USING THE HEAT OF CIRCULATING WATER AT POWER PLANTS FOR ELECTRICITY GENERATION

I. N. Butakov

Honored Worker of Science and Technology, Professor, Doctor

As is known, the design temperature of the cooling water at its inlet into the condenser of steam turbines is usually taken to be 15°C. This was an uncritical transfer to our country of Western European experience, despite the difference in climate. A water temperature of 15–18°C is the average for most rivers of our country for only three summer months, whereas for almost seven months this temperature remains at 0–2°C. If, in the condenser of a turbine designed for \(p_2 = 0.04—0.05\) ata and \(t_2 = 15^\circ\mathrm{C}\), the vacuum, owing to colder circulating water, proves deeper than the design value, then the final expansion of the steam will occur already beyond the limits of blading in the turbine exhaust branch pipe, i.e., in the best case it will be completely unused. If, however, one takes into account that, in order to create the necessary flow areas in the blades of the last stages at large steam volumes, even at \(p_2 = 0.04\) ata these flow areas have to be designed “cut off,” allowing high steam velocities in order to avoid various design complications up to increasing the number of casings, then it becomes clear that, as the vacuum is deepened, the specific steam consumption in the turbine may even increase as a result of deterioration of the relative internal efficiency. To this must be added the lowering of the condensate temperature because of the deeper vacuum, so that a turbine built for a vacuum of \(p_2 = 0.04—0.05\) ata will, at a deeper vacuum, give an excess fuel consumption due to overcooling of the condensate. It has been established from experience that a reduction in condensate temperature by \(1^\circ\mathrm{C}\) causes an excess fuel consumption of approximately 0.15%. To avoid this, one resorts to reducing the cooling ratio in order to obtain the normal vacuum \(p_2 = 0.04—0.05\) ata; this, however, worsens the heat-transfer coefficient in the condenser because of the reduction of the water velocity in its tubes. Therefore it is often preferred to raise the temperature of the cooling water by diverting part of the water from the discharge canal into the intake well, or else to plug the lower row of condenser tubes. In all such cases one has to forgo the possibility of using the deep vacuum provided by the climatic and geographical conditions of the USSR. From what has been said it is clear that, under the conditions of the greater part of the USSR, with such a sharp difference in the temperatures of water bodies during the year, one can profitably utilize the gift of our harsh nature—the low water temperatures—for about seven months of the year only in a special installation operating precisely during this seven-month period, i.e., for about 4500–5000 h/year, which coefficient of utiliza-

Fig. 1

Visible labels in the figure:

• Current generator
• Pump motor
• Supply of circulating water at \(25^\circ\text{C}\)
• Water at \(15^\circ\text{C}\), supplied from the evaporators to the condensers of the main turbines
• Washing with water at \(0.2^\circ\text{C}\) of the wetting surfaces
• Weak solution of ammonia salt, washed and collected at [[unclear: temperature value]]
• Surfaces of the final [[unclear: heater / heating]]
• Length \(23\,220\)
• Diameter \(660\)

… should be considered quite sufficient to justify the existence of such a special installation, all the more so since the additional capacity obtained precisely during this period is especially necessary, as it covers the autumn-winter peak load.

If it is assumed that the station’s main turbines must receive circulating water year-round at a temperature of \(15^\circ\mathrm{C}\), in order to have calculated conditions for their operation, then the temperature of the water at the outlet from the condenser will be of the order of \(25^\circ\mathrm{C}\), and consequently, for 4500–5000 hours per year, the special installation will have at its disposal approximately the same temperature difference as in the Claude and Boucherot installation, intended for utilizing the temperature difference in tropical seas between the surface and a depth of 1000 meters. It follows from this that, in the case under consideration as well, it is advisable to use an installation of the Claude and Boucherot type for utilizing the heat of the spent circulating water.

As regards the structural implementation of the installation described, the design developed for the first such installation, intended to operate according to the Claude and Boucherot principle and mounted on the steamship Tunis (Fig. 1), is of great interest. The total capacity of this installation is 2200 kW, to be produced by eight single-disk turbines of 275 kW each, operating on a common shaft, at the ends of which are mounted the electric generator and the compressor of the refrigerating installation, which was intended for producing ice, so necessary for tropical countries where it was planned to start the installation (Cuba Island). The eight turbines are enclosed in a common drum 6 m in diameter and 23 m long, with wall thickness of 12 mm, inside which the necessary vacuum must be maintained, since the evaporator chambers and mixing condensers are located there. Some of the latter were made in the form of surface condensers.

Fig. 2

Consideration of Fig. 1 leads to the conclusion that the construction of such an installation does not present great difficulties. It does not necessarily require the participation of special turbine-building plants, where, of course, it could be carried out with still greater success. This concerns the thermomechanical part. Electric generators, naturally, should best be obtained from specialized plants.

If \(d_{\mathrm{e}}\) is the weighted-average specific steam consumption, kg/kWh, for the condensing part of the station and \(m\) is the cooling ratio, then the amount of circulating water is \(q'_{\mathrm{water}} = d_{\mathrm{e}} m\) kg/kWh. The relationship between \(q'_{\mathrm{water}}\) and \(d_{\mathrm{e}}\) for different \(m\) is given in Fig. 2 as a bundle of straight lines.

Circulating water at a temperature of \(25^\circ\mathrm{C}\) is sent to the evaporator of the low-pressure turbine; after part of the water has evaporated, the main mass of it, upon leaving the evaporator (the waste water), has a tem—

[[unclear: beginning of line]] temperature of 15.5°C¹); a temperature of 15.5°C corresponds to a heat of evaporation \(r = 58\) cal., and, consequently, from 1 kg of circulating water entering the evaporator there will be obtained steam in the amount \((25 - 15.5):587 = 9.5/587 = 0.016\) kg, or, for each kWh of the main turbines,

\(D = 0.016 d_{\mathrm{e}} m\) kg. Figure 3 shows the dependence of \(D\) on \(d_{\mathrm{e}}\) in the form of a bundle of straight lines for different values of \(m\). A water temperature of 15.5°C in the evaporator corresponds to a steam pressure of 0.018 ata, so that, at a pressure in the condenser of 0.01 ata, we have the specific steam consumption in the low-pressure turbine

\(d_{\mathrm{e}}^{\mathrm{l.p.}} = 860/0.55 \cdot 17 = 90\) kg/kWh. Here,

Fig. 3

for the sake of caution, the relative electrical efficiency has been taken as only \(\eta_{\mathrm{e}} = 0.55\).

