Enthalpy Explained: All You Need to Know

[muslach]

Hello , I don’t think enthalpy makes sense, I am here because I want a clear explanation for what it really is.

It has been said it is the internal energy and the volume it takes as in work (W) but why? Why do we even need enthalpy what is the problem with delta U!

It is equal heat and pressure times volume it describes the work it has done to achieve the volume and the heat energy (average kinetic energy) there is no need for anything else to describe the energy change except delta U.

 

[Dale]

Enthalpy and internal energy are different things. Neither can replace the other.

In particular, whether a process is exothermic or endothermic depends on the change in enthalpy, not the change in internal energy. Internal energy only considers the internal energy and not the work that must be done on the external environment.

 

[LCSphysicist]

The enthalpy is relatively more useful than U when the environment is at P constant, that is why we use it. For example, suppose you want to "create" some system, i don't know, maybe make a cake. You give him the energy necessary to it "be a cake", is that enough? Well, normally the cake will need to increase its volume right? Do you think it will take energy from where? All energy you give it is to make him a "cake", it has no more energy to do anything. You need to give more energy to it expand and become in fact a "cake". So it need to do work on the environment. If it were a gas, the work would be Pdv... Wow, wait. If it were a gas, that is, if you "want" to create a gas, so you would need to give it its internal energy, plus Pdv to its expand, that is H = U + PV. Call it enthalpy and be happy.

An idiot example, but hope you get it.

 

[haushofer]

Internal energy only considers the internal energy...

Do you mind if I put this on a tile and hang it in my toilet?

 

[jtbell]

The enthalpy is relatively more useful than U when the environment is at P cte

I suppose "P cte" means "constant pressure". I've never seen it used in US English.

 

[Auto-Didact]

I suppose "P cte" means "constant pressure". I've never seen it used in US English.

It sounds almost French to me.

 

[256bits]

Summary:: prove to me that enthalpy can replace delta U

there is no need for anything else to describe the energy change except delta U.

So you are saying there is NO difference between a constant volume process and a constant pressure process( or any process that involves a change in volume)?
The properties of state can be combined in a manner to form another property of state.
Enthalpy is one such property that has been found to very useful in thermodynamic calculations.

Just as an example,
Steam tables even give you tabulations of internal energy, enthalpy and entropy for saturated liquid, saturated vapour and the change form one to the other.
http://homepages.wmich.edu/~cho/ME432/Appendix1Udated_metric.pdf
Table A-4, A-5
and you can study and play around with the numbers to get a better feel of enthalpy.

 

[DrStupid]

Why do we even need enthalpy what is the problem with delta U!

We want to be lazy efficient. Determing the change of internal energy at constant pressure and without non-volumetric work requires the measurement of heat and work but for the change of enthalpy we need to measure the heat only.

PS: At constant volume we would only need the heat for the change of internal energy. However, keeping the pressure constant is easier. Under normal consitions the atmosphere does this job for us.

 

[LCSphysicist]

I suppose "P cte" means "constant pressure". I've never seen it used in US English.

Hahaha cte is "constante", it is actually portuguese notation. Sometimes i didn't notice i mix both language :P

 

[vanhees71]

All the thermodynamical potentials are related by Legendre transformations and each has "natural independent" thermodynamic variables. E.g., one usually starts with the internal energy (for simplicity in the canonical ensemble of a gas, i.e., a fixed number of particles in therma contact with a heat bath at temperature $T$). Then one has
$$\mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V,$$
i.e., the "natural variables" for $U$ are $S$ and $V$ and you have the relations
$$T=\partial_S U(S,V), \quad p=-\partial_V U(S,V).$$
Sometimes one writes
$$T=(\partial_S U)_{V}, \quad p=-(\partial_V U)_S,$$
with the subscripts indicating which variable has to be kept constant when taking the partial derivative with respect to the other independent variable. It's clear that the internal energy is particularly convenient to use for isochoric processes, i.e., when $\mathrm{d} V=0$.

