Concepts of Thermodynamics
Prof. Suman Chakraborty Department of Mechanical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 28
First Law of Thermodynamics for
Unsteady Processes in a Control Volume So, far we have discussed about the application
of first law for a control mass system and for a control volume undergoing a steady state
steady flow process. But all processes are not steady and it is important that we also
understand how to address the problems where the state within the control volume at least
changes as a function of time. So, we will start today’s discussion on
first law applied to unsteady state problems. So, what kind of situation we are considering?
Let us just draw a box which is the control volume and the properties within the box.
So, by properties within the box, this box has numerous physical locations and at each
physical location you can have a separate set of properties and physics does not disallow
that, what is disallowed is from the concept of thermodynamic equilibrium consideration.
What is that? If we assume that this entire control volume is in thermodynamic equilibrium,
then we cannot, you know, distinguish the property set here from property set here,
from property set here and so on. So, everywhere the property set, by property set we mean
a pressure, volume, temperature or other relevant properties like internal energy and enthalpy,
we will learn a few more properties as we advance in this course. So, all these properties,
these properties if they are uniform over this control volume then only this control
volume can remain in equilibrium, thermodynamic equilibrium.
So, whenever we say that we are talking about equilibrium thermodynamics, we are implicitly
considering that all these points are at uniform state, but there can always be a deviation
from equilibrium and any natural process actually is deviated from equilibrium. So, that is
how you have a little bit of conflict in conceptual paradigm between equilibrium thermodynamics
and a more practical consideration of heat transfer within the system.
So, if you have heat transfer within the system, you are likely to have temperature gradients.
If you do not have temperature gradients how can you have heat transfer within the system?
But equilibrium thermodynamics assumes that you are having a local equilibrium, but that
local equilibrium state can change with time because of may be a global heat transfer to
or from the system or any work done by the system or work done on the system. But whenever
the change takes place the change takes place in such a manner that the entire system quickly
comes to equilibrium in response to that, whether quickly or slowly it is a matter of
the properties of the system, but eventually it comes to equilibrium ok.
So, but so that is not actually impractical, the system can eventually come to equilibrium.
What this equilibrium thermodynamics is not able to capture is how by function of time
this system came from one equilibrium state to another equilibrium state. It simply can
identify the equilibrium states, equilibrium state 1 changed to equilibrium state 2, but
that does not occur instantaneously. How over a period of time, through different transients,
this has occurred. That detailing is not within the perview of this macroscopic or classical
thermodynamics. So, this is a brief discussion which I thought
is required, because you know you may have a conflict of understating that if you have
heat transfer then how can you have equilibrium. So, this we have to keep in mind that when
you have heat transfer instantaneously the equilibrium state is disturbed, the system
takes some time to come to a new equilibrium state. In between what happens the dynamics
that is not within the perview of macroscopic thermodynamics. Macroscopic thermodynamics
will bother about only you know the change of state from one equilibrium state to another
equilibrium state, not you know the how I mean the system dynamics changed in between.
So, this is a control volume. So, let me just draw a little bit more clearly, there is a
generic inlet i, there can be large number of inlets. So, each will be symbolically denoted
by i, there will be generic exit e again there could be multiple exits, there is a heat transfer
q, again this is generic, if it is from the surrounding to the system then q is positive,
if it is from system to surrounding then q is negative ok. So, then next is work done
again very generically represented in this diagram, if work is done by the control volume
or energy transfer due to work is from control volume to the surrounding then it is positive,
if it is the other way then it is negative. In the mean time the state within the control
volume because it is unsteady it changes from one equilibrium state 1 to another equilibrium
state 2, this is the new thing that is beyond the previous consideration, the previous the
steady state steady flow problems this 1 and 2 states were identical equal. So, there was
no change in state within the control volume. Now, the state within the control volume changes
from state 1 to state 2. State 1 uniform over the control volume, state 2 uniform over the
control volume, this is the basic picture that we are keeping in mind. Now we will apply
the Reynolds transport theorem to fit this picture. So, let us write the Reynolds transport
theorem just for recapitulation. So, just to recapitulate, capital n is an
extensive property, small n is capital n per unit mass, Vr is the velocity of the fluid
relative to the control volume, theta is a unit vector outward normal to the surface
across which the inflow or outflow is taking place. And you know this CS is the control
surface, the surface of the control volume over which these flow boundaries are there.
