{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6:Laminar and turbulent boundary layers "
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.2, Page number: 284"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=300; #air temperature,K\n",
"v=1.5; #air velocity, m/s\n",
"t=0.5; #thickness, m\n",
"u=1.853*math.pow(10,-5); #dynamic viscosity,kg/(m*s)\n",
"v1=1.566*math.pow(10,-5); #kinematic viscosity,m**2/s\n",
"\n",
"#Calculations\n",
"Rex=v*t/v1; #reynolds no. is low enough to permit the use of laminar flow analysis.\n",
"b=4.92*t/(math.sqrt(Rex))*100; # b.l. thickness, cm\n",
"#in this case b/x=1.124/50=0.0225 so laminar flow is valid.\n",
"v2=0.8604*math.sqrt(v1*v/t);\n",
"#since v2 grows larger as x grows smaller, the condition v2<u is not satisfied very near the leading edge.\n",
"#in this case del/thickness is 0.0225.\n",
"x=0.8604*math.sqrt(v1*v/t); #velocity,m/s\n",
"y=x/t;\n",
"\n",
"#Results\n",
"print \"Boundary layer thickness is :\",round(b,3),\"cm\\n\"\n",
"print \"Since velocity grows larger as thickness grows smaller, the condition x<<u is not satisfied very near the leading edge. therefore the BI approximation themselves breakdown.\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Boundary layer thickness is : 1.124 cm\n",
"\n",
"Since velocity grows larger as thickness grows smaller, the condition x<<u is not satisfied very near the leading edge. therefore the BI approximation themselves breakdown.\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.3, Page number: 291"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"l=0.5; #total length of surface,m\n",
"Cf=0.00607; #overall friction coefficient\n",
"\n",
"#Calculations\n",
"tw=1.183*(2.25)*Cf/2; # wall shear, kg/(m*s**2)\n",
"a=0.5; #ratio of wall shear at x=l and average wall shear\n",
"#tw(x)=twavg where 0.664/(x**0.5)=1.328/(47,)893, x=1/8 m thus the wall shear stress plummets to twavg one fourth of the way from the leading edge and drops only to one half of twavg in the remaining 75 percent plate.x<600*1.566*10**(-5)/1.5=0.0063 m.\n",
"# preceding analysis should be good over almost 99 percent of the 0.5 m length of the surface.\n",
"\n",
"#Results\n",
"print \"Overall friction coefficient is :\",Cf,\"\\n\"\n",
"print \"Wall shear is :\",tw,\"kg/(m*s^2)\\n\"\n",
"print \"The preceding analysis should be good over almost 99 percent of the 0.5m length of the surface.\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Overall friction coefficient is : 0.00607 \n",
"\n",
"Wall shear is : 0.00807841125 kg/(m*s^2)\n",
"\n",
"The preceding analysis should be good over almost 99 percent of the 0.5m length of the surface.\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.4, Page number: 296"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"l=0.06; #length of heater, m\n",
"p=15; #pressure of heater, atm\n",
"T1=440; #temperature of heater, K\n",
"v=2; #free stream velocity,m/s\n",
"T2=460; #constant temperature of heater, K\n",
"T3=450; #mean temperature of heater, K\n",
"k=0.674; #thermal conductivity ,W/(m.K)\n",
"\n",
"#Calculations\n",
"q=2*(0.332)*(k/l)*math.sqrt(v*l/(1.72*10**-7))*(T2-T1)/1000;\t\t #(from Fig 6.10)formula for heat flux is q=2*(0.664)*k/l*(Rel**0.5)*(T2-T1), in kW/m^2\n",
"\n",
"#Results\n",