The power of the low-pressure turbine is

\[ N = \frac{0.016 \cdot d_{\mathrm{e}} \cdot m}{90} = 1.8 \cdot 10^{-4} \cdot d_{\mathrm{e}} \cdot m \]

kW for each kWh of the main turbines, or, if we take, as often happens, \(m = 60\) and \(d_{\mathrm{e}} = 4.5\), then we obtain

\(N = 1.8 \cdot 60 \cdot 10^{-4} \cdot 4.5 = 0.05\) kW,

i.e., the power of the low-pressure turbine in this case amounts to 5% of the power of the main turbines. The dependence of \(N\) on \(d_{\mathrm{e}}\) is shown in Fig. 4. It follows from what has been said that a low-pressure turbine using the heat of spent circulating water can in many cases cover the station’s own-use electricity consumption, since the latter, for peat-fired and oil-fired stations and for stations with chain grates, most often amounts to 4–5% (Shatura GRES in 1939—3.53%).

The quantity of cooling water per kWh of the low-pressure turbine with mixing condensation is determined from the equation

\[ 598 \cdot d_{\mathrm{e}}^{\mathrm{l.p.}} + q_{\mathrm{e}}'' = 6.6 \left(d_{\mathrm{e}}^{\mathrm{l.p.}} + q_{\mathrm{e}}''\right),^{2} \]

¹ The temperature of the discharged water has been taken as 15.5°C, since, to make up for the loss of circulating water due to the evaporation of part of it in the evaporator and due to possible leaks, it is necessary to add spent water from the condenser of the low-pressure turbine at a temperature of 6.6°C. In addition, one must take into account some cooling of the water on its way to the condensers of the main turbines.

² At a pressure in the condenser of 0.01 ata, the water temperature has been taken as 6.6°C, i.e., it is assumed that the partial pressure of the air will be so small that it may be neglected. As we shall see below, such an assumption will not differ greatly from reality.

whence

\[ q_{\mathrm{e}}''=\frac{d_{\mathrm{e}}^{\mathrm{tnd}}(598-6.6)}{5.6}=107\ \text{kg}, \]

or, per kWh of the main turbines,

\[ q_{\mathrm{water}}''=107\cdot 0.016\cdot d_{\mathrm{e}}\cdot m =1.7\,d_{\mathrm{e}}\,m\ \text{kg}. \]

The dependence of \(q_{\mathrm{water}}''\) on \(d_{\mathrm{e}}\) is shown in Fig. 5. The question arises of the expenditure of energy for station service in pumping large quantities of water in connection with the installation of the low-pressure turbine.

Fig. 4

The power expended on pumping cooling water for the low-pressure turbine per 1000 kWh of the main turbines is determined by assuming that the head to be overcome, \(H''\), in meters of water column, includes the leveling head \(H_1\), frictional and local—

Fig. 5

—losses of the turbines; determine, assuming that the head to be overcome \(H''\), in m water column, includes the leveling head \(H_1\), frictional and local—

…with the resulting deterioration of the vacuum and the possible reduction in the expenditure of energy for pumping air out of the condensers of the main turbines. This is also one of the advantageous aspects of using the installation described. Thus, in all, for 1000 kWh of the main turbines we have air \(q^{\circ}_{\text{air}} = 0.0825\, d_{\mathrm{e}} m\) kg. Of course, not all of this quantity of air has to be extracted, since part of it remains in the water discharged into the river and sent to the condensers of the main turbines. On the basis of Klot’s experiments in Ugra with a semi-industrial installation, it may be assumed that

Fig. 8

only \(3/4\) of all the air need be removed, so that subject to extraction is

\[ q = 0.75 \cdot 0.0825\, d_{\mathrm{e}} m = 0.062\, d_{\mathrm{e}} m, \]

and it must be understood that the coefficient \(3/4\) also allows for the admission of air through leaks, which further increases the safety margin in the calculation; therefore we finally take

\[ q = 0.06\, d_{\mathrm{e}} m \frac{\text{kg}}{1000\ \text{kWh}}. \]

In Figs. 8 and 9, bundles of straight lines are given for determining the quantities of air per 1000 kWh of the main turbines for various values of the cooling ratio.

powers \(N'\) and \(N''\) for different values of \(q'_{\mathrm{water}}\) and \(q''_{\mathrm{water}}\) and for different heads \(H'\) and \(H''\). Thus, in all, \(N=N'+N''\) kW is spent on pumping the water. But before the utilization installation was built, \(N_0\) kW was expended; therefore, the additional kW \(N_{\mathrm{add}}=N-N_0\) must be charged to the utilization installation. Incidentally, it should be noted that discharge of water from the mixing condenser of the low-pressure turbine into the river will occur by gravity if the leveling height proves to be not less than 10 m and the end of the downflow pipe is lowered below the water level in the river.

The question of air extraction is much more serious. The latter enters the condenser, first, with the steam from the evaporator, which carries away gases from the circulating water of the main turbines; secondly, by being released from the cooling water of the low-pressure turbine; thirdly, through leaks in the entire part of the path that is under vacuum. In Fig. 7 a curve is given for the values of the coefficient of dissolution of air in water as a function of its temperature [1]. The coefficient of dissolution is the ratio of the weight of air contained in 1 cu. m of water at the given temperature to the weight of the same volume of atmospheric air in contact with this water (Henry’s law). Therefore, for cold water at a temperature of \(1^\circ\mathrm{C}\), entering the condenser of the low-pressure turbine, we have the quantity of dissolved air

Fig. 7

\[ 1.293\cdot 0.028\cdot 273/234=0.036\ \mathrm{kg/m^3}, \]

i.e., for every 1000 kWh of the main turbines,

\[ q''_{\mathrm{air}}=0.036\cdot 1.7\cdot d_{\mathrm{e}}\cdot m =0.06\cdot d_{\mathrm{e}}\cdot m\ \mathrm{kg}. \]

