THERMAL CALCULATION OF A VELOX BOILER

G. I. Fuks and N. A. Novikov

The thermal calculation of a Velox boiler has the special feature that it is necessary to determine and coordinate with one another the magnitudes of the power of the gas turbine and of the gas and air compressors.

On the basis of knowing these powers, the magnitude of the required supplementary power will be determined.

The power of the gas turbine is limited, on the one hand, by the temperature of the gases before the gas turbine (not more than \(500^\circ\)C). On the other hand, it is necessary to have a feed-water heater downstream of the gas turbine. These stringent conditions limit the thermal drop of the gas turbine. Therefore, in most cases this heat drop is insufficient to create the power necessary for operation of the compressors. It is necessary to install additional sources of power—an electric motor or a steam turbine.

This supplementary engine serves not only for the constant addition of its work to the work of the gas turbine, but also for starting during boiler start-up, when there is no possibility of using the exhaust gases, and also for regulating the boiler. All the work expended on the compressors ultimately goes to the heating of the air and gas entering the furnace.

Ignorance of the necessary supplementary power therefore does not make it possible to lay out the balance of the Velox boiler as in ordinary boiler systems. To determine the necessary supplementary power per \(1\) kg \((\mathrm{Nm}^3)\) of fuel, at given loss magnitudes, \(n\), we proceed from the balance equation

\[ Q_{n}^{p}+\frac{860\,N\cdot\eta}{B} = \frac{D\cdot\Delta i}{B}+Q_{2}+Q_{3}+Q_{4}+Q_{5} \tag{1} \]

and the efficiency equation

\[ \eta_{y} = \frac{\dfrac{D\cdot\Delta i}{B}} {Q_{n}^{p}+\dfrac{860\,N\cdot\eta}{B}}. \tag{2} \]

Here the notation is:

\[ Q_{n}^{p} \]
— calorific value of the fuel, \(\dfrac{\mathrm{kcal}}{\mathrm{Nm}^{3}}\) \(\left(\dfrac{\mathrm{kcal}}{\mathrm{kg}}\right)\);

\[ N \]
— supplementary power in kW;

\[ \eta \]
— total efficiency of the compressors
\[ \left(\eta=\frac{\text{power in the gas}}{\text{power in the engine}}\right); \]

\[ B \]
— fuel consumption, \(\dfrac{\mathrm{Nm}^{3}}{\mathrm{hour}}\) \(\left(\dfrac{\mathrm{kg}}{\mathrm{hour}}\right)\);

\[ Q_{2} \]
— boiler loss with the exhaust gas, \(\dfrac{\mathrm{kcal}}{\mathrm{Nm}^{3}}\) \(\left(\dfrac{\mathrm{kcal}}{\mathrm{kg}}\right)\);

\(Q_3\) — boiler loss due to chemical incomplete combustion;

\(Q_4\) — boiler loss due to mechanical underburning;

\(Q_5\) — boiler loss to the surroundings;

\(\Delta i\) — heat expenditure for producing steam, \(\dfrac{\mathrm{cal}}{\mathrm{kg}}\).

We introduce

\[ n=\frac{860\cdot N\cdot \eta}{B}\ \frac{\mathrm{cal}}{\mathrm{Nm}^3}\left(\frac{\mathrm{cal}}{\mathrm{kg}}\right) \]

and determine from equations (1) and (2) the fuel consumption and the additional power

\[ B=\frac{D\cdot \Delta i}{Q_2+Q_3+Q_4+Q_5}\cdot \frac{1-\eta_v}{\eta_v}, \tag{3} \]

\[ n=\frac{Q_2+Q_3+Q_4+Q_5}{1-\eta_v}-Q_n^r. \tag{4} \]

From relation (2) it is evident that exceeding the additional power above that required is disadvantageous, since this leads to a reduction in the efficiency of the installation, i.e., ultimately, to an overconsumption of fuel.

Equations (3) and (4) make it possible to construct the graphs

\[ B=f_1(\eta_v)\quad \text{and}\quad n=f(\eta_v) \]

The power \(n_0\) required to drive the compressors (air and gas) is determined with the aid of a \(T-S\) diagram constructed for air and fuel gas (Fig. 1).

