5.7 – Reference States and State Properties¶
5.7.0 – Learning Objectives¶
By the end of this section you should be able to:
- Understand reference states and their importance.
- Understand the steam table chart and its reference point.
5.7.1 – Introduction¶
Often, we do not know the absolute value of internal energy or enthalpy (\(\hat{U}, \hat{H}\)). To give us a way of measuring these values, a reference state is chosen to represent 0 and the values are determined as a difference to the reference state.
This notebook will go over the concept of reference state and state properties.7
5.7.2 – Internal energy of Water¶
Let’s look at water at 0 C\(^\circ\) and 1 atm. If asked to find the specific energy \(U\) of this water, you might think right away that you can use \(C_p\Delta T\) but this is wrong. We cannot use this formula because we need to consider the \(\Delta T\) which refers to a change in temperature.
What we can do is use the 0 C\(^\circ\) 1 atm water as a reference and compare it to changes in energy.
We can take liquid water at 50 C\(^\circ\) 1 atm and say that in reference to 0 C\(^\circ\) water, 100 C\(^\circ\) water has \(210 \space \frac{kJ}{kg}\) more energy than 0 C\(^\circ\) liquid water.
Likewise, 100 C\(^\circ\) liquid water at 1 atm has \(420 \space \frac{kJ}{kg}\) more energy than water at 0 C\(^\circ\).
By setting water at 0 C\(^\circ\). We can use the referenced values as variables known as state variables also known as \(\hat{U}\) and \(\hat{H}\).
A state variable is a variable that its property only depends on its initial and final state.
In [8]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.read_excel('../figures/Module-5/isoT600k.xlsx', sheet_name='Sheet1', index_col=0, na_values=['NA'])
df
df.truncate(before=0, after=10)
Out[8]:
| Temperature (K) | Pressure (MPa) | Volume (l/mol) | Internal Energy (kJ/mol) | Enthalpy (kJ/mol) | Entropy (J/mol*K) | |
|---|---|---|---|---|---|---|
| 0 | 600 | 0.0 | infinite | 51.415 | 56.404 | 500.13 |
| 1 | 600 | 1.0 | 4.8861 | 51.123 | 56.009 | 130.11 |
| 2 | 600 | 2.0 | 2.3891 | 50.811 | 55.589 | 123.82 |
| 3 | 600 | 3.0 | 1.5548 | 50.477 | 55.142 | 119.88 |
| 4 | 600 | 4.0 | 1.1361 | 50.119 | 54.664 | 116.88 |
| 5 | 600 | 5.0 | 0.8834 | 49.734 | 54.151 | 114.36 |
| 6 | 600 | 6.0 | 0.71355 | 49.318 | 53.599 | 112.11 |
| 7 | 600 | 7.0 | 0.59087 | 48.864 | 53.000 | 110.03 |
| 8 | 600 | 8.0 | 0.49741 | 48.366 | 52.345 | 108.04 |
| 9 | 600 | 9.0 | 0.42316 | 47.811 | 51.620 | 106.06 |
| 10 | 600 | 10.0 | 0.36195 | 47.183 | 50.802 | 104.05 |
If you recall the steam tables” from module 4: NIST steam tables from module 4, these values are all state variables referenced to water at its triple point.
5.73 – Example question¶
Water is partially condensed out of a gas stream containing 66.9 mole% water vapor and the balance nitrogen.
Figure 1.
Step 1 - Perform all required material balances¶
- While none are needed, it might be helpful to fill in the missing mole fractions for nitrogen
Step 2 - Write the appropriate form of the energy balance (open or closed) and reduce to simplest form (get rid of any negligible terms)¶
Step 3 - Choose a reference state for the constituents¶
Water
\(l\), 25 °C, \(1 \space atm\),
Nitrogen
\(g\), 25 °C, \(1 \space atm\)
Step 4 - For a closed constant- volume system, construct a table with columns for initial and final amounts of each species (\(m_i\) or \(n_i\)) and specific internal energies relative to the chosen reference states. For an open system, construct a table with columns for inlet and outlet stream component flow rates (\(\dot{m_i}\) or \(\dot{n_i}\)) and specific enthalpies relative to the chosen reference states.¶
| Substance | \(n_{in}\) | \(\hat{H_{in}}\) | \(n_{out}\) | \(\hat{H_{out}}\) |
|---|---|---|---|---|
| H2O(v) | 0.669 | \(\hat{H_{1}}\) | 3.35 | \(\hat{H_{3}}\) |
| H2O(l) | 63.55 | |||
| N2(v) | 0.331 | \(\hat{H_{2}}\) | 33.1 | \(\hat{H_{4}}\) |