Water entering the evaporator at a temperature of \(25^\circ\mathrm{C}/273+25=293\,K\) should contain air

\[ 1.293\cdot 0.019\cdot 273/296=0.0225\ \mathrm{kg/m^3}, \]

or, per 1000 kWh of the main turbines,

\[ q_{\mathrm{air}}=0.0225\,d_{\mathrm{e}}\,m\ \mathrm{kg}. \]

But this water, being directed from the evaporator to the condensers of the main turbines, then returns from there again to the evaporator, undergoing continuous degassing; therefore the air content in it should be much smaller, since penetration of air into this water is possible only through leaks, and the most vulnerable section here is the part that is under vacuum. Having noted this, we shall nevertheless adopt in the present case, for reliability of the calculation, the full quantity of air, but in what follows we shall not take into account the inflow of air through leaks in the system, owing, on the one hand, to the difficulty of allowing for it in the absence of experimental indications applicable to the installation described, and, on the other hand, also because the admission of air through leaks is of great importance in surface condensers of normal turbines when the amount of dissolved gases in the condensate is negligible by comparison. In the installation being analyzed, however, the suction of air may be neglected in comparison with the enormous quantity of dissolved gases, especially in cold water when condensation by mixing is present. Since degassing of the circulating water will occur in the evaporator of the low-pressure turbine, in the condensers of the main turbines the danger of air penetration into these condensers through possible leaks will decrease within the limits of these condensers through possible leaks—

resistance \(H_2\) in m water column of the water conduit, but reduced by the amount \(H_3\), m water column, of the difference between the barometric column and the column corresponding to the pressure in the condenser of the low-pressure turbine, so that

\(H'' = H_1 + H_2 - H_3\) in m water column.

Therefore the power expended, in kW, is

\[ N'' = \frac{1000 \cdot q''_{\text{water}} \cdot H''} {3600 \cdot 75 \cdot 1.36 \cdot 0.65} = 0.0042\, q''_{\text{water}} H'', \]

where the efficiency of the pump unit, including the electric motor, is \(\eta_{\text{p}} = 0.65\). Similarly, for pumping the outgoing water from the evaporator through the condenser of the main turbines and back to the evaporator, the power expended per 1000 kWh of the main turbines is

Fig. 6

\[ N' = \frac{1000 \cdot q'_{\text{water}} \cdot H'} {3600 \cdot 1.36 \cdot 75 \cdot 0.65} = 0.0042\, q'_{\text{water}} H', \]

where \(H'\) is understood to mean the frictional and local resistances, in m water column, of the entire path along which the water passes. Fig. 6 makes it possible to determine

Now we can proceed to determine the partial pressure of air in the mixing condenser of the low-pressure turbine, which, as is known, can be found from the formula:

\[ P_b=\frac{G_b \cdot R_b \cdot T_b}{D \cdot V_{\text{н}} \cdot 10000} =\frac{29{,}27 \cdot \gamma \cdot T_b}{10000 \cdot V_{\text{н}} \cdot x}. \]

Here

\[ \gamma = G_b/D=\frac{0{,}06 \cdot d_{\text{э}} \cdot m}{16 \cdot d_{\text{э}} \cdot m}=0{,}0038 \]

is the ratio of the amount of air to the amount of steam,

\[ T_b = T_{\text{п}} = 279{,}6\ \text{K} \]

is the temperature of the steam–air mixture,

\(V_{\text{н}}\) is the specific volume of steam in \(\text{m}^3/\text{kg}\), and the steam content \(x=1{,}0\).

Fig. 9

Since at \(P_2=0{,}01\) ata \(V_{\text{н}}=131{,}6\), then

\[ P_b=\frac{29{,}27 \cdot 0{,}0038 \cdot 279{,}6}{131{,}6 \cdot 10^4} =0{,}0000238\ \text{ata}. \]

The value \(P_b\) of the pressure is so small that it is quite permissible to neglect it in the calculations. Here we also obtain the specific volume of the air at its partial pressure in the condenser of the low-pressure turbine:

\[ V_b=\frac{29.27\cdot279.6}{0.0000238}=262\ \text{m}^3/\text{kg}, \]

i.e., it proves to be twice as large as the specific volume of the steam, so that, during extraction, for one volume of air there will be two volumes of steam.

After determining the quantity of air to be extracted, we turn to determining the power required for this. In order to reduce the expenditure of the latter, the principle of fractional removal of air should be used, as applied by Claude in his installation, consisting in preliminary degassing near the tops of the barometric columns of cold and warm water, as shown in the diagram of Fig. 10, where \(B\)—evaporator, \(C\)—condenser; \(D\)—deaerators; \(R\)—cooler; \(T\)—turbine; 1—cold water; 2—water from the condenser; 3—water leaving the evaporator; 4—warm water; \(a, b, c, d\), and \(e\)—pipelines for the mixture of air and steam at

Fig. 10

temperatures \(1^\circ\)–\(2^\circ\text{C}\) \((a,b)\), \(6.6^\circ\text{C}\) \((c)\), \(25^\circ\text{C}\) \((d,e)\), and corresponding pressures for mercury columns of 65 mm; 35 mm; 7.35 mm; 35 mm; and 65 mm. Thus, three stages of extraction are obtained, and in all five mixture takeoffs are obtained. As a result, the work of extraction will be reduced, since, as is known, in isothermal compression it is proportional to the logarithm of the ratio of atmospheric pressure to the pressure at the point where the mixture is taken off; and, consequently, when taking off the mixture at a pressure of 0.1 ata, extraction work proportional to the argument \(\log \frac{1.0}{0.1}\), or 1.0, is required, whereas this work after the condensation chamber is proportional to \(\log \frac{1.00}{0.01}\), i.e., 2.0 and even 3 times greater, if one takes into account the additional volumes of steam that strongly rarefy each volume of air in the condenser. Therefore, the power required for extracting gases from the barometric columns will be approximately 3–4 times less than only from

condenser of the low-pressure turbine. In the specifications for the Cloo extractor it was stipulated that 60% by weight of the air should be extracted from barometric columns at a pressure of 65 mm Hg (0.088 ata), 25% at a pressure of 35 mm Hg (0.048 ata), and only 15% at condenser pressure together with 4 volumes of steam [2]. As a result of applying the principle of fractional air removal, the operation of gas extraction was reduced, so that the power ultimately expended amounted to no more than 7–8% of the power of the low-pressure turbine installation [3]. Let us also note that, in the extractor built according to the design of Academician Rateau, in order to avoid compression

Fig. 11

to atmospheric pressure of the steam mixed with air, provision is made for compressing the steam to pressures in the intermediate stages of the extractor that communicate with coolers, where the steam is partially condensed and thus no longer participates in the subsequent compression.

