) # in mm\n",
"r1=r1/1000.0 # in m\n",
"r2=(25.0+3.0)/1000.0 # m\n",
"\n",
"\n",
"#Calculation\n",
"import math\n",
"rm1=(r2-r1)/math.log(r2/r1) \n",
"k1=45.0 #W/(m.K)\n",
"R1=(r2-r1)/(k1*(2*math.pi*rm1*L)) # Thermal resistance of wall pipe[K/W]\n",
"\n",
"#For inner lagging:\n",
"k2=0.08 #W/(m.K)\n",
"ri1=0.028 #m\n",
"ri2=(ri1+r1) # m\n",
"rmi1=(ri2-ri1)/math.log(ri2/ri1)\n",
"R2=(ri2-ri1)/(k2*2*math.pi*rmi1*L) #Thermal resistance of inner lagging [K/W]\n",
"\n",
"#For outer lagging:\n",
"k3=0.04 #W/(m.K)\n",
"ro1=0.053 #m\n",
"ro2=(ro1+0.04) # m\n",
"rmo1=(ro2-ro1)/math.log(ro2/ro1)\n",
"R3=(ro2-ro1)/(k3*2*math.pi*rmo1*L) #Thermal resistance of outer lagging\n",
"\n",
"R=R1+R2+R3\n",
"Ti=550.0 #K #inside\n",
"To=330.0 #K # outside\n",
"dT=Ti-To #Temperature difference\n",
"Q=dT/R\n",
"\n",
"\n",
"#Result\n",
"print\"Rate of heat loss per metre of pipe,Q=\",round(Q,1),\"W/m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Rate of heat loss per metre of pipe,Q= 62.7 W/m\n"
]
}
],
"prompt_number": 126
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.3,Page no:2.19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat Loss in pipe\n",
"import math\n",
"\n",
"#Variable declaration\n",
"r1=44.0 #i [mm]\n",
"r1=r1/1000.0 #i[m]\n",
"r2=0.094 #i [m]\n",
"r3=0.124 #i [m]\n",
"T1=623.0 #iTemperature at outer surface of wall in[K]\n",
"T3=313.0 #iTemperature at outer surface of outer insulation [K]\n",
"k1=0.087 #iThermal conductivity of insulation layer 1..in [W/m.K]\n",
"k2=0.064 #iThermal conductivity of insulation layer 2 [W/m.K]\n",
"l=1 #i Length of pipe [m]\n",
"\n",
"#Calculation\n",
"rm1=(r2-r1)/math.log(r2/r1) #imath.log mean radius of insulation layer 1 [m]\n",
"rm2=(r3-r2)/math.log(r3/r2) #imath.log mean radius of insulation layer 2[m]\n",
"#iPutting values in following eqn:\n",
"Q= (T1-T3)/((r2-r1)/(k1*2*math.pi*rm1*l)+(r3-r2)/(k2*2*math.pi*rm2*l)) \n",
"\n",
"#Result\n",
"print\"Heat loss per meter pipe is\",round(Q,1),\"W/m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss per meter pipe is 149.2 W/m\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.4,Page no:2.21"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss in interface\n",
"\n",
"#Variable declaration\n",
"A=1 #Heat transfer area [sq m]\n",
"x1=0.229 # thickness of fire brick in [m]\n",
"x2=0.115 # thickness of insulating brick in [m]\n",
"x3=0.229 # thickness of building brick in [m]\n",
"k1=6.05 #thermal conductivity of fir brick [W/(m.K)]\n",
"k2=0.581 #thermal conductivity of insulating brick [W/m.K]\n",
"k3=2.33 #thermal conductivity of building brick [W/m.K]\n",
"T1=1223 # inside temperature [K]\n",
"T2=323 # Outside temperature[K]\n",
"\n",
"#Calculation\n",
"dT=T1-T2 #Overall temp drop [K]\n",
"R1=(x1/k1*A) #thermal resistance 1\n",
"R2=(x2/k2*A) # Thermal resistance 2\n",
"R3=(x3/k3*A) #Thermal resistance 3\n",
"Q=dT/(R1+R2+R3) #w/SQ m\n",
"Ta=-((Q*R1)-T1) #from Q1=Q=(T1-Ta)/(x1/k1*A)\n",
"#Similarly\n",
"Tb=(Q*R3)+T2 \n",
"\n",
"#Result\n",
"print\"Interface temperature:\\ni-Between FB-IB=\",round(Ta),\" K \\nii-Between IB-PB=\",round(Tb,1),\"K\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Interface temperature:\n",
"i-Between FB-IB= 1121.0 K \n",
"ii-Between IB-PB= 587.8 K\n"
]
}
],
"prompt_number": 130
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.5,Page no:2.23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss per unit area\n",
"\n",
"#Variable declaration\n",
"A=1 #let [sq m]\n",
"x1=0.23 #thickness of fir brick layer[m]\n",
"x2=0.115 # [m]\n",
"x3=0.23 #[m]\n",
"T1=1213.0 #Temperature of furnace [K]\n",
"T2=318.0 #Temperature of furnace [K]\n",
"dT=T1-T2 #[K]\n",
"k1=6.047 #W/(m.K) (fire brick)\n",
"k2=0.581 #W/(m.K) (insulating brick)\n",
"k3=2.33 #W/(m.K) (building brick)\n",
"\n",
"#Calculation\n",
"Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3)) #Heat lost per unit Area in Watt\n",
"Q_by_A=round(Q_by_A,1)\n",
"R1=(x1/k1) #Thermal resistance\n",
"R2=(x2/k2)\n",
"R3=(x3/k3)\n",
"R1=round(R1,2)\n",
"R2=round(R2,1)\n",
"R3=round(R3,1)\n",
"Ta=T1-((dT*R1)/(R1+R2+R3))\n",
"Ta=round(Ta)\n",
"Tb=((dT*R3)/(R1+R2+R3))+T2\n",
"\n",
"\n",
"#Result\n",
"print\"Heat loss per unit area is\",Q_by_A,\"W=\",Q_by_A,\"J/s\\n\"\n",
"print\"Ta=\",Ta,\"K(APPROX) =Temperature at the interface between fire brick and insulating brick\"\n",
"print\"Tb=\",round(Tb),\"K Temperature at the interface between insulating and building brick\"\n",
"print\"\\nNOTE:Tb is wrongly calculated in the book as 565 K\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss per unit area is 2674.2 W= 2674.2 J/s\n",
"\n",
"Ta= 1108.0 K(APPROX) =Temperature at the interface between fire brick and insulating brick\n",
"Tb= 581.0 K Temperature at the interface between insulating and building brick\n",
"\n",
"NOTE:Tb is wrongly calculated in the book as 565 K\n"
]
}
],
"prompt_number": 159
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.7,Page no:2.26"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss in wall\n",
"#Part-(a)\n",
"\n",
"#Variable declaration\n",
"A=1.0 # sq metre\n",
"x1=114.0 # mm\n",
"x1=x1/1000.0 # metre\n",
"k1=0.138 # W/(m.K)\n",
"R1= x1/(k1*A)\n",
"x2=229.0 #mm\n",
"x2= x2/1000.0 # metre\n",
"k2=1.38 # W/m.K\n",
"R2=x2/(k2*A)\n",
"dT=1033.0-349.0\n",
"\n",
"#Calculation\n",
"#Heat loss\n",
"Q1=dT/(R1+R2)\n",
"#Part(b)\n",
"#contact resistance=cr\n",
"cr=0.09 #K/W\n",
"R=R1+R2+cr\n",
"Q=dT/R\n",
"\n",
"#Result\n",
"print\"Heat loss from 1 sq metre wall=\",round(Q1,1),\"W\"\n",
"print\"Heat loss from 1 sq metre when resistance present=\",round(Q,1),\"W\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss from 1 sq metre wall= 689.5 W\n",
"Heat loss from 1 sq metre when resistance present= 632.1 W\n"
]
}
],
"prompt_number": 161
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.8,Page no:2.27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Loss per area\n",
"\n",
"#Variable declaration\n",
"x1=0.02 #[m]\n",
"x2=0.01 #[m]\n",
"x3=0.02 #[m]\n",
"k1=0.105 #W/(m.k)\n",
"k3=k1 #W/(m.K)\n",
"k2=0.041 #W/(m.K)\n",
"T1=303\n",
"T2=263\n",
"\n",
"#Calculation\n",
"dT=T1-T2 #[K]\n",
"Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3))\n",
"R=0.625 #K/W\n",
"Tx=293 #K\n",
"Rx=0.9524 #K/W\n",
"x=R*(T1-Tx)/(dT*Rx)\n",
"x=x*100 #mm\n",
"\n",
"#Result\n",
"print\"The temperature of 293 K will be reached at point\",round(x,1),\"mm from the outermost wall surface of the ice-box\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The temperature of 293 K will be reached at point 16.4 mm from the outermost wall surface of the ice-box\n"
]
}
],
"prompt_number": 163
