This is a diagram of a turbine which we use
in thermodynamics and the turbine is used
to generate shaft work, and it does so by
extracting energy from a high pressure, high
temperature stream of fluid that enters the
turbine, it expands within the turbine and
then it exits at a lower temperature and a
lower pressure and the difference between
the high energy of the fluid flowing into
it and the low energy of the fluid leaving
it, that net difference is the shaft work
that the turbine generates, to make that happen
the insides of the turbine have to be incredibly
intricate but from a thermodynamics standpoint
we don't care about what goes on within this
black box, all we care about is the flow of
energy, what we would say is that if this
system is running under steady conditions
the total rate of all forms of energy entering
the system is equal to the total rate of all
forms of energy leaving. So I'll draw a control
volume around the turbine and what we're concerned
with are all the different ways energy can
flow into it, obviously energy will flow in
from the high pressure, high temperature steam
entering the turbine, it'll leave as the lower
temperature, lower pressure steam, it'll also
leave as shaft work which we desire, so there's
various ways energy can get into the turbine,
we can have internal energy entering, there
can be kinetic energy of the fluid entering
and potential energy of the fluid entering,
so these three forms of energy enter the turbine
at this location. One form of energy that's
really easy to forget, and it's not a terribly
intuitive, is known as flow energy or flow
work, and it's the work required by the upstream
fluid to push the downstream fluid into the
turbine, to conceptualize this I've deleted
the majority of the feed and I've put a little
piston/cylinder device here and the syringe
needs to push the downstream fluid into the
turbine and that requires a downward force
on the plunger, that force I'll call F1, and
that force is equal to the pressure at point
one multiplied by the cross sectional area
of the pipe, A1 and over a period of time
this plunger has displaced a distance L1 and
I can define the amount of flow work, Wflow,
as equal to the force times the length, F1
L1 and you can show that this is equal to
P1 times the displaced volume and the rate
at which flow work is entering the system
Wflow dot, is equal to the derivative d dt
of P1 V1, if the pressure is constant that's
equal to P1 dV1 dt, or I can say P1 dmV1hat
dt, and since V1hat, the specific volume at
state one is a constant I'll end up with mdot
times P V1hat and this is the rate at which
flow work enters the system or we'd say the
rate at which PV work enters the system so
to demonstrate this the upstream fluid is
pushing the fluid down into the turbine and
it's doing flow work, not only does the fluid
have to push fluid into the turbine, the fluid
leaving the turbine has also got to push its
downstream fluid out against some pressure P2, so
we look at the flow work at state 2 the exhaust
of the turbine, some work is leaving the system,
so here we see fluid pushing this plunger
along as it exhausts the turbine. So again,
if we look at our energy balance we've got
internal energy entering associated with the
molecular interactions and in addition to
kinetic and potential energy we also need
to account for this flow energy entering,
so here are the four forms of energy entering
the system, at steady state that has to equal
the total rate at which all forms of energy
leave the control volume, those four are equal
to the internal energy leaving, the flow energy
leaving, or that the system is doing, plus
any kinetic or potential energies leaving,
there's also one other form of energy leaving
so I need to add that into my energy balance,
so I'll say plus shaft work, and with a turbine
this is what we're primarily interested in,
and let me write this symbolically, in a nutshell
these four forms of energy are large entering
the turbine relative to the four forms of
energy leaving the turbine, that means that
this turbine can produce a great deal of shaft
work, in other words if it can extract a lot
of this energy it can do more shaft work,
and we can simplify this a bit by saying that
the mass flow entering the system has to equal
the mass flow leaving the system at steady
state, and let's just drop the subscripts
and just call it mdot, and I'll rearrange
this expression, again I factored out the
mass flows. One thing to look at and to keep in
mind is that these terms, they're often seen
in analysis of thermodynamic systems and they're
seen so much in fact that for a matter on
convenience we'll say that this U1 plus p1
Vhat1, we're going to call it H1, and we'll
give that H a name, we'll call it enthalpy.
So I'm going to solve for Wout now and make
that substitution for H, an assumption we
can often make is that these terms H1 and
H2 are usually fairly large compared to the
kinetic and potential energy differences and
they're so large in fact that it may only
make a percent or two difference if we're
trying to calculate W out we might get by
by just saying that W out is equal to the
mass flow rate through the turbine times the
enthalpy of the stream entering minus H2,
in analyzing this system, if I come back to
the diagram what is often available to you
is the temperature and the pressure at the
inlet or at point one and I can use those
to look up H1 and often times I've got a T2
and P2 and I can use those to look up H2 and
when I want to find the work that the turbine
is generating, if I know the mass flow rate
I know H1 and I know H2, the amount of enthalpy
extracted from the stream is effectively equal
to the shaft work that the turbine is producing.
Lastly, I want to note that I don't need to
use enthalpy to solve these problems, as long
as I've got a conceptual feel for what internal
energy means and a conceptual feel for what
flow work means I can use this relationship
H1 is equal to U1 plus P1 times the specific
volume at state 1 and that's all I need if
I go down I've got a piece of a table for
superheated steam and if I look at this,
here's superheated steam at a pressure of
one megapascal, so let's say that P1 is equal
to one megapascal and if I knew that the temperature,
for example, is equal to 500 degrees C I want
to show you that I don't even need this enthalpy
column, I know that H is equal to U plus P
times the specific volume, if I plug in numbers
for U P and V, and when I do this calculation
what I come up with is a value of 3479.1 kilojoules
per kilogram, and that is the exact same value
for enthalpy, and this column for H is completely
redundant, I don't need it whatsoever but
it's really nice because otherwise I would
have to make this calculation over and over
again, and because we see it so often in thermodynamic
analysis for convenience they've added this
extra column.