"print \"Heat flux is :\",round(q,3),\"kW/m^2\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat flux is : 124.604 kW/m^2\n",
"\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.5, Page number: 308"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=293; #air temperature,K\n",
"v=15; #air velocity,m/s\n",
"T2=383; #temperature of plate,K\n",
"l=0.5; #length of plate,m\n",
"w=0.5; #width of plate,m\n",
"Pr=0.707; #prandtl no.\n",
"k=0.02885; #thermal conductivity of ,W/(m.K)\n",
"\n",
"#Calculations\n",
"Rel=v*l/(0.0000194); #reynolds no.\n",
"Nul=0.664*(Rel)**0.5*Pr**(1/3); \t #nusset no.\n",
"h1=367.8*(k)/l; #average convection coefficient, W/(m**2*K)\n",
"Q=h1*l**(2)*(T2-T1); #heat transferred,W\n",
"h2=h1/2 #convection coefficient at trailing , W/(m**2*K)\n",
"a1=4.92*l/math.sqrt(Rel)*1000 #hydrodynamic boundary layer,m\n",
"a2=a1/(Pr)**(1/3); #thermal boundary layer,mm\n",
"\n",
"#Results\n",
"print \"Average heat trensfer coefficient is :\",round(h1,3),\"W/m^2/K\\n\"\n",
"print \"Total heat transferred is\",round(Q,3),\"W\\n\"\n",
"print \"Convection coefficient at trailing is :\",round(h2,3),\"W/(m^2*K)\\n\"\n",
"print \"Hydrodynamic boundary layer is : \",round(a1,3),\"m\\n\"\n",
"print \"Thermal boundary layer is : \",round(a2,3),\"mm\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average heat trensfer coefficient is : 21.222 W/m^2/K\n",
"\n",
"Total heat transferred is 477.496 W\n",
"\n",
"Convection coefficient at trailing is : 10.611 W/(m^2*K)\n",
"\n",
"Hydrodynamic boundary layer is : 3.956 m\n",
"\n",
"Thermal boundary layer is : 4.441 mm\n",
"\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.6, Page number: 310"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=288; # air temperature,K\n",
"v=1.8; # air velocity,m/s\n",
"l=0.6; # length of panel, m\n",
"Q=420; # power per unit area, m**2\n",
"T2=378; # maximum temperature of surface, K\n",
"k=0.0278; #thermal conductivity of ,W/(m.K)\n",
"Pr=0.709; #Prandtl no.\n",
"\n",
"#Calculations\n",
"T3=Q*l/(k)/(0.453*(l*v/(1.794*10**-5))**(0.50)*(Pr)**(1/3));#maximum temperature difference \n",
"Twmax=T1+T3; #Twmax comes out to be 106.5 C, this is very close to 105 C,if 105 is at all conservative, Q = 420 should be safe.\n",
"T4=0.453/0.6795*T3; #average temperature difference,K\n",
"Twavg=T1+T4; #average wall temperature,K\n",
"Twa=Twavg-273; #average wall temp. in \u00b0C\n",
"\n",
"#Results\n",
"print \"Average wall temperature is :\",round(Twa,3),\"C\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average wall temperature is : 75.975 C\n",
"\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.8, Page number: 313"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"v=15; #air velocity,m/s\n",
"T2=383; # temperature of plate,K\n",
"l=0.5; # length of plate,m\n",
"w=0.5; # width of plate,m\n",
"Pr=0.707; # prandtl no.\n",
"rho=1.05; #Air desnity, kg/m^3\n",
"#Calculations\n",
"Rel=v*l/(0.0000194); #reynolds no.\n",
"Nul=0.664*math.sqrt(Rel)*Pr**(1/3); # nusset no.\n",
"Cf=2*Nul/(Rel*Pr**(1/3)); #friction coefficient\n",
"s=Cf*0.5*rho*v**2; #drag shear, kg/(m*s**2)\n",
"f=s*0.5**2; #drag force, N\n",
"\n",
"#Results\n",
"print \"Drag force on heat transfer surface is :\",round(f,5),\"N\\n\"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Drag force on heat transfer surface is : 0.06307 N\n",
