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"source": [
"# Chapter 06:Incompressible viscous flow: A brief review"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.1:pg-226"
]
},
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"Introduction to heat transfer by S.K.Som, Chapter 6, Example 1\n",
"Umax in m/s is\n",
"Umax= 1.6\n",
"The shear stress T in N/m**2\n",
"T= 64.0\n",
"(dp/dx) in N/m**3 is\n",
"X= -19200.0\n",
"The Shear stress at a distance of 0.002m from the lower plate in N/m**2\n",
"t= -57.6\n",
"The shear stress at a distance of 0.002m from the upper plate in N/m**2\n",
"t= 57.6\n",
"The opposite signs in t represents the opposite directions.The plus sign is in the direction of flow and the minus sign is in the direction opposite to the flow \n",
"The pressure drop over a distance of 2m in N/m**2 is\n",
"deltaP= 38400.0\n"
]
}
],
"source": [
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 1\"\n",
" #Oil of specific gravity 0.90 and dynamic viscosity (mu=0.1Pa*s) flows between two fixed plates kept 2*b=10mm apart,So b=5mm.\n",
"#The average velocity is Uav=1.60m/s\n",
"Uav=1.60;\n",
"mu=0.1;\n",
"b=0.005; #in metre\n",
" #Umax is maximum velocity\n",
" Umax=(3/2)*Uav\n",
"print\"Umax in m/s is\"\n",
"Umax=(3/2)*Uav\n",
"print\"Umax=\",Umax\n",
" #The shear stress at the plate is given by T=2*Âµ*(Umax/b)\n",
"print\"The shear stress T in N/m**2\"\n",
"T=2*mu*(Umax/b) \n",
" #The shear sress at a distance from plate is given by t=y*(dp/dx)\n",
"#(dp/dx)=X=-3*mu*(Uav/b**2)\n",
"print\"T=\",T\n",
"print\"(dp/dx) in N/m**3 is\"\n",
"X=-3*mu*(Uav/b**2)\n",
" #Taking modulus of X by multipying it with negative sign.\n",
"print\"X=\",X\n",
"print\"The Shear stress at a distance of 0.002m from the lower plate in N/m**2\"\n",
"y=b-0.002;\n",
"t=y*(X) #NOTE:Answer given in the book is incorrect (Calculation mistake)\n",
"print\"t=\",t\n",
"print\"The shear stress at a distance of 0.002m from the upper plate in N/m**2\"\n",
"t=-y*(X) #NOTE:Answer given in the book is incorrect (Calculation mistake)\n",
"print\"t=\",t\n",
"print\"The opposite signs in t represents the opposite directions.The plus sign is in the direction of flow and the minus sign is in the direction opposite to the flow \"\n",
" #deltaP is the pressure drop\n",
"print\"The pressure drop over a distance of 2m in N/m**2 is\"\n",
" #Since pressure drop is considered at a distance of 2m so L=2m\n",
"L=2;\n",
"deltaP=(-X)*L\n",
"print\"deltaP=\",deltaP"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.3:pg-229"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
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"name": "stdout",
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"text": [
"Introduction to heat transfer by S.K.Som, Chapter 6, Example 3\n",
"The rate of change of pressure with respect to length in N/m**3\n",
"X= 2000\n",
"Flow rate(Q) in m**3/s is)\n",
"Q= 0.00333333333333\n",
"The viscosity of oil(mu)in kg/(m*s)\n",
"mu= 0.0920388472731\n"
]
}
],
"source": [
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 3\"\n",
" #Oil of specific gravity (sg)=0.90 is discharged at a rate(mdot)=3kg/s under a pressure difference dp=10KN/m**2 over a length dz=5m of a pipe having a diameter(D) of 50mm.\n",
"dp=10*10**3; #in N/m**2\n",
"dz=5;\n",
"D=0.05; #in metre\n",
"mdot=3;\n",
"sg=0.90;\n",
" #X=dp/dz is the rate of change of pressure\n",
"print\"The rate of change of pressure with respect to length in N/m**3\"\n",
"X=dp/dz\n",
"print\"X=\",X\n",
" #Flow rate is Q\n",
"print\"Flow rate(Q) in m**3/s is)\"\n",
"Q=mdot/(sg*10**3)\n",
"print\"Q=\",Q\n",
" #The viscosity of oil is mu=(pi*D**4*X)/(128*Q*dz)\n",
"print\"The viscosity of oil(mu)in kg/(m*s)\"\n",
"mu=(math.pi*D**4*X)/(128*Q)\n",