The extractor [4], built according to the design of Academician Rateau, consists of 17 wheels divided into 3 groups, rotating at a common speed of 4900 rpm, driven by an electric motor through a reduction gear. The first two wheels have a diameter of 720 mm and a peripheral speed of 335 m/s. After these two wheels the mixture is exhausted at a pressure of 0.048 ata pro-

...goes to cooler No. 1. The third and fourth wheels, with a diameter of 580 mm, pass not only the air from cooler No. 1, but also a mixture from the preliminary degassers at a pressure of 0.048 ata. The fifth wheel has a diameter of 680 mm. An additional mixture of air and steam from the second pair of degassers at a pressure of 0.088 ata is supplied here. The exhaust after the fifth wheel, at a pressure of 0.132 ata, is directed to cooler No. 2. The first three wheels of the medium-pressure group, with a diameter up to 560 mm, exhaust at 0.27 ata into cooler No. 3. The last three wheels of this group, with a reduced diameter down to 380 mm, expel the air at a pressure of 0.52 ata into cooler No. 4. The wheels of the first two groups are of the type with radial blades. These radial wheels have a great advantage: they produce no axial force, so that the first two groups have no balancing device. The high-pressure group consists of six wheels with diameters decreasing from 360 mm to 320 mm. They rotate at more moderate peripheral speeds (177–188 m/sec). These wheels compress the air from 0.52 ata to 1.05 ata, in order to discharge it to the atmosphere without intermediate cooling. The wheels of the third group are of the old type—cellular; consequently an axial pressure toward the suction side is produced here. Therefore a balancing piston is provided for the third group. In Monteuil’s article a longitudinal section is given of an assembly of three casing-groups, whose total length, including the electric motor, is 8.4 m; the shaft connection between the casings is semi-elastic. The three shaft exits are completely isolated from the surrounding atmosphere, so that the penetration of air into the machine is precluded. This result is achieved by the provision of closed bearings. To avoid the entry of lubricant into the compressor, a balancing channel is made in the lower part of the bearing, where an intermediate pressure is established between the pressures existing at both ends of the closed bearing.

The following table gives the design values of the suction and exhaust pressures at the ends of each group of wheels for the case of the utilization installation that we are considering.

As is known, the most advantageous ratio \(x\) between the final compression pressure and the initial pressure in the stages of multistage compressors, from the point of view of the power expended, is obtained when

\[ x=\sqrt[n]{\frac{P_n+1}{P_1}}, \]

where \(n\) is the number of separate compressions. In our case, consequently,

\[ x=\sqrt[5]{1.05/0.01}\simeq 2.54. \]

But it turns out that the excess of the required power is insignificant even with large deviations, of the order of 20%, from the theoretical intermediate pressure: it is expressed in fractional parts of a percent, usually not exceeding 0.5% [5]. Therefore there is no need to strive for the exact fulfillment of the most advantageous ratio. Here it is useful, of course, also to take into account the experience of constructing the first extractor according to the design of Academician Rateau. We shall take the distribution, by weight, over the extraction points of the air being extracted as follows: 15% in the condenser, 25% at a pressure of 0.03 ata, and 60% at a pressure of 0.1 ata.

We now turn to determining the power expended in compressing the mixture of air and steam, taking into account the fractional principle and the presence of intercoolers. Three hypotheses may be taken as the basis of the calculation here: 1) isothermal compression; 2) adiabatic compression; 3) polytropic compression. The theoretical powers expended on compression will be unequal, namely \(N_{\text{isot}} < N_{\text{adiab}} < N_{\text{polit}}\). To pass from theoretical power to actual power, it is necessary to introduce, for the three processes indicated above, three different experimental coefficients, so that

\[ N_{\text{actual}} = \frac{N_{\text{isot}}}{\eta_{\text{isot}}} = \frac{N_{\text{adiab}}}{\eta_{\text{adiab}}} = \frac{N_{\text{polit}}}{\eta_{\text{polit}}}. \]

Assuming that compression of the mixture will occur according to a polytropic law with polytropic exponent \(n = 1.3\), since in the present case steam will participate in the process together with the air, we find the required theoretical power from the formula

\[ L = \frac{n}{n - 1}\cdot P_1 V_1 \left[ \left(P_2/P_1\right)^{\frac{n-1}{n}} - 1 \right], \]

or, in kW,

\[ N = \frac{1.3\,P_1\,V_1}{0.3\cdot 3600\cdot 75\cdot 1.36} \left[ \left(P_2/P_1\right)^{0.23} - 1 \right], \]

if \(V_1\) is given in m³/hr. To facilitate the calculations, one may use the graph (Fig. 11), putting \(P_1/10000 = 1\) atm. and \(V_1 = 1\) m³. Plotting \(P_2/P_1\) on the abscissa axis and the power on the ordinate axis, we obtain the power in kW/m³ for an initial pressure of 1 atm. This power for the first pair of wheels, at \(P_2/P_1 = 3\), will be equal to

\[ L_1 = \frac{0.03250 + 0.03580}{2} = 0.0339\ \text{kW/m}^3. \]

The quantity of air extracted from the condenser is \(0.15 q\) kg/1000 kWh, and its specific volume is 262 m³/kg, which, owing to the presence of steam, is doubled, since, as indicated above, in the condenser there are two volumes of steam per one volume of air; hence at every moment a double weight of steam, relative to the weight of air, will be extracted from the condenser. Therefore the required total power for the first two wheels is

\[ N_1 = 0.0339\cdot 0.15\cdot q\cdot 262\cdot 2\cdot 0.01 = 0.0266\,q = 0.0016\,d_{\text{э.м}}\ \text{kW/1000 kWh}. \]

