Do I use Enthalpy or Internal Energy (or flip a coin)?

[jpblomberg]

Hello all,

When I am solving a thermodynamics (energy of compressed gas/liquid) problem, which cases should I use Enthalpy and which cases for Internal Energy? It has been 11 years since I took thermodynamics and I don't remember which one should be used.

Things I do remember: The Internal Energy of a substance includes the rotational and translation energy of the atoms/molecules that make up the substance. Enthalpy includes the Internal Energy + (Pressure * Volume). The units for both of these are the same, energy per mass.

I am not clear as to why some calculations use Internal Energy and some use Enthalpy. When is Internal Energy used and when is Enthalpy used?



Credo ut Intelligam -St. Augustine

 

[iainuts]

I think the conventional way of writing the first law is perhaps the biggest confusion when trying to figure out which to use. Rather than use what's normally found in text books, I've found that expanding the first law slightly and applying it to a control volume helps a lot.

When doing a thermo problem, I'd suggest starting by drawing a control volume around that portion of the system you're doing the analysis on, then applying the following form of the first law*:

dU = Hin - Hout + Qin - Qout + Win - Wout

Note that this follows the convention of energy into a control volume being positive (adding energy to the CV) and energy leaving a control volume being negative (subtracting energy from the CV).

When drawing your control surface around the volume, you need to identify what is entering or exiting this volume. Any fluid entering or leaving brings with it or takes away some enthalpy depending on it's temperature and pressure. Note that the enthalpy, just like internal energy, is simply a state defined by the fluid's temperature and pressure or other information you have on it. Typically, two bits of info are needed. For example, a fluid may be two phase, so in that case knowing the pressure and quality (sometimes called vapor fraction) of the fluid gives you the enthalpy.

Similarly, heat and work may cross the control surface also. The equation above simply shows that the sum total of all those forms of energy equate to the difference in the control volume's internal energy.

One could for example, draw a control volume around a valve. If we assume the valve is well insulated (no heat transfer) and since there is no work done by or on the valve, then if the control volume is under steady state conditions, the internal energy does not change either. In this very simple case, we find the enthalpy into the valve is equal to the enthalpy out of the valve (Hin = Hout).

Another example would be to draw a control surface around a constant volume cylinder being pressurized with a gas. Here we find that dU = Hin. This helps explain why cylinders warm up as they are being filled, thus there may also be heat flow out of the cylinder, so dU = Hin - Qout.

For a compressor or pump for example, we might draw a control volume around it and we'd find similarly that dU = 0 (no energy stored inside the CV), and Hout = Hin + Win (assuming other values are zero). If the compressor was rejecting heat, we would have Hout = Hin + Win - Qout.

*Note that potential and kinetic energy changes are neglected in this equation.

PS: If in doubt about what to use, flipping a coin always works too.

 

[UmeshMathur]

In general, most process calculations are done using the enthalpy H rather than the internal energy U. Therefore, all commercial process simulators report stream enthalpies rather than internal energies.

All commercial modeling for unit operations is based on use of enthalpy, e.g., the steady or unsteady flow energy equation (1st law of thermodynamics) or for systems involving chemical reactions or phase equilibrium. This seems to be a standard "convention" that I have never seen violated. Of course, one could do everything with internal energy, but this would create an incompatibility with everybody else's numbers and, further, add zero additional value.

 

[gtsim]

I also use internal energy for non-steady flow processes (e.g. internal combustion engine modelling) and enthapy for steady flow processes (e.g. modelling gas and steam turbines).

 

[thermcool]

Change in the energy contents of the system undergoing some process at constant volume is change in internal energy
In this case, the system volume is constant and any change in the system energy will directly reflect the change in the internal energy.

In the other case, if the pressure is kepy constant during the process, the system may do some work also. Therefore the energy contents of the system is not the reflection of the changes in the internal energy but it is the sum of the internal energy plus the work done by or on the system.