The enthalpy is defined as the potential with "natural variables" $S$ and $p$. It's given by the corresponding Legendre-transform of $U$, i.e.,
$$H=U+p V.$$
Indeed, then one finds
$$\mathrm{d} H=\mathrm{d} U + V \mathrm{d} p + p \mathrm{d} V=T \mathrm{d} S+ V \mathrm{d} p.$$
Now indeed the "natural independent variables" for $H$ are $S$ and $p$, $H=H(p,V)$, and you have the relations
$$T=(\partial_S H)_{p}, \quad V=(\partial_p H)_{S}.$$
It's also clear that $H$ is convenient for isobaric processes, i.e., for $\mathrm{d} p=0$, and that's why it's particularly often used in chemistry when considering reactions at constant (lab) pressure.

 

[Chestermiller]

Enthalpy is just a very convenient property to work with in analyzing many kinds of thermodynamic systems. As you get more experience, you will see how it comes into play. Could we get along without enthalpy and just work entirely with internal energy and entropy in analyzing thermodynamic systems? Yes. But, as I said, working with enthalpy is very frequently more concise and convenient.

 

[Sailor Al]

Steam tables even give you tabulations of internal energy, enthalpy and entropy for saturated liquid, saturated vapour and the change form one to the other.
http://homepages.wmich.edu/~cho/ME432/Appendix1Udated_metric.pdf
Table A-4, A-5
and you can study and play around with the numbers to get a better feel of enthalpy.

I'm still confused. Those Steam TablesA-4 and A-5 quote enthalpy as kJ/kg. Isn't Enthalpy measured in Joules (or kilo Joules), not Joules/kg?

 

[Chestermiller]

I'm still confused. Those Steam TablesA-4 and A-5 quote enthalpy as kJ/kg. Isn't Enthalpy measured in Joules (or kilo Joules), not Joules/kg?

Enthalpy is an extensive property, so if you divide it by the mass of the material you have, you get the enthalpy per unit mass. Since the steam tables don't know how much mass you have, they give the enthalpy per unit mass.. You just multiply this by the amount of mass of material you have to get the enthalpy.

 

[Sailor Al]

Enthalpy is an extensive property, so if you divide it by the mass of the material you have, you get the enthalpy per unit mass. Since the steam tables don't know how much mass you have, they give the enthalpy per unit mass.. You just multiply this by the amount of mass of material you have to get the enthalpy.

Still trying to get my head around enthalpy. In what way is it relevant to the behaviour of an ideal gas? Can you give a simple example please?

 

[Sailor Al]

I should have been more specific: Is the consideration of enthalpy relevant in the study of aerodynamics of aircraft wings and yacht sails, i.e. in (dry) air at wind speeds of 0 to 300 kts, in temps of 0 to 25°C in the normal atmosphere below say 10,000 ft? Is enthalpy more relevant to exo/endo-thermic reactions?

 

[Chestermiller]

I should have been more specific: Is the consideration of enthalpy relevant in the study of aerodynamics of aircraft wings and yacht sails, i.e. in (dry) air at wind speeds of 0 to 300 kts, in temps of 0 to 25°C in the normal atmosphere below say 10,000 ft?

Yes.

Is enthalpy more relevant to exo/endo-thermic reactions?

More relevant than what?

You seem to be coming at this at a very elementary level. Please summarize your background in Thermodynamics study so that we can tailor our answer your questions at the appropriate level.

 

[Sailor Al]

Yes.

More relevant than what?

You seem to be coming at this at a very elementary level. Please summarize your background in Thermodynamics study so that we can tailor our answer your questions at the appropriate level.

I have a degree in mathematical physics Birmingham UK 1967, and failed thermodynamics! I have re-connected with the topic over the past two years in an attempt, as a racing yachtsman, to understand the source of the aerodynamic force from a sail.
I have done a lot of reading, including Marchaj, Fossati, Gentry and I have studied Eiffel's research report in his book. I am now three weeks into reading Anderson to see if he can provide some answers and have come unstuck at Section 7.2 when for some unexplained reason, he introduces:
"A related quantity is the specific enthalpy, denoted by h and defined as h = e + pv (7.3)"
I have returned to my Physics text (Starling and Woodall) to find enthalpy introduced in Ch 13 "Equations of state" and read that it's involved in the work separating molecules (p284).
Now the aerodynamics of sails is not about state changes: Air is a simple and pretty ideal gas in all stages of sailing (as it also is in domestic flying). No state changes, no molecular attraction to overcome, so I'm confused.
Maybe you can help?
[EDIT] Why is it important? Because I'm trying to understand the "air can be considered incompressible below Mach 0.3" assumption which he "proves" in an incredibly long and complex argument which, inter alia, includes, and thus relies upon, the differentiation of the enthalpy equation at (7.19). [/EDIT]