The third wheel is traversed by a mixture of air and steam from intercooler No. 1, connected to the first pair of wheels. Although part of the exhaust steam from this pair condenses in cooler No. 1, nevertheless steam remains in the mixture in an amount corresponding to the saturation temperature in cooler No. 1. It is customary to cool gas in intercoolers to a temperature approximately 10 degrees higher than the initial temperature (Hütte, 1936, vol. III, p. 608). Since the initial temperature of the cooling water under our conditions is \(1^\circ\)—\(2^\circ\text{C}\), allowing its heating in the cooler to \(10^\circ\text{C}\), it will be sufficient to assume cooling of the mixture to \(15^\circ\text{C}\). This temperature corresponds to a partial pressure of steam \(P_{\text{steam}} = 0.0174\) ata. Therefore the partial pressure of air is

\[ P_{\text{air}} = 0.0300 - 0.0174 = 0.0126\ \text{ata}. \]

The volume occupied by the air after cooler No. 1 will be

\[ V_{\text{air}} = \frac{G_{\text{air}}\cdot R_{\text{air}}\cdot T}{P_{\text{air}}} = \frac{29.27\cdot 0.15\cdot q\cdot 288}{126} = 10.1\,q = 0.62\,d_{\text{э.м}}\ \text{m}^3/1000\ \text{kWh}. \]

This air, in the saturated state at a temperature of 15°C, will bring moisture from cooler No. 1 (Fig. 12)

\[ \frac{12.8 \cdot 0.62 \cdot d_{\mathrm{э}} \cdot m}{1000} = 0.008\,d_{\mathrm{э}} \cdot m \ \text{kg}/1000\ \text{kWh}, \]

or, in cubic meters per 1000 kWh,

\[ 0.008 \cdot 78 \cdot d_{\mathrm{э}} \cdot m = 0.62\,d_{\mathrm{э}} \cdot m, \]

so that per 1 volume of air there is 1 volume of steam. After the first two impellers, \(0.15 \cdot 262 \cdot q\ \mathrm{m^3}\) of air and the same amount of steam entered the cooler (No. 1), whose weight is

\[ \frac{0.15 \cdot q \cdot 262}{131.6} = 0.018\,d_{\mathrm{э}} \cdot m \]

kg/1000 kWh.

Thus, there will remain in the cooler

\[ (0.018 - 0.008)\,d_{\mathrm{э}}m = 0.01\,d_{\mathrm{э}} \cdot m \ \text{kg}/1000\ \text{kWh}. \]

The power expended on the third impeller is

\[ N'_1 = 0.0136 \cdot 0.03 \cdot 2 \cdot 0.62 \cdot d_{\mathrm{э}} \cdot m = 0.0005\,d_{\mathrm{э}} \cdot m \]

\[ \text{kW}/1000\ \text{kWh}. \]

After the third impeller, air from the preliminary-deaeration pipelines enters in an amount of \(0.25q\) at a pressure of 0.03 ata. The partial pressure of the steam in the cold column is 0.0067 ata. The partial pressure of the air is

\[ P_{\mathrm{air}} = 0.05 - 0.0067 = 0.0433\ \text{ata}. \]

Volume of air:

\[ V_{\mathrm{air}} = \frac{29.27 \cdot 0.187 \cdot q \cdot 275}{433} = 0.21\,d_{\mathrm{э}} \cdot m \]

The amount of steam for saturation of the air at a temperature of 1°C (Fig. 12) is

\[ \frac{5.6 \cdot 0.21 \cdot d_{\mathrm{э}} \cdot m}{1000} = 0.00117\ \text{kg}/1000\ \text{kWh}, \]

and, consequently, per \(1\ \mathrm{m^3}\) of air there is \(1\ \mathrm{m^3}\) of steam.

The amount of air from the warm column is \(0.25 \cdot 0.75 \cdot q'_{\mathrm{air}}\). The partial pressure of steam is \(P_{\mathrm{steam}} = 0.0323\) ata, so that the partial pressure of the air is

\[ P_{\mathrm{air}} = 0.05 - 0.0323 = 0.0117\ \text{ata}. \]

The volume of air is

\[ V_{\mathrm{air}} = \frac{29.27 \cdot 0.75 \cdot q'_{\mathrm{air}} \cdot 298 \cdot 0.25}{177} = \]

\[ = 0.208\,d_{\mathrm{э}} \cdot m\ \mathrm{m^3}/1000\ \text{kWh}. \]

Fig. 12

The weight of steam carried away by the air at a temperature of 25°C (Fig. 12) is

\[ \frac{23 \cdot 0.208 \cdot d_{\mathrm{э}} \cdot m}{1000} = 0.0048\,d_{\mathrm{э}} \cdot m\ \text{kg}/1000\ \text{kWh}. \]

The temperature of the compressed mixture after the third wheel (Fig. 11) is \(35+15=50^\circ\mathrm{C}\). Therefore, after the mixtures from the two preliminary-deaeration pipelines have entered before the fourth wheel, the temperature of the mixture will be:

\[ \begin{aligned} &50(0.15\cdot 0.06\cdot d_{\mathrm{e}}\cdot m+2\cdot 0.008\cdot d_{\mathrm{e}}\cdot m) +25(0.25\cdot 0.75\cdot 0.0225\cdot d_{\mathrm{e}}\cdot m+ \\ &\quad +2\cdot 0.0048\cdot d_{\mathrm{e}}\cdot m) +1(0.25\cdot 0.75\cdot 0.06\cdot d_{\mathrm{e}}\cdot m+2\cdot 0.00117\cdot d_{\mathrm{e}}\cdot m)= \\ &=x(0.15\cdot 0.06\cdot d_{\mathrm{e}}\cdot m+2\cdot 0.008\cdot d_{\mathrm{e}}\cdot m +0.25\cdot 0.75\cdot 0.0225\cdot d_{\mathrm{e}}\cdot m+ \\ &\quad +2\cdot 0.0048\cdot d_{\mathrm{e}}\cdot m +0.25\cdot 0.75\cdot 0.06\cdot d_{\mathrm{e}}\cdot m+2\cdot 0.00117\cdot d_{\mathrm{e}}\cdot m), \end{aligned} \]

whence \(x=28^\circ\mathrm{C}\). The volume of the mixture from the cold column, when heated from \(1^\circ\mathrm{C}\) to \(28^\circ\mathrm{C}\), will increase by

\[ 2\cdot 0.21\cdot d_{\mathrm{e}}\cdot m\cdot 301/274=0.46\cdot d_{\mathrm{e}}\cdot m. \]

