{ "cells": [ { "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" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.6" } }, "nbformat": 4, "nbformat_minor": 0 }