 

[vanhees71]

A very concise book at an introductory level is

R. Becker, Theory of Heat, Springer (1955)

As a mathematical physicist it should be easy for you to understand that behind all the thermodynamic potentials are just Legendre transformations to get from the internal energy, $U(S,V,N)$ to other equivalent descriptions of the system with other "natural independent variables". It's all derived from the differential
$$\mathrm{d} U=T \mathrm{d} S -P \mathrm{d} V+\mu \mathrm{d}N.$$
Here $T$ are the temperature, $S$ the entropy, $P$ the pressure, $V$ the volume, $\mu$ the chemical potential (for the conserved particle number or charge $N$).

To get a potential with "natural independent variables" $S$, $P$, and $N$ you do the Legendre transformation to the enthalpy,
$$H=U+p V,$$
which indeed leads to
$$\mathrm{d} H = \mathrm{d} U + P \mathrm{d} V + V \mathrm{d} P = T \mathrm{d} S + V \mathrm{d} P + \mu \mathrm{d} N.$$

 

[Chestermiller]

As long as the temperature of the air does not change much due to air compression and viscous frictional heating at the surface of the wing or sail, you can treat the flow as isothermal, and you don't need to address enthalpy. Enthalpy comes in when you need to consider the occurrence of temperature changes (like in re-entry of a space vehicle). So in your cases, you can use the isothermal flow differential equations of fluid mechanics.

 

[Steve4Physics]

Still trying to get my head around enthalpy. In what way is it relevant to the behaviour of an ideal gas? Can you give a simple example please?

At the risk of being accused of being too simplistic in this erudite environment, try this...

Suppose you have a sample of an ideal gas at absolute zero.

You want to raise its temperature to some value T - and introduce this heated gas into a constant-pressure environment with pressure P.

To do this, you have to supply two lots of energy:

1) the internal energy (U) to raise the temperature from 0 to T; this energy is the kinetic energy of the (ideal) gas particles at temperature T;

2) sufficient energy to push the constant-pressure environment outwards (because the gas volume changes from zero to some value V = nRT/P); this requires an amount of work W = PΔV = P(V-0) = PV.

The total energy is the enthalpy: H = U + PV

Changes in the enthalpy during a process reflect not only changes to internal energy but also work done on/by the environment.

[Minor edit made for clarity.]

 

[Sailor Al]

As long as the temperature of the air does not change much due to air compression and viscous frictional heating at the surface of the wing or sail, you can treat the flow as isothermal, and you don't need to address enthalpy. Enthalpy comes in when you need to consider the occurrence of temperature changes (like in re-entry of a space vehicle). So in your cases, you can use the isothermal flow differential equations of fluid mechanics.

But the temperature of the air will change due to compression - it doesn't matter if the change large or small, the process (air over a wing or sail) is not isothermal.
I don't want to use isothermal equations.

The aerodynamic process is not exothermic or endothermic and there is no change of state (or phase), between gas, liquid or solid, so what's enthalpy doing in the equations?

I want to understand why Anderson introduced enthalpy.

 

[Sailor Al]

A very concise book at an introductory level is

R. Becker, Theory of Heat, Springer (1955)

As a mathematical physicist it should be easy for you to understand that behind all the thermodynamic potentials are just Legendre transformations to get from the internal energy, $U(S,V,N)$ to other equivalent descriptions of the system with other "natural independent variables". It's all derived from the differential
$$\mathrm{d} U=T \mathrm{d} S -P \mathrm{d} V+\mu \mathrm{d}N.$$
Here $T$ are the temperature, $S$ the entropy, $P$ the pressure, $V$ the volume, $\mu$ the chemical potential (for the conserved particle number or charge $N$).

To get a potential with "natural independent variables" $S$, $P$, and $N$ you do the Legendre transformation to the enthalpy,
$$H=U+p V,$$
which indeed leads to
$$\mathrm{d} H = \mathrm{d} U + P \mathrm{d} V + V \mathrm{d} P = T \mathrm{d} S + V \mathrm{d} P + \mu \mathrm{d} N.$$

I have a copy of Becker, Theory of Heat.
It references enthalpy on 9 pages of 370 but nowhere does it provide a definition other than the formula: Free Enthalpy G=E -TS+pV which is not explained.
My question is why Anderson would use of enthalpy in the discussion of the aerodynamics of wings and sails.

 

[Sailor Al]

At the risk of being accused of being too simplistic in this erudite environment, try this...

Suppose you have a sample of an ideal gas at absolute zero.

You want to raise its temperature to some value T - and introduce this heated gas into a constant-pressure environment with pressure P.

To do this, you have to supply two lots of energy:

1) the internal energy (U) to raise the temperature from 0 to T; this energy is the kinetic energy of the (ideal) gas particles at temperature T;

2) sufficient energy to push the constant-pressure environment outwards (because the gas volume changes from zero to some value V = nRT/P); this requires an amount of work W = PΔV = P(V-0) = PV.

The total energy is the enthalpy: H = U + PV

Changes in the enthalpy during a process reflect not only changes to internal energy but also work done on/by the environment.

[Minor edit made for clarity.]

No, please don't risk being too simplistic. That's what has led me to this point, but simplicity has to be logically valid.
You can't start even a simplistic explanation with "Suppose you have a sample of an ideal gas at T=0 (i.e. absolute zero)" because at absolute zero theoretically a gas has a volume of zero (in practice of course it would be a solid)
So, I'm afraid your explanation doesn't help answer my question.

 

[Steve4Physics]

You can't start even a simplistic explanation with "Suppose you have a sample of an ideal gas at T=0 (i.e. absolute zero)"

I would argue that for a classical ideal gas, it is perfectly reasonable to consider the limiting state where the temperature is zero. The corresponding volume will, of course, also be zero because an ideal gas consists of point particles (each having zero volume).

But if preferred, you could change the argument in Post #20 so that the initial temperature and volume are arbitrarily small - e.g. $T_i = 10^{-20} K$ and $V_i =10^{-20}m^3$. It doesn't change the explanation.

[Edited]

 

[Sailor Al]

I would argue that for a classical ideal gas, it is perfectly reasonable to consider the limiting state where the temperature is zero. The corresponding volume will, of course, also be zero because an ideal gas consists of point particles (each having zero volume).

But if preferred, you could change the argument in Post #20 so that the initial temperature and volume are arbitrarily small - e.g. $T_i = 10^{-20} K$ and $V_i =10^{-20}m^3$. It doesn't change the explanation.Formation of the solid state requires inter-particle forces. But an ideal gas has no inter-particle forces. But for a real gas, you are correct.

OK, so you get a value for enthalpy, but what do you do with it?
If you add heat to a gas you will increase its enthalpy in both of its terms. U will increase and so will PV will increase.
I can find no "rules" for its use. No-one is saying there is a law of conservation of enthalpy.
What use is it in aerodynamics?

 

[Chestermiller]

But the temperature of the air will change due to compression - it doesn't matter if the change large or small, the process (air over a wing or sail) is not isothermal.
I don't want to use isothermal equations.

The change will be very small, and you'll be killing yourself to include this for no reason. In any event, you can treat the flow as isothermal and then confirm the small temperature rise (and its predicted small effects) a'postiori.

The aerodynamic process is not exothermic or endothermic and there is no change of state (or phase), between gas, liquid or solid, so what's enthalpy doing in the equations?

I want to understand why Anderson introduced enthalpy.

Enthalpy (per unit mass) is just a very convenient function of temperature, equal to the internal energy per unit mass plus pressure times specific volume. This combination occurs frequently in thermodynamic situations, particularly those involving steady flow. It is present in the energy balance equation (say for steady flow over a wing or sail) to enable you to calculate the temperature distribution. Of course, we don't need to work in terms of enthalpy, and can work in terms of internal energy instead, but, like I said, the combination u + pv frequently occurs in the flow equations and is easier to work with.

The problems come in when we start to try to assign physical interpretation to enthalpy. This invariably causes confusion to beginners for no reason. It is better just to consider enthalpy as a convenient parameter to work with.

 

[Sailor Al]

Enthalpy (per unit mass) is just a very convenient function of temperature, equal to the internal energy per unit mass plus pressure times specific volume. This combination occurs frequently in thermodynamic situations, particularly those involving steady flow. It is present in the energy balance equation (say for steady flow over a wing or sail) to enable you to calculate the temperature distribution. Of course, we don't need to work in terms of enthalpy, and can work in terms of internal energy instead, but, like I said, the combination u + pv frequently occurs in the flow equations and is easier to work with.

