diff options
Diffstat (limited to 'Mechanical_Engineering_Thermodynamics/Chapter_3.ipynb')
| -rwxr-xr-x | Mechanical_Engineering_Thermodynamics/Chapter_3.ipynb | 240 |
1 files changed, 240 insertions, 0 deletions
diff --git a/Mechanical_Engineering_Thermodynamics/Chapter_3.ipynb b/Mechanical_Engineering_Thermodynamics/Chapter_3.ipynb new file mode 100755 index 00000000..f5793d78 --- /dev/null +++ b/Mechanical_Engineering_Thermodynamics/Chapter_3.ipynb @@ -0,0 +1,240 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3 - Temperature and Heat"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1 - Pg 33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the mass of water required per pound of iron \n",
+ "import math\n",
+ "#Initialization of variables\n",
+ "T1=500 #F\n",
+ "T2=100 #F\n",
+ "Tf=75 #F\n",
+ "cpi=0.120 #B/lb F\n",
+ "cpw=1.0 #B/lb F\n",
+ "#calculations\n",
+ "Qw=1*cpw*(T2-Tf)\n",
+ "Qi=-1*cpi*(T2-T1)\n",
+ "mw=Qi/Qw\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Mass of water = \",mw,\"lb water/lb iron\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass of water = 1.92 lb water/lb iron\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2 - Pg 34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the net heat transferred\n",
+ "import math\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "#Initialization of variables\n",
+ "m=5 #lb\n",
+ "T1=1540+460 #R\n",
+ "T2=540+460 #R\n",
+ "#calculations\n",
+ "def q(T):\n",
+ "\tcp=m*(0.248+0.448*math.pow(10,-8) *T*T)\n",
+ "\treturn cp;\n",
+ "\n",
+ "Q=scipy.integrate.quad(q,T1,T2)\n",
+ "#results\n",
+ "print '%s %d %s' %(\"Heat transferred =\",Q[0],\"Btu\")\n",
+ "print '%s' %(\"The answer is a bit different due to rounding off error in textbook\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transferred = -1292 Btu\n",
+ "The answer is a bit different due to rounding off error in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3 - Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the heat required for the process\n",
+ "#Initialization of variables\n",
+ "Tm=235 #F\n",
+ "Tb=832 #F\n",
+ "T=70 #F\n",
+ "cps=0.18 #B/lb F\n",
+ "cpl=0.235 #B/lb F\n",
+ "Lf=15.8 #B/lb\n",
+ "Lv=120 #B/lb\n",
+ "m=10 #lb\n",
+ "#calculations\n",
+ "Qa=m*cps*(Tm-T)\n",
+ "Qb=m*Lf\n",
+ "Qc=m*cpl*(Tb-Tm)\n",
+ "Qd=m*Lv\n",
+ "Q=Qa+Qb+Qc+Qd\n",
+ "#results\n",
+ "print '%s %.1f %s' %(\"Heat required =\",Q,\"Btu\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat required = 3057.9 Btu\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4 - Pg 36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the mass of ice required for the process\n",
+ "#Initialization of variables\n",
+ "T1=22 #F\n",
+ "T2=32 #F\n",
+ "T3=40 #F\n",
+ "T4=70 #F\n",
+ "cps=0.501 #B/lb F\n",
+ "cpw=1 #B/lb F\n",
+ "Lf=143.3 #B/lb\n",
+ "m=40 #lb\n",
+ "#calculations\n",
+ "Qa=cps*(T2-T1)\n",
+ "Qb=Lf\n",
+ "Qc=cpw*(T3-T2)\n",
+ "Qd=m*cpw*(T3-T4)\n",
+ "mi=-Qd/(Qa+Qb+Qc)\n",
+ "#results\n",
+ "print '%s %.2f %s' %(\"Mass of ice required =\",mi,\"lb ice\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass of ice required = 7.68 lb ice\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5 - Pg 37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the extra mass of ice required\n",
+ "#Initialization of variables\n",
+ "T1=22 #F\n",
+ "T2=32 #F\n",
+ "T3=40 #F\n",
+ "T4=70 #F\n",
+ "cps=0.501 #B/lb F\n",
+ "cpw=1 #B/lb F\n",
+ "Lf=143.3 #B/lb\n",
+ "m=40 #lb\n",
+ "cp=0.092\n",
+ "mc=10\n",
+ "#calculations\n",
+ "Qa=cps*(T2-T1)\n",
+ "Qb=Lf\n",
+ "Qc=cpw*(T3-T2)\n",
+ "Qe=mc*cp*(T3-T4)\n",
+ "mi=-Qe/(Qa+Qb+Qc)\n",
+ "#results\n",
+ "print '%s %.3f %s' %(\"Extra Mass of ice required =\",mi,\"lb ice\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extra Mass of ice required = 0.177 lb ice\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |