i-t Diagram for Combustion Products
Abstract Generated abstract
This paper presents a universal enthalpy, temperature diagram for combustion products, intended to determine the enthalpy of gas mixtures from temperature and composition, or conversely to estimate temperature from enthalpy and composition. The construction is based on expressing mixture enthalpy as the enthalpy of diatomic gases plus corrections for carbon dioxide and water vapor fractions, with graphical summation performed using auxiliary curves and 45 degree lines. For the inverse problem, the paper describes a successive approximation construction and explains why the readings converge. The diagram is computed from Timofeev’s heat capacity tables, fitted by least squares for 0 to 2000 degrees Celsius, and the expected calculation error is estimated not to exceed about 1.5 percent when gas composition is treated as complete, including water vapor.
Full Text
i–t Diagram for Combustion Products
G. I. Fuks
The existing i–t diagrams may be divided into two groups:
- Special diagrams, for example, intended for calculations with a specified fuel.
- Universal diagrams.
The proposed diagram belongs to the latter type. It is constructed according to the relation
\[ iV_{\mathrm{g}} = i_2 V_2 + i_{\mathrm{CO_2}} V_{\mathrm{CO_2}} + i_{\mathrm{H_2O}} V_{\mathrm{H_2O}}, \tag{1} \]
where \(i, i_2, i_{\mathrm{CO_2}}\), and \(i_{\mathrm{H_2O}}\) are the enthalpies, in \(\mathrm{n m^3}\), of the combustion products, diatomic gases, \(\mathrm{CO_2}\), and \(\mathrm{H_2O}\), respectively; \(V_{\mathrm{g}}, V_2, V_{\mathrm{CO_2}}\), and \(V_{\mathrm{H_2O}}\) are the volumes, in total \(\mathrm{n m^3}\), of the diatomic gases, \(\mathrm{CO_2}\), and \(\mathrm{H_2O}\), respectively. Since
\[ V_{\mathrm{g}} = V_2 + V_{\mathrm{CO_2}} + V_{\mathrm{H_2O}}, \tag{2} \]
then
\[ iV_{\mathrm{g}} = i_2 V_2 + (i_{\mathrm{CO_2}} - i_2)V_{\mathrm{CO_2}} + (i_{\mathrm{H_2O}} - i_2)V_{\mathrm{H_2O}}. \tag{3} \]
If it is taken into account that
\[ \frac{V_{\mathrm{CO_2}}}{V_{\mathrm{g}}} = r_{\mathrm{CO_2}}, \tag{4} \]
\[ \frac{V_{\mathrm{H_2O}}}{V_{\mathrm{g}}} = r_{\mathrm{H_2O}}, \]
then one may obtain
\[ i = i_2 + r_{\mathrm{CO_2}}(i_{\mathrm{CO_2}} - i_2) + r_{\mathrm{H_2O}}(i_{\mathrm{H_2O}} - i_2). \tag{5} \]
The principal role is played by the diatomic gases. The numerical values of the enthalpy of the diatomic gases as a function of temperature are plotted on the horizontal straight line. The values of the quantities
\[ i_{\mathrm{CO_2}} - i_2 \]
and
\[ i_{\mathrm{H_2O}} - i_2 \]
are plotted upward and downward from the main straight line at the corresponding temperature abscissas. Points with identical values of \(r_{\mathrm{CO_2}}\) and \(r_{\mathrm{H_2O}}\) are connected by curves with the corresponding inscriptions. Summation according to relation (5) is carried out by means of straight lines with a slope of \(45^\circ\) (see calculation example 1, plotted on the diagram).
The solution of the inverse problem, i.e., determination of the gas temperature from a given enthalpy and composition, can be performed on the universal diagram only by trial. In this case the method of successive approximation is used, which consists in the sequence of constructions abcdefghit (see calculation example 2, plotted on the diagram). In this procedure, each time the temperature axis intersects a vertical, an increasingly accurate value of the sought gas temperature is read off.
The correctness of this procedure can be proved as follows. If one assumes that, in the temperature interval where the construction is carried out,
\[ i_{\mathrm{H_2O}} - i_2 = idem, \tag{a} \]
…the line with the corresponding notation \(r_{\mathrm{H_2O}}\) will run horizontally. In this case, the very first construction will already make it possible to read off the desired temperature, as follows directly from the direct problem. Since in fact
\[ i_{\mathrm{H_2O}} - i_2 \]
increases with temperature, the indicated construction will give a somewhat underestimated value. Continuing this construction in the form of a set of lines \(abcefg\), we would also obtain the true value of the desired temperature if condition (a) were satisfied in the temperature interval from the previously determined value to the true one. In reality, however, owing to the increase of \(i_{\mathrm{H_2O}} - i_2\) with temperature, a somewhat higher temperature is read off. The further construction \(ghit\) again gives a somewhat underestimated temperature, etc. At the same time, the subsequent readings become closer and closer, so that further constructions become unnecessary.
