# Entropy change with elevation

1. Jul 11, 2014

### nayanm

I have seen in a number of thermodynamics lectures that the entropy change of a system as it falls approximately isothermally from some height h to the ground is: ΔS = mgh/T

(The proof basically has you conceive of a reversible process between the same two states where some upwards force acts against the force of gravity in order to maintain quasi-mechanical equilibrium. Once the object hits the ground in this theoretical reversible process, it must be heated by Q=mgh to account for the work done by the object on the additional force. With the heat added, the final state of the system in the reversible process matches that in the actual process. The change in entropy is thus Q_rev/T = mgh/T. I can explain this in more detail if needed.)

My question: how come we don't take change in entropy associated with change in height into account in the open system entropy balance? Every textbook I have seen accounts for only entropy inflow due to mass flowing in, entropy inflow due to heat transfer in, and entropy outflow due to mass flowing out.

But why is there no mgh/T term for change in elevation?

Last edited: Jul 11, 2014
2. Jul 11, 2014

### WannabeNewton

Are you asking about the absence in textbook discussions of entropy flow due to change in elevation in general open thermodynamic systems? If not, then your question read verbatim is quite unclear to me.

If you are referring to your specific example then the entire point of the calculation is to derive the entropy flow due to change in elevation in a gravitational field so clearly the $\Delta S = mgh/T$ term is present in the end result.

3. Jul 11, 2014

### nayanm

Yes, that's exactly what I'm asking.

For example: http://www.learnthermo.com/T1-tutorial/ch08/lesson-B/pg04.php
Every other thermodynamics textbook I've seen has some variation of that equation (with no explicit elevation term).

Since in the open system energy balance, the same textbooks do include a Δ(potential energy), they aren't assuming no change in elevation either.

So then why doesn't this change in elevation give rise to an explicit term in the open system? Is it perhaps implicitly included in the sin - sout terms?

4. Jul 11, 2014

### Staff: Mentor

Are you referring to the change in entropy of a parcel of gas that moved from altitude h to ground level, or an object that behaves differently than a gas?

In any event, I don't like the way your books and lectures have treated the change in entropy. Entropy is a function of state (a property of the material), so it only depends on the two end points (i.e., the pressures and temperatures at the end points), and not on the process path that is used. So the fact that the object is moved from a higher altitude to the ground is not relevant to calculating the entropy change, except for determining the pressure at altitude compared to the 1 atm at ground level. The pressure at altitude can be determined from the hydrostatic equation. Once the pressures and temperatures at the two end points are known, the entropy change per unit mass is determined. This calculation is easy to do for an ideal gas.

Chet

5. Jul 12, 2014

### nayanm

I am trying to reconcile the result I'm obtaining for a rigid body in a gravitational field (entropy changes with height) with what I know about flowing streams from my Engineering Thermodynamics (Smith and Van Ness) textbook (entropy depends only on pressure and temperature).

Since the entropy of a rigid body varies with height, a rigid body at the reference level and the same rigid body 5 meters above the ground are clearly not at the same state. So then is a parcel of air at the reference level and the same parcel of air 5 meters above the ground (at the same temperature and pressure) at the same state?

In other words: how can a parcel of air at some pressure P and some temperature T be at the same entropy IRRESPECTIVE of height when, at the same time, the entropy of a rigid body does depend upon height?

6. Jul 12, 2014

### nayanm

Another way to phrase the question that I just thought of (based on a different thread you replied to):

I know for a simple system where there is a single force-displacement conjugate pair (pressure-volume for instance), 2 properties of the system are sufficient to specify the state.

But in an open system where there is an additional mechanical force (m*g) acting to change a coordinate (height), doesn't it logically follow that we need 3 properties to specify the state (e.g. V, T, and z)?

