I teach undergraduate thermodynamics and I was quite ashamed that I couldn't explain to a student, the following. I thought I'd bring it to physics.SE in hope of providing my student a good explanation.

The question was concerning the energy equation for an open system:

$$ \underbrace{\frac{\mathrm{d} E}{\mathrm{d}t}}_{\begin{array}{c}\text{Rate of change}\\ \text{of total energy}\\ \text{in the system}\end{array}} = \underbrace{\delta \dot{Q}}_{\begin{array}{c}\text{Rate of}\\ \text{heat transfer}\end{array}} - \underbrace{\delta \dot{W}}_{\begin{array}{c}\text{Work}\\ \text{extracted/input}\end{array}} + \underbrace{\dot{m} \left(h_1 + \frac{V_1^2}{2} + g z_1 \right)}_\text{energy of inlet stream} - \underbrace{\dot{m} \left(h_1 + \frac{V_1^2}{2} + g z_1 \right)}_\text{energy of outlet stream}$$

The young lady asked me that for steady state operations, the rate of change of total energy, $dE/dt$ is zero. So why are $\delta \dot{Q}$ and $\delta \dot{W}$ are not zero as they are also rates of heat exchange and work generation/input. It is just obvious to me but I don't know how to explain this to a 17 year old.

I'd appreciate it if someone could help me out on this.

  • $\begingroup$ It would help if you explain what the terms are. I don't know about everyone else, but it is not at all obvious to me what any of the symbols in the third and fourth terms on the right hand side mean. Are they supposed to exactly cancel, as you've written? Anyway, the steady state solution $\frac{dE}{dt}=0$ just means that the average amount of energy in the system is not changing with time. So presumably any amount of heat that is transferred to the system is completely converted to work by whatever process within the system that you're modelling, meaning $\delta \dot{Q} = -\delta\dot{W}$. $\endgroup$ Feb 26 '13 at 22:25
  • $\begingroup$ @MarkMitchison Well this was my exact explanation! It apparently isn't obvious to 17-18 year old freshmen/sophomores! :( $\endgroup$
    – dearN
    Feb 26 '13 at 22:43
  • $\begingroup$ Well that is the answer! The best you can do is explain it clearly, she will work it out for herself eventually. Why don't you try asking her why she thinks that the rates of heat and work flow should be zero? Another useful thing is to think of an example that would be familiar to her which demonstrates this principle. I would try and suggest something, but you never actually explained what on Earth we are talking about here! Failing that, try to concoct your own example. The best demonstration of a physical principle is usually provided by Nature herself :) $\endgroup$ Feb 26 '13 at 22:51
  • $\begingroup$ @MarkMitchison Yes, I will try and turn to nature. My explanation (which should have been included) was to show that the dE/dt term is an accumulation rate which is zero for steady state. $\endgroup$
    – dearN
    Feb 26 '13 at 22:53
  • $\begingroup$ Sorry but I have no idea what that means! What is the physical system, what is coming in the 'inlet stream', what is $\dot{m}$, $V_1$ etc. etc.?? I appreciate you might not be bothered to include these things, but if you have time it would be helpful to future users of this site with a similar question if you were to edit this information into your post. Also, other people might have some better suggestions than me once they know what you are talking about. It really is not clear to anyone except maybe a specialist in whatever field of physics we are supposed to be talking about here. $\endgroup$ Feb 26 '13 at 22:58

The key to explaining this lies in two pieces that you have correctly written in the equation but have to make the student appreciate:

1) The difference between a $d$ and a $\delta$

$d$ means a small differential change, while $\delta $ just means a small quantity($\delta$ is used to denote a change in variational calculus, but that is different). Therefore in your equation $\delta Q/dt$ is just a small quantity per unit time (or per small change in time).

