Thus far, we have thought about the internal energy (U) of a substance. Enthalpy is a very close cousin, and enthalpy is used in many engineering calculations, unlike internal energy.

While we have not physically described it, the internal energy is (or is a manifestation of) the molecular energies within a substance. This would include kinetic energy, potential energy, vibrational energy, spin energy and so on.

Enthalpy is internal energy * plus*.

*is what this post is about.*

**Plus what**We recall that for a material, the differential change in a substances internal energy is codified as:

dU = dQ – dW

Where dU is internal energy change, dQ is heat flow into or out of the system, and dW is work done by or to the system. You will note that I am now writing dW as a negative term. Why? Because as heat is added to a material, the material will, if unconstrained, tend to expand and do work on the atmosphere, thereby lessening the internal energy somewhat. This is the typical form of this expression.

We also recall that this expression can be rewritten as:

dU = dQ – p * dV

Rather than try to start by describing what enthalpy * is*, let’s do a calculation and see how it

*from internal energy first.*

**differs**Let us imagine a glass jar holding about 13 cubic feet of air (Why 13 cubic feet? ‘Cause that weighs about 1 pound.) Because the volume of the jar does not change, the work term in our formula goes away and we are left with

dU = dQ

Now, it turns out that the internal energy of a material and it’s temperature are directly related via the specific heat of the material (see earlier post if you want a little background.) Thus, we may rewrite our equation as:

dU = Cv * dT = dQ

Change in internal energy equals specific heat at constant volume times temperature change, which is equal to the energy flow Q. Remember, we are not allowing any work to be done because the volume is fixed.

Well, lets do the math. Under normal temperature and pressure conditions, the * constant volume specific heat* of air is about 0.17 Btu per pound-degree-Rankine. You will recall that for

*changes*in temperature, Fahrenheit degrees may be substituted for Rankine.

Let us imagine we add enough heat to raise the temperature of the air by 50 degrees. The change in internal energy would be:

dU = 1 pound-air * 50 degrees * 0.17 Btu/Lb-degR = 8.5 Btu. So the internal energy has increased by 8.5 Btu.

Now it turns out that when we did this, the enthalpy of the air also changed. The enthalpy is calculated with the * constant pressure specific heat*. In the case of air, Cp is approximately 0.24 Btu per pound-degree-Rankine. Let us designate enthalpy as h. Then:

dh = 1 pound-air * 50 degrees * 0.24 Btu/Lb-degR = 12 Btu. 40% more than the change in internal energy!

So what’s up with enthalpy? And what do we do with it?

Well, the usual description is as follows:

Most thermodynamic processes of interest do not occur in balloons and jars, they occur in systems that have materials flowing through them. For example, a cooling coil has water continuously flowing through it.

Sticking with our coil example, it takes energy (work) to induce flow through a coil. The enthalpy (and Cp) accounts for this flow work.

You may recall that work is a force times a distance, and that this can be algebraically manipulated to become a product of changes to volume and pressure. For the flow work, this work is:

dW = d(pV)

For reasons we won’t get into in this blog (but may be sort of intuitive), this flow work expression can be rewritten as:

dW = p * dV + V * dP

Now, I said above enthalpy and internal energy differ by the flow work, so we can express enthalpy as follows:

dh = dU + p * dV + V * dp.

But wait, dU = dQ – p * dV, so

dh = dQ – p * dV + p * dV + V * dp

or

dh = dQ + V * dp

Compared with internal energy:

dU = dQ – p * dV

We can see by inspection that enthalpy changes will be bigger than internal energy changes when heat is added or removed from a material – just look at the sign of the work term and think about what adding or removing heat will do. In the case of our jar of air, we observe that heating the air included pressure energy that would allow the air to be transported were it in an “open” system. Open systems allow materials to flow through them.

To me, internal energy is a bit more intuitive than the enthalpy, but that may just be me. In any event, enthalpy, not internal energy, is generally used when making calculations.

Note, by the way, that if you have a material such as water sitting in a jar at some pressure p, and you have a hose running with the same pressure p, the temperature of the water in the jar should be higher. The intuitive reasoning is that the flowing water has to expend some energy to move through the hose, so that deducts from the internal energy and, therefore, the temperature.

More formally, we realize that if they are at the same pressure, the internal energy of the water in the jar equals the enthalpy of the flowing water. Therefore, the internal energy must be higher in the jar of water, and therefore the temperature must be higher too.

Now let’s start finishing up.

We have talked around specific heats a little bit, but now we can get a little more detailed.

Cv, specific heat at constant volume, captures the **change in internal energy relative to the change in the material’s temperature**.

Cp, specific heat at constant pressure, describes the * change in enthalpy relative to the change in the material’s temperature*.

Engineering calculations use Cp, which makes sense.

***

Some of these prior posts have been a little heavy for what is intended as a general blog, but the pieces are almost in place for us to really think and wonder about energy in intelligent ways.