Obviously,

\[ n_0=n_{gm}+n', \]

where \(n_{gm}\) is the power of the gas turbine,

\(n'\) is the required additional power.

To determine the power delivered by the gas turbine, it is necessary to specify the operation of the tail section of the boiler. As a rule, it is economically advantageous to use the maximum temperature drop on the economizer heating surface, leaving to the boiler itself, if possible, only the generation of steam. The only obstacle in this respect may be an inadmissibly high temperature in the gas turbine.

We determine the quantity of heat going to heat the water in the water economizer

\[ Q_{ez}=\frac{D\cdot \Delta t_e}{B}+Q_5^e, \]

where \(\Delta t_e=t''_{nut}-t'_{nut}\),

\(Q_5^e\) — loss to the surroundings by the economizer, \(\dfrac{\mathrm{cal}}{\mathrm{Nm}^3}\left(\dfrac{\mathrm{cal}}{\mathrm{kg}}\right)\).

We lay off the value \(Q_{ez}\) on the \(I-T\) diagram constructed for the given composition of flue gases, on the ordinate of the temperature of the exiting gases. By this procedure we find the temperature before the water economizer, i.e., the temperature of the gases after the gas turbine,

\[ T_{ez}=T''_{gm}. \]

We repeat this calculation for several values of \(\eta_v\) (i.e., for the corresponding values of fuel consumption), at the corresponding temperatures before the water economizer.

We specify a practical pressure drop in the gas turbine and, with the aid of the \(T-S\) and \(I-T\) diagrams for flue gases, determine the adiabatic temperature drop at various \(\eta_v\) (corresponding to various \(T''_{gm}\)).

The power on the turbine shaft is expressed as:

\[ n_{2m}=AL_{ad}^{2m}\cdot \eta_{oi}\cdot \eta_m\ \frac{\mathrm{cal}}{\mathrm{Nm}^3}\left(\frac{\mathrm{cal}}{\mathrm{kg}}\right). \]

And, knowing the power of the gas turbine and the required power, we find the magnitude of the supplementary power as a function of \(\eta_y\).

We construct the curve \(n'=f(\eta_y)\) (Fig. 2). The point of intersection of the curves \(n=f(\eta_y)\) and \(n'=f(\eta_y)\) corresponds to a definite boiler operating regime, in which the power of the gas turbine, the power of the supplementary energy source, and the selected calculated efficiency of the installation are coordinated with one another. Thus the preliminary calculation of the Velox boiler is completed.

EXAMPLE OF CALCULATION

Determine the magnitude of the supplementary power and the efficiency of a Velox boiler with capacity

\[ D=90\ \frac{\mathrm{tons}}{\mathrm{h}}, \]

with steam parameters \(P=17\ \mathrm{ata}\);

\[ T_n=375^\circ\mathrm{C}, \]

operating on peat gas with

\[ Q_n^p=3030\ \frac{\mathrm{cal}}{\mathrm{Nm}^3}. \]

Previously determined are

\[ V_0=3.01\ \frac{\mathrm{Nm}^3}{\mathrm{Nm}^3};\qquad V_g=3.976\ \frac{\mathrm{Nm}^3}{\mathrm{Nm}^3}. \]

1. We assign the values

\[ \eta_y=0.9;\ 0.93;\ 0.94;\ 0.95 \]

and calculate for these values the following quantities:

\[ B;\ n;\ n_0;\ n_{2m};\ Q_{vz};\ n'. \]

We show the calculation for one value, \(\eta_y=0.9\); the remaining calculations are collected in a table and shown in Fig. 3.

1) Fuel consumption

\[ B=\frac{D\cdot \Delta i}{Q_2+Q_3+Q_4+Q_5}\cdot \frac{1-\eta_y}{\eta_y}. \]

Previously, by the ordinary method, it was determined that

\[ Q_2=107\ \frac{\mathrm{cal}}{\mathrm{Nm}^3}, \]

\[ Q_5=91.4\ \frac{\mathrm{cal}}{\mathrm{Nm}^3}, \]

\[ Q_3=Q_4=0\ \frac{\mathrm{cal}}{\mathrm{Nm}^3}, \]

\[ B=\frac{9000\cdot 695}{107+91.4}\cdot \frac{1-0.9}{0.9} =35080\ \frac{\mathrm{Nm}^3}{\mathrm{h}}. \]

2) Supplementary power from the balance equation:

\[ n=\frac{Q_2+Q_3+Q_4+Q_5}{1-\eta_y}-Q_n^p =\frac{107+91.4}{1-0.9}-3030= \]

\[ =-1046\ \frac{\mathrm{cal}}{\mathrm{Nm}^3}\sim -265\ \frac{\mathrm{cal}}{\mathrm{Nm}^3\ \mathrm{g}}. \]

3) With the aid of the constructed \(T-S\) diagram for peat gas and air, we determine the power of the air and gas compressors.

Air-Blower Power

We have, as given,

\[ P_1 = 1 \ \text{ata}; \quad t_1 = 20^\circ\text{C}; \]

\[ P_2 = 3 \ \text{ata}. \]

We plot the air-blowing process on the \(T-S\) diagram for air (points \(1-2\)) and find the value of the adiabatic compression work

\[ AL_{ad} = 33 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{of air}}. \]

Taking \(\eta_{ad} = 0.72\), we find

\[ AL_{\text{actual}} = \frac{33}{0.72} = 45.7 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{air}}. \]

We convert the compression work to \(1 \ \text{Nm}^3\) of flue gases:

\[ AL_{ov} = \frac{AL_{\text{actual}} \cdot V_0}{V_2} = \frac{45.7 \cdot 3.01}{3.975} = 34.8 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{f.g.}}. \]

Gas-Blower Power

\[ P_1 = 1.03 \ \text{ata}; \quad t_1 = 20^\circ\text{C}; \quad P_2 = 3 \ \text{ata}. \]

We plot the gas-blowing process on the \(T-S\) diagram for fuel gas (points \(1-2\)) and find

\[ AL_g = \frac{40}{0.72 \cdot 3.976} = 14 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{f.g.}}. \]

Total Power

\[ n_0 = AL'_{da} + AL_g = 34.8 + 14 = 48.8 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{f.g.}}. \]

We reduce the compressor power to the shaft power:

\[ n_0 = \frac{48.8}{\eta_m} = \frac{48.8}{0.95} = 51.4 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{f.g.}}. \]

4) Heat used to preheat the water in the economizer,

\[ Q'_{ea} = \frac{D \cdot \Delta t_e}{V_2 \cdot B} + \frac{Q_5^e}{V_2} = \frac{90000 \cdot 81}{3.976 \cdot 35080} + \frac{17.1}{3.976} = 56.6 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{f.g.}}. \]

5) From the \(I-T\) diagram for flue gases we find

\[ T'_{vz} = T'_{zt} = 240^\circ\text{C}. \]

6) We set the pressure drop in the gas turbine as \(\Delta p_{zt} = 1.4 \ \text{ata}\) and determine the adiabatic heat drop with the aid of the \(T-S\) and \(I-T\) diagrams for flue gases.

We find the value of \(n_{zt}\):

\[ n_{zt} = 24.5 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{f.g.}}. \]

7) Additional power

\[ n' = n_0 - n_{zt} = 51.4 - 24.5 = +26.9 \ \frac{\text{kcal}}{\text{Nm}^3 \ \text{f.g.}}. \]

We plot the curves (Fig. 2)

\[ 1) \quad n = f(\eta_y) \]

\[ 2) \quad n' = f'(\eta_y) \]

Fig. 1.

Fig. 2.

The point of intersection of the curves corresponds to the value of the additional power

\[ n = 10 \frac{\mathrm{kcal}}{\mathrm{Nm^3}\ \mathrm{d.g.}} \]

and the efficiency \(\eta_y = 0.935\).

Converting to kilowatts, we obtain the additional power

\[ N = \frac{B \cdot n \cdot V_2}{860} = \frac{19830 \cdot 10 \cdot 3.976}{860} = 915\ \mathrm{kW} \]

Fig. 3.

Table 1.