The volume of the mixture from the warm column will also increase when the temperature is raised from \(25^\circ\mathrm{C}\) to \(28^\circ\mathrm{C}\):

\[ 2\cdot 0.203\cdot d_{\mathrm{e}}\cdot m\cdot 301/298=0.42\cdot d_{\mathrm{e}}\cdot m. \]

The volume of the mixture from the third wheel, however, will decrease when the temperature falls from \(50^\circ\mathrm{C}\) to \(28^\circ\mathrm{C}\), and will be

\[ 2\cdot 0.62\cdot d_{\mathrm{e}}\cdot m\cdot 301/323=1.16\cdot d_{\mathrm{e}}\cdot m. \]

Thus the total volume of the mixture of air and steam before the fourth wheel is

\[ (0.46+0.42+1.16)d_{\mathrm{e}}\cdot m=2.04\cdot d_{\mathrm{e}}\cdot m\ \text{cu. m}/1000\ \text{kWh}. \]

The compression work in the fourth and fifth wheels from Fig. 11, at \(P_2/P_1=\)

\[ =0.10/0.05=2,\quad \text{will be}\quad \frac{0.0217+0.0190}{2}=0.0203\ \text{kW}/\text{m}^3. \]

The total required power is

\[ N'_2=0.0203:0.05\cdot 2.04\cdot d_{\mathrm{e}}\cdot m =0.00207\cdot d_{\mathrm{e}}\cdot m\ \text{kW}/1000\ \text{kWh}. \]

After the sixth wheel the mixture enters cooler No. 2, to which the mixture of air and steam from the two preliminary-deaeration pipelines at a pressure of \(0.1\) ata is also directed. We take the temperature in the cooler as \(20^\circ\mathrm{C}\), and the amount of air will be \(q=0.06\cdot d_{\mathrm{e}}\cdot m\).

The partial pressure of steam at \(20^\circ\mathrm{C}\) is \(P_{\text{steam}}=0.0238\) ata, and the partial pressure of air is \(P_{\text{air}}=1.0-0.0238=0.0762\) ata. The volume of air is

\[ V_{\text{air}}=\frac{29.27\cdot 298\cdot 0.06\cdot d_{\mathrm{e}}\cdot m}{762} =0.69\cdot d_{\mathrm{e}}\cdot m. \]

The amount of steam (Fig. 12) carried away by the air is

\[ \frac{17\cdot 0.69\cdot d_{\mathrm{e}}\cdot m}{1000} =0.01175\cdot d_{\mathrm{e}}\cdot m\ \text{kg}/1000\ \text{kWh}, \]

which corresponds to a steam volume of \(0.01175\cdot 57.8=0.69\cdot d_{\mathrm{e}}\cdot m\); consequently, one volume of steam corresponds to one volume of air. The compression work (Fig. 11) at \(P_2/P_1=0.25/0.1=2.5\) will be \(0.0281\ \frac{\text{kW}}{\text{m}^3}\). The total compression work in the group of wheels 6–8 is

\[ N_3=0.0281\cdot 0.1\cdot 2\cdot 0.69\cdot d_{\mathrm{e}}\cdot m =0.00387\cdot d_{\mathrm{e}}\cdot m\ \text{kW}/1000\ \text{kWh}. \]

After the eighth wheel, cooler No. 3 has a mixture temperature of 20°C, so that the partial vapor pressure \(P_{\text{vap}} = 0.0238\) ata and the partial air pressure \(P_{\text{air}} = 0.25 - 0.0238 = 0.2262\) ata.

The volume of air

\[ V_{\text{air}}=\frac{29.27\cdot q\cdot 293}{2262}=0.228\cdot d_{\text{e}}\cdot m\ \text{m}^3/1000\ \text{kWh}. \]

The amount of vapor carried away (Fig. 12)

\[ \frac{17\cdot 0.228\cdot d_{\text{e}}\cdot m}{1000} =0.00388d_{\text{e}}\cdot m\ \text{kg}/1000\ \text{kWh}. \]

The volume of vapor is \(0.00388\cdot d_{\text{e}}\cdot m\cdot 57.8=0.226\cdot d_{\text{e}}\cdot m\). With a compression ratio \(0.5/0.25=2\), the total compression work in the group of wheels 9–11 is

\[ N_4=0.0203\cdot 0.25\cdot 2\cdot 0.228\cdot d_{\text{e}}\cdot m =0.00232\cdot d_{\text{e}}\cdot m \]

In cooler No. 4 we shall take the temperature to be 15°C. Then \(P_{\text{vap}}=0.0174\) ata and \(P_{\text{air}}=0.5-0.0174=0.4826\) ata. The volume of air is

\[ V_{\text{air}}=\frac{29.27\cdot q\cdot 288}{4826}=0.105\cdot d_{\text{e}}\cdot m. \]

The amount of vapor carried away is

\[ \frac{13\cdot 0.105\cdot d_{\text{e}}\cdot m}{1000} =0.001360\ d_{\text{e}}\cdot m\ \text{kg}/1000\ \text{kWh}. \]

The compression ratio is \(1.05/0.5=2.1\). The total compression work in the last group of wheels 12–17 will be

\[ N_5=0.022\cdot 0.5\cdot 2\cdot 0.105\cdot d_{\text{e}}m =0.00232\cdot d_{\text{e}}m. \]

Thus the power that must be expended to drive the extractor will be

\[ \begin{aligned} N_1 &= 0.0016\cdot d_{\text{e}}\cdot m\\ N'_2 &= 0.0005\cdot d_{\text{e}}\cdot m\\ N''_2 &= 0.0021\cdot d_{\text{e}}\cdot m\\ N_3 &= 0.0039\cdot d_{\text{e}}\cdot m\\ N_4 &= 0.0023\cdot d_{\text{e}}\cdot m\\ N_5 &= 0.0023\cdot d_{\text{e}}\cdot m\\ \overline{N} &= 0.0127\cdot d_{\text{e}}\cdot m. \end{aligned} \]

If we take \(\eta_{\text{polyt}}=0.6\), then the power actually expended is

\[ N_{\text{actual}}=\frac{0.0127\cdot d_{\text{e}}\cdot m}{0.6} =0.0212\cdot d_{\text{e}}\cdot m\ \text{kW}/1000\ \text{kWh}. \]

In Fig. 13 a bundle of straight lines is constructed for the dependence of \(N_{\text{actual}}\) on \(d_{\text{e}}\) for different values of \(m\).