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.9,Page no:2.28"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss\n",
"\n",
"#Variable declaration\n",
"import math\n",
"ID=50.0 #Internal diameter[mm]\n",
"dT=(573.0-303.0) \n",
"r1=ID/2.0 #mm\n",
"r1=r1/1000.0 # metres\n",
"OD=150.0 #Outer diameter[mm]\n",
"r2=OD/2.0 # mm\n",
"r2=75.0/1000.0 # m\n",
"#Thermal conductivity\n",
"k=17.45 # W/(m.K) \n",
"\n",
"#Calculation\n",
"#Q/A=dT/(r2-r1)/k\n",
"A1=4*math.pi*(r1**2) \n",
"A2=4*math.pi*(r2**2) \n",
"A=math.sqrt(A1*A2)\n",
"Q=(A*k*dT)/(r2-r1)\n",
"\n",
"#Result\n",
"print\"Heat loss=Q=\",round(Q),\"W\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss=Q= 2220.0 W\n"
]
}
],
"prompt_number": 164
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.10,Page no:2.29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat Passed\n",
"\n",
"#Variable declaration\n",
"A= 1.0 #sq m\n",
"x1=0.15\n",
"x2=0.01\n",
"x4=0.15\n",
"T1=973.0 #[K]\n",
"T2=288.0 #[K]\n",
"dT=T1-T2 #[K]\n",
"#Thermal conductivities\n",
"k1=1.75 \n",
"k2=16.86\n",
"k3=0.033\n",
"k4=5.2\n",
"\n",
"#Calculation\n",
"#in absence of air gap,sum of thermal resistances \n",
"sR=(x1/k1*A)+(x2/k2*A)+(x4/k4*A)\n",
"round(sR,3)\n",
"sR=0.1153 #approximate\n",
"Q= dT/sR\n",
"\n",
"#When heat loss,Q=1163,then new resistance =sR1\n",
"Q1=1163.0 #[W/sq m]\n",
"sR1=dT/Q1\n",
"#width of air gap be w then\n",
"w=(sR1-sR)*k3*A # [m]\n",
"w=w*1000 #in [mm]\n",
"\n",
"#Result\n",
"print\"Heat lost per sq meter is\",round(Q),\"W/sq m\"\n",
"print\"Width of air gap is\",round(w,1),\"mm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat lost per sq meter is 5941.0 W/sq m\n",
"Width of air gap is 15.6 mm\n"
]
}
],
"prompt_number": 184
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.11,Page no:2.30"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Heat loss in Insulated pipe\n",
"import math\n",
"\n",
"#Variable declaration\n",
"d1=300.0 #[mm]\n",
"r1=d1/2.0 # [mm]\n",
"r1=r1/1000.0 #[m]\n",
"r2=r1+0.05 #[m]\n",
"r3=r2+0.04 #[m]\n",
"x1=0.05 #[m]\n",
"x2=0.04 #[m]\n",
"k1=0.105 #W/(m.K)\n",
"k2=0.07 #W/(m.K)\n",
"\n",
"#Calculation\n",
"rm1= (r2-r1)/math.log(r2/r1) # [m]\n",
"rm2=(r3-r2)/math.log(r3/r2) #[m]\n",
"L=1 #let\n",
"A1=math.pi*rm1*L #let L=1\n",
"R1=x1/(k1*A1) \n",
"A2=math.pi*rm2*L\n",
"R2=x2/(k2*A2)\n",
"T1=623.0 #[K]\n",
"T2=323.0 #[K]\n",
"dT=T1-T2 #[K]\n",
"\n",
"#Part a\n",
"Q_by_L= dT/(R1+R2) #Heat loss\n",
"\n",
"#Part b:\n",
"P=2*math.pi*(r1+x1+x2) #[m]\n",
"Q_by_L_peri=Q_by_L/P # [W/sq m]\n",
"R1=x1/(k1*A1) \n",
"sR=0.871+0.827\n",
"dT1=dT*R1/sR\n",
"\n",
"\n",
"#Result\n",
"#Part a\n",
"print\"Heat loss is:\",round(Q_by_L,1),\"W/m\" \n",
"#Part b:\n",
"print\"Heat lost per sq meter of outer insulation is\",round(Q_by_L_peri),\"W/sq m\"\n",
"print\"Temperature between two layers of insulation=\",round((T1-dT1)),\"K\"\n",
"\n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss is: 176.3 W/m\n",
"Heat lost per sq meter of outer insulation is 117.0 W/sq m\n",
"Temperature between two layers of insulation= 469.0 K\n"
]
}
],
"prompt_number": 187
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.12,Page no:2.31"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss in Composite brick\n",
"\n",
"#Variable declaration\n",
"x1=0.01 #[m]\n",
"x2=0.15 #[m]\n",
"x3=0.15 #[m]\n",
"T1=973.0 #[K]\n",
"T2=423.0 #[K]\n",
"dT=T1-T2 \n",
"#Thermal conductivities\n",
"k1=16.86 #[W/m.K]\n",
"k2=1.75 #[W/m.K]\n",
"k3=5.23 #[W/m.K]\n",
"k_air=0.0337 # [W/m.K]\n",
"A=1 #[sq m]\n",
"\n",
"#Calculation\n",
"sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A))\n",
"Q=dT/sigma_R #Heat flow in [W\n",
"Tm= Q*x3/k3 #Temperature drop in magnesite brick\n",
"#Interface temperature=iT\n",
"iT=T2+Tm #[K]\n",
"sigma_xbyk= A*dT/1163 #with air gap for reducing heat loss to 1163 per sq m\n",
"x_by_k=sigma_xbyk-sigma_R #x/k for air\n",
"t=x_by_k*k_air\n",
"t=t*1000 \n",
"\n",
"#Calculation\n",
"print\"Width of the air gap is\",round(t,2),\"mm\"\n",
"\n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Width of the air gap is 12.06 mm\n"
]
}
],
"prompt_number": 189
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.13,Page no:2.32"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat flow in a pipe\n",
"import math\n",
"#Variable declaration\n",
"L=1 #assume [m]\n",
"k1=43.03 #[W/(m.K)\n",
"k2=0.07 #(W/m.K)\n",
"T1=423 #inside temperature [K]\n",
"T2=305 # [K]\n",
"r1=0.0525 #[mm]\n",
"r2=0.0575 #[m]\n",
"r3=0.1075 #[m]\n",
"\n",
"#Calculation\n",
"#r3=r3/1000 #[m]\n",
"Q=(2*math.pi*L*(T1-T2))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2)) #Heat loss per metre \n",
"#Part 2\n",
"#T=Temperature of outer surface\n",
"T=T1-(Q*math.log(r2/r1))/(k1*2*math.pi*L) \n",
"#Part iii\n",
"id=0.105 #inside diametre in [m]\n",
"A=math.pi*id*1 #inside area in [sq m]\n",
"C=Q/(A*(T1-T2)) #conductance per length\n",
"\n",
"#Result\n",
"print\"Heat flow per metre of pipe is\",round(Q,2),\"W/m\"\n",
"#Part 2\n",
"print\"Temperature at outer surface of steel pipe:\",round(T,2),\"K\" \n",
"#Part iii\n",
"print\"Conductance per m length based on inside area is\",round(C,1),\"W/K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat flow per metre of pipe is 82.93 W/m\n",
"Temperature at outer surface of steel pipe: 422.97 K\n",
"Conductance per m length based on inside area is 2.1 W/K\n"
]
}
],
"prompt_number": 192
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.15,Page no:2.35"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Thickness of insulation \n",
" \n",
"#Variable declaration\n",
"A=1 # [sq m]\n",
"x1=0.1 #m\n",
"x2=0.04\n",
"k1=0.7\n",
"k2=0.48\n",
"\n",
"#Calculation\n",
"sigma=x1/(k1*A)+x2/(k2*A) #K/W\n",
"#Q=4.42*dT\n",
"#Q=dT/sigma\n",
"#with rockwool insulation added,Q_dash=0.75*Q\n",
"k3=0.065 # W/(m.K)\n",
"#Q_dash=dT/sigma+x3/k3*A\n",
"#On solving Q and Q_dash we get\n",
"x3=((1/(0.75*4.42))-sigma)*k3 #[m]\n",
"x3=x3*1000 # [mm]\n",
"\n",
"#Result\n",
"print\"Thickness of rockwool insulation required=\",round(x3,2),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of rockwool insulation required= 4.91 mm\n"
]
}
],
"prompt_number": 193
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.16,Page no:2.36"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Reduction in heat loss in insulated pipe\n",
"d1=40.0 # Diameter of pipe[mm]\n",
"r1=(d1/2.0)/1000.0 #Outside radius in [m]\n",
"t1=20.0 #Insulation 1 thickness in [mm]\n",
"t1=t1/1000 #[m]\n",
"t2=t1 #Insulation 2 thickness in[m]\n",
"r2=r1+t1 #radius after 1st insulation in [m]\n",
"r3=r2+t2 #Radius after second insulation in [m]\n",
"\n",
"#Calculation\n",
"import math\n",
