What is Enthalpy and How is it Used in Thermodynamics?

[clevermetal]

Hi, new forumer here.

I've found this site a couple times before from google searches and whatnot but I've always been too scared to post here as everyone seems much too intelligent for my questions.

However I can't find a decent answer to this question, perhaps there isn't one, however I remain hopeful! Anyway, enough chitchat.

What is enthalpy?
I get that h = u + Pv
but when you're calculating heat added in a boiler or work done in a turbine it's always H2-H1 which leads me to believe it's the internal energy of the gas/vapour. But then that's U. So I'm confused as to what enthalpy actually IS

Cheers
please don't flame too hard, I'm only a second year undergraduate!

edit: my english skill is...well... lacking.

 

[Acut]

You may view the difference in enthalpy as the heat liberated by a process that happens under constant pressure.

Some non-rigorous high-school (and even college-level!) textbooks say that enthalpy is energy, which I find awkwardly confusing and unnecessary for the student. When I first studied it, I had precisely the same doubt you had. U is the real energy. H is just a parameter which happens to be easier to use in many applications - because many real-world processes happen at relatively constant pressure (1 atm).

Now, notice that only changes in H (H2 - H1) are meaningful here. Unlike real energy (U), absolute values of enthalpy (H) have no real or intuitive meaning. In fact, many chemistry textbooks arbitrarily (well, not so arbitrarily) choose to set the enthalpy of certain substances at certain conditions to zero to make calculations easier.

 

[clevermetal]

You may view the difference in enthalpy as the heat liberated by a process that happens under constant pressure.

Ahhh that's a much easier definition to understand. So it really has no relevance as an absolute quantity. Cheers for that.

However now another question springs to mind: if it's modelling a constant pressure process then why can we use it to find the work done in a turbine in which there is a significant pressure decrease?

 

[Studiot]

Hello clevermetal and welcome to Physics Forums.

As preveiously stated enthalpy is much used by engineers to make calculations easier, but there you can go further than the forumula you found.

Enthapy (H) is also called Total Heat or Heat Content and is introduced by the sum

H = U +pv

It can be directly measured by calorimetric methods if some state is arbitrarily chosen as zero.

The term pv is the work done in expanding the substance from zero to volume v at constant pressure p or alternatively the work done in pumping a fluid into an enclosure (constV) at a steady pressure.

H = U + pv

dH = dU + pdv + vdp

But dU + pdv = δq (First Law)

so

dH = δq + vdp

The condition for adiabatic expansion is that δq = 0

Thus for adiabatic expansion

dH = vdp

The total heat for a finite adiabatic expansion is the area under a pv graph and is obtained by integration

H_a - H_b = ∫vdp = area ABCD in my sketch

Engineers call this quantity the adiabatic heat drop.

This is the turbine work you were asking about.

The next step (again by engineers) is to extend this to the equation for steady flow of fluids.

 

[Darwin123]

Hi, new forumer here.

I've found this site a couple times before from google searches and whatnot but I've always been too scared to post here as everyone seems much too intelligent for my questions.

However I can't find a decent answer to this question, perhaps there isn't one, however I remain hopeful! Anyway, enough chitchat.

What is enthalpy?
I get that h = u + Pv
but when you're calculating heat added in a boiler or work done in a turbine it's always H2-H1 which leads me to believe it's the internal energy of the gas/vapour. But then that's U. So I'm confused as to what enthalpy actually IS

The change in enthalpy of a system is the heat energy that is either transferred to the system or created under conditions of constant pressure.
The implicit assumption in your examples is that the external pressure of the turbine and boiler are both constant in time. Thus, the heat energy that is given off by the system in both cases is H2-H1. So under conditions of constant external pressure,
dH=TdS,
where H is the enthalpy, T is the temperature and S is the entropy.
The phrase "heat energy" has an language ambiguity that most teachers don't address. There is no specific type of energy that can be called heat energy. "Heat energy" can be optical, infrared, vibrational or kinetic. So by definition, the change in "heat energy" is the temperature times the change in entropy.
The language ambiguity comes about because there are two components to the change in entropy. I get around this ambiguity by imagining that the entropy is like an indestructible gas, while temperature is the pressure of that gas. You can handle it any way you like.
There are two components in the change of entropy. There is entropy that is being transferred and the entropy that is being created. I like to write this as,
dS=dS_Rev+dS_Irrev,
where dS_Rev is the entropy being transferred and dS_Irrev is the entropy that is created. The transfer of entropy only occurs from high temperature to low temperature.
Consider a boiler and a turbine that are under constant pressure but adjusted for optimal efficiency. Let there be an increase of enthalpy of "H2-H1" for each.
A boiler does not have moving parts, so the frictional forces are zero. Furthermore, at equilibrium the temperature of the boiler is the same as the temperature of the heat source.
dS_Irrev=0,
dS_Rev>0,
H2-H1=TdS_Rev
Consider a turbine that is doing work but has friction.The turbine has moving parts, so friction is significant. In fact, all the work of the turbine is being used somewhere else. The energy transferred outside the turbine by work is completely used up by some form of friction. Therefore,
dS_Rev=0,
dS_Irrev>0,
H2-H1=TdS_Irrev
In the turbine case, dS_Irrev may be proportional to the amount of work the turbine is doing. If the turbine weren't doing work, there would be no friction.