For compiling the diagram, the heat-capacity tables prepared by V. N. Timofeev [1] were used. Processing these tables by the method of least squares led to the following formulas for the mean heat capacity from \(0^\circ\) to \(t^\circ\mathrm{C}\) at constant pressure in the interval \(0^\circ\)–\(2000^\circ\mathrm{C}\).
- Diatomic gases:
\[ c = 0.310 + 0.0000241\,t \ \frac{\mathrm{kcal}}{\mathrm{nm}^3\,\mathrm{deg}.} \tag{6} \]
with a maximum error of \(1\%\).
- \(\mathrm{CO_2}\):
\[ c = 0.430 + 0.0000769\,t \ \frac{\mathrm{kcal}}{\mathrm{nm}^3\,\mathrm{deg}.} \tag{7} \]
with a maximum error of \(4\%\).
- \(\mathrm{H_2O}\):
\[ c = 0.354 + 0.0000572\,t \ \frac{\mathrm{kcal}}{\mathrm{nm}^3\,\mathrm{deg}.} \tag{8} \]
with a maximum error of \(1.3\%\).
With the same error, the values of the enthalpy of the gases were calculated. Taking into account that the main part of the combustion products consists of the nitrogen of the air supplied for combustion, it can be shown that the maximum error in calculation by these relationships, and consequently by the diagram as well, should not exceed \(1.5\%\).
When using the diagram it is necessary to take into account that the composition of the gases must be calculated with allowance for water vapor, i.e., for “complete” and not for “dry” gases. Knowing the volumes of the dry gases \(V_{c2}\) and of the water vapor \(V_{\mathrm{H_2O}}\), the total volume is calculated as
\[ V_z = V_{c2} + V_{\mathrm{H_2O}} . \tag{9} \]
If the percentage content of \(\mathrm{O_2}\) in the dry gases is \(\mathrm{CO_2}\%\), then
\[ r_{\mathrm{CO_2}} = \mathrm{CO_2}\frac{V_{c2}}{V_z}. \tag{10} \]
For calculating \(r_{\mathrm{H_2O}}\) we have
\[ r_{\mathrm{H_2O}} = \frac{V_{\mathrm{H_2O}}}{V_z}. \tag{11} \]
Further,
\[ r_z = 100 - (r_{\mathrm{CO_2}} + r_{\mathrm{H_2O}}). \tag{12} \]
REFERENCES
- Timofeev V. N.—Physical constants of gases, Izvestiya VTI, 6, 1946.
\(i\)-\(t\) Diagram for Combustion Products
Enthalpy, kcal/Nm³
Temperature, °C
Curves and markings visible on the diagram include:
- \(20\%\ \mathrm{CO_2}\)
- \(10\%\)
- \(30\%\)
- \(20\%\)
- \(40\%\ \mathrm{H_2O}\)
Enthalpy, kcal/Nm³
Temperature, °C
Curves and markings visible on the diagram include:
- \(20\%\ \mathrm{CO_2}\)
- \(10\%\)
- \(30\%\)
- \(20\%\)
- \(40\%\ \mathrm{H_2O}\)
Examples
Given
\(t = 940^\circ\mathrm{C}\)
\(r_{\mathrm{CO_2}} = 13\%\)
\(r_{\mathrm{H_2O}} = 22\%\)
Find \(i\).
Solution
\(a-b-c-d\)
\(i = 356\ \mathrm{kcal/Nm^3}\)
Given
\(i = 822\ \mathrm{kcal/Nm^3}\)
\(r_{\mathrm{CO_2}} = 10\%\)
\(r_{\mathrm{H_2O}} = 21\%\)
Find \(t\).
Solution
\(e-f-c-d-e\)
\(p-g-h-i\)
\(t = 2000^\circ\mathrm{C}\)
Enthalpy, kcal/Nm³
Temperature, °C
Curves and markings visible on the diagram include:
- \(20\%\ \mathrm{CO_2}\)
- \(10\%\)
- \(20\%\)
- \(40\%\ \mathrm{H_2O}\)