2) Steady state is when a system's properties don't change with time: Heat is not a systems property it is a transfer of energy. This is precisely why one never says $dQ$. Same for work, or matter. Now matter is a funny one it can be a change and a transfer. So technically if one is doing an atom balance one says \begin{align*} dN=\delta N \end{align*} or if your system leaks a particular molecules while also produces it by a chemical reaction one would write, \begin{align*} dN_i = \delta N_{i, leak}+\delta N_{i, reaction} \end{align*} Also energy transfers are always path dependent since they are not a system's property but depend on the process path. This is why $\delta $ also reminds us of that path dependence.

Essentially in steady state no changes to system properties occur but the quantity of heat or work is not zero it is still what it is.

  • $\begingroup$ This is a good point that $\delta$ represents an inexact differential and that $\mathrm{d}$ is an exact differential. The fact that $\delta \dot{Q}$ and $\delta \dot{W}$ are path dependent but $\mathrm{d}\dot{E}$ is state dependent is certainly the clue to this answer. $\endgroup$
    – dearN
    Feb 27 '13 at 9:56
  • $\begingroup$ I will also point out that $\delta \dot{Q}$ is not wrong but can be confusing for a student. It gives the impression that it is the change in a time derivative or at the very least a derivative while its neither. Usually in batch systems one tends to write $dE=\delta Q+\delta W$ (ignore the time ) or in steady-flow case one tends to highlight the by writing out $\delta Q/dt$ $\endgroup$
    – Sankaran
    Feb 27 '13 at 13:16

The expression for the total variation in energy is

$$\frac{dE}{dt} = \frac{dQ}{dt} + \frac{dW}{dt} + \frac{dE_\mathrm{matter}}{dt}$$

where the last terms accounts for variations in energy due to matter flows in and/or out the open system. It is better to leave this last term generic, just as the heat and work terms, instead picking a specific form for the flow of matter, which will be only valid for specific cases.

Note that I am using exact differentials for both work and heat. This is right. In classical thermodynamics heat and work have to be represented by inexact differentials $\delta Q$ and $\delta W$, but once you extend the classical thermodynamics space of variables with the variable time, you can use exact differentials. A good discussion of why heat and work are exact differentials $dQ(t)$ and $dW(t)$ when you include time is given in the well-known textbook by Prigogine & Kondepudi: «Modern thermodynamics: from heat engines to dissipative structures». If you cannot obtain this book, I discuss the same topic in a recent work which is free (open access): International Journal of Thermodynamics, Vol 16, No 3, 102-108

Now the main question: a stationary state (SS) is one for which the properties and flows are constant. This implies

$$\left(\frac{dE}{dt}\right)_\mathrm{SS} = 0$$


$$\left(\frac{dQ}{dt}\right)_\mathrm{SS} = \mathrm{constant}$$

$$\left(\frac{dW}{dt}\right)_\mathrm{SS} = \mathrm{constant}$$

$$\left(\frac{dE_\mathrm{matter}}{dt}\right)_\mathrm{SS} = \mathrm{constant}$$

The definition says that the flows have to be constant, they can be zero or non-zero. This contrast with the definition of equilibrium (EQ) as the state for which the flows are zero and the properties constant

$$\left(\frac{dE}{dt}\right)_\mathrm{EQ} = 0$$


$$\left(\frac{dQ}{dt}\right)_\mathrm{EQ} = 0$$

$$\left(\frac{dW}{dt}\right)_\mathrm{EQ} = 0$$

$$\left(\frac{dE_\mathrm{matter}}{dt}\right)_\mathrm{EQ} = 0$$

Note that zero is a constant, therefore technically an equilibrium state is just a particular case of a stationary state. As DeGroot & Mazur remark in their celebrated textbook on «Non-equilibrium thermodynamics»:

stationary states can be either equilibrium or non-equilibrium states depending on the boundary conditions, which are imposed on the system.


I think she is not looking for a detailed theoretical answer. Just give an example. May be you can even start with the question: can you imagine the machine which constantly gets heat but still it's state doesn't change (as it produces some work or due to mass exchange)? If you can, that is the answer, why δQ and δW are not zero. In fact any engine is such a machine. But its' construction is quite complicated. The most simple example I can imagine is a pipe with water flow, which is heated in one point. It always gets heat, but that state may last unchanged for ever (so dE=0).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.