"print\"mu=\",mu"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.7:pg-250 "
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
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"outputs": [
{
"name": "stdout",
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"text": [
" Introduction to heat transfer by S.K.Som, Chapter 6, Example 7\n",
"The maximum length of plate in m is \n",
"L= 2.5\n",
"The average skin friction coefficient is\n",
"cfL= 1.328\n",
"Drag force on one side of plate in N is\n",
"Fd= 21.5136\n"
]
}
],
"source": [
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 7\"\n",
" #A flat plate B=1.2m wide and of length L is kept parallel to a uniform stream of air of velocity Uinf=3m/s in a wind tunnel.\n",
"Uinf=3;\n",
"B=1.2;\n",
" #If it is desired to have a laminar boundary layer only on the plate \n",
"#Assume that the laminar flow exists up to a reynold number(ReL)=5*10**5.Take density of air as rhoair=1.2kg/m**3 and viscosity of air as nuair=1.5*10**-5 m**2/s.\n",
"nuair=1.5*10**-5;\n",
"rhoair=1.2;\n",
"ReL=5*10**5;\n",
" #For maximum length of the plate reynolds number is ReL=Uinf*L/nuair\n",
"#so L=ReL*nuair/Uinf\n",
"print\"The maximum length of plate in m is \"\n",
"L=ReL*nuair/Uinf\n",
"print\"L=\",L\n",
" #The average skin friction coefficient is cfL=1.328/(ReL)**(1/2)\n",
"print\"The average skin friction coefficient is\"\n",
"cfL=1.328/(ReL)**(1/2)\n",
"print\"cfL=\",cfL\n",
" #Fd is drag force\n",
"print\"Drag force on one side of plate in N is\"\n",
"Fd=cfL*(rhoair*Uinf**2/2)*B*L\n",
"print\"Fd=\",Fd"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex6.10:pg-268 "
]
},
{
"cell_type": "code",
"execution_count": 17,
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"outputs": [
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"name": "stdout",
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"text": [
"Introduction to heat transfer by S.K.Som, Chapter 6, Example 10\n",
"Wind velocity(Uinf)in m/s is\n",
"Uinf= 10\n",
"Reynolds number is\n",
"ReL= 4000000.0\n",
"Friction coefficient is\n",
"CbarfL= 0.0735645\n",
"Drag force on one side of the plate per unit metre width in Newton is \n",
"FD= 26.48322\n",
"The turbulent boundary layer thickness at the trailing edge in metre is \n",
"delta= 2.274\n"
]
}
],
"source": [
"import math\n",
" \n",
"print\"Introduction to heat transfer by S.K.Som, Chapter 6, Example 10\"\n",
" #Wind at a speed of U=36km/hr blows over a flat plate of length,L=6m .If the density and kinematic viscosity of air are rho=1.2kg/m**3 and mu=1.5*10**-5m**2/s respectively.\n",
"U=36;\n",
"L=6;\n",
"rho=1.2;\n",
"mu=1.5*10**-5;\n",
" #Wind velocity in m/s is Uinf\n",
"print\"Wind velocity(Uinf)in m/s is\"\n",
"Uinf=U*1000/3600\n",
"print\"Uinf=\",Uinf\n",
" #Reynolds number is given by ReL=L*Uinf/mu\n",
"print\"Reynolds number is\"\n",
"ReL=L*Uinf/mu\n",
"print\"ReL=\",ReL\n",
" #We consider that transition of boundary layer takes place from laminar to turbulent takes place at ReL=5*10**5.\n",
"#Therfore the corresponding friction coefficient is given by CbarfL=(0.074-ReL**(1/5))-(1742/ReL)\n",
"print\"Friction coefficient is\"\n",
"CbarfL=(0.074/ReL**(1/5))-(1742/ReL)\n",
"print\"CbarfL=\",CbarfL\n",
" #Drag force on one side of the plate per unit metre width is given by FD=CbarfL*rho*Uinf**2*L/2\n",
"print\"Drag force on one side of the plate per unit metre width in Newton is \"\n",
"FD=CbarfL*rho*Uinf**2*L/2\n",
"print\"FD=\",FD\n",
" #The turbulent boundary layer thickness at the trailing edge is given by delta=L*(0.379/ReL**(1/5))\n",
"print\"The turbulent boundary layer thickness at the trailing edge in metre is \"\n",
"delta=L*(0.379/ReL**(1/5))\n",
"print\"delta=\",delta"
]
}
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