7. Jul 12, 2014

### Staff: Mentor

Hi nayanm.
I can see why this is so confusing to you. Before we proceed further, though, let's take a step backward for a moment and ask the question as to whether these lectures and books of yours have calculated the entropy change correctly for a rigid body whose elevation changes. The analysis in Smith and Van Ness is correct, in that the entropy is a function of pressure and temperature. Suppose that we consider the case of (a) an parcel of ideal gas and (b) a rigid body that is moved to different heights at different pressures. In Smith and Van Ness, they give an equation for the partial derivative of entropy with respect to pressure at constant temperature (for an arbitrary material). What is that equation? We can use that equation to determine the change in entropy in both these cases, if the temperature is held constant, but the parcel is moved from a given altitude (where the pressure is lower) to the surface of the earth (where the pressure is higher).

Chet

8. Jul 12, 2014

### nayanm

Chet,

I believe the equation you are referring to is: $dS = \frac{Cp}{T}dT - (\frac{dV}{dT})_P dP$ (imposed with the constant temp. condition)

However, this equation (and the Maxwell relationships from which it is obtained) presupposes that entropy can be expressed solely as a function of two other properties only, which is precisely what I'm trying to convince myself in the first place.

Rather, I'm trying to evaluate the partial derivative $(\frac{dS}{dh})_P$ for an arbitrary material.

I'm finding it a little difficult to type up this proof. Would you mind if I hand-wrote it out and scanned it?

9. Jul 12, 2014

### Staff: Mentor

The above equation is the one I was referring to. Before proceeding further, I will ask the question, How can entropy be a function of altitude, if it is a physical property of the material?

I'm going to continue with the derivation. For an ideal gas, $(\frac{∂V}{∂T})_P=\frac{R}{P}$

So, $dS = - \frac{R}{P}dP=-RdlnP$

If we integrate this from altitude h to the surface, we obtain:

$$ΔS = -Rdln\left(\frac{P(0)}{P(h)}\right)$$

Using the hydrostatic equation, we can show that, at constant temperature:
$$P(h)=P(0)e^{-\frac{Mgh}{RT}}$$
where M is the molecular weight of air. If we substitute this into the equation for ΔS, we get:

$$ΔS = -\frac{Mgh}{T}$$

This is the entropy change per mole. If m is the mass of the gas packet, then

$$ΔS = -\frac{mgh}{T}$$

I'll give you a little time to digest this, before I continue.

Chet

10. Jul 12, 2014

### Staff: Mentor

The result I presented in my previous post agrees with the entropy change that you presented in your original post. Note however that this relationship applies exclusively to the case of an ideal gas parcel. For a solid object, as we will show below, a different relationship applies. Note also that the height h only comes into the equation in a very tangential way, by considering how the air pressure varies with altitude, as described by the hydrostatic equation. The entropy is really just a function of the pressures at the two altitudes, and is not directly and inherently related to the altitude h.

Next, lets consider the case of a solid object. In this case, the starting equation is still:
$$dS=-\left(\frac{∂V}{∂T}\right)_PdP$$
This equation is first re-written as:
$$dS=-V\left(\frac{∂lnV}{∂T}\right)_PdP=-αVdP$$
where α is the volumetric coefficient of thermal expansion. If we make the reasonable approximation that, for a solid, the solid parcel volume V and the coefficient of thermal expansion are nearly constant, we obtain:
ΔS=-Vα(P(0)-P(h))
The solid parcel volume is related to the mass of the particle by V = mv, where v is the specific volume of the solid and m is the mass of the solid parcel. So:
ΔS=-mvα(P(0)-P(h))
If we plug in the hydrostatic pressures P(0) and P(h) into this relationship, we obtain:
$$ΔS=-mvαP(0)(1-e^{-\frac{Mgh}{RT}})$$
In the limit of small values of h, this equation reduces to:
$$ΔS=-\frac{mgh}{T}\left(\frac{vαP(0)M}{R}\right)$$
So the entropy change for a solid parcel moved from altitude h to altitude 0 is very different from that for an ideal gas. The term in parenthesis in the equation will typically be very small. Try some numbers for the term in parenthesis to see that this is the case.