"\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.9, Page number: 328"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=297; # river water temp.,K\n",
"T2=283; # ocean water temp., K\n",
"n=5; # no. of knots\n",
"k=0.5927; # thermal conductivity,W/(m*K)\n",
"a=998.8; #density of water, kg/m**3\n",
"Cp=4187; # heat capacity, J/kg/K\n",
"Pr=7.66; #Prandtle no.\n",
"x=1; #distance from forward edge,m\n",
"v=1.085*10**-6; # kinematic viscosity, m**2/s\n",
"u=2.572; # velocity of knot,m/s\n",
"\n",
"#Calculations\n",
"T3=(T1+T2)/2; # avg. temp.,K\n",
"Rex=u/v # reynolds no.\n",
"Cf=0.455/(math.log(0.06*Rex))**2 # friction coefficient\n",
"h=k/x*0.032*(Rex)**(0.8)*Pr**(0.43); # heat transfer coefficient,W/(m**2*K)\n",
"h1=a*Cp*u*Cf/2/(1+12.8*(Pr**0.68-1)*math.sqrt(Cf/2)); #heat transfer coefficient,W/(m**2*K)\n",
"\n",
"#Results\n",
"print \"Friction coefficient is :\",round(Cf,5),\"\\n\"\n",
"print \"Convective heat transfer coefficient at a distance of 1 m fom the forward edge is :\",round(h,3),\"W/(m^2*K)\\n\"\n",
"print \"Heat transfer coefficient by another method is :\",round(h1,3),\"W/(m^2*K)\\n\"\n",
"print \"The two values of h differ by about 18 percent, which is within the uncertainity \\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Friction coefficient is : 0.00323 \n",
"\n",
"Convective heat transfer coefficient at a distance of 1 m fom the forward edge is : 5728.966 W/(m^2*K)\n",
"\n",
"Heat transfer coefficient by another method is : 6843.244 W/(m^2*K)\n",
"\n",
"The two values of h differ by about 18 percent, which is within the uncertainity \n",
"\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 6.10, Page number: 329"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"l=2; # length of plate,m\n",
"p=1000; # power density,W/m**2\n",
"u=10; # air velocity,m/s\n",
"T1=290; # wind tunnel temp.,K\n",
"p2=1; # pressure,atm\n",
"v=1.578*10**-5; # kinematic viscosity, m**2/s\n",
"k=0.02623; # thermal conductivity,W/(m*K)\n",
"Pr=0.713; # prandtl no.\n",
"Rel=u*l/v; # reynolds no. at 10 m/s\n",
"Nul=1845; # nusselt no.\n",
"Re=400000; #Reynolds no. at the end of turbulent transition engine\n",
"\n",
"#Calculations\n",
"h=Nul*k/l; #convection coefficient,W/(m**2*K)\n",
"Tavg=T1+p/h; #Average temperature of plate, K\n",
"\n",
"#to take better account of the transition region, we can use churchill eqn.\n",
"x=Rel*Pr**(2/3)/(math.sqrt(1+(0.0468/Pr)**(2/3))); \n",
"x1=1.875*x*Re/Rel;\n",
"Nul1=0.45+0.6774*math.sqrt(x)*math.sqrt(1+((x/12500)**(3/5))/(1+(x1/x)**3.5)**0.4);\n",
"H=Nul1*k/l; #convection coefficient,W/(m**2*K)\n",
"Tw=290+1000/H; #average temperature of plate,K\n",
"\n",
"#Results\n",
"print \"Average temperature of plate is :\",round(Tavg,2),\" K\\n\"\n",
"print \"Average temperature of plate is :\",round(Tw,2),\" K , thus in this case, the average heat transfer coefficient is 33 percent higher when the transition regime is included.\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average temperature of plate is : 331.33 K\n",
"\n",
"Average temperature of plate is : 321.54 K , thus in this case, the average heat transfer coefficient is 33 percent higher when the transition regime is included.\n",
"\n"
]
}
],
"prompt_number": 23
}
],
"metadata": {}
}
]
}