Most of the heat transfer process in practical life takes place at constant pressure (which is atmospheric pressure) so we make use of the enthalpy.

remember
Energy contents at constant volume = Internal energy
Energy contents at constant pressure = Enthalpy = Internal energy plus work done by (or on) the system.

 

[UmeshMathur]

thermocool:

I regret having to adopt a negative tone in this response to your post as it seems to me to disseminate some erroneous concepts.

In the vast majority of industrial processes, internal energy changes even when volume is not kept constant, just as enthalpy changes when pressure is not constant. Examples are all compressors, turbines, etc. that handle compressible fluids and either deliver or consume any significant amount of power. Simply heating a fluid 10 degrees C in a heat exchanger (i.e., with neither T nor P held constant) will certainly increase BOTH the U and H unless you are determined to construct a pathological case for the sake of argument.

Also, I disagree with the blanket assertion that "Most of the heat transfer process in practical life takes place at constant pressure (which is atmospheric pressure) so we make use of the enthalpy", as this is patently untrue for virtually all chemical processes. There are excellent thermodynamic reasons to choose enthalpy over internal energy for flow processes (steady or unsteady flow), with or without chemical reaction, that have nothing to do with pressure remaining constant.

Finally, both points at the end of your post are at best incomplete and quite misleading. The mere fact that U and H can be calculated using Cv and Cp, respectively (with appropriate corrections for departure from ideality), has ABSOLUTELY NOTHING to do with volume or pressure being constant in ANY process, nor with whether or not work was done on or by the system.

 

[thermcool]

UmeshMathur,

Aim of putting that general statement was to give an idea of the enthalpy and internal energy, and how are they compared to each other.

Wheneven you can choose enthaply over internal energy or vice versa, the distinction is made by the choice of the control volume. this is why you may use internal energy or enthalpy in the systems like compressors or turbines. Whenever you make a choice, it is with respect to the control volume. Open system, close system are all relative terms that decide the choice of energy parameters.

For example, When the change in internal energy is accompanied by the expansion or contraction of the gases or vapour produced during phase change process, this represents the change in enthalpy.

Obviously , change in enthalpy reflect the change in the internal energy of the system plus the useful work done by the system. it think it is very much clear from the equation i gave before. If the system is not allowed to do some useful work, then the mere change in the energy is only change in the internal energy. Also, change in enthalpy is reflected when the energy crosses system boundary which may be due to expansion or flow of heat which is ultimately converted to some useful work.

 

[UmeshMathur]

You can always compute U from H and vice versa, at any point in any system, from the definition: H = U + PV

Both are fundamental properties of any substance at a given condition. Which one you use is up to you, really, and is basically a matter of convention. However, you would be foolhardy to insist on using U for flow processes, as work done here is generally related to enthalpy far more simply than to internal energy. Similarly, for closed systems (as gtsim has noted) you are better off using internal energy to do the work/energy accounting.

This material is so elementary (covered in the first few lectures of an introductory thermo course) that I feel reluctant to pursue it any further.

 

[mechprocess]

Use internal energy for closed systems becuase you have no flow work. The energy equation is derived from (u+pv)*w (mass flow) rate excluding specific kenetic and potential energies and work and heat addtion.

I assume you are performing an energy balance on a fluid with known physical properties. If not, then you can measure flow, temperature and pressure drop at upstream and downstream conditions. Remember that the pressuer energy equation is derived from the energy equation.

w(u1 + p1v1 + V1^2/2gc + (g/gc)Z1) = w( u2+ p2v2 + V2^2/2gc + (g/gc)Z2) - W + Q

and has units of ft-lbf/sec

This turns into the pressure energy equation by dividing by the mass flow rate w so that the units are ft-lbf/lbm

Note that (u2 - u1) = your losses and can be defined by the Darcy Equation = (fLV^2)/2dgc again with units of ft-lbf/lbm

Don't forget to include heat Q/w if your process is gaining or losing heat.

You can derive the Bernoulli Equation this way.