Note that, in the case of a solid parcel, the height h again only comes into the equation in a very tangential way, by considering how the air pressure varies with altitude, as described by the hydrostatic equation.

Chet

11. Jul 13, 2014

### nayanm

Chet,

Thank you very much for the detailed derivation. I rewrote the entirety of both posts out by myself and the logic makes sense to me.

One small point of simplification: you’ll notice I was specifically considering a rigid body which, by definition, must be incompressible. Thus, alpha = 0, and your equation again reduces to -mgh/T.

I understand conceptually that entropy is a physical property and thus should be independent of height, but I’m still having trouble mathematically proving it to myself, independent of working substance.

I’ve attached a variation of the entropy vs. height proof I was taught. It'll probably make more sense to look at the attachment first. I’ll attempt to explain my reasoning here as well in case my writing is difficult to understand.

An important qualification: Let’s assume that the following process takes place in a vacuum where external pressure = 0 at every height.

Process we are trying to analyze:
Consider a rigid, thermally insulated block of weight mg that falls from some height h to the ground.
Applying an energy balance, we see ΔU = +mgh.

Hypothetical reversible path:
In order to calculate change in entropy between two state points, we can always conceive of a reversible process between the state points and calculate the heat transfer in that process.

In order for a process to be reversible, all generalized forces must balance. Thus, we set up a pulley system where a block of equal mass is lifted up in the surroundings and the resulting tension balances the force of gravity acting on our system. At the conclusion of this (albeit infinitely slow) process, we notice that ΔU = 0. Thus, we are not yet at out final state.

In order to reach the final state, we heat the body by Q = mgh so that the internal energy rises by mgh, bringing us to our final state.
Thus (assuming negligible change in temperature): ΔS = Qrev/T = mgh/T.

We made no mention of external pressure in this derivation (we could easily be in a vacuum/a planet with no atmosphere), yet the entropy STILL changes by an amount mgh/T. This contradicts your statement that the entropy change with height is purely a consequence of pressure changing with height.

Is there something I have assumed wrong in this derivation?

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Last edited: Jul 13, 2014
12. Jul 13, 2014

### Staff: Mentor

Actually, no. If α=0, then, from the equation, ΔS=0.

Yes. You left out a term in the first law energy balance equation. You can verify this by checking out the complete form of the first law equation given in Smith and Van Ness. Actually, for the change you described, ΔU = 0. The really interesting question is "What happens after the rigid body collides inelastically with the ground?"

Bird, Stewart, and Lightfoot, Transport Phenomena, does a great job of discussing the relationship between the overall energy balance equation, the mechanical energy balance equation, and the thermodynamic energy balance equation. This is really important stuff that most thermo books omit. In this book, they also show a derivation of the differential entropy balance equation (in an exercise at the end of one of the chapters). This elucidates specifically how viscous dissipation and irreversible heat conduction determine the amount of entropy generation.

When I have time later, I'll look over the derivations you attached.

Chet

13. Jul 13, 2014

### nayanm

My apologies. Of course, you're right. But, I'm afraid this leaves me even more confused than before.

I was using the form: Δ(K.E.) + Δ(P.E.) + ΔU = Won + Qin.

For the actual process, I reasoned Q=0 (body is thermally insulated), W=0 (no external forces), and Δ(K.E.)=0 (block is at rest in both the initial and final state). Also, since we are in a vacuum with no air resistance, the full amount of the potential energy is converted to internal energy when the block collides with the ground.

I'll look for the relevant sections in BSL as you suggested.

14. Jul 13, 2014

### Staff: Mentor

OK. I see. So, the instant before the object hits the ground , delta U is still zero. Now you are assuming that an inelastic collision takes place between the object and the ground. But, because the object is rigid, all the Inelastic deformation takes place in the ground, This is where the heat is generated. Some of this heat may temporarily make its way into the object by conduction, but eventually all the heat will be conducted away by the ground. So, in the end, the temperature of the object will be back at its original value, and the object will have suffered no internal energy or entropy change. If , somehow, one were able to remove the object from contact with the ground before all the small amount of heat that had diffused into the object had a chance to diffuse out again, the object would have suffered a small change in internal energy, but nothing like mgh.

Chet

15. Jul 13, 2014

### nayanm

The way I see it, as the object is falling, its atoms have a net directionality of movement downwards. When this is abruptly brought to a halt by collision with the ground, the vibrational energy of these molecules inevitably increase, increasing internal energy of our system.

We have assumed perfect thermal insulation of the object, so none of this energy is conducted out as heat.

But even if the internal energy increases by some amount < mgh (let's just call it ΔU), my original claim seems to remain valid. In the hypothetical reversible path, we must still heat the object by Qrev = ΔU in order for the reversible process to emulate the actual process. And thus, entropy still changes despite the fact that we are in a vacuum/no atmosphere and pressure hasn't changed.

16. Jul 13, 2014

### Staff: Mentor

I have a different perspective for you to consider.

Your original question asked if we should be considering entropy as a function of height. You yourself have now shown that, up to the point where the object makes contact with the ground, its internal energy as well as its entropy has not changed. So you have now demonstrated that entropy is not a function of height.

Now I would like you to consider your object as experiencing two separate processes, rather than a single overall process.

Process 1: The object falls from a height h until it is about to make contact with the ground. Its potential energy has decreased, but its kinetic energy has increased. But, as we have seen, its internal energy and its entropy have not changed. So, for process 1, the internal energy and the entropy of the body are constant.
Process 2: An object traveling at a velocity $\sqrt{2gh}$ collides with a much larger object (the ground). The collision may be purely elastic, purely inelastic, or partially elastic.

Note that collision Process 2 can be considered to occur independent of how the initial velocity of the object was established. So, here again, the height h is only incidental to Problem 2.

A rigid body is just a limiting case of a deformable body that has a very high stiffness. So let's allow the body be a little deformable. If the body is perfectly elastic and the ground is very rigid, the collision will be purely elastic, and body will rebound with a velocity equal to its downward velocity prior to the collision. In this case then, the internal energy change and the entropy change for collision Process 2 (as well as for the overall combination of Processes 1 and 2) will be zero.

If the body is not perfectly elastic, part of the kinetic energy of the falling body will be converted to internal energy, and its entropy will increase.

If the collision is perfectly in-elastic, all the object's kinetic energy will be converted to internal energy. Under these circumstances, the change in internal energy will be mgh. And the temperature of the object will rise by mgh/C, where C is the heat capacity. If the object is somehow insulated from the ground, this temperature rise will remain constant within the object. If the temperature rise is negligible compared to the initial absolute temperature of the object, then the entropy increase will be approximately mgh/T0, where T0 is the initial temperature. If the temperature rise in not negligible compared to the initial absolute temperature of the object, the entropy change will be $Cln(1+\frac{mgh}{CT_0})$. These results give the maximum that the entropy can increase. In reality, for a partially elastic collision, the entropy change will lie somewhere between this value and zero.

In any event, our results show that, for the combination of Processes 1 and 2, the entropy is not an explicit function of height.

Chet

17. Jul 14, 2014

### nayanm

This last post of yours very elegantly clarified my confusion.

My mistake, as I can see now, was to calculate change in entropy for a fall + perfectly elastic collision process and then erroneously accredit gravity for this entropy change.

Thank you very much for sticking with me through this.

I'm not really satisfied by the way thermodynamics is being taught to me as a second year undergrad. We need more people like you who can effectively clear up conceptual confusions that students may have.

Take care,
Nayan

Last edited: Jul 14, 2014