"#Since python does not handles symbolic constants,we will assume some values:\n",
"#(1)\n",
"#Let the layer M-1 be nearer to the surface\n",
"L=1.0 #[m]\n",
"T1=10.0 #Temperature of inner surface of pipe [K]\n",
"T2=5.0 #Temperature of outer surface of insulation [K]\n",
"k=1.0 #Thermal conductivity\n",
"k1=k #For M-1 material\n",
"k2=3*k #For material M-2\n",
"Q1=(T1-T2)/(math.log(r2/r1)/(2*math.pi*L*k1)+math.log(r3/r2)/(2*math.pi*L*k2))\n",
"#(2)\n",
"#Let the layer of material M-2 be nearer to the surface\n",
"Q2=(T1-T2)/(math.log(r2/r1)/(2*math.pi*L*k2)+math.log(r3/r2)/(2*math.pi*L*k1))\n",
"#For dummy variables unity...\n",
"#For any value of k,T1 and T2,Q1 is always less than Q2\n",
"\n",
"#Result\n",
"print\"M-1 near the surface is advisable(i.e Arrangement one will result in less heat loss)\"\n",
"per_red=(Q2-Q1)*100/Q2\n",
"print\"Percent reduction in heat loss is\",round(per_red,1),\"percent\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"M-1 near the surface is advisable(i.e Arrangement one will result in less heat loss)\n",
"Percent reduction in heat loss is 23.2 percent\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.17,Page no:2.37"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss in a pipe\n",
"import math\n",
"\n",
"#Variable declaration\n",
"T1=523 #[K]\n",
"T2=323 #[K]\n",
"r1=0.05 #[m]\n",
"r2=0.055 #[m]\n",
"r3=0.105 #[m]\n",
"r4=0.155 #[m]\n",
"k1=50 #[W/(m.K)]\n",
"k2=0.06 #[W/(m.K)]\n",
"k3=0.12 #W/(m.K)\n",
"\n",
"#Calculation\n",
"#CASE 1\n",
"Q_by_L1=2*math.pi*(T1-T2)/((math.log(r2/r1))/k1+(math.log(r3/r2))/k2+(math.log(r4/r3))/k3) #[W/m]\n",
"#Case 2\n",
"Q_by_L2=2*math.pi*(T1-T2)/((math.log(r2/r1))/k1+(math.log(r3/r2))/k3+(math.log(r4/r3))/k2)\n",
"perct=(Q_by_L2-Q_by_L1)*100/Q_by_L1\n",
"\n",
"#Result\n",
"#CASE 1\n",
"print\"Heat loss=\",round(Q_by_L1,1),\"W/m\"\n",
"#Case 2\n",
"print\"If order is changed then heat loss=\",round(Q_by_L2,2),\"W/m\" \n",
"print\"Loss of heat is increased by\",round(perct,2),\" percent by putting material with higher thermal conductivity near the pipe surface\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss= 89.6 W/m\n",
"If order is changed then heat loss= 105.76 W/m\n",
"Loss of heat is increased by 18.04 percent by putting material with higher thermal conductivity near the pipe surface\n"
]
}
],
"prompt_number": 198
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.18,Page no:2.38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Arrangements for heat loss\n",
"\n",
"#Variable declaration\n",
"#Assume:\n",
"L=1.0 #[m]\n",
"r1=0.10 #[m] Outside radius od pipe\n",
"ia=0.025 #inner insulaiton [m]\n",
"import math \n",
"r2=r1+ia #Outer radius of inner insulation\n",
"r3=r2+ia #Outer radius of outer insulation\n",
"\n",
"#Calculation\n",
"#CASE 1:'a' near the pipe surface\n",
"#let k1=1\n",
"k1=1.0 #Thermal conductivity of A[W/m.K]\n",
"#and k2=3k1=3\n",
"k2=3.0 #Thermal conductivity of B[W/m.K]\n",
"#Let dT=1\n",
"dT=1.0\n",
"Q1=dT/(math.log(r2/r1)/(2*math.pi*k1*L)+math.log(r3/r2)/(2*math.pi*k2*L))\n",
"#CASE 2:'b' near the pipe surface \n",
"Q2=dT/(math.log(r2/r1)/(2*math.pi*k2*L)+math.log(r3/r2)/(2*math.pi*k1*L))\n",
"\n",
"#Result\n",
"print\"ANSWER-(i)\\nQ1=\",round(Q1,2),\"W \\nQ2=\",round(Q2,2),\" W \"\n",
"print\"Q1 is less than Q2.i.e arrangement A near the pipe surface and B as outer layer gives less heat loss\\n\"\n",
"percent=(Q2-Q1)*100/Q1 #percent reduction in heat loss\n",
"print\"ANSWER-(ii) \\nPercent reduction in heat loss (with near the pipe surface)=\",round(percent,1),\"%(approx)\" \n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"ANSWER-(i)\n",
"Q1= 22.13 W \n",
"Q2= 24.48 W \n",
"Q1 is less than Q2.i.e arrangement A near the pipe surface and B as outer layer gives less heat loss\n",
"\n",
"ANSWER-(ii) \n",
"Percent reduction in heat loss (with near the pipe surface)= 10.6 %(approx)\n"
]
}
],
"prompt_number": 206
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.19,Page no:2.39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Insulation thickness\n",
"\n",
"#Variable declaration\n",
"x1=0.224 # m\n",
"k1=1.3 # W/(m.K)\n",
"k2=0.346 # W/(m.K)\n",
"T1=1588.0 # K\n",
"T2= 299.0 # K\n",
"QA=1830.0 # W/ sq metre #heat loss\n",
"\n",
"#Calculation\n",
"#Q/A=(T1-T2)/x1/k1+x2/k2\n",
"x2=k2*((T1-T2)*1/(QA)-(x1/k1))\n",
"x2=x2*1000 \n",
"\n",
"#Result\n",
"print\"Thickness of insulation=\",round(x2),\"mm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of insulation= 184.0 mm\n"
]
}
],
"prompt_number": 209
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.20,Page no:2.39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss in furnace\n",
"\n",
"#Variable declaration\n",
"\n",
"#for clay\n",
"k1=0.533 #[W/(m.K)]\n",
"#for red brick\n",
"k2=0.7 #[W/m.K]\n",
"\n",
"\n",
"#Calculation\n",
"\n",
"#Case 1\n",
"A=1 #Area\n",
"x1=0.125 #[m]\n",
"x2=0.5 #[m]\n",
"#Resistances\n",
"r1=x1/(k1*A) #Res of fire clay [K/W]\n",
"r2=x2/(k2*A) #Res of red brick[K/W]\n",
"r=r1+r2\n",
"#Temperatures\n",
"T1=1373 #[K]\n",
"T2=323 #[K]\n",
"Q=(T1-T2)/r #[W/sq m]\n",
"Tdash=T1-Q*r1 #[K]\n",
"\n",
"#Case2\n",
"# Heat loss must remain unchanged,Thickness of red brick also reduces to its half\n",
"x3=x2/2 #[m]\n",
"r3=x3/(k2*A) #[K/W]\n",
"Tdd= T2+(Q*r3) #[K]\n",
"#Thickness of diatomite be x2,km be mean conductivity\n",
"Tm=(Tdash+Tdd)/2 #[K]\n",
"km=0.113+(0.00016*Tm) #[W/(m.K]\n",
"x2=km*A*(Tdash-Tdd)/Q #[m]\n",
"x2=x2*1000 #[mm]\n",
"\n",
"#Result\n",
"print\"Thickness of diatomite layer=\",round(x2),\"mm\"\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of diatomite layer= 93.0 mm\n"
]
}
],
"prompt_number": 210
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.21,Page no:2.41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Rate of heat loss in pipe\n",
"\n",
"#Variable declaration\n",
"k1=0.7 #common brick W/((m.K)\n",
"k2=0.48 #gypsum layer [W/(m.K)\n",
"k3=0.065 #Rockwool [W/m.K]\n",
"#Heat loss with insulatiob will be 20% of without insulation\n",
"A=1 #sq m\n",
"x1=0.1 #[m]\n",
"x2=0.04 #[m]\n",
"\n",
"#Calculation\n",
"R1=x1/(k1*A) #K/W\n",
"R2=x2/(k2*A) #K/W\n",
"R=R1+R2 #K/W\n",
"#R3=x3/(k3*A)\n",
"QbyQd=0.2\n",
"sigRbyRd=QbyQd\n",
"x3=(R/QbyQd-R)/15.4 #m\n",
"x3=x3*1000 #[mm]\n",
"\n",
"#Result\n",
"print\"Thickness of rockwool insulation\",round(x3),\"mm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of rockwool insulation 59.0 mm\n"
]
}
],
"prompt_number": 211
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.22,Page no:2.44"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss from insulated steel pipe\n",
"\n",
"#Variable declaration\n",
"Ts=451.0 #Steam temperature in [K]\n",
"Ta=294.0 #Air temperature in [K]\n",
