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| A '''mass balance''', also called a '''material balance''', is an application of [[conservation of mass]] to the analysis of physical systems. By accounting for material entering and leaving a system, [[mass flow]]s can be identified which might have been unknown, or difficult to measure without this technique. The exact [[conservation law]] used in the analysis of the system depends on the context of the problem, but all revolve around mass conservation, i.e. that [[matter]] cannot disappear or be created spontaneously.<ref name="himmelblau">{{cite book | last = Himmelblau | first = David M. | title = Basic Principles and Calculations in Chemical Engineering | edition = 2nd | year = 1967 | publisher = [[Prentice Hall]] | url = http://www.pearsonhighered.com/educator/product/Basic-Principles-and-Calculations-in-Chemical-Engineering/9780132346603.page}}</ref>{{rp|59-62}}
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| Therefore, mass balances are used widely in [[engineering]] and [[Environmental analysis|environmental analyses]]. For example, mass balance theory is used to design [[chemical reactor]]s, to analyse alternative processes to produce chemicals, as well as to model [[pollution]] dispersion and other processes of physical systems. Closely related and complementary analysis techniques include the [[population balance equation|population balance]], [[Energy accounting|energy balance]] and the somewhat more complex [[entropy]] balance. These techniques are required for thorough design and analysis of systems such as the [[refrigeration cycle]].
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| In environmental monitoring the term '''budget calculations''' is used to describe mass balance equations where they are used to evaluate the monitoring data (comparing input and output, etc.) In biology the [[dynamic energy budget]] theory for metabolic organisation makes explicit use of mass and energy balances.
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| ==Introduction==
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| The general form quoted for a mass balance is ''The mass that enters a system must, by conservation of mass, either leave the system or accumulate within the system ''.
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| Mathematically the mass balance for a system without a chemical reaction is as follows:<ref name="himmelblau"/>{{rp|59-62}}
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| <div align="center"><math> \mathrm{Input} = \mathrm{Output} + \mathrm{Accumulation} \, </math></div>
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| Strictly speaking the above equation holds also for systems with [[chemical reaction]]s if the terms in the balance equation are taken to refer to total mass, i.e. the sum of all the chemical species of the system. In the absence of a chemical reaction the amount of any chemical species flowing in and out will be the same; this gives rise to an equation for each species present in the system. However, if this is not the case then the mass balance equation must be amended to allow for the generation or depletion (consumption) of each chemical species. Some use one term in this equation to account for chemical reactions, which will be negative for depletion and positive for generation. However, the conventional form of this equation is written to account for both a positive generation term (i.e. product of reaction) and a negative consumption term (the reactants used to produce the products). Although overall one term will account for the total balance on the system, if this balance equation is to be applied to an individual species and then the entire process, both terms are necessary. This modified equation can be used not only for reactive systems, but for population balances such as arise in [[particle mechanics]] problems. The equation is given below; note that it simplifies to the earlier equation in the case that the generation term is zero.<ref name="himmelblau"/>{{rp|59-62}}
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| <div align="center"><math> \text{Input} + \text{Generation} = \text{Output} + \text{Accumulation} \ + \text{Consumption} </math></div>
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| *In the absence of a [[nuclear reaction]] the number of [[atoms]] flowing in and out must remain the same, even in the presence of a chemical reaction.
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| *For a balance to be formed, the boundaries of the system must be clearly defined. <!-- this statement is unclear, but it means to say you must have a good knowledge of where the inlets and outlets to your system are -->
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| *Mass balances can be taken over physical systems at multiple scales. <!-- Needs an example which the reader can relate to. I tried the idea of a watercourse having a balance at the river level as well as in the water cycle, but it seemed to ill-defined -->
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| *Mass balances can be simplified with the assumption of [[steady state]], in which the accumulation term is zero.
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| ==Illustrative example==
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| [[Image:mass bal clarifier.svg|thumb|350px|right|Diagram showing clarifier example]]
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| A simple example can illustrate the concept. Consider the situation in which a [[slurry]] is flowing into a [[settling tank]] to remove the solids in the tank. Solids are collected at the bottom by means of a [[conveyor belt]] partially submerged in the tank, and water exits via an overflow outlet.
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| In this example, there are two substances: solids and water. The water overflow outlet carries an increased concentration of water relative to solids, as compared to the slurry inlet, and the exit of the conveyor belt carries an increased concentration of solids relative to water.