We now have all the data needed to determine the energy consumption for the station service of the utilization installation for any condensing station having an operating capacity \(N_{\text{main tur}}\) of the main turbines. Let us take as an example such a station with a capacity \(N_{\text{main tur}}=14000\ \text{kW}\), \(d_{\text{e}}=6.25\ \text{kg}/\text{kWh}\) and \(m=60\). Then:

From Fig. 2: \(q'_{\text{water}}=375\ \text{kg}/\text{kWh}\);

From Fig. 3: \(D=6.0\ \text{kg}/\text{kWh}\).

From Fig. 4: \(N = 0.067\ \text{kW/kWh}\)

From Fig. 5: \(q''_{\text{water}} = 620\ \text{kg/kWh}\).

From Fig. 6 we have: 1) with a tract resistance through the condensers of the main turbines \(H' = 10.25\), we obtain \(N' = 16\ \text{kW}/1000\ \text{kWh}\); 2) for supplying cooling water to the condenser of the low-pressure turbine, at \(H'' = 10\ \text{m}\), we obtain \(N'' = 25\ \text{kW}/1000\ \text{kWh}\). Altogether, then, \(14(25 + 16) = 575\). The power absorbed by the pumps for supplying circulating water to the main turbines is \(N_0 = 420\ \text{kW}\), so that the required additional power is \(N_{\text{add}} = 575 - 420 = 155\ \text{kW}\). The power required for the air extractor (Fig. 13) is

\[ N_{\text{act}} = 110\ \text{kW}. \]

Altogether, the energy for the auxiliary needs of the utilization installation is \(N_{\text{aux}} = 155 + 110 = 265\ \text{kW}\). The total power of the utilization installation (Fig. 4) is \(14000 \cdot 0.067 = 940\ \text{kW}\), so that the energy consumption for auxiliary needs is \(265 \cdot 100 / 940 = 28\%\). To cover the other auxiliary needs of the power station, \(940 - 265 = 675\ \text{kW}\) remains, i.e. \(675 \cdot 100 / 14000 = 4.8\%\). The energy consumption for air extraction is \(110 \cdot 100 / 940 = 11.7\%\), whereas in Claude’s case it was determined to be \(7\text{–}8\%\). This excess consumption must apparently be explained by the fact that in the utilization installation a deeper vacuum of \(0.01\ \text{ata}\) is adopted, as against \(0.019\) in Claude’s. In fact—

Fig. 13

...it is possible to expect a reduction in the energy expenditure for air extraction, for, as indicated above, the water leaving the evaporator and directed to the condenser of the main turbines will to a considerable extent be deaerated. Some reduction in the power required for air extraction will probably be achieved by a closer approach to the most advantageous ratio \(X\) between the final compression pressure and the initial pressure in the stages of the multistage compressor; finally, one should not lose sight of the fact that the actual power of the low-pressure turbine itself was calculated with great caution; in reality it is permissible to expect a higher value of the relative efficiency of this turbine, and consequently also a reduction in the percentage of energy spent on auxiliary needs.

In conclusion it is necessary to emphasize once again that, with the use of the utilization installation, savings in fuel consumption will be achieved owing to the elimination of the possibility of condensate supercooling¹, and also owing to the reduction in the steam consumption for the ejectors of the condensers of the main turbines as a result of deaeration of the circulating water, which should reduce the entry of air through leaks in the condenser, with a simultaneous increase in the vacuum owing to an increase in the heat-transfer coefficient, which in turn will contribute to fuel economy. If, in addition, the evaporated make-up of the circulating water is sent to chemical treatment, then, in addition to everything stated above, complications in the boiler plant from water leaks in the condensers of the main turbines will be eliminated. Moreover, in this way it will undoubtedly be possible to reduce the number of washings and cleanings of the condensers of the main turbines, the fouling of which causes deterioration of the vacuum and, consequently, also leads to overconsumption of fuel, often disrupts the uninterrupted operation of stations, and often causes a short supply of electric power to consumers². As is known, the problem of maintaining condensers in proper order is becoming more and more difficult because of the growing pollution of cooling-water sources by sewer waste waters from industry and large population centers. One further probable element of economy should also be discussed here, namely the potential reduction in the consumption of electric power for the power station’s own needs owing to the possibility of generating direct current in the utilization installation, as is the case at many stations in the USA. Direct current permits economical speed regulation over wide limits, which is important for the electric motors of many auxiliary mechanisms of power stations and may be of special significance with the forthcoming mass introduction of automation. But the main reduction in the specific fuel consumption will, of course, be obtained from the additional generation of kWh in the low-pressure turbine.

As indicated above, part of the condensers of the Claude and Boucherot installation was designed as surface condensers, which was intended to obtain distillate for the preparation of ice. For our conditions this possibility of obtaining condensate in a utilization installation of the Claude and Boucherot type may have special significance, since it can free us from the necessity of using, for make-up in the event of condensate loss at the station, chemically purified water for 6–7 months of the year, during the periods of the most intensive operation of power stations.

This distillate is degassed, which is also very valuable. In a number of cases, the very possibility of obtaining such condensate for make-up may prove decisive, and then it will not be necessary to attach great importance to the inevitable reduction in the capacity of the utilization plant as a consequence of the worsening of the vacuum when surface condensers are used.

If it is assumed that the utilization plant will give only 5% savings in specific fuel consumption, then, for the private example considered above, the annual monetary saving, with the cost of standard fuel at 50 rubles per ton and the specific consumption of such fuel at 1.2 kg/kWh, will be

$14000 \cdot 4500 \cdot 1.2 \cdot 0.05 \cdot 50/1000 = 190000$ rubles.

With 10% deductions according to the capitalization factor, this makes it possible to capitalize the sum of 1,900,000 rubles in order, without monetary loss to the national economy, to obtain a 5% saving in specific fuel consumption. Hence the maximum permissible value of the cost of the installed kW is

$1900000/940 = 2000$ rubles.

The corresponding calculations, made for the conditions of the Shatura GRES with its specific consumption of standard fuel of 0.54 kg/kWh and the cost of this fuel at 91 rubles/t, give a maximum permissible cost of the installed kW of the utilization plant of about the same 2000 rubles, on the assumption that the fuel saving will be only 4%. Both these figures show that the maximum permissible amount of capital investment per installed kW of the utilization plant should prove sufficient to realize such installations for obtaining savings in fuel consumption.

