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hello everyone and welcome to another
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lesson
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in this video we will be discussing the
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process of performing a material balance
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after watching this video it is expected
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that you will be able to describe
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the relationships between the inputs and
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outputs of single unit processes
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let's begin in its most basic definition
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we say that a material balance is a
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direct application
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of the law of conservation of mass to
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the design of processes
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it's like we are applying the law of
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conservation of mass to a specific
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process
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at steady state we say that this is
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characterized as
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mass input is equal to mass output
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majority of the problems that we will be
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solving are steady states
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problems the definition of steady states
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will be discussed
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later in general we say that the law of
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conservation of mass
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is stated as mass can neither be created
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nor destroyed
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in any physical or chemical process
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mathematically speaking we can represent
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it with this
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general balance equation that is mass
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inputs
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plus mass generated minus mass
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output minus mass consumed is equal to
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the accumulation of mass
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this general material balance equation
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can be simplified depending on the type
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of the process
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throughout our analysis of the material
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balance process we will be dividing our
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systems into two categories
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those are reactive systems and
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non-reactive systems
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two out of five terms in our general
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mass balance equation pertains to
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reactive processes only
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we have generation and consumption
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generation and consumption are
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determined by stoichiometry
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and they are exclusive with reactive
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systems accumulation is another special
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term in our mass balance equation
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in such a way that it is present
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whenever there is an
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imbalance between the inputs and the
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output
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for now we will be sticking with
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non-reactive processes
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without accumulation those are
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non-reactive steady state process
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and with that we have to discuss what
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are the different types of processes
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the type of process or our assumption on
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the processes will have an
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impact on how do we proceed with our
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material balance equation
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you can generally classify processes
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into three types those are
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the batch process which we also call
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transient process
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the continuous or steady states
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processes this are majority of our
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examples
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and we also have semibatch or fed batch
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processes
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the semi-batch process is a hybrid
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between the batch and the continuous
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process
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for now let's focus on the simpler
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continuous processes
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for the continuous processes in which
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there is no chemical reaction
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we can remove the terms of generation
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accumulation
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and consumption in our general material
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balance
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therefore for a continuous process we
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can simplify that by writing in
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is equal to out
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this is our primary equation that we
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will be applying to almost
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all of our examples basically when you
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say that the process is steady states
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and there is no chemical reaction then
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you can use this very simple mass
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balance
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the input is equal to output equation
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can be applied to streams
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and it can also be applied to components
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of streams
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and we will illustrate that with our
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many examples showing how the material