 

[clevermetal]

Cheers for the responses! It's much clearer now.

 

[Couchyam]

Enthalpy can also be used in situations where "pressure" (as it is normally understood) does not appear: for instance, in many substances magnetization changes depending on the applied magnetic field. This creates a magnetic analogue to pressure with a corresponding "enthalpy". The mathematics is the same, but the interpretation is just slightly different.

 

[Darwin123]

Hi, new forumer here.

I've found this site a couple times before from google searches and whatnot but I've always been too scared to post here as everyone seems much too intelligent for my questions.

However I can't find a decent answer to this question, perhaps there isn't one, however I remain hopeful! Anyway, enough chitchat.

What is enthalpy?
I get that h = u + Pv
but when you're calculating heat added in a boiler or work done in a turbine it's always H2-H1 which leads me to believe it's the internal energy of the gas/vapour. But then that's U. So I'm confused as to what enthalpy actually IS

Cheers
please don't flame too hard, I'm only a second year undergraduate!

edit: my english skill is...well... lacking.

The enthalpy of a reaction is the "heat energy" given off by the reaction in a system
subject to an external pressure that is constant in time. When the pressure on a system is constant,
dH=TdS,
where dH is the change in enthalpy, dS is the change in entropy, and T is the temperature. When the volume of a system is constant,
dU=T dS,
where dU is the change in internal energy.
The tricky part of this description is the word "heat". The word "heat" sometimes refers to energy transferred by conduction, sometimes refers to entropy transferred by conduction, and sometimes refers to the entropy that is created by an irreversible process.
The heat energy, dQ, of a reaction is,
dQ=T dS,
where
dQ is the "heat energy" that is either transferred or created,
dS is the change in entropy,
T is the temperature of the system.
The problem is that Q is ambiguously defined. "Q" varies with the schedule of
heating even in reversible reactions. However, "S" is an equation of state. Therefore, "S" is always unambiguous when the reaction is reversible. Note that entropy can either be transferred or created, it can not be destroyed.

On the more practical side, the heat energy in a bomb calorimeter is the energy that leaves the container (i.e., the system) by conduction after a reaction has taken place. The heat energy of the bomb calorimeter is always equal to TdS. However, there is a difference in the heat energy depending on whether the contents of the container is kept at a constant temperature or a constant pressure.
The enthalpy (dH) is the heat energy given off by a reaction in a bomb calorimeter when the pressure on the system is kept constant. The internal energy (dU) is the heat given of by a reaction in a bomb calorimeter when the volume of the system is kept constant.
Very often, a system of interest is either kept at constant volume or kept at constant pressure. For most reactions, the two values (dH and dU) are different. However, they (dH and dU) can sometimes be the same.
Be careful around the word "heat" or the phrase "heat energy". They are deceptive. The dictionaries say that "heat" is a form of energy. However, there is no form of energy that can be unambiguously defined as heat. The least ambiguous definition of heat energy is,
dQ=T dS.
Heat is not always infrared energy. Heat is not always the kinetic energy of a molecule. Heat is not always vibrational energy. Heat is always,
dQ=T dS.
Entropy is the well-defined concept. Not heat energy.

 

[Zeppos10]

In the above structure and some specifics are lacking: I will take the response by ACUT to elaborate:

You may view the difference in enthalpy as the heat liberated by a process that happens under constant pressure..

But how do you connect ∆H=Q with H=U+pV ??
we know ∆U = Q + W = Q + W' + W" where W" = useful work = shaft work = technical work. W' is work done against the atmosphere = -p'∆V, if p' is the external=atmospheric pressure. Now define H=U+p'V (!) and the first law becomes: ∆H= Q+W" .
If no useful work is done (W"=0: this not the case for most turbines) it follows: ∆H=Q.
Note how important the 'constant pressure' condition is: it makes p=p', if you didn't do that at in the definition of H. (nb: solids have enthalpy as property, but not an internal pressure).

Some non-rigorous high-school (and even college-level!) textbooks say that enthalpy is energy, which I find awkwardly confusing and unnecessary for the student.

So enthalpy is a form of energy, ie the sum of internal energy U and the energy of displacement p'V. I do not see anything awkward.

Now, notice that only changes in H (H2 - H1) are meaningful here. Unlike real energy (U), absolute values of enthalpy (H) have no real or intuitive meaning. In fact, many chemistry textbooks arbitrarily (well, not so arbitrarily) choose to set the enthalpy of certain substances at certain conditions to zero to make calculations easier.

In ANY balance equation for a conserved quantity ∆X = X(out)-X(in) and this is true whether X is absolute or relative: ie it does not matter.
You can calculate absolute enthalpy values if you want to, but there is no need to do so and the convention is not to do so.

see also: https://www.physicsforums.com/showthread.php?t=338573

 

[Zeppos10]

see also attachment at
https://www.physicsforums.com/showthread.php?t=88987&referrerid=219693