The problems come in when we start to try to assign physical interpretation to enthalpy. This invariably causes confusion to beginners for no reason. It is better just to consider enthalpy as a convenient parameter to work with.

You say it is a very convenient function...but convenient for who? I understand the equation H = U +PV, but what use is it? If heat is applied, or work added (i.e. compression) to a quantity of (ideal) gas, there's no phase change, no exo/endothermic reaction, the value of H increase in both of the terms. There's no law of conservation of enthalpy.
Why does enthalpy appear in aerodynamics of sails or wings at subsonic, atmospheric conditions?

 

[Chestermiller]

You say it is a very convenient function...but convenient for who? I understand the equation H = U +PV, but what use is it? If heat is applied, or work added (i.e. compression) to a quantity of (ideal) gas, there's no phase change, no exo/endothermic reaction, the value of H increase in both of the terms. There's no law of conservation of enthalpy.

Incorrect. What you are saying is that you have no idea how the enthalpy function is applied in practice to solve for the temperature distribution in a fluid flow (and its feedback effect on the flow). So you have some learning to do.

The first step is to get an understanding of how enthalpy comes into play in the derivation of the open system version of the first law of thermodynamics, which considers fluid flow into- and out of a system, in addition to heat and work. You can learn about this in Fundamentals of Engineering Thermodynamics by Moran et al. (Chapter 4) The enthalpy function enters naturally in this derivation in place of internal energy for a closed system.

Your next step should be to read Chapter 11 of Transport Phenomena by Bird, Stewart, and Lightfoot : The Equations of Change for non-Isothermal Systems. This looks at situations where the temperature and other parameters are changing both with spatial position and time using partial differential equations. See in particular Table 11.4 which gives the energy balance equation in terms of enthalpy as well as in terms of CpdT, which is the differential of enthalpy.

Why does enthalpy appear in aerodynamics of sails or wings at subsonic, atmospheric conditions?

After completing the studies I just recommended, you will realize the answer to this question.

 

[Sailor Al]

Incorrect.

That's a bit blunt!
Could you expand a little please as to which of those statements is incorrect?
1) H = U +PV
2) If heat is applied, or work added (i.e. compression) to a quantity of (ideal) gas, there's no phase change, no exo/endothermic reaction,
3) the value of H increase in both of the terms.
4)There's no law of conservation of enthalpy.

 

[Chestermiller]

That's a bit blunt!
Could you expand a little please as to which of those statements is incorrect?

2) If heat is applied, or work added (i.e. compression) to a quantity of (ideal) gas, there's no phase change, no exo/endothermic reaction,

So what. Just adding heat and doing work (including viscous work) causes enthalpy and temperature to change: dH=CpdT

4)There's no law of conservation of enthalpy.

Yes there is. Just master the material that I recommended for you and you will how this equation is derived.

 

[Sailor Al]

Yes there is. Just master the material that I recommended for you and you will how this equation is derived.

I know it's not a perfect tool, but I get exactly zero hits in a Google search for "conservation of enthalpy"

 

[Chestermiller]

I know it's not a perfect tool, but I get exactly zero hits in a Google search for "conservation of enthalpy"

Look. I gave you the exact materials you need to learn to accomplish what you want. Do you have the determination to learn these materials or not? I'm not going to spoon feed this to you. You need to study and do lots of problems. I'm sorry if I'm being blunt. I'm giving you this advice based on over 50 years of engineering experience.

 

[Sailor Al]

My question was pretty simple: How is the consideration of enthalpy relevant in the study of aerodynamics of aircraft wings and yacht sails, i.e. in (dry) air at wind speeds of 0 to 300 kts, in temps of 0 to 25°C in the normal atmosphere below say 10,000 ft?
Aerodynamics is about the understanding of the aerodynamic force that results in lift and drag over a wing.
I have asked why H = U + PV has any bearing on this study?
You are saying "useful to solve for the temperature distribution in a fluid flow (and its feedback effect on the flow)." but that doesn't seem to have anything to do with the subject at hand - especially when in a separate post you are saying that " you can treat the flow as isothermal".
I'm not being deliberately obtuse, and I am reading most of the material being offered, but none of it explains my problem.

 

[Chestermiller]

My question was pretty simple: How is the consideration of enthalpy relevant in the study of aerodynamics of aircraft wings and yacht sails, i.e. in (dry) air at wind speeds of 0 to 300 kts, in temps of 0 to 25°C in the normal atmosphere below say 10,000 ft?

My answer was that I don't think it is relevant, and that you can solve this isothermally..

Aerodynamics is about the understanding of the aerodynamic force that results in lift and drag over a wing.
I have asked why H = U + PV has any bearing on this study?

In your situation, temperature can have a negligible effect of the forces, but if you want to include this effect, you also need to include a heat balance equation (based on enthalpy) and the effect of temperature on the fluid mechanics equations. Have you had a course in fluid dynamics, including the Navier Stokes equations?

You are saying "useful to solve for the temperature distribution in a fluid flow (and its feedback effect on the flow)." but that doesn't seem to have anything to do with the subject at hand - especially when in a separate post you are saying that " you can treat the flow as isothermal".

If the flow is isothermal (negligible effect of temperature), then you don't need to solve for the temperature distribution. You just solve the isothermal fluid dynamics equations. You can then substitute the calculated flow velocity distribution into the enthalpy balance equation to check to see whether the calculated temperature distribution is indeed negligible.

I'm not being deliberately obtuse, and I am reading most of the material being offered, but none of it explains my problem.

In my judgment, you're "in over your head."

 

[Sailor Al]

My answer was that I don't think it is relevant, and that you can solve this isothermally..

If the flow is isothermal (negligible effect of temperature), then you don't need to solve for the temperature distribution. You just solve the isothermal fluid dynamics equations. You can then substitute the calculated flow velocity distribution into the enthalpy balance equation to check to see whether the calculated temperature distribution is indeed negligible.

In my judgment, you're "in over your head."

If you cast your mind back to the reason I asked the question in post #17, you will note I said:
"I have done a lot of reading, including Marchaj, Fossati, Gentry and I have studied Eiffel's research report in his book. I am now three weeks into reading Anderson to see if he can provide some answers and have come unstuck at Section 7.2 when for some unexplained reason, he introduces:
"A related quantity is the specific enthalpy, denoted by h and defined as h = e + pv (7.3)""
And I asked "why it is important?"
I 100% concur with your answer " I don't think it is relevant", and is 100% correct. But now I have to ask why the heck did Anderson include it in his book entitled "Fundamentals of Aerodynamics"?
Also as I said in #17 "Because I'm trying to understand the "air can be considered incompressible below Mach 0.3" assumption which he "proves" in an incredibly long and complex argument which, inter alia, includes, and thus relies upon, the differentiation of the enthalpy equation at (7.19)"
If the introduction of enthalpy is not relevant, but is a part of the justification of the <M0.3 argument, doesn't that invalidate the whole argument?
I am ensuring that I don't go "in over my head" by working painstakingly through Anderson's line of reasoning.
Your response clearly confirms my conclusion that his reasoning is deeply flawed.
I know I'm not ready to "solve the isothermal fluid dynamics equations", and before I do I want to have some confidence I need to. If Anderson's reasoning is so flawed, and he's one of the top dudes in the field, then I'm pretty sure I can save myself a lot of time and frustration by staying well away.

 

[Chestermiller]

Please consider delving into the references I recommended.

As far as the books you have been studying, if they have problems at the end of the chapters, please consider solving a major fraction of these problems. Reading is good, but only solving problems can solidify your understanding and give you a working knowledge of the subject.

 

[Sailor Al]

Please consider delving into the references I recommended.

As far as the books you have been studying, if they have problems at the end of the chapters, please consider solving a major fraction of these problems. Reading is good, but only solving problems can solidify your understanding and give you a working knowledge of the subject.

But if the references are explaining enthalpy, and enthalpy is not relevant to aerodynamics, what's the point of that if my study is aerodynamics. Don't you see the contradiction?
I have my answer, you have confirmed enthalpy is irrelevant to aerodynamics.

 

[Motore]

I have my answer, you have confirmed enthalpy is irrelevant to aerodynamics.

Read post #19!

 

[Sailor Al]

Read post #19!

Yes, I know, it is a conundrum. But when I press him for why he qualifies his answer with "As long as the temperature of the air does not change much" he ducks for the undergrowth .
What is your view: is enthalpy useful in describing the process in #17?