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balance process is carried out
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let us first establish the basics the
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basics of how to make
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or how to perform a material balance we
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will be following the five steps
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enumerated here
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but we will do this with an example so
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that you can follow along
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let's consider this example an aqueous
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solution of sodium hydroxide contains
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20 sodium hydroxide it is desired to
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produce
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an 8 sodium hydroxide solution by
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diluting a stream of the 20
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solution with a stream of pure water
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determine the feed rates of the 20
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solution
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and diluting water needed to produce 2
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310 bounce mass per minutes of the 8
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solution let's first have number one
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identify the process process unit and
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process streams
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identifying the type of process and the
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type of process unit
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would greatly help us in determining
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what type of solution will we pursue
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with these types of problems
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first what is the process from our
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understanding of the problem
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we are taking one stream that is the 20
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sodium hydroxide solution
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and we want to generate an 8 sodium
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hydroxide solution
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basically we want to dilute the 20
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solution by adding
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water by doing so what we are doing is a
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mixing
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process so in short the process in this
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case is mixing
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basically we are mixing two different
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streams in order to come up with
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one stream and we want to come up with
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an eight percent sodium hydroxide
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solution
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so let's write here the process is
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mixing
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next let's identify what is our process
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unit
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if you will remember the process unit is
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an apparatus or equipment that is
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used to carry out the process so what
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equipment do you need
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if the process is mixing the answer is
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you need a mixer
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so therefore the process unit is a mixer
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finally we identify what are our process
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streams
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the process streams are the inputs and
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outputs for the system
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the first process stream is the 20
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sodium hydroxide solution
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this is process stream number one and
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let's classify if it's an inputs or an
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output from our understanding this is an
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input
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to the process what is our next process
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stream
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from the wording of the problem it is
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desired to produce
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an eight percent sodium hydroxide
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solution so the eight percent sodium
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hydroxide solution
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is our process stream number two and
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this is considered
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as an output because we want to produce
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this stream coming from an input which
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is the 20
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sodium hydroxide solution okay now what
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do we need to mix with the 20
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sodium hydroxide solution in order to
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lower its concentration
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of course we will be mixing water and
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that is another process stream
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we want to produce the eight percent
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sodium hydroxide solution
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by diluting a stream of the twenty
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percent solution which is one of our
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inputs
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with a stream of pure water water
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is our process stream number three and
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this is considered as
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another input so by dissecting the
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problem we have determined that
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the process is mixing the process units
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is a mixer
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and we have three process streams we
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have two
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input streams and one output stream
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we are done with step number one with
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the information gathered from step
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number one
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we then proceed to step number two draw
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and label the process
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using block diagrams label the given
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values with correct
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units also do not forget to properly
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place your inputs
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and outputs inputs are arrows
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going toward the process unit and
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outputs are arrows
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going away from the process units here's
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the same example and let's attempt to
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draw our flowchart to represent the
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system
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with the knowledge we obtained from step
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number one
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we know that the process unit is a mixer
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so let's draw a box
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let's label this as our mixer
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and let's represent our process streams
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we determine that we have two input
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streams
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our first stream is the 20 sodium
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hydroxide solution
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i will call this stream a and let us
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write what do we know about stream a
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well first we know that stream a is an
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aq's solution it's an ages mixture
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that is composed of 20 sodium hydroxide
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and presumably the remaining 80 percent
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is water because this
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is an aq solution meaning it has to
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contain water
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okay so that is our first process stream
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what is our second process stream let us
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recall what is our second input
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in order for us to lower the
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concentration of stream a we need to mix
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it with stream b
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which is pure water
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this is the diluent in this case okay
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remember that in the previous step we
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identified two
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inputs and one output so we are done
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with the two inputs let us now
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characterize our outputs
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our output is what we want to produce we
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want to produce
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an eight percent sodium hydroxide
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solution
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then that should be our output stream i
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will call the output stream as
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c and stream c contains eight percent
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sodium hydroxide and presumably the
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remaining 92 percent is water because we
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only have two components in this problem
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the sodium hydroxide and the water this
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is the diagram that represents the
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problem
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it is from this diagram that will be
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basing all of our remaining solution
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so it's very important that you get this
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part correct
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you need to make sure that all of the
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given in the problem is represented
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properly in your diagram
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if you cannot properly draw a diagram
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then it would be questionable whether
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you can answer the rest of the problem
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okay so representing the problem in a
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diagram is the most important step
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in the material balance process now
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let's browse through the rest of the
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problem and see
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what other information can we get that
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would be helpful in solving the problem
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we are done with the compositions of the
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stream so we have
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stream a stream b and stream c all
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having their compositions
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but there is another given in the
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problem that is 2
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310 pounds mass per minute of the 8
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solution this parameter is a mass flow
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rate because of the unit's pound mass
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per minute
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and this mass flow rate pertains to the
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flow of the products it was specified
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that it is for the eight percent
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solution
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so we say here that our stream c has a
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mass flow rate of two three one
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pound mass per minute and that is our
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complete
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diagram always make sure that your
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entire problem is represented in the
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diagram
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now let's proceed to step number three
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step number three is
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select a basis we have talked about
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selecting a basis in our previous lesson
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and in this case you may want to select
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a basis that would benefit your problem
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solving approach
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for example if you are given a flow rate
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b that may
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a mass flow rate volumetric rate or
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molar flow rate a very sensible basis is
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to assume whatever is the denominator
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units of that flow rate for example
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if you are given two three one zero
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pounds mass per minutes
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one of our best bases could be we just
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assume that we have one minute of
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operation or one minute of mixing
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right because if you assume that your
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basis is one minute
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then you already have the mass of stream
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c that is 2
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310 pounds mass in fact that is what we
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are going to do
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we say that our basis is one
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minute of operation
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this basis allows us to have the
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simplest material balance solution
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you can assume any other value aside
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from one minute let's say for example
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that your basis is one hour of operation
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you can also do that
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but then there would be a couple of
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other extra steps that you need to do
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in order to adjust your given to your
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chosen basis
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so choose the simplest basis whenever
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possible and that's step three
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let's proceed to step four select your
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system boundaries
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and perform material balances starting
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with the overall material balance
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then follow with the individual balances
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of the components so let's do that piece
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by piece
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what is a boundary a boundary is simply
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a dotted line
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that we draw around certain systems in
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which we are considering
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what streams are going in and what
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streams are going out
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of my system boundary for processes with
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only a single unit you are left with no
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choice but to select your unit as your
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boundary
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so for clarification let me draw the
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boundary here i
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say that we are setting the boundary
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around the mixer
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in our future lessons when we will
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consider processes with several unit
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processes
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then we have to specify our boundaries
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okay but for now it is enough to say
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that for a single unit your boundary is
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around
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that unit before proceeding with the
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balances let's
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first determine what is required from
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this problem or what is being asked from
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the problem
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the problem statement is given here
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determine the feed rates of the 20
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solution and diluting water needed to
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produce 2
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310 pounds mass per minute of the eight
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percent solution
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our required in this case is a and
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b remember that a represents the flow
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rate
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of the twenty percent sodium hydroxide
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solution while b
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represents the flow rate of the pure
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water stream
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and we just want to determine their feed
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rates or simply the flow rates in which
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they are going into the mixer
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now let's perform the omb omb
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stands for an overall material balance
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or overall mass
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balance this omb is a simple equation
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that simply states input is equal to
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output
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that is our law of conservation of mass
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for non-reactive steady state processes
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so let's just determine what streams are
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going in and what streams are going out
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and they should be equal
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for our problem here we have two streams
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going in that's stream a and stream b
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so we say that a plus b
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this is the summation of your input
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streams
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and this is equal to the summation of
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your
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output streams we only have one output
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stream so we can say that a and b
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is equal to c and that is your omb
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or that is your overall material balance
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it's that simple
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at this point you can now substitute
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whatever values are given for either a
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b or c we are given the flow rate of
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stream
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c so we can substitute that as a
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plus b is equal to two three one
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zero we just have to remember that all
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of the terms in this overall material
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balance
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are in pound units if you will notice
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this equation contains two unknowns
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that's a
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and b and we cannot solve an equation
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with two unknowns
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unless we find another equation with the
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same
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unknowns and we will get our second
00:14:23
unknown
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with what we call the component balances
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for the component balances you will take
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a look at a particular
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component of the system let's define
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what is a component
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a component is a substance that can be
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defined separately
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for our problem in our example here we
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only have two components the first
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component is sodium hydroxide
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and the second component is water if you
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take a look at the systems
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stream a contains two components naoh
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and water
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stream c also contains two components
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naoh and water
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but stream b only contains one component
00:15:04
that's water because
00:15:05
stream b does not contain any sodium
00:15:08
hydroxide so we say that for the entire
00:15:10
system we have two components
00:15:12
and therefore you have the capability to
00:15:14
perform two
00:15:15
component balances one for sodium
00:15:18
hydroxide
00:15:18
and the other for water but in this case
00:15:21
we only need
00:15:22
one of the component balances because we
00:15:24
already have our equation number one
00:15:26
which is our overall material balance so
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let me illustrate
00:15:30
our sodium hydroxide balance for the
00:15:33
component balances you will still be
00:15:35
applying the law of conservation of mass
00:15:37
that's input is equal to output for
00:15:39
non-reactive steady-state systems
00:15:41
but we only have to focus on that
00:15:43
particular component
00:15:45
our mass balance in this case i will
00:15:47
write it as naoh
00:15:49
in is equal to naoh
00:15:52
out we are simply saying since there is
00:15:56
no chemical reaction and there is no
00:15:57
accumulation that
00:15:58
whatever sodium hydroxide comes in it
00:16:00
must also be the sodium hydroxide that
00:16:02
comes out
00:16:03
right and we just have to figure out how
00:16:05
to mathematically represent that
00:16:07
so let's write our equation for the
00:16:10
sodium hydroxide balance
00:16:12
for our sodium hydroxide in we can have
00:16:14
two sources
00:16:15
stream a and stream b however we know
00:16:18
that stream b
00:16:19
does not contain any sodium hydroxide
00:16:21
because it's pure
00:16:22
water so we will not be including that
00:16:24
in our balance
00:16:25
okay stream a contains 20
00:16:28
sodium hydroxide we are assuming that
00:16:30
this is 20 by mass
00:16:32
because this is a liquid system how do
00:16:34
we mathematically write that
00:16:36
well we write the amount of sodium
00:16:38
hydroxide coming in
00:16:39
as 0.2 times a
00:16:42
0.2 is the mass fraction of sodium
00:16:45
hydroxide in stream a
00:16:47
and letter a represents the total mass
00:16:49
flow rate of stream a so when you
00:16:50
multiply the total mass flow rate with
00:16:52
the mass fraction of the component
00:16:54
you would get the mass of the component
00:16:57
so we say that this
00:16:58
is the mass of
00:17:01
sodium hydroxide in stream a
00:17:07
to complete our material balance on the
00:17:09
other side of the equation we represent
00:17:10
the mass of sodium hydroxide coming out
00:17:13
of the system and that is represented by
00:17:15
our single output which is stream c
00:17:18
in stream c we have eight percent sodium
00:17:20
hydroxide
00:17:21
so we write that as 0.08 times
00:17:25
c 0.08 is the mass fraction of sodium
00:17:28
hydroxide and c
00:17:29
represents the total mass flow rate of
00:17:31
the output stream
00:17:33
now if you take a look at this equation
00:17:35
we don't know what's the value of a but
00:17:36
we do know the value of c so we can
00:17:38
substitute
00:17:39
we write 0.2 a is equal to 0.08 times
00:17:44
the value of c
00:17:45
2 310 pounds
00:17:48
mass we are using mass units only
00:17:51
instead of mass flow rate units because
00:17:52
of our bases
00:17:54
one minute operation solving this
00:17:56
equation we can look for the value of a
00:17:59
substituting that is two three one zero
00:18:03
times point zero eight divided by points
00:18:06
two this is the mass of stream a nine
00:18:09
hundred
00:18:10
twenty four pounds mass
00:18:14
since our basis is one minute operation
00:18:16
you can also say that the mass flow rate
00:18:18
of stream a
00:18:19
is 924 pound mass per
00:18:22
minute now that we have the mass of
00:18:26
stream a
00:18:27
we can substitute this back to our
00:18:28
overall material balance or omb
00:18:31
in order to determine the value of
00:18:32
stream b
00:18:34
rearranging our omb we have the equation
00:18:38
b is equal to c minus a
00:18:41
and we calculate that as two three one
00:18:43
zero
00:18:44
minus two hundred ninety four stream b
00:18:48
has a mass of one three eight six
00:18:51
pounds mass in terms of flow rates we
00:18:55
can
00:18:56
write that as 1386
00:18:59
pounds mass per minute and just like
00:19:02
that we were able to determine the feed
00:19:04
rates of streams a
00:19:06
and b our interpretation of this
00:19:08
solution is
00:19:09
if you want to obtain 2 310
00:19:12
pound mass per minutes of an 8 sodium
00:19:15
hydroxide solution
00:19:17
you need to mix 924 pound mass per
00:19:20
minutes
00:19:21
of a 20 sodium hydroxide solution
00:19:24
with 1 386 pound mass per minutes
00:19:28
of pure water okay that is our
00:19:31
interpretation
00:19:32
and that's how you perform a material
00:19:34
balance
00:19:35
at the end of the day this all boils
00:19:37
down on how well can you interpret a
00:19:40
problem
00:19:40
so you have to practice your reading
00:19:42
comprehension skills because that
00:19:44
factors a lot in understanding
00:19:46
what the worded problem is trying to
00:19:48
tell us remember that your
00:19:50
representation of the problem
00:19:52
is based on your understanding of the
00:19:53
problem and your final answer
00:19:56
is based on the material balance which
00:19:58
is based on your interpretation of the
00:20:00
problem
00:20:01
so if you cannot understand the problem
00:20:03
you cannot draw a proper diagram
00:20:05
you cannot perform a proper material
00:20:07
balance you would arrive with the wrong
00:20:09
answer
00:20:09
the key for mastering this skill is
00:20:11
constant practice you have to solve as
00:20:13
many problems as you can in order for
00:20:15
you to get acquainted with the process
00:20:17
of material balances
00:20:19
now that we know the process let's try
00:20:21
another example
00:20:26
the oxygen in air is usually sufficient
00:20:28
to sustain
00:20:29
combustion but it can be enriched to
00:20:32
make combustion more efficient
00:20:34
in a certain process it is required to
00:20:36
produce a 50
00:20:37
50 mixture of oxygen and nitrogen by
00:20:40
mixing air
00:20:41
and pure industrial oxygen how much in
00:20:44
kilo moles per day
00:20:46
of the 50 50 mixture can be produced
00:20:48
from 250 cubic meters per day
00:20:51
of pure oxygen at stp stp stands for
00:20:55
standard temperature and pressure that
00:20:57
is zero degree celsius and one
00:20:58
atmosphere
00:21:00
since we are dealing with gases in this
00:21:02
problem we are going to assume that the
00:21:04
given 50 50 mixture pertains to
00:21:06
50 mole percent oxygen and 50 mole
00:21:09
percent nitrogen
00:21:11
let's perform our step-by-step process
00:21:13
first one is identify the process
00:21:16
the process unit and the process streams
00:21:18
the process in this case again
00:21:20
is mixing because we want to produce an
00:21:22
enriched version of
00:21:24
air in which you will be mixing air and
00:21:26
pure
00:21:27
oxygen since the process is mixing we
00:21:30
know that the process unit is a mixer
00:21:32
and we also know what are the process
00:21:34
streams the 50 50 mixture is a process
00:21:37
stream
00:21:37
air is a process stream and pure oxygen
00:21:40
is another process stream
00:21:42
now that we have that knowledge let us
00:21:44
represent the problem
00:21:45
in a diagram our process unit
00:21:49
is a mixer
00:21:53
and based on the problem we have three
00:21:55
process streams
00:21:56
we have two input streams and
00:21:59
one output stream for this problem let
00:22:03
us name the input streams as
00:22:04
f1 and f2
00:22:08
and let's name the product as p for the
00:22:11
drawing of the diagram you are not
00:22:12
limited on how you want to represent
00:22:15
your input and output streams you can
00:22:17
use whatever variables you want to use
00:22:19
i will call stream f1 as our air stream
00:22:23
and air typically contains 79
00:22:27
nitrogen gas and 21 oxygen
00:22:30
gas stream f2 is our pure
00:22:33
oxygen stream this is 100 percent oxygen
00:22:38
and our product stream is what we want
00:22:40
to achieve we want to produce a 50
00:22:43
50 mixture of oxygen and nitrogen that
00:22:46
is 50 percent
00:22:47
nitrogen and 50 oxygen
00:22:53
another given in this problem is we want
00:22:55
to use
00:22:56
250 cubic meters per day of pure oxygen
00:22:59
at
00:23:00
stp so in our f2 we have here 250
00:23:05
cubic meters per day at stp
00:23:10
that is a volumetric flow rate and we
00:23:12
may want to convert that to a molar flow
00:23:14
rate
00:23:14
to suit our material balance problem in
00:23:17
this problem our unit is not in mass
00:23:20
it is in moles because we are dealing
00:23:22
with gases
00:23:23
for non-reactive processes the mole
00:23:25
balance is the same as the mass balance
00:23:27
the number of moles coming in
00:23:29
is equal to the number of moles coming
00:23:31
out just make sure to understand that
00:23:33
this only works for
00:23:35
non-reactive process when we go to the
00:23:37
reactive processes
00:23:38
you cannot perform the omb as a mole
00:23:41
balance
00:23:43
in this problem we are required to solve
00:23:45
what is
00:23:46
the molar flow rate of the product
00:23:48
stream particularly
00:23:50
in kilomoles per day or the given feed
00:23:52
rate of
00:23:53
stream f2 now that we have illustrated
00:23:56
our problem let's proceed to step three
00:23:58
which is choose your basis
00:24:02
as i have mentioned if you are given a
00:24:04
flow rate be that may mass
00:24:06
volumetric or molar flow rate you can
00:24:09
use the denominator unit of the flow
00:24:11
rate as the basis
00:24:12
in this case we are given 250 cubic
00:24:14
meters per day
00:24:15
of stream f2 the denominator unit is day
00:24:19
therefore a sensible basis is we use one
00:24:22
day of
00:24:23
operation by the application of our
00:24:25
basis it is our understanding that
00:24:27
stream f2
00:24:28
now has a volume of 250 cubic
00:24:32
meters and this is pure oxygen
00:24:35
before we proceed to the material
00:24:36
balance we have to convert
00:24:38
everything to either mole or mass units
00:24:41
250 cubic meters is in molar units
00:24:44
you cannot use that in the overall mole
00:24:47
balance
00:24:47
because it's a volume right so we have
00:24:50
to convert from volume to moles and
00:24:52
since this is a gas
00:24:53
we can assume that oxygen behaves
00:24:55
ideally and we can use the ideal gas
00:24:57
equation
00:24:58
so we want to convert from volume to
00:25:01
mole
00:25:02
we use the ideal gas equation pv is
00:25:04
equal to nr
00:25:06
t we solve for the number of volume of
00:25:08
oxygen
00:25:09
this becomes pv over rt
00:25:14
remember that we are using stp
00:25:16
conditions that
00:25:17
is zero degree celsius and one
00:25:19
atmosphere of pressure
00:25:20
let's substitute to solve for the number
00:25:22
of moles of oxygen
00:25:24
pressure that's 101 325 pascals
00:25:28
multiplied by volume 250 cubic meters
00:25:31
divided by the universal gas constant r
00:25:34
in standard si units that's 8.314
00:25:37
pascal cubic meters per mole per kelvin
00:25:40
and
00:25:41
we multiply that by the absolute
00:25:42
temperature zero degree celsius that's
00:25:45
273
00:25:46
point 15 kelvin the number of moles of
00:25:49
oxygen is 11
00:25:51
154 points 38
00:25:54
moles o2 or in kilomoles
00:25:58
this becomes 11.15
00:26:04
this is basically the value of stream 2
00:26:07
and we can use this with our material
00:26:09
balance
00:26:10
our omb or overall material balance in
00:26:14
this case it also
00:26:15
pertains to overall mole balance is now
00:26:18
inputs f1 plus f2
00:26:21
is equal to outputs that's p
00:26:25
substituting what is already known we
00:26:28
have f1
00:26:29
plus 11.15 is equal to p
00:26:32
always remember that these are in
00:26:34
kilomole units
00:26:37
since we have two unknowns f1 and b we
00:26:40
have to look for
00:26:41
other equations to describe the system
00:26:43
in order for us to solve the unknowns
00:26:45
after the omb we proceed to the
00:26:47
component balance
00:26:49
let me take this opportunity to discuss
00:26:51
what we call the thai
00:26:53
components the thai component if present
00:26:56
is what allows us to simplify our
00:26:58
material balances for a particular
00:27:00
problem
00:27:01
the thai component is defined as a
00:27:03
component
00:27:04
that only enters through one stream and
00:27:07
only exits
00:27:09
through one stream let us examine our
00:27:11
components and take a look if they
00:27:13
satisfy the definition of the thai
00:27:15
component
00:27:16
based on our problem we only have two
00:27:18
components oxygen
00:27:19
and nitrogen let's take a look at oxygen
00:27:22
oxygen
00:27:23
enters at f1 and at f2
00:27:26
and exits at p it enters at two streams
00:27:30
and
00:27:30
exits at one stream therefore oxygen is
00:27:33
not
00:27:34
a thai component a thai component can
00:27:36
only enter and exit
00:27:38
on one stream each let's take a look at
00:27:40
nitrogen
00:27:41
nitrogen enters at f1 and nitrogen exits
00:27:45
at p it enters at one stream and exits
00:27:49
at one stream
00:27:50
therefore nitrogen is a thai component
00:27:54
our thai component balance is 0.79
00:27:57
of f1 is equal to 0.5 of
00:28:02
p this means that stream f1 contains 79
00:28:06
nitrogen and the stream p contains 50
00:28:10
nitrogen gas there is no nitrogen in
00:28:13
stream f2 so that does not appear in our
00:28:15
nitrogen balance
00:28:17
rearranging this equation this becomes
00:28:19
f1 is equal to
00:28:21
0.5 over 0.79
00:28:24
times p i'll call this equation 2 and we
00:28:27
can substitute equation 2
00:28:29
from our omb
00:28:32
our omb will become 0.5
00:28:35
over 0.79 times p
00:28:38
plus 11.15 is equal to
00:28:41
p equation null has only one unknown and
00:28:46
that is the product
00:28:47
p from here we can rearrange the
00:28:49
equation and solve for the value of p
00:28:51
rearranging this becomes negative 11.15
00:28:57
divided by point five over
00:29:00
point seventy nine minus one
00:29:06
p is thirty point thirty seven
00:29:10
kilo moles
00:29:13
since our basis is one day of operation
00:29:16
we can say that the flow rate of our
00:29:17
product is
00:29:19
30.37 kilomoles per
00:29:22
day that is our final answer
00:29:26
the problem only asks for the molar flow
00:29:28
rate of the product but if you want you
00:29:30
can also determine the molar flow rate
00:29:32
of
00:29:32
stream f1 which is one of our feed
00:29:34
streams you can use equation number two
00:29:37
that is 0.5 divided by 0.79
00:29:41
divided by p 30.37
00:29:44
f1 is 19.22 kilomoles per day
00:29:49
you can also verify from our overall
00:29:50
material balance if the values you
00:29:52
obtained are mathematically sensible
00:29:55
for example one way of checking if you
00:29:57
are correct is you simply add f1
00:29:59
and f2 and their value should be equal
00:30:01
to p so we have the value of
00:30:03
f1 19.22 and to this we add f2
00:30:06
which is 11.15 this should be equal to
00:30:12
30.37 which is the value of our product
00:30:15
we can say that our calculations are
00:30:17
correct
00:30:18
by now you are getting a feel on how
00:30:20
material balance problems are
00:30:22
tackled or how they are processed just
00:30:24
follow our five steps and you will be
00:30:26
okay
00:30:28
let's solve another example
00:30:31
an experiment on the growth rate of
00:30:33
certain organism
00:30:34
requires an environment of humid air
00:30:37
enriched in oxygen
00:30:39
three input streams are fed into an
00:30:41
evaporation chamber to produce an
00:30:43
output stream with the desired
00:30:44
composition we have our three streams
00:30:47
here stream a
00:30:48
is liquid water fed at a rate of 20
00:30:51
cubic centimeters per minute
00:30:53
stream b is air composed of 21
00:30:56
oxygen and 79 percent nitrogen stream c
00:31:00
is pure oxygen with the molar flow rate
00:31:03
one-fifth
00:31:04
of the molar flow rate of stream b the
00:31:07
output gas is analyzed and is found to
00:31:09
contain
00:31:10
1.5 mole percent water draw and label
00:31:13
the flow chart of the process
00:31:15
and calculate all unknown stream
00:31:17
variables
00:31:18
so let us apply our material balance
00:31:21
steps on this problem
00:31:23
first step is to identify what is the
00:31:25
process the process is essentially
00:31:27
mixing but it was specified that our
00:31:29
unit in this case
00:31:30
is an evaporation chamber it just so
00:31:33
happens that one of our streams is
00:31:34
liquid water and we want to convert
00:31:36
liquid water
00:31:37
to water vapor because that is the
00:31:39
specification of the problem
00:31:41
so the process is mixing with a little
00:31:42
bit extra and that is including the
00:31:45
evaporation our process unit is an
00:31:47
evaporation
00:31:48
chamber let us draw our diagram for this
00:31:52
problem
00:31:53
so we have our evaporation chamber let's
00:31:55
call that ec
00:31:57
and we have our three input streams
00:32:01
those are a b and c
00:32:05
for our product stream i will be calling
00:32:08
that a stream b
00:32:10
and we are given the compositions of our
00:32:12
input streams
00:32:13
stream a is pure water that's 100
00:32:16
h2o with a liquid volumetric flow rate
00:32:21
of 20
00:32:22
cubic centimeters per minute
00:32:28
b is air so that is composed of 79
00:32:31
nitrogen and 21 oxygen
00:32:37
stream c is pure oxygen
00:32:42
and it was stated that the molar flow
00:32:44
rate of stream c
00:32:45
is one-fifth the molar flow rate of
00:32:48
stream b we write that relationship
00:32:50
as c is equal to one-fifth of
00:32:53
b and before we proceed
00:32:56
let us analyze the composition of stream
00:32:59
b or the product stream
00:33:00
we know that stream p is simply a
00:33:02
combination of the three input streams
00:33:05
and our three input stream states that
00:33:07
we have three components
00:33:08
those are water oxygen
00:33:11
and nitrogen from the problem it was
00:33:15
stated
00:33:16
that the output gas is analyzed to
00:33:18
contain 1.5 mole percent
00:33:21
water so we know that this is 1.5
00:33:26
and for our required we are asked to
00:33:28
solve for all unknown stream variables
00:33:31
those are the flow rates of a b and c
00:33:34
before proceeding to the material
00:33:35
balance part
00:33:36
let us first agree to use either mole or
00:33:39
mass
00:33:39
balances in this case since we are
00:33:41
dealing with gases and since we are
00:33:43
given a mole percent it would be easier
00:33:45
for us to perform mole balances
00:33:47
so everything must be represented as
00:33:50
either molar flow rates or mole
00:33:52
percents with that we have to convert
00:33:54
our stream a which is represented as a
00:33:56
volumetric flow rate that's 20 cubic
00:33:58
centimeters per minute
00:34:00
so let's convert 20 cubic centimeters
00:34:03
per minutes
00:34:05
two moles per minute which is a molar
00:34:07
flow rate
00:34:08
to do that we have to convert from
00:34:10
volume to mass and then from mass
00:34:12
to mole from volume to mass we'll use
00:34:14
density from mass to mole we'll use
00:34:16
molecular weight
00:34:17
we are using liquid water and its
00:34:19
density is one gram per cubic centimeter
00:34:21
so that is our first
00:34:22
factor for every one cubic centimeter
00:34:26
it weighs one gram and then finally for
00:34:29
the conversion to moles
00:34:31
we say that for every 18 grams of water
00:34:34
that is equivalent to
00:34:35
one mole of water that is the molecular
00:34:37
weight of water
00:34:39
take a look at the cancellation of units
00:34:40
cubic centimeter would cancel
00:34:42
grams would cancel and you will be left
00:34:45
with the units
00:34:46
mole in the numerator and in the
00:34:48
denominator it's
00:34:50
minutes this is the molar flow rate of
00:34:53
water
00:34:54
calculating that is 20 divided by 18
00:34:58
that's 1.11 moles per minute
00:35:01
we can now write that here stream a is
00:35:04
1.11
00:35:05
mole per minute this is the value that
00:35:07
we're going to use for our material
00:35:09
balances
00:35:10
let's begin with the material balance
00:35:12
proper let us set
00:35:14
our bases our basis in this case is
00:35:17
one minute operation
00:35:20
because our flow rate is given as moles
00:35:23
per minute
00:35:24
our overall material balance in this
00:35:26
case this is a mole balance
00:35:28
is a plus b plus
00:35:31
c is equal to b the summation of our
00:35:34
inputs
00:35:35
is equal to the summation of our outputs
00:35:38
we know the value of a so we can
00:35:40
substitute that this becomes 1.11
00:35:43
plus b plus c is equal to b
00:35:46
we call this equation one now for our
00:35:49
component balance let's identify if we
00:35:51
have a thai
00:35:52
component and that would simplify our
00:35:54
solution
00:35:56
if you take a look at water it enters at
00:35:58
stream a
00:35:59
only and exits at stream p only so water
00:36:02
is a thai
00:36:03
component let's have the water balance
00:36:08
this is 1.11
00:36:11
moles coming from stream a is equal to
00:36:16
0.015 times b
00:36:19
that's because stream p contains 1.5
00:36:22
mole percent
00:36:23
water from this thai component balance
00:36:25
you can see that we are already able to
00:36:27
solve for
00:36:28
p that's simply 1.11 divided by 0.015
00:36:34
the value of p is 74.07
00:36:39
moles now that we have the value of p
00:36:43
you can write our overall material
00:36:46
balance or omb
00:36:47
as 1.11 plus b
00:36:50
plus c is equal to 74.06
00:36:54
how can we solve for the values of b and
00:36:56
c remember that we have
00:36:58
a relationship for b and c we say that
00:37:01
the flow rate of c
00:37:02
is equal to one-fifth of b we can
00:37:05
substitute that to our overall material
00:37:07
balance this becomes
00:37:08
1.11 plus b
00:37:11
plus c is equal to one-fifth of b
00:37:14
and that's equal to 74.07
00:37:18
rearranging and substituting we are able
00:37:20
to solve for b as 74.07
00:37:23
minus 1.11 transpose from the other side
00:37:26
of the equation
00:37:27
divided by 1.2 that is one plus
00:37:31
one fifth the flow rate of b
00:37:34
is 60.8
00:37:37
moles per minute because our basis is
00:37:41
one minute
00:37:41
and from this we can solve for the flow
00:37:43
rate of c because we know that c is
00:37:45
equal to one fifth of
00:37:46
b so we simply divide b by 5.
00:37:50
the flow rate of c is 12.16
00:37:54
moles per minute and that's it we were
00:37:57
able to look for the flow rates of b
00:37:59
and c two of our missing inputs plus the
00:38:01
flow rate of our product
00:38:03
p if you look back at our diagram we
00:38:07
were able to solve for the flow rate of
00:38:08
b
00:38:09
and the flow rate of c and the flow rate
00:38:11
of p
00:38:12
however we are still missing the
00:38:14
composition of p
00:38:15
and that is what are the mole fractions
00:38:18
of oxygen and nitrogen in our product
00:38:21
stream
00:38:21
that's also important so let's try to
00:38:23
solve for that
00:38:24
for that let us represent the mole
00:38:26
fraction of
00:38:28
oxygen in stream b as the variable x
00:38:32
and for nitrogen i can represent that as
00:38:34
variable y because we are simply
00:38:36
representing with variables
00:38:37
however a more sensible approach to
00:38:39
representing the mole fraction of
00:38:41
nitrogen in p
00:38:42
would be 1 minus x
00:38:45
minus 0.015 we are simply exploiting the
00:38:49
fact that the summation of the mole
00:38:51
fractions of
00:38:52
all the components in p is equal to 1.
00:38:56
so let's solve for the value of x
00:39:02
to determine the value of x we simply
00:39:04
perform another component balance
00:39:06
if you remember we are done with the
00:39:08
overall material balance
00:39:10
and we have also used the water balance
00:39:12
let's now use the oxygen
00:39:14
balance the oxygen balance is there is
00:39:18
no oxygen in a so we will not include
00:39:20
that
00:39:20
in stream b that is 21 oxygen so we
00:39:24
write that as
00:39:25
point 21 b and then from stream c that's
00:39:30
pure oxygen so we simply write
00:39:32
c and finally on the product stream
00:39:35
we don't know the mass fraction of
00:39:37
oxygen so we write that
00:39:38
as x multiplied by p x
00:39:42
is the mole fraction of oxygen in stream
00:39:44
p and
00:39:45
p is the total molar flow rate of the
00:39:48
stream we already know the values of b
00:39:50
c and p we simply substitute those and
00:39:52
we can solve for the value of
00:39:54
x substituting that is point 21
00:39:58
times b 60.8
00:40:01
plus c 12.16
00:40:04
and the result is divided by p 74.07
00:40:10
the mole fraction of oxygen or x is
00:40:14
0.3365
00:40:16
or this is also 33.65 percent
00:40:20
oxygen and of course since you now know
00:40:22
the value of
00:40:23
x you can solve for the percentage of
00:40:25
nitrogen
00:40:26
that's simply 1 minus
00:40:29
x 0.3365
00:40:32
minus 0.015 this is the mole fraction of
00:40:36
water in stream p
00:40:38
nitrogen has a mole fraction of 0.6485
00:40:43
or that is 64.85
00:40:46
nitrogen and with that we have also
00:40:50
solved for the composition
00:40:52
of stream p and that's the nature of
00:40:54
material balances
00:40:55
you just have to exploit every possible
00:40:58
material balance you can form
00:40:59
that is first the overall material
00:41:02
balance and then after that you can
00:41:03
perform individual component balances
00:41:07
after this lesson you will be given more
00:41:09
examples detailing the proper material
00:41:11
balance solution
00:41:12
for different types of process units
00:41:15
certain process units have particular
00:41:17
modes of solution and
00:41:18
it's important that we discuss all of
00:41:20
them but for now this is the end of this
00:41:23
video
00:41:24
i hope you learned something thank you
00:41:25
for listening and as always
00:41:27
keep safe
00:41:37
[Music]