"Di=25.0 #Internal diameter of pipe [mm]\n",
"Di=Di/1000 #[m]\n",
"od=33.0 #Outer diameter of pipe [mm] \n",
"od=od/1000 #[m]\n",
"hi=5678.0 #Inside heat transfer coefficient [W/(m**2.K)]\n",
"ho=11.36 #Outsideheat transfer coefficient [W/(sq m.K)]\n",
"\n",
"#Calculation\n",
"xw=(od-Di)/2 #Thickness of steel pipe [m]\n",
"k2=44.97 #k for steel in W/(m.K)\n",
"k3=0.175 #k for rockwool in W/(m.K)\n",
"ti=38.0/1000 #thickness of insulation in [m]\n",
"r1=Di/2 #[m]\n",
"r2=od/2 #[m]\n",
"rm1=(r2-r1)/math.log(r2/r1) #[m]\n",
"r3=r2+ti #[m]\n",
"rm2=(r3-r2)/math.log(r3/r2) #[m]\n",
"Dm1=2*rm1 #[m]\n",
"Dm2=2*rm2 #[m]\n",
"import math\n",
"#Rate of heat loss = dT/(sigma_R)\n",
"L=1 #[m]\n",
"R1=1/(hi*math.pi*Di*L) #[K/W]\n",
"R1=round(R1,4)\n",
"R2=xw/(k2*math.pi*Dm1*L)\n",
"R2=round(R2,6)\n",
"R3=(r3-r2)/(k3*math.pi*Dm2*L)\n",
"R3=round(R3,3)\n",
"R3=1.086\n",
"Do=(od+2*ti) #[mm]\n",
"R4=1/(ho*math.pi*Do*L) #[m]\n",
"R4=round(R4,3)\n",
"sigma_R=R1+R2+R3+R4 \n",
"#Heat loss\n",
"dT=Ts-Ta #[K]\n",
"Q=dT/sigma_R #Heat loss [W/m]\n",
"\n",
"#Result\n",
"print\"Rate of heat loss is\",round(Q,2),\"W/m\"\n",
"print\"\\nNOTE:Slight variation in final answer due to mistake in calculation of sigma_R in textbook.\\nIn book is is taken as 1.366\\n \"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Rate of heat loss is 116.63 W/m\n",
"\n",
"NOTE:Slight variation in final answer due to mistake in calculation of sigma_R in textbook.\n",
"In book is is taken as 1.366\n",
" \n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.23,Page no:2.46"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat loss from furnace\n",
"\n",
"#Variable declaration\n",
"T1=913.0 #[K]\n",
"T=513.0 #[K]\n",
"T2=313.0 #[K]\n",
"\n",
"#Calculation\n",
"\n",
"#Q=(T1-T)/(x/(k*A))\n",
"#Q=(T-T2)/(1/(h*A))\n",
"#x=2k/h\n",
"#Q=(T1-T2)/(x/(kA)+1/(h*A))\n",
"#Therefore,Q=hA/3*(T1-T2)\n",
"#With increase in thickness(100%)\n",
"#x1=4*k/h\n",
"#Q2=(T1-T2)/(x1/k*A+1/(h*A))\n",
"#Q2=(h*A)/5)*(T1-T2)\n",
"#Now\n",
"h=1.0 #Assume\n",
"A=1.0 #Assume for calculation\n",
"Q1=(h*A/3)*(T1-T2)\n",
"Q2=((h*A)/5)*(T1-T2)\n",
"percent=(Q1-Q2)*100/Q1 #Percent reduction in heat loss\n",
"\n",
"#Result\n",
"print\"Percentage reduction in heat loss is\",percent,\"%\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage reduction in heat loss is 40.0 %\n"
]
}
],
"prompt_number": 100
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.24,Page no:2.47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Rate of heat loss\n",
"\n",
"#Variable declaration\n",
"L=1.0#m\n",
"thp=2.0#Thickness of pipe in mm\n",
"thi=10.0#Thickness of insulation in mm\n",
"T1=373.0#K\n",
"T2=298.0#K\n",
"id=30.0#mm\n",
"r1=id/2#mm\n",
"\n",
"#Calculation\n",
"r2=r1+thp#mm\n",
"r3=r2+thi#mm\n",
"#In S.I units\n",
"r1=r1/1000 #m\n",
"r2=r2/1000#m\n",
"r3=r3/1000#m\n",
"k1=17.44#W/(m.K)\n",
"k2=0.58#W/(m.K)\n",
"hi=11.63#W/(sq m.K)\n",
"ho=11.63#W/(sq m.K)\n",
"import math\n",
"Q=(2*math.pi*L*(T1-T2))/(1/(r1*hi)+(math.log(r2/r1))/k1+((math.log(r3/r2))/k2)+(1/(0.02*ho)))\n",
"\n",
"#Result\n",
"print\"Rate of heat loss,Q=\",round(Q,1),\"W\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Rate of heat loss,Q= 43.5 W\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.25,Page no:2.48"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Thickness of insulation .\n",
"from scipy.optimize import fsolve\n",
"import math\n",
"\n",
"#Variable declaration\n",
"h=8.5 #[W/sq m.K]\n",
"dT=175 #[K]\n",
"r2=0.0167 #[m]\n",
"\n",
"#Calculation\n",
"Q_by_l=h*2*math.pi*r2*dT #[W/m]\n",
"k=0.07 #For insulating material in [W/m.K]\n",
"#for insulated pipe--50% reduction in heat loss\n",
"Q_by_l1=0.5*Q_by_l #[w/m]\n",
"def f(r3):\n",
" x=Q_by_l1-dT/((math.log(r3/r2))/(2*math.pi*k)+1/(2*math.pi*r3*h))\n",
" return(x)\n",
"#by trial and error method we get:\n",
"r3=fsolve(f,0.05)\n",
"t=r3-r2 #thickness of insulation in [m]\n",
"\n",
"#Result\n",
"print\"Required thickness of insulation is\",round(t[0],4),\"m=\",round(t[0]*1000,1),\" mm or\",round(t[0]*1000),\"m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Required thickness of insulation is 0.0188 m= 18.8 mm or 19.0 m\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.26,Page no:2.49"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate heat loss per metre length\n",
"import math\n",
"\n",
"#Variable declaration\n",
"id=0.1 #internal diameter in[m]\n",
"od=0.12 #outer diameter in [m]\n",
"T1=358 #Temperature of fluid [K]\n",
"T2=298 #Temperature of surrounding [K]\n",
"t=0.03 #thickness of insulation [m]\n",
"k1=58 #[W/m.K]\n",
"k2=0.2 #W/(m.K) insulating material\n",
"h1=720 #inside heat transfer coeff [W/sq m .K]\n",
"h2=9 #W/sq m.K\n",
"r1=id/2 #[m]\n",
"r2=od/2 #[m]\n",
"r3=r2+t #[m]\n",
"\n",
"#Calculation\n",
"#Heat loss per meter=Q_by_L\n",
"Q_by_L=(T1-T2)/(1/(2*math.pi*r1*h1)+math.log(r2/r1)/(2*math.pi*k1)+math.log(r3/r2)/(2*math.pi*k2)+1/(2*math.pi*r3*h2)) #W/m\n",
"\n",
"#Result\n",
"print\"Heat loss per metre length of pipe=\",round(Q_by_L,2),\"W\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss per metre length of pipe= 114.49 W\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.27,Page no:2.50"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Mineral wool insulation\n",
"from scipy.optimize import fsolve\n",
"\n",
"#Variable declaration\n",
"import math\n",
"T1=573 #[K]\n",
"T2=323 #[K]\n",
"T3=298 #[K]\n",
"h1=29 # Outside heat transfer coefficients [W/sq m.K]\n",
"h2=12 #[W/sq m.K]\n",
"r1=0.047 #Internal radius [m]\n",
"r2=0.05 #Outer radius[m]\n",
"k1=58 #[W/m.K]\n",
"k2=0.052 #[W/m.K]\n",
"\n",
"#Calculation\n",
"#Q=(T1-T2)/(1/(r1*h1)+math.log(r2/r1)/k1+math.log(r3/r2)/k2)=(T2-T3)/(1/(r3*h2))\n",
"def f(r3):\n",
" x=(T1-T2)/(1/(r1*h1)+math.log(r2/r1)/k1+math.log(r3/r2)/k2)-(T2-T3)/(1/(r3*h2))\n",
" return(x)\n",
"\t#by trial and error method :\n",
"r3=fsolve(f,0.05)\n",
"t=r3-r2 #Thickness of insulation in [m]\n",
"#Q=h2*2*math.pi*r3*L*(T2-T3)\n",
"Q_by_l=h2*2*math.pi*r3*(T2-T3) #[W/m]\n",
"\n",
"#Result\n",
"print\"Thickness of insulation is\",round(t*1000),\"mm\"\n",
"print\"Rate of heat loss per unit length is\",round(Q_by_l[0],1),\"W/m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of insulation is 32.0 mm\n",
"Rate of heat loss per unit length is 154.1 W/m\n"
]
}
],
"prompt_number": 28
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.28,Page no:2.51"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate heat loss per sq m and temperature of outside surface\n",
"\n",
"#Variable declaration\n",
"A=1 #assume [sq m]\n",
"x1=0.006 #[m]\n",
"x2=0.075 #[m]\n",
"x3=0.2 #[m]\n",
"k1=39.0 #[W/m.K]\n",
"k2=1.1 #[W/m.K]\n",
