Prof. S. K. Konyukhov.

Critique of Methods for Testing Steam Boilers

Steam technology is, by its very nature, chiefly a technology of an evolutionary character. Slowly but surely it spreads its nets over the entire earth, embracing the most diverse branches of industry. Railways, steamships, mines, plants, factories, water-supply systems, and various kinds of industrial and economic enterprises—all these are served mainly by steam. There are, it is true, certain areas of technology where the utilization of steam has not yielded the proper results, such as, for example, aviation and air navigation, submarines; but this phenomenon is rather accidental than regular in character, and there would be nothing surprising if these isolated areas of technology should ultimately prove to be in the possession of steam. Such thoughts involuntarily come to mind when one reads of new achievements in steam technology.

Dangerous competitors of steam installations in the form of electric motors, water turbines, wind engines, internal-combustion engines, etc., despite their scientifically grounded advantages, are not yet in a position to suppress the mighty spread of steam, all the more since the representatives of steam engineering are not sitting with folded hands. In the interesting article by Professor N. F. Charnovsky, “The Age of Machines, or the Century of Uninterrupted Successes of Technical Creativity,” there is given a brief but sufficiently vivid survey of mankind’s achievements over the past century and the beginning of the present one; moreover, the impetus for these achievements was laid, of course, by steam technology.

We should like to correct a certain inaccuracy in Professor N. F. Charnovsky’s communication.

He writes: “However, Watt’s steam engine, which first operated at the end of the eighteenth century at a pressure of 2–3 atmospheres, after twenty years of uninterrupted work on its improvement—both in design and in execution—by the beginning of the nineteenth century was already able to operate at 4 atmospheres of pressure, with a simultaneous increase in the number of revolutions from 12 to 20 per minute and an increase in its thermal efficiency.”

This statement does not correspond to the truth.

Watt’s first engine had a cylinder diameter of 18 inches, and the inaccuracy of boring reached \(3/8\) inch. The piston leaked steam badly, although, according to the patent, it had packing “of paper, cork, putty, cardboard, and felt.”

It is therefore not surprising that the subsequent engines, which had a cylinder diameter of 80 inches and a piston stroke of 10 feet, developed only 62 horsepower. Until 1800 the steam pressure did not exceed 5 pounds per square inch, and only by 1830 did it rise to 20 pounds.

Slowly but steadily the pressure rose to 45, 60, 75, etc., pounds, and quite recently a pressure of 175 pounds per square inch, or about 12 atmospheres, was considered normal. The same picture of a regular increase in the number of strokes was observed in steam engines, until steam turbines appeared, at once increasing the number of revolutions per minute from hundreds to several tens of thousands (see Zörmann, Steam Turbines, p. 14).

From a purely theoretical point of view this rapid shift toward increasing the number of revolutions did not yield especially valuable results, leaving some unfilled gulf between former achievements and the new paths of technology.

Professor Baudis in Vienna therefore qualifies the steam turbine as a slow-speed engine.

The course of Professor Baudis’s reasoning may be reduced to the following. Let us denote by \(i\) the head of the steam turbine, i.e., the difference in heat in a unit of steam before the performance of work in the turbine and after the steam leaves the turbine; let

\[ A=\frac{1}{427} \]

be the heat equivalent of a unit of work. Then one may write

\[ \frac{i}{A}=H, \]

where \(H\) denotes the working power of the steam turbine, referred to one kilogram of steam. This quantity \(H\) may be regarded as the head of the steam turbine.

If, at heat head \(H\), the turbine operates with a number of revolutions \(n\), then at another heat head \(H^{1}\) it will operate with a number of revolutions equal to \(n^{1}\).

Since the rotational speed of the rotor is directly proportional to the number of revolutions per unit time, then, using the well-known formula

\[ V=\sqrt{2gH}, \]

one may write

\[ n^{1}:n=\sqrt{H^{1}}:\sqrt{H}. \]

As regards power, it will be proportional to the cube of the number of revolutions, or

\[ N^{1}:N=H^{1}\sqrt{H^{1}}:H\sqrt{H}. \]

If in these relations we put \(H^{1}=1\ \mathrm{mtr}\), we obtain

\[ n^{1}=\frac{n}{\sqrt{H}} \]

\[ N^{1}=\frac{N}{H\sqrt{H}}. \]