So, this is 1, this is 1 like that ok. So, we will write N as the total energy of
the system, remember total energy is internal energy plus kinetic energy plus potential
energy ok. So, dN/dt is dE/dt, this is small e is capital e per unit mass, remember that
e includes internal energy kinetic energy and potential energy. Now as per the description
of first law of thermodynamics of the of a system, this is the difference between heat
transfer and rate of heat transfer and rate of work done.
So, when we say rate of work done normally it is the work associated with the control
volume, that is whatever work you can derive from the control volume plus an additional
work that you require to maintain the flow in presence of pressure right. So, that is
flow energy or flow work which we have discussed in great details in one of our previous lecture.
So, instead of you know repeating that again I am just writing the expression for the flow
work. So, this is the flow work associated with the inlet and this is the flow work associated
with the outlet right. Remember it is we derived it as p by rho because in thermodynamics for
easy interpolation we use specific volume instead of density. So, 1 by rho is v, flow
work associated with the inflow is negative and flow work associated with the outflow
is positive, there could be multiple inlets and exits.
So, there I mean you can generalize these with a summation symbol, but for you know
simplicity I am not writing it and then the most important thing is that we are assuming
that the pressure and specific volume are uniform over the respective inlet and exit
sections. So, pressure pe when we say,see e is not a
point, e is you know the distributed collection of points across this. So, we are assuming
that pressure is uniform over this which is very true for pipe flow, see typically this
inlet and exit will be pipes. So, typically over the cross section of the pipe the pressure
does not vary that much as compared to how it changes longitudinally. So, pressure will
be uniform and this is not such a large cross section that density will also change substantially.
So, for all practical purposes pressure and specific volume uniform across respectively
i and e are not very bad assumptions ok. So, then we will write Q dot. So, this W dot
is actually W dot that you can get across the control volume W dot CV, but I am just
short form in short form I am writing W dot just to avoid any know confusion; then plus
ok. Now, can you tell one thing, see how do you know you know from your common sense access
that this work is negative and this work is positive.
So, I give you a physical feel about it. So, by work transfer what is happening? Some energy
is either entering the control volume or energy is leaving the control volume, look into this
symbol of work if energy is leaving the control volume then work corresponding to that is
positive. So, here when the fluid is flowing out of the control volume it is carrying with
it the energy. So, energy is leaving the control volume and that is why that is positive.
Here fluid is entering the control volume. So, some energy is pumped into the fluid in
the form of work. So, that is why it is negative ok. So, the you know why I am telling all
these because the sign of heat and work these are very important, the algebraic signs so
far as problem solving writing algebraic expressions. So, if you get a confusion in this instead
of you know getting into the basic derivations from the scratch you can use your common sense
to figure out, whether you will put a plus to that or minus to that.
So, that is what I am emphasizing it. So, this then let us write the right hand side
ok. Now this e is what? Internal energy plus kinetic energy all per unit mass. So, all
the velocities are measured relative to the inertial reference frame so that Vr and V
will be you know the same. So, then you have to recognize that these property, internal
energy kinetic energy potential energy these may vary across the cross section.
But we are again assuming that these cross section or control surface rather which is
which may be small cross section of a pipe, this control surface is such that over which
these properties are uniform. That means, height does not change appreciably, velocity
also does not vary appreciably which is again you know not a very good assumption, why?
Because, if you have studied fluid mechanics you know that the velocity across the section
of the pipe varies. So, instead of using the local velocity you
can use average velocity here with something called as multiplied by something called as
kinetic energy correction factor, this is taught in fluid mechanics. I can understand
that there are students studying this thermodynamics course who may not have studied fluid mechanics
yet. So, I am not getting into all those details. So, we are we are desperately trying to represent
this v average square by two the kinetic energy of the system and that fortunately is possible
if the velocity is very uniform across the section and that occurs when the flow is so
called turbulent. In most of the practical engineering applications you will find typically
in the industry, you will find that there are flow rates sufficiently large to have
turbulent flow and then the velocity is uniform over this section. So, this truly represents
the kinetic energy without much you know error ok.
So, given that we can argue that you can bring this property which is uniform over the section
outside the integral and rho into Vr dot eta dA integrated is the mass flow rate, positive
for outflow, negative for inflow, so we can write, ok. So, the assumption is very very
clear these properties are uniform over i and e ok. So, this is often called as uniform
flow assumption, that you know the flow at the sections respective sections i and e are
associated with uniform property ok, this is often called as uniform flow assumption.
