2024-06-25T04:40:07.000-0400

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A lot of times, the user is confused while understanding the enthalpy calculation in AMESim. The purpose of this article is to understand the enthalpy calculation in AMESim

The enthalpy (H) is defined as the sum of a system's internal energy (E) and the product of its pressure (P) and volume (V):

**H = E + PV.**

Enthalpy is a useful concept in thermodynamics because it simplifies the description of energy changes in processes occurring at constant pressure, which is a common condition in many practical situations. When a process occurs at constant pressure, the change in enthalpy ΔH is equal to the heat absorbed or released by the system

ΔH = Difference between initial and final enthalpies = Heat Exchanges

Enthalpy changes are important in various applications, such as:

- Chemical reactions: The enthalpy change of a reaction (often referred to as the heat of reaction) indicates whether the reaction is exothermic (releases heat, ΔH<0) or endothermic (absorbs heat, ΔH>0).
- Phase transitions: The enthalpy change associated with phase transitions (such as melting, boiling, and sublimation) provides information about the energy required for these processes.
- Engineering applications: Enthalpy is used in designing and analyzing energy systems like engines, refrigerators, and heat exchangers.

Overall, enthalpy is a central concept in the study of energy transfer and transformation in physical and chemical processes.

However, I**nternal energy is a state function as it depends on the state of the system (temperature, pressure, volume) but not the path taken to reach that state. **However, we cannot define an absolute value for internal energy because there's no reference point for zero internal energy. **The absolute point for internal energy is defined as the zero point of the internal energy scale. In thermodynamics, this zero point is conventionally chosen based on a reference state, and it is not an absolute quantity that can be universally applied to all systems**. The choice of zero point for internal energy is arbitrary, but it provides a reference for measuring changes in internal energy.

In short, we cannot measure the absolute enthalpy. In most of applications, we are interested in finding the change in energy. Hence, technically, relative enthalpy makes sense in most of the applications.

How is Enthalpy calculated in AMESim?

In AMESim, we calculate the change in enthalpy with respect to the reference condition and the **reference condition will be defined in the fluid properties,**

Enthalpy example with the thermal-hydraulic library -

Let us take an example of water. The specific enthalpy of water at different temperatures is

Temperature (DegC |
0.01 |
10 |
20 |
25 |
30 |
40 |
50 |
60 |
70 |
80 |

Specific enthalpy (KJ/Kg) |
0.000612 |
42.021 |
83.914 |
104.83 |
125.73 |
167.53 |
209.34 |
251.18 |
293.07 |
335.01 |

Let us create a simple model to measure the properties of water.

if we plot specific enthalpy, then you will notice that the** specific enthalpy of water is at zero at a temperature of 20 DegC.**

But **as per the above table**, the **Specific enthalpy at this temperature is 83914 J/Kg**. This specific enthalpy may have been calculated at a reference value of 0 DegC.

*In the case of AMESim. We calculate the relative enthalpy over reference enthalpy when temperature changes. In this case, we are maintaining the same temperature as the reference temperature. Since we are using zero specific enthalpy at reference 20 DegC, hence, final results will be zero enthalpy at 20 DegC.*

As per the above table, the enthalpy of water at 20 DegC is 83914 J/Kg). Now, let us set the Specific enthalpy at the reference condition equal to 83914 J/kg.

*With this, if we run the model again, then, it should come up with the value of 83914 J/Kg, which we entered in the fluid properties at reference temperature. please note, we are using the temperature of 20 Degrees in the source.*

Let us increase the temperature from 20 to 30 degrees now.

**Now specific enthalpy is 125615 J/kg. This value is nearly the same as given in the above property table. Hence, this value can be treated as absolute enthalpy, if we define the reference at 0 DegC. Please note, in this case, the addition of 10 Deg results in an increase in enthalpy by 41701 J/kg over the reference enthalpy of 83914 J/Kg. This can be considered as relative enthalpy.**

Hence, in the above case, we can say that specific enthalpy increased by 41701 J/kg due to 10 degree increase in temperature.

Now, let us change the specific enthalpy to its default value of 0.

**In this case, your reference itself is zero, hence, your specific enthalpy values will be equal to the net change in enthalpy due to temperature change.**

This value is exactly the same as we calculated before. Hence, **it can be a good idea to set the zero specific enthalpy at a reference temperature, so that, AMESim can directly produce the enthalpy difference due to changes in temperature with respect to reference temperature.**

We can conclude that enthalpy changes as your reference state changes, but your relative enthalpy will remain the same.

Enthalpy in two-phase flow library -

A similar analogy is applicable to two-phase flow.

Let us calculate the absolute enthalpy first for water. We will try to change the density from 0.01 to 990 kg/m**3 at a constant pressure of 1 barA. Reference conditions can be set in fluid index by

We will compare the absolute enthalpy for the below three option

- default reference state (red curves): we let the "reference state" to "default"
- reference state defined at liquid saturation for a given temperature (blue curves):
- reference temperature 0 degC
- reference enthalpy 200 kJ/kg
- reference entropy 1 kJ/kg/K
- reference state defined by pressure and temperature (yellow curves):
- reference temperature -20 degC
- reference pressure 1 barA
- reference enthalpy 200 kJ/kg
- reference entropy 1 kJ/kg/K

Let us create the below model with three different fluid indexes for water and set the above reference state parameter

Now we can compare the enthalpy at these three different conditions.

**You can change the x-axis to log scale for better viewability. In this case, each reference condition is different, hence absolute enthalpy is different.**

Now let us find out the relative enthalpy with respect to pressure 3 barA and density 500 kg/m**3. To do this, we need to find out the enthalpy at this point and subtract it from the absolute enthalpy.

We can do this by adding the below model in the same sketch

Now, we can find the relative enthalpy with respect to the above-defined reference.

**In this case, relative enthalpy is the same in all the cases.**

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