{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 7: permiability" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.10: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "k1= 0.302e-7 //cm/sec\n", "k2= 0.12e-7 //cm/sec\n", "e1= 1.1\n", "e2= 0.9\n", "e= 0.75\n", "//calcualtions\n", "n= (log10((k1/k2)*((1+e1)/(1+e2))))/log10(e1/e2)\n", "C= k1/(e1^n/(1+e1))\n", "k= C*(e^n/(1+e))\n", "//results\n", "printf ('hydraulic conductivity = %e cm/sec ',k')" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.11: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "H1= 2 //m\n", "H2= 3 //m\n", "H3= 4 //m\n", "k1= 1e-4 //cm/sec\n", "k2= 3.2e-2 //cm/sec\n", "k3= 4.1e-5 //cm/sec\n", "//calculations\n", "H= H1+H2+H3\n", "Kh= (1/H)*((k1*H1)+(k2*H2)+(k3*H3))\n", "Kv= H/((H1/k1)+(H2/k2)+(H3/k3))\n", "P= Kh/Kv\n", "//results\n", "printf ('ration of equivalent hydraulic conductivity = % 2f ',P)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.12: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "H= 450 //mm\n", "h= 150 //mm\n", "k1= 1e-2 //cm/sec\n", "k2= 3e-3 //cm/sec\n", "k3= 4.9e-4 //cm/sec\n", "h1= 300 //mm\n", "//calculations\n", "Kv= H/(h*(1/k1+1/k2+1/k3))\n", "i= h1/H\n", "q= Kv*i*100*3600\n", "//results\n", "printf ('rate of water supply = % 2f cm/hr ',q)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.1: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "L= 30 //cm\n", "A= 177 //cm^2\n", "h= 50 //cm\n", "Q= 350 //cm^3\n", "t= 300 //sec\n", "//claculations\n", "k=Q*L/(A*h*t)\n", "//results\n", "printf ('hydraulic conductivity = % 3f cm/sec ',k)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.2: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "L= 203 //mm\n", "A= 10.3 //cm^2\n", "a= 0.39 //cm^2\n", "h0= 508 //mm\n", "h180= 305 //mm\n", "t= 180 //sec\n", "//calculations\n", "k= 2.303*a*L*log10(h0/h180)/(A*t)\n", "//results\n", "printf ('hydraulic conductivity of sand = % 2f in/sec ',k)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.3: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of varilables\n", "k= 3e-7 //cm/sec\n", "n= 0.0911e-4 //g*sec/cm^2\n", "dw= 1 //g/cc\n", "//calculations\n", "K= k*n/dw\n", "//results\n", "printf ('absolute premeability = % 4f cm^2 ',K)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.4: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "k= 5.3e-5 //m/sec\n", "H= 3 //m\n", "a= 0.139 //radians\n", "//calculations\n", "A= H*cos(a)\n", "i= sin(a)\n", "q= k*i*A*3600\n", "//results\n", "printf ('rate of seepage = % 4f m^3/hr/m ',q)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.5: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "L= 50 //m\n", "k= 0.08e-2//m/sec\n", "h= 4 //m\n", "H1= 3 //m\n", "H= 8 //m\n", "a= 0.139 //radians\n", "//calculations\n", "i= h*cos(a)/L\n", "A= H1*cos(a)\n", "q= k*i*A\n", "//results\n", "printf ('flow rate = % 2f m^3/sec/m ',q)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.6: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "k1= 0.02 //cm/sec\n", "e1= 0.5 \n", "e2= 0.65\n", "//calculations\n", "k2= k1*(e2^3/(1+e2))/(e1^3/(1+e1))\n", "//results\n", "printf ('hydraulic conductivity at void ratio of 0.65 = % 2f cm/sec ',k2)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.8: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "e= 0.6\n", "D10= 0.09 //mm\n", "//calculations\n", "k= 2.4622*(D10^2*(e^3/(1+e)))^0.7825\n", "//results\n", "printf ('hydraulic conductivity = % 4f cm/sec ',k)" ] } , { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7.9: solved.sce" ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [ "clc\n", "//initialisation of variables\n", "e= 0.6\n", "D10= 0.09 //mm\n", "D60= 0.16 //mm\n", "//calculations\n", "Cu=D60/D10\n", "k= 35*(e^3/(1+e))*(Cu^0.6)*(D10^2.32)\n", "//results\n", "printf ('hydraulic conductivity = % 3f cm/sec ',k)" ] } ], "metadata": { "kernelspec": { "display_name": "Scilab", "language": "scilab", "name": "scilab" }, "language_info": { "file_extension": ".sce", "help_links": [ { "text": "MetaKernel Magics", "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" } ], "mimetype": "text/x-octave", "name": "scilab", "version": "0.7.1" } }, "nbformat": 4, "nbformat_minor": 0 }