INSTRUCTOR: I'm going to describe the terms
that are important for the first law of thermodynamics
when we have an open system, which means mass can flow
into the system, mass can flow out,
and so we should look at a diagram.
And we'll call whatever's in this box our system.
And that systems gonna have some internal energy,
so you will be internal energy, and I'm gonna use
this notation total to indicate this is not internal
energy per kilogram or per mol, but it's internal energy.
And I'll use this for this system so I'm distinguishing
it from the fact we have flow in, so we have some,
and well, let's work in units of mass, so kilograms
per second flowing in.
And that mass, which can have several components,
has some enthalpy per kilogram, and that's a function
of temperature, composition, phase, et cetera.
And then we have a mass flow rate out,
it does not have to be the same as the mass flow rate in
because we're looking at general situation,
and the enthalpy in general is going to be different.
The composition can be different, lots of things
can be happening at a system.
One of the things is we're gonna be adding heat,
in general, and so, the dot would mean this is
joules, or kilojoules, per second, so it's per time.
We could be adding shaft work, for example if we have
a compressor, and again, per time.
We could add expansion, compression work.
Let's put a dot over it, we're gonna add
expansion compression work, like a piston moving up and down
on the system.
So if I wanna look at the general energy balance,
what I'm gonna say is that change in the energy
is a function of how much energy we add,
how much energy we take away, and what we're going
to do here because we're interested in applications
in typical engineering situations, we're gonna ignore
kinetic energy, potential energy.
Not saying that these terms might not be important
in some cases, but the things we're interested in
we're going to ignore these, so we're setting these terms
to zero, we could certainly include them if necessary.
So then the energy balance says, there'll be
a differential equation since the change in the
internal energy of the system, with respect to time,
'cause the temperature can be changing the composition
and can be changing.
That's due to the fact we're adding energy,
so one way we're adding energy is we have low in.
And notice this is in enthalpy, for the flow term in,
it's not internal energy, the reason is because of the
PV work term, we explain that in other screen casts.
So that's one way we're adding energy.
One way we're removing energy is we have mass flow out
and again that's an enthalpy term, that's important.
And then the other terms that we talked about,
the heat, the shaft work, and the
expansion compression work.
So what we wanna do to better understand this,
let's apply it to a few situations.
This is a general equation, but it reduces when we
have restrictions on our system, so let's look
at some examples.
So first example is for a closed system.
So I've written down the general equation,
and what we're going to do is, this is zero,
this is zero, no flow in, and then I'm gonna look
at the special case where this is also zero.
Does not have to be, we could have a propeller,
excuse me, could an impeller that's mixing something up
and adding energy, but let's look at this case.
And so, then the equation reduces.
So, first, make sure this is the total, this is not
per kilogram, so then let me write down
the reduced equation.
If I were to integrate this over some time period,
then the left side would be dealt to you total,
and that would be the total amount of Q,
and the total amount of work, that we added to the system,
We often look at this on a per kilogram basis.
So, now I'm using U to represent the kilojoules per kilogram
and let me just use Q with a mark to indicate
we're talking about now per kilograms, so all these
are per kilogram.
And put this here because this is often how we look
at it in systems, we do analysis.
So that's one case, let's look at another case.
If we have a steady state system, which is certainly
very common in chemical engineering processes.
So we're looking at steady state system, and I've written
down again, copied the general equation.
So by steady state we mean any change respect to time,
had given a location, is zero.
And we're gonna look at the case where we have shaft work
but we don't have any expansion, compression,
so for example, we're looking at a turbine.
The other important part of steady state means
the mass flow rate in, equals the mass flow rate out,
otherwise we'd be accumulating mass or depleting mass
in the system, we would not be at steady state.
So then, this equation simplifies, this term is zero,
and this term is zero, and so let me write down,
combine some terms.
And so this would be the equation, for example,
if we had, let's say we had a turbine where we're
generating work from change in enthalpy,
pressure drop, the Q term is often zero for a turbine.
So this is per time, we could also write this per kilogram.
And then it would just be H in minus H out, plus Q,
plus shaft work, where these terms, remember now
are per kilogram, and the way we define our H's,
originally, these are specific enthalpy,
so they're per kilogram.
So, easiest way, when solving problems in energy balances
so to start with our general energy balance, and then remove
terms to end up with a equation.
Might end up with a differential equation,
that we're gonna integrate, depends on what's known,
what's unknown, what's fixed, what isn't.