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{
"metadata": {
"name": "",
"signature": "sha256:2bebadfad4a18ede042a136ada8984e181f57290a23ad4ed8003a1f83445a447"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 25: Machine\u2013Network Interactions"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 25.4, Page 893"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
" #Initializing the variables\n",
"Pa_P1 = -200; # From previous Question\n",
"Q = 1.4311 ; # From previous questions.\n",
"\n",
" #Calculations\n",
"DpSys = Pa_P1 + 98.9*Q**2;\n",
"print \"System Operating point (m^3/s):\",round(DpSys,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"System Operating point (m^3/s): 2.55\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 25.7, Page 906"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"import sympy\n",
"from sympy import solve,symbols\n",
" \n",
"\n",
" #Initializing the variables\n",
"Vo = 25.3; #Outlet velocity\n",
"D = 10 ; # Mean hydraulic diameter\n",
"f = 0.008; # friction factor\n",
"X = 1000; # Length of road\n",
"P = 12600; # Absorbing power\n",
"Va = 300; # Tunnel air flow\n",
"K1 = 0.96;\n",
"K2 = 0.9;\n",
"T = 590; #Thrust\n",
"rho = 1.2; # Air density \n",
"\n",
" #Calculations\n",
"alpha = (1/D)**2;\n",
"A = math.pi*D**2/4; # Area of tunnel\n",
"Vt = Va/A;\n",
"W = Vo/Vt; #Omega\n",
"E = (1-alpha*W);\n",
"C = (1-alpha*W)*(1-E)**2 + E**2 - 1;\n",
" # Manipulating equation 25.20;\n",
"LHS = f*X*(E+1)**2/D + C + 1 ;\n",
"\n",
"n1 = symbols('n1')\n",
"result=solve(K1*(2*((alpha*W**2 + (1-alpha)*E**2-1)+(n1-1)*(alpha*W*(W-1)-C/2)))-LHS)\n",
"\n",
"n=result[0]\n",
"\n",
"\n",
" # Alternative approach using equation 25.22\n",
"n2 = (rho*((4*f*X*Vt**2)/(2*D) + 1.5*Vt**2/2))*A/(K1*K2*T); \n",
"Pt = round(n2)*P;\n",
"\n",
"print \"Number of fans required :\",round(n2)\n",
"print \"Total power consumed (KW) :\",Pt/1000"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Number of fans required : 6.0\n",
"Total power consumed (KW) : 75.6\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 25.8, Page 907"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"import sympy\n",
"from sympy import solve,symbols\n",
" #Initializing the variables\n",
"f = 0.008;\n",
"T = 290;\n",
"L = 750;\n",
"Dt = 9; # Diameter Tunnel\n",
"Df = 0.63; # Diameter fan\n",
"K1 = 0.98;\n",
"K2 = 0.92;\n",
"Vo = 27.9;\n",
"n = 10;\n",
"A=math.pi*Dt**2/4\n",
"rho=1.2\n",
"X=750\n",
" #Calculations\n",
"alpha = (Df/Dt)**2;\n",
" # equation 25.20 becomes when E = 1 nad C = 0\n",
"W=symbols('W')\n",
"omega = solve(2*K1* (alpha*W**2 +(n-1)*alpha*W*(W-1)) - 4*f*L/Dt -1)\n",
" \n",
"\n",
"for i in range(1,len(omega)): # since omega is always positive and real\n",
" if omega[i]>0:\n",
" w = round(omega[i],1);\n",
"Vt = Vo/w;\n",
"\n",
"# by equation 25.22\n",
"VT=(n*(K1*K2*T)/(A*(rho*((4*f*X)/(2*Dt) + 1.5/2))))**0.5\n",
"print \"Tunnel Velocity (m/s) :\",round(VT,2)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Tunnel Velocity (m/s) : 4.05\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 25.9, Page 914"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"\n",
" #Initializing the variables\n",
"Ws = 0.45;\n",
"Ks = 3.2;\n",
"H = 152;\n",
"h = 0;\n",
"Hatm = 10.3;\n",
"Pv = 350; #vapour pressure\n",
"g = 9.81;\n",
"rho = 1000;\n",
" \n",
" #Calculations\n",
"Ht1 = 152*(Ws/Ks)**(4/3); # the value of Ht1 is 11.12 and in book it is taken as 11.2 so there will be a difference in final answer\n",
"Hvap = round(Pv/(rho*g),3);\n",
"Z = Hatm -h -Hvap -Ht1;\n",
"print \"Elevation of pump (m):\",round(Z,3)\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Elevation of pump (m): -0.851\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 25.11, Page 927"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"import sympy\n",
"from sympy import symbols,solve\n",
"import numpy as np\n",
" \n",
"\n",
" #Initializing the variables\n",
"Co = 0;\n",
"Qc = 0.0024;\n",
"V = 5400;\n",
"c = 10;\n",
" #Calculations\n",
"#####--------------------PART(A)-----------------#######\n",
"n1=symbols('n1')\n",
"def partA(n1):\n",
" Ci = 10;\n",
" # t = infinity so e^(-nt) = 0\n",
" Q=10000*Qc/(c-Co)\n",
" n1 = Q*3600/V; \n",
" return n1\n",
"ans=partA(n1)\n",
"\n",
"print \"Part(A) : number of air changes per hour if the garage is in continuous use and the maximum permissible concentration of carbon monoxide is 0.1 per cent. :\",ans,\"\\n\"\n",
"\n",
"#####--------------------PART(B)-----------------#######\n",
"n=symbols('n')\n",
"def partB(n):\n",
" Ci = 0; \n",
" n=[1.5,1.2,0.9,1.0] \n",
" t=1 # time in hours\n",
" error=[]\n",
" mini=100\n",
" ans=0\n",
" for i in range(4): \n",
" Q = V/3600; \n",
" A = 10000*Qc/Q; # as Co=0 \n",
" error.append(abs((A*(1-math.e**(-n[i]*t))/c)-n[i]));\n",
" if(error[i]<mini):\n",
" mini=error[i]\n",
" ans=n[i]\n",
" return ans \n",
"ans=partB(n)\n",
"print \"Part(B) : number of air changes per hour if this maximum level is reached after 1 hour and the garage is out of use :\",ans,'\\n'\n",
" \n",
"#####--------------------PART(C)-----------------#######\n",
"c=symbols('c')\n",
"def partC(c):\n",
" Ci = 0;\n",
" n = 1; \n",
" t = 0.333333 # 20 minutes in hours\n",
" Q = V*n/3600;\n",
" y = (Co + 10000*Qc/Q)*(1-math.e**(-n*t)) + Ci*math.e**(-n*t) ; \n",
" return y\n",
"ans=partC(c)\n",
"print \"Part(C) :the concentration after 20 minutes (Parts per 10000) :\",round(ans,3),'\\n'\n",
"#####--------------------PART(D)-----------------#######\n",
"t=symbols('t')\n",
"def partD(t):\n",
" Ci = 10;\n",
" n = 1; \n",
" c = 0.1;\n",
" t=np.log(100) \n",
" return round(t,2)\n",
"ans=partD(t) \n",
"print \"Part(D) : time necessary to run the ventilation system at the rate calculated in (b) to reduce the concentration to 0.001 per cent (in hours) :\",ans,\"hours\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Part(A) : number of air changes per hour if the garage is in continuous use and the maximum permissible concentration of carbon monoxide is 0.1 per cent. : 1.6 \n",
"\n",
"Part(B) : number of air changes per hour if this maximum level is reached after 1 hour and the garage is out of use : 1.0 \n",
"\n",
"Part(C) :the concentration after 20 minutes (Parts per 10000) : 4.535 \n",
"\n",
"Part(D) : time necessary to run the ventilation system at the rate calculated in (b) to reduce the concentration to 0.001 per cent (in hours) : 4.61 hours\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}
|