Then we can club this term with this and this term with this double tick term with this
double term. So, u becomes u plus p v if you club this with this. So, internal energy is
replaced by enthalpy if you bring it in this side ok.
So, the physical difference is that between a control mass process and the process which
is taking place in a control volume with flow is that here you have this additional thermal
energy that the fluid must poses to maintain the flow in presence of pressure. So, the
thermal energy for a flowing fluid is represented by enthalpy and not by internal energy, this
is the physical understanding of a flow process as compared to a you know stationary, you
know non flow process. So, then let me utilize this part of the board to you know complete
the derivation which will require one or two more steps. So, Q dot plus m dot i. I am just writing
i on the left. So, so far so good, now as I told
that in equilibrium thermodynamics we are not really bothered about how the change of
state takes place in between. We are only bothered about the change from equilibrium
state 1 to equilibrium state 2. So, we will integrate this over time from
time state 1 to time state 2. So, if we do that and assume, so we make an assumption
that this inlet state has not changed over time right; see inlet state has a chance of
changing over time. So, we are assuming that inlet state has not changed over time. So,
assume states i and e locally uniformed and also not changing with time right, these are
the two things that we are already assuming. So, in that case if you integrate this equation
over time, so, this is like a constant over time, i e sorry h i v i square by 2 g zi m
dot i is d m i d t these when integrated will become m i right. So, what we will get q dot
integrated will become q so integrate over time. So, you have q plus m i, this when integrated
over time will be the difference in E from time state 2 to time state 1 within the control
volume. So, this will be E at state 2 minus E at state one, right within the control volume,
plus me he plus v square by 2 plus g ze plus W ok. W dot integrated over time will become
W. Then, in the next step so, what is this E this E is E within the control volume. So,
that is not suppose to contain the term enthalpy as a thermal energy, that will contain internal
energy as a part of thermal energy. So, this is the key you have to keep in mind
for inflow and outflow the thermal energy includes enthalpy for the state within the
control volume you know the energy thermal energy part is contributed only by internal
energy. So, Q plus mi hi plus Vi square by 2 plus g zi equal to, now the most important
assumption that at state 2 we are a or at state 1 we are assuming that the properties
are uniform. So, uniform state within the control volume so; that means.
So, m2 this we can write m2 into. So, the assumption is uniform state within control
volume ok. So, you can see the control volume is undergoing an unsteady state unsteady process,
but at each time state it is uniform over space. So, this is called as unsteady but
uniform state ok. So, this equation which you know let me make
a, so, there is a h e term which I have I have to write. So, the or me term plus me
h e plus v square by 2 plus g ze plus W. So, this equation let me make it a box. So, that
you know you can. So, q plus energy transfer at the inlet is equal to the change in state
of energy between state 1 to state 2 plus the energy coming out in terms of flow and
work. So, it is basically energy in equal to energy
out plus energy change within the control volume. So, this is energy in, through heat
through flow this is energy, this is energy out and this is the change in energy within
the control volume. So, whatever energy is getting into the control volume may be a part
of that is leaving, the remaining can increase the energy from state 1 energy to state 2
energy right. So, this is the bro, this is the broad energy balance.
Essentially after a whole lot of mathematical toil we have written an equation which we
can write simply from common sense with energy balance if we can understand it properly.
Now only one point small point that remains to be noted here is that we have included
energy balance, but we should also include mass balance consistent with the energy balance. So, what is mass balance here?. So, m dot
i minus m dot e, right, the rate at which mass enters the control volume minus the rate
at which mass leaves the control volume the difference is changing the mass of the control
volume. So, this is mass conservation. So, if we integrate this over time m i minus
m e this is m 2 minus m 1, very simple. So, these two equations together constitute the
energy balance and mass balance frame work of a process called as USUF Uniform State
Uniform Flow, the under lying assumption is that it is uniform but unsteady, but at a
given instant of time the properties within the control volume are uniform. That is why
the state thermodynamic state is uniform and the flow sections are also having uniform
properties. So, uniform state, uniform flow is the terminology that is correctly associated
with the description of this energy balance and this mass balance.
Let us stop here today. In the next class we will start working out a few problems illustrating
these basic equations. Thank you very much.
