#!/usr/bin/env python # coding: utf-8 # # Homework 16 # # In[1]: import numpy as np import matplotlib.pyplot as plt get_ipython().run_line_magic('matplotlib', 'inline') from scipy.integrate import quad from scipy.interpolate import interp1d from scipy.optimize import curve_fit, fsolve # ### Problem 1 # Use fsolve to find the two roots of the polynomial $f(x) = 2x^2 + 3x - 10$. # In[ ]: # ### Problem 2 # # Use fsolve to find the solution of the following two equations: # # $$f(x,y) = 2x^{2/3}+y^{2/3}-9^{1/3},$$ # # $$g(x,y) = \frac{x^2}{4} + \sqrt{y} - 1. $$ # # Use an initial guess of $x_0=1$, $y_0$ = 1. # In[ ]: # ### Problem 3 # # Species heat capacities are given by: # # $$c_{p,i}(T) = R_g(a_{0,i} + a_{1,i}T + a_{2,i}T^2 + a_{3,i}T^3 + a_{4,i}T^4).$$ # # | Species | $a_0$ | $a_1$ | $a_2$ | $a_3$ | $a_4$ | # |---------|-----------------|-------------------|------------------|-------------------|-------------------| # | 'CH4'| 1.28392437e+00| 1.00851346e-02| -2.66241814e-06| 4.99192762e-11| 5.62132385e-14| # | 'O2' | 2.92112373e+00| 2.25021636e-03| -1.33975172e-06| 3.97866288e-10| -4.37971416e-14| # | 'N2' | 3.30542104e+00| 3.63227936e-04| 5.26679110e-07| -3.30019712e-10| 5.24837849e-14| # | 'CO2'| 2.78249997e+00| 7.06457251e-03| -4.50912355e-06| 1.35740239e-09| -1.55514712e-13| # | 'H2O'| 3.80784936e+00| 3.17295233e-04| 1.41348052e-06| -6.61323323e-10| 8.91786378e-14| # # **NOTE, THESE ARE NOT THE SAME AS IN PREVIOUS HOMEWORK.** # # The enthalpy of the species at temperature $T$ is given by # # $$h_i(T) = h_{f,i} + \int_{T_{ref}}^{T}c_{p,i}(T)dT.$$ # * Here, $h_{f,i}$ is the heat of formation of the species at temperature $T_{ref}=298$. # # | Species | $h_{f,i}$ (J/mol) | # |---------|--------------------| # | CO2 | -393509 | # | H2O(g) | -241818 | # | O2 | 0 | # | N2 | 0 | # | CH4 | -74520 | # # **Methane combustion is given by the following reaction** # $$CH_4 + 2O_2 + 7.52N_2 \rightarrow CO_2 + 2H_2O + 7.52N_2.$$ # * The reactants are at room temperature (298 K). # * 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. # * The product temperature is called the **adiabatic flame temperature**, and without heat losses is the temperature of a stoichiometric methane (~natural gas) flame. # # **Find the adiabatic flame temperature in K and $^o$F** given by the following equation: # # $$h_{CH4}(298) + 2h_{O2}(298) + 7.52h_{N_2}(298) = h_{CO2}(T_{ad}) + 2h_{H2O}(T_{ad}) + 7.52h_{N_2}(T_{ad}).$$ # # * This is one equation in one unknown. # * $h_{CH4}(298)$, etc., refers to the enthalpy of the species at a temperature of 298. # # In[3]: # Dictionaries of the data for convenience to save typing. # You can reuse or reformat (you don't have to use a dictionary). a = { 'CH4':np.array([1.28392437e+00, 1.00851346e-02, -2.66241814e-06, 4.99192762e-11, 5.62132385e-14]), 'O2':np.array([2.92112373e+00, 2.25021636e-03, -1.33975172e-06, 3.97866288e-10, -4.37971416e-14]), 'N2':np.array([3.30542104e+00, 3.63227936e-04, 5.26679110e-07, -3.30019712e-10, 5.24837849e-14]), 'CO2':np.array([2.78249997e+00, 7.06457251e-03, -4.50912355e-06, 1.35740239e-09, -1.55514712e-13]), 'H2O':np.array([3.80784936e+00, 3.17295233e-04, 1.41348052e-06, -6.61323323e-10, 8.91786378e-14]), } hf = { 'CO2': -393509, 'H2O': -241818, 'O2' : 0, 'N2' : 0, 'CH4': -74520, } #------------------------ # ### Problem 4 # # 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. # # We have the following unknowns $Q_0$, $Q_1$, $Q_2$, $f_0$, $f_1$, $f_2$. # # We have the following six equations in these unknowns: # # $$Q_0 + Q_1 + Q_2 = Q_{tot},$$ # # $$\frac{f_0L_0\rho}{2D_0}\left(\frac{4Q_0}{\pi D_0^2}\right)^2 = \frac{f_1L_1\rho}{2D_1}\left(\frac{4Q_1}{\pi D_1^2}\right)^2,$$ # # $$\frac{f_0L_0\rho}{2D_0}\left(\frac{4Q_0}{\pi D_0^2}\right)^2 = \frac{f_2L_2\rho}{2D_2}\left(\frac{4Q_2}{\pi D_2^2}\right)^2,$$ # # $$\frac{1}{\sqrt{f_0}} = -2\log_{10}\left(\frac{\epsilon_0}{3.7D_0} + \frac{2.51\mu\pi D_0}{4\rho Q_0\sqrt{f_0}}\right),$$ # # $$\frac{1}{\sqrt{f_1}} = -2\log_{10}\left(\frac{\epsilon_1}{3.7D_1} + \frac{2.51\mu\pi D_1}{4\rho Q_1\sqrt{f_1}}\right),$$ # # $$\frac{1}{\sqrt{f_2}} = -2\log_{10}\left(\frac{\epsilon_2}{3.7D_2} + \frac{2.51\mu\pi D_2}{4\rho Q_2\sqrt{f_2}}\right).$$ # # The following quantities are given: # # $$Q_{tot} = 0.01333\,m^3/s,$$ # $$L_0 = 100\,{m},$$ # $$L_1 = 150\,{m},$$ # $$L_2 = 80\,{m},$$ # $$D_0 = 0.05\,{m},$$ # $$D_1 = 0.045\,{m},$$ # $$D_2 = 0.04\,{m},$$ # $$\epsilon_0 = 0.00024\,{m},$$ # $$\epsilon_1 = 0.00012\,{m},$$ # $$\epsilon_2 = 0.0002\,{m},$$ # $$\mu = 1.002\times 10^{-3}\,kg/(m\cdot s),$$ # $$\rho = 998\,kg/m^3.$$ # # 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. # # Notes: # * The first equation says the total flow rate is the sum of the flow rates through each pipe. # * 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. # * 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. # * The last three equations relate the friction factor in each pipe to its flow rate and pipe properties. # * 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 $4Q_0/(\pi D_0^2)$. # * In the frictions factor equations, $\mu\pi D/(4\rho Q)$ is just $1/Re$. Again, we made the substitution for convenience, but again, we could have written local variables for the $Re$'s and then used the $Re$'s in the equations. # In[ ]: