{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Homework 10\n" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "import numpy as np\n", "from scipy.optimize import root" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Problem 1\n", "Use fsolve to find the two roots of the polynomial $f(x) = 2x^2 + 3x - 10$." ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Problem 2\n", "\n", "Use `root` to find the solution of the following two equations:\n", "\n", "$$f(x,y) = 2x^{2/3}+y^{2/3}-9^{1/3},$$\n", "\n", "$$g(x,y) = \\frac{x^2}{4} + \\sqrt{y} - 1. $$\n", "\n", "Use an initial guess of $x_0=1$, $y_0$ = 1." ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Problem 3\n", "\n", "Species heat capacities are given by:\n", "\n", "$$c_{p,i}(T) = R_g(a_{0,i} + a_{1,i}T + a_{2,i}T^2 + a_{3,i}T^3 + a_{4,i}T^4).$$\n", "\n", "| Species | $a_0$ | $a_1$ | $a_2$ | $a_3$ | $a_4$ |\n", "|---------|-----------------|-------------------|------------------|-------------------|-------------------|\n", "| 'CH4'| 1.28392437e+00| 1.00851346e-02| -2.66241814e-06| 4.99192762e-11| 5.62132385e-14|\n", "| 'O2' | 2.92112373e+00| 2.25021636e-03| -1.33975172e-06| 3.97866288e-10| -4.37971416e-14|\n", "| 'N2' | 3.30542104e+00| 3.63227936e-04| 5.26679110e-07| -3.30019712e-10| 5.24837849e-14|\n", "| 'CO2'| 2.78249997e+00| 7.06457251e-03| -4.50912355e-06| 1.35740239e-09| -1.55514712e-13|\n", "| 'H2O'| 3.80784936e+00| 3.17295233e-04| 1.41348052e-06| -6.61323323e-10| 8.91786378e-14|\n", "\n", "The enthalpy of the species at temperature $T$ is given by\n", "\n", "$$h_i(T) = h_{f,i} + \\int_{T_{ref}}^{T}c_{p,i}(T)dT.$$\n", "* Here, $h_{f,i}$ is the heat of formation of the species at temperature $T_{ref}=298$. \n", "\n", "| Species | $h_{f,i}$ (J/mol) |\n", "|---------|--------------------|\n", "| CO2 | -393509 |\n", "| H2O(g) | -241818 |\n", "| O2 | 0 |\n", "| N2 | 0 |\n", "| CH4 | -74520 |\n", "\n", "**Methane combustion is given by the following reaction**\n", "$$CH_4 + 2O_2 + 7.52N_2 \\rightarrow CO_2 + 2H_2O + 7.52N_2.$$\n", "* The reactants are at room temperature (298 K).\n", "* The products have the same enthalpy as the reactants, but are much hotter because internal energy in the chemical bonds has been converted to sensible energy.\n", "* The product temperature is called the **adiabatic flame temperature**, and without heat losses is the temperature of a stoichiometric methane (~natural gas) flame.\n", "\n", "**Find the adiabatic flame temperature in K and $^oF$** given by the following equation:\n", "\n", "$$h_{CH4}(298) + 2h_{O2}(298) + 7.52h_{N_2}(298) = h_{CO2}(T_{ad}) + 2h_{H2O}(T_{ad}) + 7.52h_{N_2}(T_{ad}).$$\n", "\n", "* This is one equation in one unknown.\n", "* $h_{CH4}(298)$, etc., refers to the enthalpy of the species at a temperature of 298 K.\n", "\n", "**In solving the problem, create a function `h(T,a,hf)` that returns the enthalpy at a given temperature `T` for a given species with coefficients `a` and `hf`.**\n", "- Then, `hCH4 = h(T,aCH4,hfCH4)` etc.\n" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "aCH4 = np.array([1.28392437e+00, 1.00851346e-02, -2.66241814e-06, 4.99192762e-11, 5.62132385e-14])\n", "aO2 = np.array([2.92112373e+00, 2.25021636e-03, -1.33975172e-06, 3.97866288e-10, -4.37971416e-14])\n", "aN2 = np.array([3.30542104e+00, 3.63227936e-04, 5.26679110e-07, -3.30019712e-10, 5.24837849e-14])\n", "aCO2 = np.array([2.78249997e+00, 7.06457251e-03, -4.50912355e-06, 1.35740239e-09, -1.55514712e-13])\n", "aH2O = np.array([3.80784936e+00, 3.17295233e-04, 1.41348052e-06, -6.61323323e-10, 8.91786378e-14])\n", "\n", "hfCO2 = -393509\n", "hfH2O = -241818\n", "hfO2 = 0\n", "hfN2 = 0\n", "hfCH4 = -74520\n", "\n", "#------------------------\n", "\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Problem 4 \n", "\n", "We have a parallel pipe network with three pips. We want to find the flow rates $Q$ through each pipe. The flow rates are written in terms of the friction factors $f$ for each pipe.\n", "\n", "We have the following unknowns $Q_0$, $Q_1$, $Q_2$, $f_0$, $f_1$, $f_2$. \n", "\n", "We have the following six equations in these unknowns:\n", "\n", "$$Q_0 + Q_1 + Q_2 = Q_{tot},$$\n", "\n", "$$\\frac{f_0L_0\\rho}{2D_0}\\left(\\frac{4Q_0}{\\pi D_0^2}\\right)^2 = \\frac{f_1L_1\\rho}{2D_1}\\left(\\frac{4Q_1}{\\pi D_1^2}\\right)^2,$$\n", "\n", "$$\\frac{f_0L_0\\rho}{2D_0}\\left(\\frac{4Q_0}{\\pi D_0^2}\\right)^2 = \\frac{f_2L_2\\rho}{2D_2}\\left(\\frac{4Q_2}{\\pi D_2^2}\\right)^2,$$\n", "\n", "$$\\frac{1}{\\sqrt{f_0}} = -2\\log_{10}\\left(\\frac{\\epsilon_0}{3.7D_0} + \\frac{2.51\\mu\\pi D_0}{4\\rho Q_0\\sqrt{f_0}}\\right),$$\n", "\n", "$$\\frac{1}{\\sqrt{f_1}} = -2\\log_{10}\\left(\\frac{\\epsilon_1}{3.7D_1} + \\frac{2.51\\mu\\pi D_1}{4\\rho Q_1\\sqrt{f_1}}\\right),$$\n", " \n", "$$\\frac{1}{\\sqrt{f_2}} = -2\\log_{10}\\left(\\frac{\\epsilon_2}{3.7D_2} + \\frac{2.51\\mu\\pi D_2}{4\\rho Q_2\\sqrt{f_2}}\\right).$$\n", "\n", "The following quantities are given:\n", "\n", "$$Q_{tot} = 0.01333\\,m^3/s,$$\n", "$$L_0 = 100\\,{m},$$\n", "$$L_1 = 150\\,{m},$$\n", "$$L_2 = 80\\,{m},$$\n", "$$D_0 = 0.05\\,{m},$$\n", "$$D_1 = 0.045\\,{m},$$\n", "$$D_2 = 0.04\\,{m},$$\n", "$$\\epsilon_0 = 0.00024\\,{m},$$\n", "$$\\epsilon_1 = 0.00012\\,{m},$$\n", "$$\\epsilon_2 = 0.0002\\,{m},$$\n", "$$\\mu = 1.002\\times 10^{-3}\\,kg/(m\\cdot s),$$\n", "$$\\rho = 998\\,kg/m^3.$$\n", "\n", "It is a good idea to make your guesses consistent. That, make sure your guesses for $Q_0$, $Q_1$, and $Q_2$ add up to the given $Q_{tot}$. Reasonable guesses for $f_0$, $f_1$, $f_2$ are 0.01. \n", "\n", "Notes: \n", "* The first equation says the total flow rate is the sum of the flow rates through each pipe. \n", "* The second equation says that the pressure drop through pipe 0 is the same as that through pipe 1, since pipes 0 and 1 are connected at their ends.\n", "* The third equation says that the pressure drop through pipe 1 is the same as that through pipe 2, since pipes 1 and 2 are connected at their ends.\n", "* The last three equations relate the friction factor in each pipe to its flow rate and pipe properties.\n", "* In the second two equations, $4Q/(\\pi D^2)$ is just the velocity in the pipe. But we substituted $v=4Q/(\\pi D^2)$ for convenience even though it makes the equations a bit harder to read. An alternative would be to define local terms like $v_0 = 4Q_0/(\\pi D_0^2)$, and then use $v_0$ in the equations above instead of $4Q_0/(\\pi D_0^2)$.\n", "* In the friction factor equations, $\\mu\\pi D/(4\\rho Q)$ is just $1/Re$. We made the substitution for convenience, but we could have written local variables for the $Re$'s and then used the $Re$'s in the equations." ] }, { "cell_type": "code", "execution_count": null, "metadata": { "collapsed": true }, "outputs": [], "source": [] } ], "metadata": { "hide_input": false, "kernelspec": { "display_name": "base", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.14.7" }, "varInspector": { "cols": { "lenName": 16, "lenType": 16, "lenVar": 40 }, "kernels_config": { "python": { "delete_cmd_postfix": "", "delete_cmd_prefix": "del ", "library": "var_list.py", "varRefreshCmd": "print(var_dic_list())" }, "r": { "delete_cmd_postfix": ") ", "delete_cmd_prefix": "rm(", "library": "var_list.r", "varRefreshCmd": "cat(var_dic_list()) " } }, "types_to_exclude": [ "module", "function", "builtin_function_or_method", "instance", "_Feature" ], "window_display": false } }, "nbformat": 4, "nbformat_minor": 1 }