For geometrically similar turbines operating at one and the same head, the power of the turbines is proportional to the square of the rotor diameters; on the other hand, the number of revolutions is inversely proportional to this diameter. In such a case the relation is easily established:

\[ N^{1}=\frac{n_s^{2}}{n_1^{2}};\quad n_s=n_1\sqrt{N^{1}}. \]

This relation is easily reduced to the following:

\[ n_s=\frac{n}{H}\sqrt{\frac{N}{\sqrt{H}}}. \]

Example: Let a steam turbine operate at a pressure of 12 atmospheres, and let the superheated-steam temperature be \(300^\circ\mathrm{C}\). Further, let the vacuum be \(95\%\). The power of the turbine is 10,000 P.S. The turbine is directly connected to an eight-pole alternating-current generator operating at 50 periods.

According to the steam tables, the heat head is \(i=220\) heat units.

In that case

\[ H = 427 \times 220 = 93940 \text{ meters}. \]

This steam turbine must be compared, in its operation, with a water turbine operating under a head almost equal to 9.4 kilometers.

If the generator operates at 50 periods, then the number of pole reversals per minute will be

\[ 2 \times 50 \times 60 = 6000 \]

Therefore the number of revolutions \(n\) per minute will be

\[ n = \frac{6000}{8} = 750, \]

If \(n_1\) and \(N^1\) are calculated, we obtain

\[ n_1 = \frac{750}{\sqrt{93940}} = 2.46 \text{ revolutions} \]

\[ N^1 = \frac{10000}{93940\sqrt{93940}} = 0.00035 \text{ P.S.} \]

The specific number of revolutions of the turbine will be

\[ n_s = 2.46\sqrt{0.00035} = 0.046/\text{min.} \]

If these values are compared with the analogous values for a water turbine, interesting conclusions are obtained. Let us take the water turbine serving the electric station at Follie (Sweden). It operates with a head

\[ H = 1650 \]

meters; its power is \(N = 3000\) P.S. and its number of revolutions is \(n = 500\) per minute.

For this turbine we obtain

\[ n_1 = \frac{500}{\sqrt{1650}} = 12.4/\text{minute} \]

\[ N^1 = \frac{3000}{1650\sqrt{1650}} = 0.0445 \text{ P.S.} \]

\[ n_s = 12.4\sqrt{0.0445} = 2.6/\text{minute} \]

Comparison of the quantities \(n_1\) and \(n_s\) for the water turbine with the same values for the steam turbine shows that the steam turbine is much slower-running than the water turbine.

The same leap is now being observed in the transition from the former low steam pressure of 12 atmospheres to pressures of 60–100—120–224 atmospheres. This transition to high pressures gives reason to expect that the efficiency of steam installations will become equal to the efficiency of internal-combustion engines, and thus the long-expected funeral of steam technology will be postponed for an indefinitely long time. “The rumors of its death have been somewhat exaggerated,” to use Mark Twain’s words.

There is no doubt that a new development in the use of considerably higher-pressure steam will again reveal an unfilled place in the theory of steam engines.

First of all, it will be necessary to decide the question of what the vapor-like medium represents at high pressures. Perhaps it is no longer pe-

ceased to be steam and turned into gas. It may be more advantageous not to use water in engines, brought to the vaporous state at high pressure, but to use detonating gas (hydrogen and oxygen), which possesses enormous thermal energy.

According to Lemathelier and Mallard, the specific heat of water vapor is expressed by the following formula:

\[ S = 0.34 + 0.00030\,t. \]

But one cubic meter of hydrogen releases 2613 calories on combustion. If we ask to what temperature these 2613 calories can raise 1 cubic meter of water vapor, then we must use the equation

\[ 2613=\int_0^t (0.34+0.00030\,t)\,dt =0.34\,t+0.00015\,t^2. \]

Solving this quadratic equation, we obtain \(t = 3191^\circ\).

The transition to the use of high-pressure steam entails the necessity of a new design of steam boilers, and with this, new approaches to testing steam generators.

Here two questions arise: 1) how such tests are carried out, and 2) how such tests ought to be carried out.

Questions of this kind were once posed by the professor of applied mechanics Juté (Saint-Étienne), and were resolved quite satisfactorily.

A rough method for assessing the working capacity of a boiler is reduced to determining the quantity of steam corresponding to one kilogram of burned coal, or of fuel in general. Taking into account the calorific capacity of the fuel and the steam pressure, we obtain a basis for judging the working capacity of the boiler. This method represents a practical, rough means of making a gross estimate of the efficiency coefficient of the boiler.

