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{
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{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 1 Introduction"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 1.1"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Rate of heat transfer per unit area is 3.7 MW/sq meter\n"
]
}
],
"source": [
"#Example Number 1.1\n",
"# HOW MUCH HEAT IS TRANSFERRED THROUGH THE PLATE\n",
"\n",
"#VARIABLE DECLARATION\n",
"\n",
"k = 370 \t\t\t # [W/m] at 250 degree celsius\n",
"dt = 100-400 \t\t\t#[degree celsius] temperature difference\n",
"dx = 3*10**(-2) \t\t#[m] thickness of plate\n",
"\n",
"#CALCULATION\n",
"\n",
"q = -k*dt/dx \t\t\t#[MW/square meter]\n",
"\n",
"#RESULTS\n",
"\n",
"print \"Rate of heat transfer per unit area is\", q/1000000 ,\"MW/sq meter\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 1.2"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Rate of heat transfer is 2.156 kW\n"
]
}
],
"source": [
"#Example Number 1.2\n",
"# CALCULATE THE HEAT TRANSFER\n",
"\n",
"#Variable declaration\n",
"\n",
"Twall = 250 \t\t\t#[degree celsius] wall temperature\n",
"Tair = 20 \t\t\t#[degree celsius] air temperature\n",
"h = 25 \t\t\t\t#[W/square meter] heat transfer coefficient\n",
"l = 75*10**(-2) \t\t#[m] length of plate\n",
"b = 50*10**(-2) \t\t#[m] width of plate\n",
"area = l*b \t\t\t#[square meter] area of plate\n",
"dt = 250-20 \t\t\t#[degree celsius]\n",
"\n",
"\n",
"#Calculation\n",
"\n",
"q = h*area*dt \t\t\t# [W] from newton's law of cooling\n",
"\n",
"\n",
"#Result\n",
"\n",
"print\"Rate of heat transfer is\",round(q/1000,3),\"kW\" \n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 1.3"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Inside plate temperature is 253.05 degree C\n"
]
}
],
"source": [
"#Example Number 1.3\n",
"# Calculate the inside plate temperature.\n",
"\n",
"#variable declaration\n",
"\n",
"Qconv = 2156 \t # [W] from previous problem\n",
"Qrad = 300\t\t # [W] given\n",
"dx = 0.02\t\t # [m] plate thicknesss\n",
"l = 0.75 \t\t # [m] length of plate \n",
"w = 0.5 \t\t # [m] width of plate\n",
"k = 43\t\t\t #[W/m] from table 1.1\n",
"area = l*w\t\t #[square meter] area of plate\n",
"\n",
"#Calculation\n",
"\n",
"Qcond = Qconv+Qrad \t # [W]\n",
"dt = Qcond*dx/(k*area) \t # [degree celsius] temperature difference\n",
"Ti = 250+dt \t\t # inside temperature\n",
"\n",
"#Results\n",
"\n",
"print\"Inside plate temperature is\",round(Ti,2),\"degree C\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 1.4"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Heat transfer is: 22.0 W\n",
" This is equal to the electric power which must be applied\n"
]
}
],
"source": [
"#Example Number 1.4\n",
"# Calculate electric power to be supplied to the wire \n",
"\n",
"#Variable declaration\n",
"\n",
"d = 1*10**(-3) \t\t#[m] diameter of wire\n",
"l = 10*10**(-2) \t#[m] length of wire\n",
"Sarea = 22*d*l/7 \t#[square meter] surface area of wire\n",
"h = 5000 \t\t#[W/square meter] heat transfer coefficient\n",
"Twall = 114 \t\t# [degree celsius]\n",
"Twater = 100\t # [degree celsius]\n",
"\n",
"#total convection loss is given by equation(1-8)\n",
"\n",
"#Calculation\n",
"\n",
"Q = h*Sarea*(Twall-Twater) # [W]\n",
"\n",
"#Results\n",
"\n",
"print\"Heat transfer is:\",Q,\"W\" \n",
"print\" This is equal to the electric power which must be applied\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 1.5"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Heat transfer per unit area is: 69.03 kW/sq meter\n"
]
}
],
"source": [
"#Example Number 1.5\n",
"# radiation heat transfer\n",
"# Calculate the heat transfer per unit area\n",
"\n",
"#Variable declaration\n",
"\n",
"sigma = 5.669*10**(-8) \t\t#[W/square meter*k^(4)] universal constant\n",
"T1 = 273+800 \t\t\t# [k] first plate temperature\n",
"T2 = 273+300 \t\t\t# [k] second plate temperature\n",
"\n",
"#equation(1-10) may be employed for this problem\n",
"\n",
"#Calculation\n",
"Q = sigma*(T1**4-T2**4) \t# [W/square meter]\n",
"\n",
"#Results\n",
"\n",
"print\"Heat transfer per unit area is:\",round(Q/1000,2),\" kW/sq meter\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exa 1.6"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total heat loss is: 55.67 W/m\n"
]
}
],
"source": [
"#Example Number 1.6\n",
"# total heat loss by convection and radiation\n",
"\n",
"#Variable declaration\n",
"d = 0.05 #[m] diameter of pipe\n",
"Twall = 50 #[degree celsius] \n",
"Tair = 20 #[degree celsius]\n",
"emi = 0.8 #emissivity\n",
"h = 6.5 #[W/square meter] heat transfer coefficient for free convection\n",
"import math\n",
"Q1 = h*math.pi*d*(Twall-Tair) #[W/m] convection loss per unit length\n",
"sigma = 5.669*10**(-8) # [W/square meter*k^(4)] universal constant\n",
"T1 = 273+Twall # [k]\n",
"T2 = 273+Tair # [k]\n",
"Q2 = emi*math.pi*d*sigma*((T1**(4))-(T2**(4))) # [W/m] heat loss due to radiation per unit length\n",
"Qtotal = Q1+Q2 # [W/m] total heat loss per unit length\n",
"\n",
"print\"Total heat loss is:\",round(Qtotal,2),\"W/m\"\n"
]
}
],
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