 

[Motore]

What is your view: is enthalpy useful in describing the process in #17?

In that specific scenario in my opinion, it complicate things because the change in temperature is small, but it is useful if you want a precise way to compute things. But you said it's irrelevant to aerodynamics which is not the case as people solving aerodynamics questions use it often.
https://www.grc.nasa.gov/www/k-12/airplane/enthalpy.html

 

[Sailor Al]

Thanks for that link.
It relates to a very special situation where heat is added to a closed system in which pressure remains constant.
But in aerodynamics heat is never added to the system, and I don't think you can remove heat from a system without changing its pressure.

I'm not sure what it has to do with aerodynamics.

 

[vanhees71]

Particularly in fluid dynamics enthalpy plays an important role. In ideal fluid dynamics (adiabatic motion) it's the right thermodynamic potential to use. Euler's equation reads
$$\rho \mathrm{D}_t \vec{v}=\rho (\partial_t \vec{v} + (\vec{v} \cdot \vec{\nabla}) \vec{v})=-\vec{\nabla} P.$$
With $h$ the enthalpy per unit mass of the fluid and thus
$$\mathrm{d}h=T \mathrm{d} S + \frac{1}{\rho} \mathrm{d} P$$
you get, for adiabatic motion, $\vec{\nabla} P=\rho \vec{\nabla} h$ and thus
$$\mathrm{D}_t \vec{v}=-\nabla h.$$
Another very valuable source is vol. 6 of Landau and Lifshitz (fluid dynamics).

 

[Sailor Al]

Particularly in fluid dynamics enthalpy plays an important role.

But I'm not necessarily interested in fluid dynamics.
I'm interested in the aerodynamics of yachts' sails. Where is enthalpy useful in the aerodynamics of sails or wings?

 

[Motore]

But I'm not necessarily interested in fluid dynamics.
I'm interested in the aerodynamics of yachts' sails.

Wikipedia:
In physics and engineering, fluid dynamics is a subdiscipline of fluid mechanics that describes the flow of fluids—liquids and gases. It has several subdisciplines, including aerodynamics (the study of air and other gases in motion) and hydrodynamics (the study of liquids in motion).

 

[vanhees71]

But I'm not necessarily interested in fluid dynamics.
I'm interested in the aerodynamics of yachts' sails. Where is enthalpy useful in the aerodynamics of sails or wings?

Aerodynamics is a subfield of fluid dynamics.

 

[Chestermiller]

But if the references are explaining enthalpy, and enthalpy is not relevant to aerodynamics, what's the point of that if my study is aerodynamics. Don't you see the contradiction?
I have my answer, you have confirmed enthalpy is irrelevant to aerodynamics.

It is irrelevant only if you are assuming that the flow is isothermal (i.e., non-isothermal effects are negligible).

 

[Chestermiller]

But I'm not necessarily interested in fluid dynamics.
I'm interested in the aerodynamics of yachts' sails. Where is enthalpy useful in the aerodynamics of sails or wings?

It is useful only when the amount of viscous frictional heating and compressional heating are sufficient (i.e., high Mach numbers) for the temperature to change significantly. For the range of operating conditions you have enumerated, it is not useful.

 

[vanhees71]

Then the free enthalpy (or Gibbs energy), $G=U-ST+PV=H-ST$ is the right potential:
$$\mathrm{d} F=\mathrm{d} U -S \mathrm{d} T -T \mathrm{d} S + P \mathrm{d} V + V \mathrm{d} P = -S \mathrm{d} T + V \mathrm{d} P.$$

 

[Steve4Physics]

I want to understand why Anderson introduced enthalpy.

You could read-ahead through the text in Anderson’s book after he defines enthalpy - to see how he then uses enthalpy. Or have you already done that and found it is not used?

 

[Sailor Al]

You could read-ahead through the text in Anderson’s book after he defines enthalpy - to see how he then uses enthalpy. Or have you already done that and found it is not used?

Ha! Yes, and that is the nub.
I have read through the text, and he does use it, specifically Section 7.4.2 where he uses it to derive (7.19)

Then via a very long and complex route* he "proves" that the errors in using air as compressible below M 0.3 are negligible.

*I'm working on a way to show the argument, but it needs work:

Here's a link to the analysis, please feel free to comment.