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    "# 2.1 – Reactive Balances\n",
    "\n",
    "---"
   ]
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    "## 2.1.0 – Learning Objectives\n",
    "\n",
    "By the end of this section you should be able to:\n",
    "\n",
    "1. Understand two of the three reactive balance methods.\n",
    "2. Solve two of the three reactive balance methods.\n",
    "\n",
    "---"
   ]
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    "## 2.1.1 – Introduction\n",
    "\n",
    "Often when we evaluate processes, __chemical reactions__ take place. These chemical reactions __add complexity__ to our system by introducing new unknowns. To reduce these complexities, we can use __molecular species balances__, __atomic balances__ or the __extent of reactions__ to help solve problems.\n",
    "\n",
    "---"
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    "## 2.1.2 – The Production of Ethylene\n",
    "\n",
    "Ethylene is a major chemical product primarily used to form the plastic polyethylene. It is also a natural chemical released by fruits to ripen. You can separate your bananas to help them ripen faster. \n",
    "\n",
    "Let's consider the production of ethylene in steady-state. In production, the dehydrogenation of Ethane forms Ethylene:\n",
    "\n",
    "$$\n",
    "    C_2H_{6 \\space (g)} \\longrightarrow C_2H_{4 \\space (g)} + H_{2 \\space (g)}\n",
    "$$\n",
    "\n",
    "![](../figures/Module-2/Flow-2.1-2.svg)\n",
    "\n",
    "Attribution: Said Zaid-Alkailani, Ngai To Lo & UBC [CC BY 4.0 de (https://creativecommons.org/licenses/by/4.0/)]\n",
    "\n",
    "---"
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    "## 2.1.3 – Molecular Species Balance\n",
    "\n",
    "The __molecular species balances__ are the balances of all chemical compounds in this reaction. For example:\n",
    "\n",
    "$$C_2 H_6 \\space \\text{balance}: \\space Accumulation = Input -Output +Generation - Consumption $$\n",
    "\n",
    "Since the system is in steady state and the reaction does not produce Ethane: \n",
    "\n",
    "$$ Accumulation = 0 , \\space Generation = 0 $$\n",
    "\n",
    "Rearranging and applying similar logic to the all of the species gives us 3 equations:\n",
    "\n",
    "### $C_2 H_6$ balance: $Input = Output + Consumption$\n",
    "\n",
    "$$ 10 \\space \\frac{kmol}{h} = \\dot{n}_{(1, \\space C_2 H_6)} + Consumption_{C_2 H_6} $$\n",
    "\n",
    "### $C_2 H_4$ balance: $Generation = Output$\n",
    "\n",
    "$$ Generation_{C_2 H_4} = \\dot{n}_{(2, \\space C_2 H_4)} $$\n",
    "\n",
    "### $H_2$ balance: $Generation = Output$\n",
    "\n",
    "$$ Generation_{H_2} = 4 \\space \\frac{kmol}{h} $$\n",
    "\n",
    "Since the reaction above has stoichiometric ratio of 1:1:1, we can determine $Generation_{H_2} = Generation_{C_2 H_4} = Consumption_{C_2 H_6}$. Therefore,\n",
    "\n",
    "$$ \\dot{n}_{(2, \\space C_2 H_4)} = 4 \\space \\frac{kmol}{h} $$\n",
    "\n",
    "$$ \\dot{n}_{(1, \\space C_2 H_6)} = 6 \\space \\frac{kmol}{h} $$\n",
    "\n",
    "---"
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    "## 2.1.4 – Atomic Species Balance\n",
    "\n",
    "The __atomic species balances__ are the balances of all the elemental species in this reaction. The Two species in the above reaction are Hydrogen and Carbon. Atomic species **must** follow the law of conservation of mass, thus both species take on the equation:\n",
    "\n",
    "### Carbon (C) balance: $Input = Output$\n",
    "\n",
    "$$ 10 \\space \\frac{kmol \\space C_2 H_6}{h} \\times \\frac{2 \\space kmol \\space C}{1 \\space kmol \\space C_2 H_6} = \\dot{n}_{(1, \\space C_2 H_6)} \\times \\frac{2 \\space kmol \\space C}{1 \\space kmol \\space C_2 H_6} + \\dot{n}_{(2, \\space C_2 H_4)} \\times \\frac{2 \\space kmol \\space C}{1 \\space kmol \\space C_2 H_4} $$\n",
    "\n",
    "$$ 10 \\space \\frac{kmol \\space C}{h} = \\dot{n}_{(1, \\space C_2 H_6)} + \\dot{n}_{(2, \\space C_2 H_4)} $$\n",
    "\n",
    "### Hydrogen (H) balance: $Input = Output$\n",
    "\n",
    "$$ 10 \\space \\frac{kmol \\space C_2 H_6}{h} \\times \\frac{6 \\space kmol \\space H}{1 \\space kmol \\space C_2 H_6} = 4 \\space \\frac{kmol \\space H_2}{h} \\times \\frac{2 \\space kmol \\space H}{1 \\space kmol \\space H_2} + \\dot{n}_{(1, \\space C_2 H_6)} \\times \\frac{6 \\space kmol \\space H}{1 \\space kmol \\space C_2 H_6} + \\dot{n}_{(2, \\space C_2 H_4)} \\times \\frac{6 \\space kmol \\space H}{1 \\space kmol \\space C_2 H_4} $$\n",
    "\n",
    "$$ 52 \\space \\frac{kmol \\space C}{h} = 6 \\space \\dot{n}_{(1, \\space C_2 H_6)} + 4 \\space \\dot{n}_{(2, \\space C_2 H_4)} $$\n",
    "\n",
    "$$ \\therefore \\space \\dot{n}_{(1, \\space C_2 H_6)} = 6 \\space \\frac{kmol}{h} $$\n",
    "\n",
    "$$ \\therefore \\space \\dot{n}_{(2, \\space C_2 H_4)} = 4 \\space \\frac{kmol}{h} $$"
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