#!/usr/bin/env python
# coding: utf-8

# <!--NOTEBOOK_HEADER-->
# *This notebook contains course material from [CBE20255](https://jckantor.github.io/CBE20255)
# by Jeffrey Kantor (jeff at nd.edu); the content is available [on Github](https://github.com/jckantor/CBE20255.git).
# The text is released under the [CC-BY-NC-ND-4.0 license](https://creativecommons.org/licenses/by-nc-nd/4.0/legalcode),
# and code is released under the [MIT license](https://opensource.org/licenses/MIT).*

# <!--NAVIGATION-->
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# # Adipic Acid Flowsheet

# ## Summary
# 
# This [Jupyter notebook](http://ipython.org/notebook.html) demonstrates the formulation and solution of material balances for a hypothetical adipic acid process described by Murphy (2005, Example 2.15) using the [symbolic algebra package Sympy](http://sympy.org/en/index.html).

# ## Problem Statement

# [Adipic acid](http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=196) (C<sub>6</sub>H<sub>10</sub>O<sub>4</sub>) (also called hexanedioic acid) rarely occurs in nature, but approximately 2.5 billion kilograms are produced per year from petrochemical feedstocks primarily for the production of nylon.
# 
# <img src="http://upload.wikimedia.org/wikipedia/commons/thumb/7/7e/Adipic-acid-3D-balls.png/320px-Adipic-acid-3D-balls.png" width = 200px/>
# 
# Recently there has been interest in [producing adipic acid from renewable resources](http://www.ihs.com/products/chemical/technology/pep/bio-based-adipic-acid.aspx). [For example, starting with glucose](http://pubs.acs.org/doi/abs/10.1021/ja00080a057) (C<sub>6</sub>H<sub>12</sub>O<sub>6</sub>), muconic acid (C<sub>6</sub>H<sub>6</sub>O<sub>4</sub>) is produced through fermentation with a genetically modified e. coli. via the reaction
# 
# <sup>7</sup>&frasl;<sub>3</sub> C<sub>6</sub>H<sub>12</sub>O<sub>6</sub> + <sup>17</sup>&frasl;<sub>2</sub> O<sub>2</sub> &#10141; C<sub>6</sub>H<sub>6</sub>O<sub>4</sub> + 8 CO<sub>2</sub> + 11 H<sub>2</sub>O
# 
# that is subsequently hydrogentated to form adipic acid
# 
# C<sub>6</sub>H<sub>6</sub>O<sub>4</sub> + 2 H<sub>2</sub>  &#10141; C<sub>6</sub>H<sub>10</sub>O<sub>4</sub>
# 
# Murphy (2005, Example 2.15) outlines a hypothetical flowsheet for this process:
# 
# <img src = "https://raw.githubusercontent.com/jckantor/CBE20255/master/images/adipic_acid_flowsheet.png" width=800px>
# 
# Neglecting the _E. coli_, solve for the flowrates necessary to produce 12,000 kg/hour of adipic acid assuming that glucose is available as a 10 mg/mL solution.
# 

# ## Solution

# ### Process Variables

# We begin by relabeling the process flowsheet with stream numbers, stream variables, and extents of reaction. There are chemical species are abbreviated as follows:
# 
# * A: adipic acid
# * C: carbon dioxide
# * G: glucose
# * H: hydrogen
# * M: muconic acid
# * N: nitrogen
# * O: oxygen
# * W: water
# 
# and where X<sub>1</sub> and X<sub>2</sub> denote the extents of reactions 1 and 2, respectively. The stream variables denote chemical flowrates in units of kg/hour. The extents of reaction will be in units of kg-mol/hour.
# 
# <img src = "https://raw.githubusercontent.com/jckantor/CBE20255/master/images/adipic_acid_flowsheet_2.png" width=800px>

# In[1]:


# Import the symbolic algebra package sympy
import sympy as sym

# Extents of reactions 1 and 2
sym.var('X1 X2')

# Stream variables
sym.var('O1 N1')           # Stream 1
sym.var('G2 W2')           # Stream 2
sym.var('H3')              # Stream 3
sym.var('N4 C4')           # Stream 4
sym.var('W5')              # Stream 5
sym.var('A6')              # Stream 6
sym.var('O7 N7 G7 W7')     # Stream 7
sym.var('N8 W8 C8 M8')     # Stream 8
sym.var('M9')              # Stream 9
sym.var('H10 M10')         # Stream 10


# Because the flowsheet includes reactions, it will be necessary to include molecular weights in the mass balance expressions. For this purpose we gather the molecular weights of all species into a python dictionary indexed by the chemical species.

# In[2]:


MW = {
    'A': 146.14,
    'C': 44.01,
    'G': 180.16,
    'H': 2.02,
    'M': 142.11,
    'N': 14.01,
    'O': 16.00,
    'W': 18.02
}


# ### Specifications

# In[3]:


specs = [
    sym.Eq(A6, 12000),
    sym.Eq(N1/MW['N'], (0.79/0.21)*(O1/MW['O'])),
    sym.Eq(G2, 0.01*W2)
]


# ### Material Balances

# In[4]:


mixer1 = [
    sym.Eq(0, O1 - O7),
    sym.Eq(0, N1 - N7),
    sym.Eq(0, G2 - G7),
    sym.Eq(0, W2 - W7)
]

reactor1 = [
    sym.Eq(0,  O7 - MW['O']*(17/2)*X1),
    sym.Eq(0,  N7 - N8),
    sym.Eq(0,  G7 - MW['G']*(7/3)*X1),
    sym.Eq(0, -C8 + MW['C']*8*X1),
    sym.Eq(0, -M8 + MW['M']*X1),
    sym.Eq(0,  W7 - W8 + MW['W']*11*X1)
]

separator = [
    sym.Eq(0, N8 - N4),
    sym.Eq(0, C8 - C4),
    sym.Eq(0, M8 - M9),
    sym.Eq(0, W8 - W5)
]

mixer2 = [
    sym.Eq(0, M9 - M10),
    sym.Eq(0, H3 - H10)
]

reactor2 = [
    sym.Eq(0, H10 - MW['H']*2*X2),
    sym.Eq(0, M10 - MW['M']*X2),
    sym.Eq(0, -A6 + MW['A']*X2)
]

mbals = mixer1 + reactor1 + separator + mixer2 + reactor2


# In[5]:


specs + mbals


# ### Solution

# In[6]:


soln = sym.solve(mbals + specs)
soln


# In[7]:


for key in soln.keys():
    print("{:3s}    {:10.1f}".format(str(key), float(soln[key])))


# In[ ]:





# <!--NAVIGATION-->
# < [Separating Milk](http://nbviewer.jupyter.org/github/jckantor/CBE20255/blob/master/notebooks/03.04-Separating-Milk.ipynb) | [Contents](toc.ipynb) | [Material Balances](http://nbviewer.jupyter.org/github/jckantor/CBE20255/blob/master/notebooks/04.00-Material-Balances.ipynb) ><p><a href="https://colab.research.google.com/github/jckantor/CBE20255/blob/master/notebooks/03.05-Adipic-Acid-Flowsheet.ipynb"><img align="left" src="https://colab.research.google.com/assets/colab-badge.svg" alt="Open in Colab" title="Open in Google Colaboratory"></a>