"k3=0.66 #[W/m.K]\n",
"h0=65.0 #W/sq m .K\n",
"T1=900.0 #K\n",
"T2=300.0 #K\n",
"\n",
"#Calculation\n",
"sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A)) \n",
"\n",
"#To calculate heat loss/sq m area\n",
"Q=(T1-T2)/sigma_R #[W/sq m]\n",
"#Q/A=T-T2/(1/h0), where T=Temp of outside surface\n",
"#So, T=T2+Q/(A*h0)\n",
"T=Q/(A*h0)+T2 #[K]\n",
"\n",
"#Result\n",
"print\"Heat loss per sq metre area is:\",round(Q,1),\"W/sq m\" \n",
"print\"Temperature of outside surface of furnace is:\",round(T,1),\"K (\",round(T-273,1),\"degree C)\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss per sq metre area is: 1551.4 W/sq m\n",
"Temperature of outside surface of furnace is: 323.9 K ( 50.9 degree C)\n"
]
}
],
"prompt_number": 42
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.29,Page no:2.52"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine necessary thickness of insulation brick\n",
"\n",
"#Variable declaration\n",
"A=1.0 #Assume [sq m]\n",
"x1=0.003 #[m]\n",
"x3=0.008 #[m]\n",
"k1=30.0 #[W/m.K]\n",
"k2=0.7 #[W/m.K]\n",
"k3=40.0 #[W/m.K]\n",
"T1=363.0 #[K]\n",
"T=333.0 #[K]\n",
"T2=300.0 #[K]\n",
"h0=10.0 #W/sq m.K\n",
"\n",
"#Calculation\n",
"#Q=(T1-T2)/(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A))\n",
"#Also,Q=(T-T2)/(1/(h0*A))\n",
"#So, (T1-T2)/((x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A))=(T-T2)/(1/(h0*A))\n",
"#or,x2=k2*A((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3/(k3*A))\n",
"x2=k2*A*((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3/(k3*A)) #[m]\n",
"\n",
"#Result\n",
"print\"Thickness of insulating brick required is\",round(x2*1000,1),\"mm\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thickness of insulating brick required is 63.4 mm\n"
]
}
],
"prompt_number": 44
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.30,Page no:2.53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat flow through furnace wall\n",
"\n",
"#Variable declaration\n",
"hi=75.0 #[W/sq m.K)\n",
"x1=0.2 #m\n",
"x2=0.1 #[m]\n",
"x3=0.1 #[m]\n",
"T1=1943 #[K]\n",
"k1=1.25 #W/m.K\n",
"k2=0.074 #/W/m.K\n",
"k3=0.555 #W/m.K\n",
"T2=343.0 #K\n",
"A=1.0 #assume [sq m]\n",
"\n",
"#Calculation\n",
"sigma_R=1.0/(hi*A)+x1/(k1*A)+x2/(k2*A)+x3/(k3*A) \n",
"#Heat loss per i sq m\n",
"Q=(T1-T2)/sigma_R #[W]\n",
"#if T=temperature between chrome brick and koalin brick then \n",
"#Q=(T1-T)/(1/(hi*A)+x1/(k1*A))\n",
"#or T=T1-(Q*(1/(hi*A)+x1/(k1*A)))\n",
"T=T1-(Q*(1.0/(hi*A)+x1/(k1*A))) #[K]\n",
"#if Tdash=temperature at the outer surface of middle layer,then\n",
"#Q=(Tdash-T2)/(x3/(k1*A))\n",
"#or Tdash=T2+(Q*x3/(k3*A))\n",
"Tdash=T2+(Q*x3/(k3*A)) #[K]\n",
"\n",
"#Result\n",
"print\"Heat loss per sq m is:\",round(Q,1),\"W\"\n",
"print\"Temperature at inner surface of middle layer=\",round(T,1),\"K(\",round(T-273,1),\"degree C)\"\n",
"print\"Temperature at outer surface of middle layer=\",round(Tdash,1),\"K (\",round(Tdash-273,1),\"degree C)\" \n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss per sq m is: 938.5 W\n",
"Temperature at inner surface of middle layer= 1780.3 K( 1507.3 degree C)\n",
"Temperature at outer surface of middle layer= 512.1 K ( 239.1 degree C)\n"
]
}
],
"prompt_number": 51
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.31,Page no:2.54"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate:(a) Heat loss per unit length \n",
"#(b)Reduction in heat loss\n",
"import math\n",
"\n",
"#Variable declaration\n",
"hi=10 #W/sq m.K\n",
"h0=hi #W/sq.m.K\n",
"r1=0.09 #m\n",
"r2=0.12 #m\n",
"t=0.05 #thickness of insulation [m]\n",
"k1=40 #W/m.K\n",
"k2=0.05 #W/m.K\n",
"T1=473 #K\n",
"T2=373 #K\n",
"\n",
"#Calculation\n",
"Q_by_L=2*math.pi*(T1-T2)/(1/(r1*hi)+math.log(r2/r1)/k1+1/(r2*h0)) #W/m\n",
"#After addition of insulation:\n",
"r3=r2+t #radius of outer surface of insulaiton\n",
"Q_by_L1=2*math.pi*(T1-T2)/(1/(r1*hi)+math.log(r2/r2)/k1+math.log(r3/r2)/k2+1/(r3*h0)) # W\n",
"Red=Q_by_L-Q_by_L1 #Reduciton in heat loss in [W/m]\n",
"percent_red=(Red/Q_by_L)*100 #% Reduction in heat loss\n",
"\n",
"#Result\n",
"print\"Ans (a) Heat loss=\",round(Q_by_L,2),\"W/m \"\n",
"print\"Ans (b) Percent reduction in heat loss is\",round(percent_red,1),\"%\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ans (a) Heat loss= 321.94 W/m \n",
"Ans (b) Percent reduction in heat loss is 77.5 %\n"
]
}
],
"prompt_number": 54
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.32,Page no:2.55"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine: i-Heat flux across the layers and\n",
"#ii-Interfacial temperature between the layers\n",
"\n",
"#Variable declaration\n",
"T1=798.0 #K\n",
"T2=298.0 #K\n",
"x1=0.02 #m\n",
"x2=x1 #m\n",
"k1=60.0 #W/m.K\n",
"k2=0.1 #W/m.K\n",
"hi=100.0 #W/sq m.K\n",
"h0=25.0 #W/sq m.K\n",
"\n",
"#Calculation\n",
"Q_by_A=(T1-T2)/(1.0/hi+x1/k1+x2/k2+1.0/h0) #W/sq m\n",
"#If Tis the interfacial temperature between steel plate and insulating material\n",
"#Q_by_A=(T-T2)/(x2/k2+1/h0)\n",
"T=Q_by_A*(x2/k2+1.0/h0)+T2\n",
"\n",
"#Result\n",
"print\"Ans (i)- Heat flux across the layers is\",round(Q_by_A),\"W/sq m\" \n",
"print\"Ans-(ii)-Interfacial temperature between layers is\",round(T,1),\"K(\",round(T-273,1),\"degree C)\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ans (i)- Heat flux across the layers is 1997.0 W/sq m\n",
"Ans-(ii)-Interfacial temperature between layers is 777.4 K( 504.4 degree C)\n"
]
}
],
"prompt_number": 58
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.33,Page no:2.56"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Determine Temperature at the outer surface of wall and convective conductance on the outer wall\n",
" \n",
" \n",
"#Variable declaration \n",
"T1=2273 #Temperature of hot gas:[K]\n",
"T4=318 #Ambient aur temperature[K]\n",
"Qr1_by_A=23260 #Heat flow by radiation from gases to inside surface of wall[W/sq m]\n",
"hi=11.63 #Heat transfer coefficient on inside wall:[W/sq m.K]\n",
"K=58 #Thermal conductivity of wall[W.sq m/K]\n",
"Qr4_by_A=9300 #Heat flow by radiation from external surface to ambient[W/sq m]\n",
"T2=1273 #Inside Wall temperature[K]\n",
"Qr1=Qr1_by_A #W for\n",
"A=1 #sq m\n",
"\n",
"\n",
"#Calculation\n",
"Qc1_by_A=hi*(T1-T2) #W/sq m\n",
"Qc1=Qc1_by_A #for A=1 sq m\n",
" #Thermal resistance:\n",
"R=1.0/K #K/W per sq m\n",
"#Now Q=(T2-T3)/R,i.e \n",
"#External wall temp T3=T2-Q*R\n",
"#Q entering wall=\n",
"Q_enter=Qr1+Qc1 #W\n",
"T3=T2-Q_enter*R #K\n",
"T3=673 #Approximate\n",
"#Heat loss due to convection:\n",
"Qc4_by_A=Q_enter-Qr4_by_A #W/sq m\n",
"#Qc4_by_A=h0*(T3-T4)\n",
"#or h0=Qc4_by_A/(T3-T4)\n",
"h0=Qc4_by_A/(T3-T4) #W/sq m.K\n",
"\n",
"\n",
"#Result\n",
"print\"Convective conductance is:\",round(h0,1),\"W/m^2.K\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Convective conductance is: 72.1 W/m^2.K\n"
]
}
],
"prompt_number": 60
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.34,Page no:2.60"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Critical radius of insulation\n",
"\n",
"#Variable declaration\n",
"T1=473 #[K]\n",
"T2=293 #[K]\n",
"k=0.17 #W/(m.K)\n",
"h=3 #W/(sq m.K)\n",
"h0=h #W/sq m.K\n",
"rc=k/h #m\n",
"r1=0.025 #Inside radius of insulaiton [mm] \n",
"\n",
"#Calculation\n",
"import math\n",
"q_by_l1=2*math.pi*(T1-T2)/(math.log(rc/r1)/k+1/(rc*h0)) #Heat transfer with insulation in W/m\n",
"#Without insulation:\n",
"q_by_l2=h*2*math.pi*r1*(T1-T2) #W/m\n",
"inc=(q_by_l1-q_by_l2)*100/q_by_l2 #Increase of heat transfer\n",
"k=0.04 #Fibre glass insulaiton W/(sq m.K)\n",
"rc=k/h #Critical radius of insulaiton\n",
"\n",
"#Result\n",
"print\"When covered with insulation:\\nHeat loss=\",round(q_by_l1,1),\"W\"\n",
"print\"When without insulation,\\nHeat loss=\",round(q_by_l2,1),\"W\"\n",
"print\"\\nPercent increase =\",round(inc,2),\"percent\" \n",
"print\"In this case the value of rc=\",round(rc,4),\"m is less than the outside radius of pipe (\",r1,\"m)\"\n",
"print\"So additon of any fibre glass would cause a decrease in the heat transfer\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"When covered with insulation:\n",
"Heat loss= 105.7 W\n",
"When without insulation,\n",
"Heat loss= 84.8 W\n",
"\n",
"Percent increase = 24.66 percent\n",
"In this case the value of rc= 0.0133 m is less than the outside radius of pipe ( 0.025 m)\n",
"So additon of any fibre glass would cause a decrease in the heat transfer\n"
]
}
],
"prompt_number": 65
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.36,Page no:2.61"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the heat loss per metre of pipe and outer surface temperature\n",
"\n",
"#Variable declaration\n",
"import math\n",
"k=1.0 #Thermal conductivity in [W/sq m.K]\n",
"h=8.0 #Het transfer coeff in W/sq m.K\n",
"rc=k/h #Critical radius in m\n",
"T1=473.0 #K\n",
"T2=293.0 #K\n",
"r1=0.055 #Outer radius =inner radius in [m]\n",
"\n",
"#Calculation\n",
"Q_by_L=2*math.pi*(T1-T2)/(math.log(rc/r1)/k+1.0/(rc*h))\n",
"#For outer surface\n",
"#Q_by_L=2*math.pi*(T-T2)/(1/rc*h)\n",
"# implies that, T=T2+Q_by_L/(rc*2*math.pi)\n",
"T=T2+Q_by_L/(rc*2*math.pi*h) #K\n",
"\n",
"#Result\n",
"print\"Heat loss per meter of pipe is\",round(Q_by_L),\"W/m(approx)\"\n",
"print\"Outer surface temperature is: \",round(T,1),\"K(\",round(T-273,1),\"degree C)\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss per meter of pipe is 621.0 W/m(approx)\n",
"Outer surface temperature is: 391.8 K( 118.8 degree C)\n"
]
}
],
"prompt_number": 68
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.37,Page no:2.78"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Calculate the time required for a ball to attain a temperature of 423 K\n",
"\n",
"#Variable declaration\n",
"k_steel=35.0 #W/m.K\n",
"Cp_steel=0.46 #kJ/(kg*K)\n",
"Cp_steel=Cp_steel*1000 #J/(kg*K)\n",
"h=10 #W/sq m.K\n",
"rho_steel=7800.0 #kg/cubic m\n",
"dia=50.0 #mm\n",
"dia=dia/1000 #m\n",
"R=dia/2 #radius in m\n",
"\n",
"#Calculation\n",
"import math\n",
"A=4*math.pi*R**2 #Area in sq m\n",
"V=A*R/3 #Volume in cubic meter\n",
"Nbi=h*(V/A)/k_steel\n",
"#As Nbi<0.10,internal temp gradient is negligible\n",
"T=423.0 #K\n",
"T0=723.0 #K\n",
"T_inf=373.0 #K\n",
"#(T-T_inf)/(T0-T_inf)=e**(-h*At/rho*Cp*V)\n",
"t=-rho_steel*Cp_steel*R*math.log((T-T_inf)/(T0-T_inf))/(3*h) #s\n",
"\n",
"#Result\n",
"print\"Time required for a ball to attain a temperature of 423 K is\",round(t),\"s=\",round(t/3600,2),\"h\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time required for a ball to attain a temperature of 423 K is 5818.0 s= 1.62 h\n"
]
}
],
"prompt_number": 70
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.38,Page no:2.78"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Time take in Steel ball quenched\n",
"\n",
"#Variable declaration\n",
"dia=50.0 #mm\n",
"dia=dia/1000 #m\n",
"r=dia/2 #radius in m\n",
"h=115.0 #W/sq m.K\n",
"rho=8000.0 #kg/cubic m\n",
"Cp=0.42 #kJ/kg.K\n",
"Cp=Cp*1000 #J/(kg*K)\n",
"\n",
"#Calculation\n",
"import math\n",
"A=4*math.pi*r**2 #Area in sq m\n",
"V=A*r/3 #Volume in cubic m\n",
"T=423.0 #K\n",
"T_inf=363.0 #K\n",
"T0=723 #K\n",
"#(T-T_inf)/(T0-T_inf)=e**(-3ht/(rho*Cp*r))\n",
"t=-rho*Cp*r*math.log((T-T_inf)/(T0-T_inf))/(3*h) #Time in seconds\n",
"\n",
"#Result\n",
"print\"Time taken by centre of ball to reach a temperature of 423 K is\",round(t,2),\"s or\",round(t/60,2),\"minutes\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time taken by centre of ball to reach a temperature of 423 K is 436.25 s or 7.27 minutes\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.39,Page no:2.79"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#HT in a Ball plunged in a medium\n",
"\n",
"#Variable declaration\n",
"h=11.36 #W/sq m.K\n",
"k=43.3 #w/(m.K)\n",
"r=25.4 #radius in mm\n",
"r=r/1000 # radius in m\n",
"\n",
"#Calculation\n",
"import math\n",
"A=4*math.pi*r**2 #Area of sphere [sq m]\n",
"V=A*r/3 #Volume in [cubic m]\n",
"rho=7849.0 #kg/cubic m\n",
"Cp=0.4606*10**3 #J/kg.K\n",
"t=1.0 #hour\n",
"t=t*3600 #seconds\n",
"T_inf=394.3 #[K]\n",
"T0=700.0 #[K]\n",
"# (T-T_inf)/(T0-T_inf)=e**(-3*h*t/rho*Cp*V)\n",
"T=T_inf+(T0-T_inf)*(math.exp((-h*A*t)/(rho*Cp*V))) \n",
"\n",
"#Result\n",
"print\"Temperature of ball after 1 h=\",round(T),\"K (\",round(T-273),\"degree C)(APPROXIMATE)\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Temperature of ball after 1 h= 475.0 K ( 202.0 degree C)(APPROXIMATE)\n"
]
}
],
"prompt_number": 74
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.40,Page no:2.80"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Slab temperature suddenly lowered\n",
"\n",
"#Variable declaration\n",
"import math\n",
"rho=9000.0 #kg/cubic m\n",
"Cp=0.38 #kJ/(kg.K)\n",
"Cp=Cp*1000 #J/(kg.K)\n",
"k=370.0 #W/m.K\n",
"h=90.0 #W/sq m.K\n",
"l=400.0 #mm\n",
"l=l/1000 #length of copper slab\n",
"t=5.0/1000 #thickness in [m]\n",
"\n",
"#Calculation\n",
"A=2*l**2 #Area of slab\n",
"V=t*l**2 #Volume in [cubic m]\n",
"L_dash=V/A #[m]\n",
"#for slab of thickness 2x\n",
"#L_dash=x\n",
"L_dash=0.025 #[m]\n",
"Nbi=h*L_dash/k #< 0.10\n",
"var=h*A/(rho*Cp*V)\n",
"#As Nbi<0.10,we can apply lumped capacity analysis\n",
"T=363.0 #[K]\n",
"T_inf=303.0 #[K]\n",
"T0=523.0 #[K]\n",
"t=-(math.log((T-T_inf)/(T0-T_inf)))/var\n",
"\n",
"#Result\n",
"print\"Time at which slab temperature becomes 363 K is\",round(t,1),\"s\"\n",