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| '''Assumptions'''
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| *Steady state
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| *Non-reactive system
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| '''Analysis'''
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| Suppose that the slurry inlet composition (by mass) is 50% solid and 50% water, with a mass flow of {{val|100|ul=kg|upl=min}}. The tank is assumed to be operating at steady state, and as such accumulation is zero, so input and output must be equal for both the solids and water. If we know that the removal efficiency for the slurry tank is 60%, then the water outlet will contain {{val|20|ul=kg|upl=min}} of solids (40% times {{val|100|ul=kg|upl=min}} times 50% solids). If we measure the flow rate of the combined solids and water, and the water outlet is shown to be {{val|60|ul=kg|upl=min}}, then the amount of water exiting via the conveyor belt must be {{val|10|ul=kg|upl=min}}. This allows us to completely determine how the mass has been distributed in the system with only limited information and using the mass balance relations across the system boundaries.
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| ==Mass feedback (recycle)==
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| [[Image:CoolingTower.png|frame|right|256px|Cooling towers are a good example of a recycle system]]
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| Mass balances can be performed across systems which have cyclic flows. In these systems output streams are fed back into the input of a unit, often for further reprocessing.<ref name="himmelblau" />{{rp|97-105}}
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| Such systems are common in [[Grinding (abrasive cutting)|grinding]] circuits, where materials are crushed then sieved to only allow a particular size of particle out of the circuit and the larger particles are returned to the grinder. However, recycle flows are by no means restricted to solid mechanics operations; they are used in liquid and gas flows, as well. One such example is in [[cooling tower]]s, where water is pumped through a tower many times, with only a small quantity of water drawn off at each pass (to prevent solids build up) until it has either evaporated or exited with the drawn off water.
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| <!-- Insert generic derivation of mass flow around a recycle with a recyc ratio of R, defined as returned flow/outlet flow-->
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| The use of the recycle aids in increasing overall conversion of input products, which is useful for low per-pass conversion processes (such as the [[Haber process]]).
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| ==Differential mass balances==
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| A mass balance can also be taken [[Calculus|differentially]]. The concept is the same as for a large mass balance, but it is performed in the context of a limiting system (for example, one can consider the limiting case in time or, more commonly, volume). A differential mass balance is used to generate [[differential equation]]s that can provide an effective tool for modelling and understanding the target system.
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| The differential mass balance is usually solved in two steps: first, a set of governing differential equations must be obtained, and then these equations must be solved, either analytically or, for less tractable problems, numerically.
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| The following systems are good examples of the applications of the differential mass balance:
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| # Ideal (stirred) Batch reactor
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| # Ideal tank reactor, also named [[Continuous stirred-tank reactor|Continuous Stirred Tank Reactor]] (CSTR)
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| # Ideal [[Plug flow reactor model|Plug Flow Reactor]] (PFR)
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| ===Ideal batch reactor===
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| The ideal completely mixed batch reactor is a closed system. [[Isothermal process|Isothermal conditions]] are assumed, and mixing prevents concentration gradients as reactant concentrations decrease and product concentrations increase over time.<ref name="weber">{{cite book | last = Weber | first = Walter J., Jr. | title = Physicochemical Processes for Water Quality Control | year = 1972 | publisher = [[John Wiley & Sons|Wiley-Interscience]] | isbn = 0-471-92435-0}}</ref>{{rp|40-41}} Many chemistry textbooks implicitly assume that the studied system can be described as a batch reactor when they write about reaction kinetics and [[chemical equilibrium]].
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| The mass balance for a substance A becomes
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| <div align="center"><math> \mathrm{IN} + \mathrm{PROD} = \mathrm{OUT} + \mathrm{ACC} </math></div>
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| <div align="center"><math> 0 + r_{\mathrm{A}} V = 0 + \frac{dn_{\mathrm{A}}}{dt} </math></div>
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| where ''r''<sub>A</sub> denotes the rate at which substance A is produced, V is the volume (which may be constant or not), ''n''<sub>A</sub> the number of moles (''n'') of substance A.
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| In a fed-batch reactor some reactants/ingredients are added continuously or in pulses (compare making porridge by either first blending all ingredients and then letting it boil, which can be described as a batch reactor, or by first mixing only water and salt and making that boil before the other ingredients are added, which can be described as a fed-batch reactor). Mass balances for fed-batch reactors become a bit more complicated.
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| ====Reactive example====
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| In the first example, we will show how to use a mass balance to derive a relationship between the [[Air–fuel ratio|percent excess air]] for the [[combustion]] of a hydrocarbon-base fuel oil and the percent oxygen in the combustion product gas. First, normal dry air contains {{val|0.2095|ul=mol}} of oxygen per mole of air, so there is one mole of {{chem|O|2}} in {{val|4.773|u=mol}} of dry air. For stoichiometric combustion, the relationships between the mass of air and the mass of each combustible element in a fuel oil are:
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| <div align="center"><math>\mathrm{Carbon:} \frac {\mathrm{mass \ of \ air}}{\mathrm{mass \ of \ C}} = \frac {4.773 \times 28.96} {12.01} = 11.51</math></div>
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| <div align="center"><math>\mathrm{Hydrogen:} \frac {\mathrm{mass \ of \ air}}{\mathrm{mass \ of \ H}} = \frac {\frac {1}{4} (4.773) \times 28.96} {1.008} = 34.28</math></div>
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| <div align="center"><math>\mathrm{Sulfur:} \frac {\mathrm{mass \ of \ air}}{\mathrm{mass \ of \ S}} = \frac {4.773 \times 28.96} {32.06} = 4.31</math></div>
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| Considering the accuracy of typical analytical procedures, an equation for the mass of air per mass of fuel at stoichiometric combustion is:
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| <div align="center"><math>\frac {\mathrm{mass \ of \ air}}{\mathrm{mass \ of \ fuel}} = AFR_{mass} = 11.5(wC) + 34.3(wH) + (wS - wO)</math></div>
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| where wC, wH, wS, and wO refer to the mass fraction of each element in the fuel oil, sulfur burning to SO2, and AFR<sub>mass</sub> refers to the [[air-fuel ratio]] in mass units.
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| For {{val|1|ul=kg}} of fuel oil containing 86.1% C, 13.6% H, 0.2% O, and 0.1% S the stoichiometric mass of air is {{val|14.56|u=kg}}, so AFR = 14.56. The combustion product mass is then {{val|15.56|u=kg}}. At exact stoichiometry, {{chem|O|2}} should be absent. At 15 percent excess air, the AFR = 16.75, and the mass of the combustion product gas is {{val|17.75|u=kg}}, which contains {{val|0.505|u=kg}} of excess oxygen. The combustion gas thus contains 2.84 percent {{chem|O|2}} by mass. The relationships between percent excess air and %{{chem|O|2}} in the combustion gas are accurately expressed by quadratic equations, valid over the range 0–30 percent excess air:
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| <div align="center"><math>\mathrm{% \ excess \ air} = 1.2804 \times (\mathrm{%O_2 \ in \ combustion \ gas})^2 + 4.49 \times (\mathrm{%O_2 \ in \ combustion \ gas})</math></div>
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| <div align="center"><math>\mathrm{%O_2 \ in \ combustion \ gas} = -0.00138 \times (\mathrm{% \ excess \ air})^2 + 0.210 \times (\mathrm{% \ excess \ air})</math></div>
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| In the second example we will use the law of [[mass action]] to '''derive''' the expression for a [[chemical equilibrium]] constant.
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| Assume we have a closed reactor in which the following liquid phase reversible reaction occurs:
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| <div align="center"><math> a\mathrm{A} + b\mathrm{B} \leftrightarrow c\mathrm{C} + d\mathrm{D} </math></div>
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| The mass balance for substance A becomes
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| <div align="center"><math> \mathrm{IN} + \mathrm{PROD} = \mathrm{OUT} + \mathrm{ACC} </math></div>
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|
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| <div align="center"><math> 0 + r_{\mathrm{A}} V = 0 + \frac{dn_{\mathrm{A}}}{dt} </math></div>
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| As we have a liquid phase reaction we can (usually) assume a constant volume and since <math> n_{\mathrm{A}}= V * C_{\mathrm{A}} </math> we get
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| <div align="center"><math> r_{\mathrm{A}} V = V \frac{dC_{\mathrm{A}}}{dt} </math></div>
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| or
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| <div align="center"><math> r_{\mathrm{A}} = \frac{dC_{\mathrm{A}}}{dt} </math></div>
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| In many text books this is given as the definition of [[reaction rate]] without specifying the implicit assumption that we are talking about reaction rate in a closed system with only one reaction. This is an unfortunate mistake that has confused many students over the years.
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| According to the law of [[mass action]] the forward reaction rate can be written as
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| <div align="center"><math> r_1=k_1 [\mathrm{A}]^a[\mathrm{B}]^b </math></div>
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| and the backward reaction rate as
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| <div align="center"><math> r_{-1}=k_{-1}1 [\mathrm{C}]^c[\mathrm{D}]^d </math></div>
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| The rate at which substance A is produced is thus
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| <div align="center"><math> r_{\mathrm{A}} = a ( r_{-1} - r_1 ) </math></div>
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| and since, at equilibrium, the concentration of A is constant we get
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| <div align="center"><math> r_{\mathrm{A}} = a ( r_{-1} - r_1 ) = \frac{dC_{\mathrm{A}}}{dt} =0</math></div>
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| or, rearranged
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| <div align="center"><math> \frac{k_1}{k_{-1}}=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}=K_{eq}</math></div>
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| ===Ideal tank reactor/continuously stirred tank reactor===
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| {{main|Continuous stirred-tank reactor}}
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| The continuously mixed tank reactor is an open system with an influent stream of reactants and an effluent stream of products.<ref name="weber" />{{rp|41}} A lake can be regarded as a tank reactor, and lakes with long turnover times (e.g. with low flux-to-volume ratios) can for many purposes be regarded as continuously stirred (e.g. homogeneous in all respects). The mass balance then becomes
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| <div align="center"><math> \mathrm{IN} + \mathrm{PROD} = \mathrm{OUT} + \mathrm{ACC} </math></div>
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| <div align="center"><math> Q_0\cdot C_{\mathrm{A},0} + r_A\cdot V = Q\cdot C_{\mathrm{A}} + \frac{dn_{\mathrm{A}}}{dt} </math></div>
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| where ''Q''<sub>0</sub> and Q denote the volumetric flow in and out of the system respectively and ''C''<sub>A,0</sub> and ''C''<sub>A</sub> the concentration of A in the inflow and outflow respective. In an open system we can never reach a chemical equilibrium. We can, however, reach a [[dynamic equilibrium|steady state]] where all '''state variables''' (temperature, concentrations etc.) remain constant (<math> \mathrm{ACC} = 0 </math>).
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| ====Example====
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| Consider a bathtub in which there is some bathing salt dissolved. We now fill in more water, keeping the bottom plug in. What happens?
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| Since there is no reaction, <math> \mathrm{PROD} =0</math> and since there is no outflow <math> Q=0 </math>. The mass balance becomes
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| <div align="center"><math> \mathrm{IN} + \mathrm{PROD} = \mathrm{OUT} + \mathrm{ACC} </math></div>
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|
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| <div align="center"><math> Q_0 \cdot C_{\mathrm{A},0} +0 = 0\cdot C_{\mathrm{A}} + \frac{dn_{\mathrm{A}}}{dt} </math></div>
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| or
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| <div align="center"><math> Q_0\cdot C_{\mathrm{A},0}= \frac{dC_{\mathrm{A}}V}{dt}=V \frac{dC_{\mathrm{A}}}{dt} + C_{\mathrm{A}}\frac{dV}{dt}</math></div>
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| Using a mass balance for total volume, however, it is evident that <math>\frac{dV}{dt}=Q_0</math>and that <math>V=V_{t=0}+Q_0t</math>. Thus we get
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| <div align="center"><math>\frac{dC_{\mathrm{A}}}{dt}=\frac{Q_0}{(V_{t=0}+Q_0t)}\left( C_{\mathrm{A},0}-C_{\mathrm{A}} \right) </math></div>
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| Note that there is no reaction and hence no [[reaction rate]] or [[rate law]] involved, and yet <math>\frac{dC_{\mathrm{A}}}{dt}\neq 0</math>. We can thus draw the conclusion that reaction rate can not be defined in a general manner using <math> \frac{dC}{dt} </math>. One '''must''' first write down a mass balance before a link between <math> \frac{dC}{dt} </math> and the reaction rate can be found. Many textbooks, however, define reaction rate as
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| <div align="center"><math>v= \frac{dC_{\mathrm{A}}}{dt} </math></div>
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| without mentioning that this definition implicitly assumes that the system is closed, has a constant volume and that there is only one reaction.
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| ===Ideal plug flow reactor (PFR)===
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| The idealized plug flow reactor is an open system resembling a tube with no mixing in the direction of flow but perfect mixing perpendicular to the direction of flow. Often used for systems like rivers and water pipes if the flow is turbulent. When a mass balance is made for a tube, one first considers an [[infinitesimal]] part of the tube and make a mass balance over that using the ideal tank reactor model.<ref name="weber" />{{rp|46-47}} That mass balance is then [[integral|integrated]] over the entire reactor volume to obtain:
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| <div align="center"><math> \frac{d(Q\cdot C_A)}{dV} = r_A </math></div>
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| In numeric solutions, e.g. when using computers, the ideal tube is often translated to a series of tank reactors, as it can be shown that a PFR is equivalent to an infinite number of stirred tanks in series, but the latter is often easier to analyze, especially at steady state.
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| ==More complex problems==
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| In reality, reactors are often non-ideal, in which combinations of the reactor models above are used to describe the system. Not only chemical reaction rates, but also [[mass transfer]] rates may be important in the mathematical description of a system, especially in [[heterogeneous]] systems.<ref>{{cite book | last = Perry | first = Robert H. | last2 = Chilton | first2 = Cecil H. | last3 = Kirkpatrick | first3 = Sidney D. | title = [[Perry's Chemical Engineers' Handbook|Chemical Engineers' Handbook]] | edition = 4th | year = 1963 | publisher = [[McGraw-Hill#McGraw-Hill_Education|McGraw-Hill]] | page = 4-21}}</ref>
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| As the chemical [[reaction rate]] depends on temperature it is often necessary to make both an [[Energy accounting|energy balance]] (often a heat balance rather than a full fledged energy balance) as well as mass balances to fully describe the system. A different reactor models might be needed for the energy balance: A system that is closed with respect to mass might be open with respect to energy e.g. since heat may enter the system through [[heat conduction|conduction]].
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| ==Commercial use==
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| In industrial process plants, using the fact that the mass entering and leaving any portion of a process plant must balance, [[data validation and reconciliation]] algorithms may be employed to correct measured flows, provided that enough redundancy of flow measurements exist to permit statistical reconciliation and exclusion of detectably erroneous measurements. Since all real world measured values contain inherent error, the reconciled measurements provide a better basis than the measured values do for financial reporting, optimization, and regulatory reporting. Software packages exist to make this commercially feasible on a daily basis.
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| ==See also==
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| * [[Bioreactor]]
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| * [[Chemical reactor]]
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| * [[Chemical engineering]]
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| * [[Chemical equilibrium]]
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| * [[Conservation of mass]]
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| * [[Continuity equation]]
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| * [[Continuous stirred-tank reactor]]
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| * [[Dilution (equation)]]
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| * [[Energy accounting]]
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| * [[Mass action]]
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| * [[Mass flux]]
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| * [[Material balance planning]]
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| * [[Data validation and reconciliation]]
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| ==References==
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| {{reflist}}
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| ==External links==
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| * [http://www.eng.uwaterloo.ca/~che100/chapt4_BALANCES.ppt Material Balance Calculations]
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| * [http://www.ncat.edu/~chemeng/WORD%20docs/CHEN%20200/C200-f07-L07.doc Material Balance Fundamentals]
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| * [http://jbrwww.che.wisc.edu/~jbraw/chemreacfun/ch4/slides-matbal.pdf The Material Balance for Chemical Reactors]
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| * [http://www.em-ea.org/Guide%20Books/Book-1/1.4%20MATERIAL%20%20AND%20ENERGY%20BALANCE.pdf Material and energy balance]
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| * [http://www.freepatentsonline.com/6751527.html Heat and material balance method of process control for petrochemical plants and oil refineries, United States Patent 6751527]
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| * {{cite book | last = Morris | first = Arthur E. | last2 = Geiger | first2 = Gordon | last3 = Fine | first3 = H. Alan | title = Handbook on Material and Energy Balance Calculations in Material Processing | url = http://www.wiley.com/WileyCDA/WileyTitle/productCd-1118065654.html | edition = 3rd | year = 2011 | isbn = 978-1-118-06565-5 | publisher = [[John Wiley & Sons|Wiley]] }}
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| [[Category:Mass]]
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| [[Category:Chemical engineering]]
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