It is interesting to note in passing that the large Claude and Boucherot installations for utilizing the temperature difference in tropical seas, according to Boucherot, were supposed to cost 1000–1500 paper francs (1926) per installed kW [6], whereas large condensing stations in Europe at that time cost 300–450 gold francs per installed kW, which in paper francs, at the exchange rate of the time, gave 2100–3100 francs [7], i.e., the installed kW of these stations was about twice as expensive as that of Claude and Boucherot.

But the interests of the national economy require of us not only a struggle to economize fuel: they also decree for us an intense struggle to economize metal, since the supply, in particular, of iron ores in the world is rather limited. Some preliminary calculations carried out in the diploma project of the student of the Tomsk Polytechnic Institute Z. A. Tikhonova show that the consumption of metal per installed kW of a utilization plant at a GRES of 14 MW capacity does not exceed that for a locomotive-type plant of the same capacity (about 200 kg/kW). Thus it cannot be said that this consumption of metal is inadmissible. Nevertheless, there is a probable possibility of considerably reducing such consumption. The fact is that the utilization plant operates under vacuum conditions, i.e., its drum, the bulkiest part, is subjected to external compression by atmospheric pressure. Under these conditions it is tempting to construct the drum from elements of stone casting, which, as is known, works excellently in compression and satisfactorily in bending, giving temporary resistance to compression, with a stellate structure, of the order of up to 10,000 kg/cm² and to bending of 550–740 kg/cm² [8]. The problem of mastering stone casting as a machine-building material must already be seriously posed before domestic machine-building, independently of the utilization plant. It is highly doubtful, however, that the large drums of utilization plants can be manufactured whole from stone casting: they will, one must suppose, be made from individual elements that will have to be joined together, achieving a state of the most complete tightness, paralyzing any possibility of pro-

penetration of air into the internal cavity of the drum. Some kind of technology analogous to the welding of separate elements as applied to stone casting must be developed. The problem is, of course, not insurmountable; sooner or later it should confront engineers enthusiastic about this field, who are inclined to blaze new trails in technology. It is then that the utilization plant will become still more attractive and will acquire yet another serious advantage.

The possibility is not excluded of making the drum of the low-pressure turbine from concrete, using dense mixes with the application of a dense plaster coat with troweling; in the present case the constant wetting of the walls inside the drum will help to achieve air-tightness. [9] The wall thickness \(\delta\) of the drum should turn out to be quite acceptable, since the outer radius of a cylindrical vessel loaded by a single external pressure \(p_n\) is determined by formula [10]

\[ r_2 \geq r_1 \sqrt{\frac{R_d}{R_d - 2p_n}}, \]

where \(R_d\) is the permissible compressive stress. Taking \(p_n = 1\) ata and, for concrete, \(R_d = 30\ \mathrm{kg/cm^2}\), we obtain \(r_2 = 310\ \mathrm{cm}\), so that \(\delta \simeq 100\ \mathrm{mm}\). In practice this thickness will have to be made somewhat larger, for in carrying out the work it is difficult to avoid small irregularities of the cylindrical part with deviation toward ovality; therefore in practice it is required only to satisfy condition [11]

\[ e = 2\,\frac{r_1 - r'_1}{r_1 + r'_1} < 0.01, \]

In any case, the wall thickness of the drum will be acceptable. If further investigation of this question confirms the advisability of using concrete, then the problem of constructing low-pressure turbines will be greatly simplified. Some difficulties may perhaps be expected here in connection with vibrations of the turbine rotors, which may lead to impairment of the tightness of the drum.

Literature

1) Chambadal. Le degazage fractionne des eaux chaude et froide dans le procede Claude—Boucherot. Chaleur et Industrie, 1934, No. 168, p. 911.

2) G. Claude. L’extraction des gaz dissous dans le procede Claude—Boucherot d’utilisation de l’energie thermique des mers tropicals. Le Genie civil, vol. C, No. 23, p. 571.

3) G. Claude. Sur l’utilisation de l’energie thermique des mers. Comptes rendus, vol. 185, p. 988.

4) C. Monteil. Description et essais en usine d’un extracteur d’air, Systeme Rateau, devant etre utilise dans les appareils Claude—Boucherot. Rev. gen. de l’electricite, 18/II 1933, pp. 217—229.

5) Hintz. Thermodynamic Foundations of Piston and Turbocompressors. 1933. Energoizdat, p. 40.

6) Boutaric. L’utilisation de l’energie thermique des oceans. Chaleur et Industrie, 1928, No. 94, p. 60.

7) Le Paige. Les Tendences actuelles dans construction des supercentrales electriques. Rev. univ. des mines, 8-e serie, vol. II, No. 6, 15/IX 1929, p. 25.

8) Construction Industry, 1935, No. 3, p. 4. According to data of Academician Budnikov (Chemical Engineering, 1940, No. 4–5, article by Volodin and Klinov), the corresponding figures are 830 and 980 kg/cm².

9) Technical Encyclopedia, vol. II, 1937, p. 738, article by V. Skramtaev on concretes.

10) Kadykov V. I. Calculation of Cylindrical Vessels. Leningrad, 1933, p. 23.

11) Bulgakov V. V. The Influence of Deviations of a Pipe’s Shape from Circular on Its Resistance to External Pressure. GTI, Moscow, 1930, p. 3.

Noticed Misprints

1. Supercooling of the condensate in some cases in winter at our stations reached 10°–20°C. At ShGES, only by introducing a number of measures was it possible to reduce this supercooling and keep it within 3°–5°C (ShGES im. V. I. Lenin. 1925–1935, p. 114). At the thermal power station of the Gorky Automobile Plant this supercooling is kept at 2°–3°C (Teplo i Sila, 1935, No. 2, article by Lavrov, p. 20). Thus, it may be assumed that supercooling of the condensate increases steam consumption by at least 0.5%. ↩

2. At Chegres, the introduction of regular cleaning of condensers with caustic soda every 6 days gave an increase in turbine efficiency of approximately 0.5% (Teplosilovoe khozyaistvo, 1940, No. 12, article by Gvozdetsky). ↩