"print\"CALCULATION MISTAKE IN BOOK IN LAST LINE\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time at which slab temperature becomes 363 K is 123.4 s\n",
"CALCULATION MISTAKE IN BOOK IN LAST LINE\n"
]
}
],
"prompt_number": 75
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.41,Page no:2.80"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Flow over a flat plate\n",
"\n",
"#Variable declaration\n",
"rho=9000.0 #kg/cubic meter\n",
"Cp=0.38 #kJ/(kg.K)\n",
"Cp=Cp*1000 #J/kg.K\n",
"k=370.0 #W/(m.K)\n",
"T0=483.0 #K\n",
"T_inf=373.0 #K\n",
"delta_T=40.0 #K\n",
"\n",
"#Calculation\n",
"T=T0-delta_T #K\n",
"t=5.0 #time in [minutes]\n",
"t=t*60 #[seconds]\n",
"#A=2A.....Two faces\n",
"#V=A.2x\n",
"#2x=thickness of slab=30 mm=0.03 m\n",
"x=0.015 #[m]\n",
"th=2*x #thickness of slab\n",
"h=-rho*Cp*x*math.log((T-T_inf)/(T0-T_inf))/t\n",
"\n",
"#Result\n",
"print\"Heat transfer coefficient is:\",round(h,1),\"W/(m^2.K)\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat transfer coefficient is: 77.3 W/(m^2.K)\n"
]
}
],
"prompt_number": 77
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.42,Page no:2.81"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#TIme required in Stainless steel rod immersed in water\n",
"\n",
"#Variable declaration\n",
"rho=7800.0 #[kg per cubic m]\n",
"h=100.0 #W/(sq m.K) Convective heat transfer coeff\n",
"Cp=460.0 #J/(kg.K)\n",
"k=40.0 #W/(m.K)\n",
"L=1.0 #[m] length ofrod\n",
"D=10.0 #mm \n",
"D=D/1000 #diameter in[m]\n",
"R=D/2 #raidus in [m]\n",
"\n",
"#Calculation\n",
"import math\n",
"#For cylindrical rod:\n",
"A=2*math.pi*R*L #Area in [sq m]\n",
"V=math.pi*R**2*L #Volume in [cubic m]\n",
"L_dash=V/A #[m]\n",
"Nbi=h*L_dash/k #Biot number\n",
"#N_bi<0.10,Hence lumped heat capavity is possible\n",
"T=473.0 #[K] \n",
"T_inf=393.0 #[K]\n",
"T0=593.0 #[K]\n",
"t=-rho*Cp*V*math.log((T-T_inf)/(T0-T_inf))/(h*A)\n",
"\n",
"#Result\n",
"print\"Time required to reach temperature\",T,\"K is\",round(t,1),\"s\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time required to reach temperature 473.0 K is 82.2 s\n"
]
}
],
"prompt_number": 79
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.43,Page no:2.82"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Time constant of Chromel alumel thermocouple \n",
"\n",
"#Variable declaration\n",
"import math\n",
"rho=8600.0 #[kg/cubic m]\n",
"Cp=0.42 #kJ/(kg.K)\n",
"Cp=Cp*1000 #J/(kg.K)\n",
"dia=0.71 #[mm]\n",
"dia=dia/1000 #[dia in m]\n",
"R=dia/2 #radius [m]\n",
"h=600.0 #convective coeff W/(sq m.K)\n",
"\n",
"#Calculation\n",
"#Let length =L=1\n",
"L=1.0 #[m]\n",
"A=2*math.pi*R*L \n",
"V=math.pi*(R**2)*L \n",
"tao=(rho*Cp*V)/(h*A) \n",
"#at1\n",
"t=tao\n",
"#From (T-T_inf)/(T0-T_inf)=e**(-t/tao)\n",
"ratio=math.exp(-t/tao) #Ratio of thermocouple difference to initial temperature difference\n",
"print\"Time constant of the thermocouple is\",round(tao,3),\"s\" \n",
"print\"At the end of the time period t=tao=\",round(tao,3),\"s\"\n",
"print\"Temperature difference b/n the thermocouple and the gas stream would be\",round(ratio,3),\"of the initial temperature difference\"\n",
"print\"It should be reordered after\",round(4*tao),\"s\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time constant of the thermocouple is 1.069 s\n",
"At the end of the time period t=tao= 1.069 s\n",
"Temperature difference b/n the thermocouple and the gas stream would be 0.368 of the initial temperature difference\n",
"It should be reordered after 4.0 s\n"
]
}
],
"prompt_number": 84
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.44,Page no:2.83"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Thermocouple junction\n",
"import math\n",
"\n",
"#Variable declaration\n",
"rho=8000.0 #kg/cubic m\n",
"Cp=420.0 #J/(kg.K)\n",
"h_hot=60.0 # for hot stream W/(sq m.K) \n",
"dia=4.0 #[mm]\n",
"t=10.0 \n",
"\n",
"#Calculation\n",
"r=dia/(2.0*1000) #radius in [m]\n",
"#For sphere\n",
"V=(4.0/3.0)*math.pi*r**3 #Volume in [cubic m]\n",
"A=4*math.pi*r**2 #Volume in [sq m]\n",
"tao=rho*Cp*V/(h_hot*A) # Time constant in [s]\n",
"ratio=math.exp(-t/tao) # %e**(-t/tao)=(T-T-inf)/(T0-T_inf)\n",
"T_inf=573.0 #[K]\n",
"T0=313.0 #[K]\n",
"T=T_inf+ratio*(T0-T_inf)\n",
"#IN STILL AIR:\n",
"h_air=10.0 #W/(sq m .K)\n",
"tao_air=rho*Cp*V/(h_air*A) #[s]\n",
"t_air=20.0 #[s]\n",
"ratio_air=math.exp(-t_air/tao_air)\n",
"T_inf_air=303.0 #[K]\n",
"T0_air=T \n",
"T_air=T_inf_air+ratio_air*(T0_air-T_inf_air)\n",
"\n",
"#Result\n",
"#ANS-[i]\n",
"print\"Time constant of thermocouple is\",round(tao,2),\"s\"\n",
"#ANS-[ii]\n",
"print\"Temperature attained by junction 20 s after removing from the hot air stream is:\",round(T_air),\"K\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time constant of thermocouple is 37.33 s\n",
"Temperature attained by junction 20 s after removing from the hot air stream is: 368.0 K\n"
]
}
],
"prompt_number": 85
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.45,Page no:2.84"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Batch reactor time taken \n",
"from scipy.optimize import fsolve\n",
"from scipy import integrate\n",
"import math\n",
"\n",
"#Variable declaration\n",
"T_inf=390.0 #[K]\n",
"U=600.0 #[W/sq m.K]\n",
"Ac=1.0 #[sq m]\n",
"Av=10.0 #Vessel area in [sq m]\n",
"m=1000.0 #[kg]\n",
"Cp=3.8*10**3 #[J/kg.K]\n",
"To=290.0 #[K]\n",
"T=360.0 #[K]\n",
"h=8.5 #[W/sq m.K]\n",
"\n",
"#Calculation\n",
"#Heat gained from the steam=Rate of increase of internal energy\n",
"#U*A*(T_inf-T)=m*Cp*dT\n",
"def f(t):\n",
" x=math.log((T_inf-To)/(T_inf-T))-U*Ac*t/(m*Cp) \n",
" return(x)\n",
"t=fsolve(f,1) #[in s]\n",
"t=round(t) #[in s]\n",
"Ts=290 \n",
"print\"Time taken to heat the reactants over the same temperature range is\",t,\"s\"\n",
"def g(T):\n",
" t1=m*Cp/(U*Ac*(T_inf-T)-h*Av*(T-Ts))\n",
" return(t1)\n",
"t1= integrate.quad(g,To,T)\n",
"def fx(Tmax):\n",
" m=U*Ac*(T_inf-Tmax)-h*Av*(Tmax-Ts)\n",
" return(m)\n",
"T_max=fsolve(fx,1)\n",
"\n",
"#Result\n",
"print\"In CASE 1\\nTime taken to heat the reactants =\",t,\"s .ie\",round(t/3600,2),\"h\" \n",
"print\"In CASE 2 \\nTime taken to heat the reactants =\",round(t1[0]),\"s\"\n",
"print\"Maximum temperature at which temperature can be raised is\",round(T_max,1),\"K\" \n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time taken to heat the reactants over the same temperature range is 7625.0 s\n",
"In CASE 1\n",
"Time taken to heat the reactants = 7625.0 s .ie 2.12 h\n",
"In CASE 2 \n",
"Time taken to heat the reactants = 8905.0 s\n",
"Maximum temperature at which temperature can be raised is 377.6 K\n"
]
}
],
"prompt_number": 92
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.46,Page no:2.95"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Heat dissipation by aluminium rod \n",
"import math \n",
"\n",
"#Variable declaration\n",
"dia=3.0 #[mm]\n",
"dia=dia/1000 #[m]\n",
"r=dia/2 #radius in[m]\n",
"k=150 #W/(m.K)\n",
"h=300 #W/(sq m.K)\n",
"T0=413 #[K]\n",
"T_inf=288 #[K]\n",
"\n",
"#Calculation\n",
"A=math.pi*(r**2) #Area in [sq m]\n",
"P=math.pi*dia #[W/sq m.K]\n",
"Q=(T0-T_inf)*math.sqrt(h*P*k*A) #Heat dissipated in [W]\n",
"\n",
"#Result\n",
"print\"Heat dissipated by the rod is\",round(Q,3),\"W\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat dissipated by the rod is 6.844 W\n"
]
}
],
"prompt_number": 94
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.47,Page no:2.96"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Aluminium fin efficiency\n",
"#Variable declaration\n",
"k=200.0 #W/(m.K)\n",
"h=15.0 #W/(sq m.K)\n",
"T0=523.0 #[K]\n",
"T_inf=288.0 #[K]\n",
"theta_0=T0-T_inf \n",
"dia=25.0 #diameter[mm]\n",
"dia=dia/1000 #diameter[m]\n",
"r=dia/2 #radius in [m]\n",
"\n",
"#Calculation\n",
"import math\n",
"P=math.pi*dia #[m]\n",
"A=math.pi*r**2 #[sq m]\n",
"#For insulated fin:\n",
"m=math.sqrt(h*P/(k*A))\n",
"L=100.0 #length of rod in [mm]\n",
"L=L/1000 #length of rod in [m]\n",
"Q=theta_0*math.tanh(m*L)*math.sqrt(h*P*k*A) #Heat loss \n",
"nf=math.tanh(m*L)/(m*L) #Fin efficiency for insulated fin\n",
"#At the end of the fin: theta/theta_0=(cosh[m(L-x)]/cosh(mL))\n",
"#at x=L, theta/theta_0=1/(cosh(mL)\n",
"T=T_inf+(T0-T_inf)*(1/math.cosh(m*L)) #[K]\n",
"\n",
"\n",
"#Result\n",
"#ANSWER-1\n",
"print\"Heat loss by the insulated rod is\",round(Q,1),\"W\"\n",
"#ANSWER-2\n",
"print\"Fin efficiency is\",round(nf*100,1),\"percent\"\n",
"#ANSWER-3\n",
"print\"Temperature at the end of the fin is\",round(T,1),\"K\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat loss by the insulated rod is 26.6 W\n",
"Fin efficiency is 96.2 percent\n",
"Temperature at the end of the fin is 509.6 K\n"
]
}
],
"prompt_number": 97
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.49,Page no:2.98"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Finding effective Pin fins\n",
"\n",
"#Variable declaration\n",
"k=300.0 #W/(m.K)\n",
"h=20.0 #W.(sq m.K)\n",
"P=0.05 #[m]\n",
"A=2.0 #[sq cm]\n",
"A=A/10000 #[sq m]\n",
"T0=503.0 #[K]\n",
"T_inf=303.0 #[K]\n",
"\n",
"\n",
"#Calculation\n",
"theta_0=T0-T_inf #[K]\n",
"import math\n",
"m=math.sqrt(h*P/(k*A))\n",
"\n",
"#CASE 1: 6 Fins of 100 mm length\n",
"L1=0.1 #Length of fin in [m]\n",
"Q=math.sqrt(h*P*k*A)*theta_0*math.tanh(m*L1) #[W]\n",
"#For 6 fins\n",
"Q=Q*6 #for 6 fins [W]\n",
"\n",
"#CASE 2: 10 fins of 60 mm length\n",
"L2=60.0 #[mm]\n",
"L2=L2/1000 #[m]\n",
"Q2=math.sqrt(h*P*k*A)*theta_0*math.tanh(m*L2) #[W]\n",
"Q2=Q2*10 #For 10 fins\n",
"\n",
"\n",
"\n",
"#Result\n",
"print\"As,Q for 10 fins of 60 mm length(\",round(Q2,2),\"W) is more than Q for 6 fins of 100 mm length (\",round(Q,2),\"W)\"\n",
"print\"The agreement-->10 fins of 60 mm length is more effective\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"As,Q for 10 fins of 60 mm length( 117.66 W) is more than Q for 6 fins of 100 mm length ( 113.75 W)\n",
"The agreement-->10 fins of 60 mm length is more effective\n"
]
}
],
"prompt_number": 101
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.50,Page no:2.98"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Metallic wall surrounded by oil and water \n",
"#Variable declaration\n",
"h_oil=180.0 #W/(sq m.K)\n",
"h_air=15.0 #W/(sq m.K)\n",
"T_oil=353.0 #[K]\n",
"T_air=293.0 #[K]\n",
"delta_T=T_oil-T_air #[K]\n",
"k=80.0 #Conductivity in [W/(m.K)]\n",
"for_section=11.0*10**-3 #[m]\n",
"L=25.0 #[mm]\n",
"L=L/1000 #[m]\n",
"W=1.0 #[m] Width,..let\n",
"t=1.0 #[mm] \n",
"t=t/1000 #[m]\n",
"A=W*t #[m]\n",
"P=2*t\n",
"Af=2*L*W #sq m\n",
"N=1.0\n",
"\n",
"#Calculation\n",
"\n",
"import math\n",
"Ab=for_section-A #[sq m]\n",
"#CASE 1: Fin on oil side only\n",
"m=math.sqrt(h_oil*P/(k*A)) \n",
"nf_oil=math.tanh(m*L)/(m*L)\n",
"Ae_oil=Ab+nf_oil*Af*N #[sq m]\n",
"Q1=delta_T/(1/(h_oil*Ae_oil)+1/(h_air*for_section)) #[W]\n",
"#CASE 2: Fin on air side only\n",
"m=math.sqrt(h_air*P/(k*A))\n",
"nf_air=math.tanh(m*L)/(m*L)\n",
"nf_air=0.928 #Approximation\n",
"Ae_air=Ab+nf_air*Af*N #[sq m]\n",
"Q=delta_T/(1.0/(h_oil*for_section)+1.0/(h_air*Ae_air)) #[W]\n",
"\n",
"#Result\n",
"print\"In oil side,Q=\",round(Q1,2),\"W\"\n",
"print\"In air side,Q=\",round(Q,2),\"W\"\n",
"print\"From above results we see that more heat transfer takes place if fins are provided on the air side\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"In oil side,Q= 9.75 W\n",
"In air side,Q= 35.56 W\n",
"From above results we see that more heat transfer takes place if fins are provided on the air side\n"
]
}
],
"prompt_number": 103
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example no:2.51,Page no:2.100"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Brass wall having fins\n",
"\n",
"#Variable declaration\n",
"k=75.0 #Thermal conductivity [W/(m.K)]\n",
"T_water=363.0 #[K]\n",
"T_air=303.0 #[K] \n",
"dT=T_water-T_air #delta T\n",
"h1=150.0 # for water[W/(sq m.K)]\n",
"h2=15.0 #for air [W/(sq m.K)]\n",
"W=0.5 #Width of wall[m]\n",
"L=0.025 #[m]\n",
"\n",
"#Calculation\n",
"Area=W**2 #Base Area [sq m]\n",
"t=1.0 #[mm]\n",
"t=t/1000 #[m]\n",
"pitch=10.0 #[mm]\n",
"pitch=pitch/1000 #[m]\n",
"N=W/pitch #[No of fins]\n",
"\n",
"\n",
"#Calculations\n",
"A=N*W*t #Total cross-sectional area of fins in [sq m]\n",
"Ab=Area-A #[sq m]\n",
"Af=2*W*L #Surface area of fins [sq m]\n",
"import math\n",
"#CASE 1: HEAT TRANSFER WITHOUT FINS\n",
"A1=Area #[sq m]\n",
"A2=A1 #[sq m]\n",
"Q=dT/(1.0/(h1*A1)+1.0/(h2*A2)) #[W]\n",
"print\"Without fins,Q=\",round(Q,2),\"W\"\n",
"#CASE 2: Fins on the water side\n",
"P=2*(t+W) \n",
"A=0.5*10**-3 \n",
"m=math.sqrt(h1*P/(k*A))\n",
"nfw=math.tanh(m*L)/(m*L) #Effeciency on water side\n",
"Aew=Ab+nfw*Af*N #Effective area on the water side [sq m]\n",
"Q=dT/(1.0/(h1*Aew)+1.0/(h2*A2)) #[W]\n",
"print\"With fins on water side,Q=\",round(Q,2),\"W\"\n",
"#CASE 3: FINS ON THE AIR SIDE\n",
"m=math.sqrt(h2*P/(k*A))\n",
"nf_air=tanh(m*L)/(m*L) #Effeciency\n",
"Aea=Ab+nf_air*Af*N #Effective area on air side\n",
"Q=dT/(1.0/(h1*A1)+1.0/(h2*Aea)) #[W]\n",
"print\"With Fins on Air side,Q=\",round(Q,1),\"W\"\n",
"#BOTH SIDE:\n",
"Q=dT/(1.0/(h1*Aew)+1.0/(h2*Aea)) #[W]\n",
"\n",
"#Result\n",
"print\"With Fins on both side,Q=\",round(Q,1),\"W\" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Without fins,Q= 204.55 W\n",
"With fins on water side,Q= 219.24 W\n",
"With Fins on Air side,Q= 800.3 W\n",
"With Fins on both side,Q= 1084.7 W\n"
]
}
],
"prompt_number": 107
}
],
"metadata": {}
}
]
}