Often the amount of heat obtained from the boiler furnace and used by the water for its heating and evaporation is calculated. This amount of heat is compared with the calorific capacity of the fuel. Good results of such calculations give utilization coefficients of 65% to 80%. The heat not received, in the amount of 35%—20%, is considered lost to radiant loss, conduction, to heat carried away by the flue gases, to the expenditure from incomplete combustion, and so on. If it is possible to calculate these heat losses separately, then the heat balance of the boiler installation is made up without difficulty.

Of course, this method is more perfect in the sense of evaluating the productivity of the industrial apparatus that a steam boiler is.

But even this method of evaluating the productivity of a boiler is not sufficiently satisfactory. If attention is paid only to the amount of heat exchanged in the machine, then from the standpoint of the theory of the work obtained this is still not enough; it is necessary to establish the temperatures of exchange. But by what are these temperatures of exchange veiled in the concepts: calorific capacity, on the one hand, and heat of vaporization, on the other? They are not there. However, the heat received by steam does not under all conditions give the same work. Here a large role is played by temperature and by the steam pressure dependent on this temperature.

The concept of calorific capacity also suffers from uncertainty in its formulation. Let us speak of the calorific capacity of the fuel at constant pressure, which, of course, is most important for the boiler installation. We are interested in the heat delivered by the fuel to the steam during combustion at constant pressure. If the pressure does not change, then the amount of heat will always be the same, whatever the path of transition from the initial state to one and the same final state. Let the combustion proceed adia—

... adiabatically at temperature \(X\). Subsequently the products of combustion are cooled to the temperature of the surrounding medium \(T_0\). In that case, the calorific capacity of the fuel represents the heat given up during this cooling from \(X\) to \(T_0\). It is quite obvious that different parts of this heat are given up at different temperatures, and therefore, from the point of view of their work, they cannot be compared with one another.

Hence arises the need to change the methods for evaluating productivity. Let us suppose that a steam boiler produces, with one kilogram of coal, \(9.46\ \mathrm{kg}\) of dry saturated steam at a temperature of \(273^\circ + 181^\circ = 454\), which corresponds to an absolute pressure of \(10.5\ \mathrm{kg}/\mathrm{sq.\ cm}\). Let, further, the temperature of the condenser be \(273^\circ + 30 = 303^\circ\).

The total heat of vaporization will be, according to the latest tables for saturated steam, \(666.6 - 30 = 636.6\) calories.

Thus, \(9.46\) kilograms of steam will contain

\[ 636.6 \times 9.46 = 6022.24 \text{ calories.} \]

Let us cause this steam to operate according to the Rankine cycle.

As is known, the coefficient of useful action of this cycle may be expressed by the following formula

\[ \eta_r = \frac{ q_1 - q_2 - T_2 \operatorname{Lgn}\frac{T_1}{T_2} + \frac{r_1 (T_1 - T_2)}{T_1} }{ q_1 - q_2 + r_1 }. \]

But the difference between \(q_1 - q_2\) is approximately equal to \(T_1 - T_2\). Consequently, in simplified form we obtain

\[ \eta_r = \frac{ (T_1 - T_2)\left(1 + \frac{r_1}{T_1}\right) - T_2 \operatorname{Lgn}\frac{T_1}{T_2} }{ T_1 - T_2 + r_1 }. \]

In our case

\[ T_1 = 454^\circ;\quad T_2 = 303;\quad r_1 = 482.95. \]

For these data the coefficient of useful action is expressed by the fraction \(0.295\). Consequently, the work delivered by the Rankine cycle may be expressed as \(1777\) large calories.

This, then, will be the productivity of the boiler, where both the heat obtained by the steam and the heat of the boiler are taken into account.

Let us denote this productivity by \(P\). Let us now clarify the expenditures of the boiler. To determine the mechanical value of the combustible, let us assume, following Zeuner, that the perfect process of combustion occurs at constant pressure without exchange of heat with the external medium. The temperature of combustion will, as before, be \(X\). Further, let these products be cooled to the temperature \(T_0\) of the external medium. During this cooling, the hot gases, changing their temperature from some \(T + dT\) to \(T\), release a certain quantity of heat. Let us direct this heat into an imaginary machine operating according to the Carnot cycle between the temperatures \(T\) and \(T_0\). Then we shall have an infinite number of ideal machines operating according to the Carnot cycle and having the lower limit \(T_0\), and the upper limit of temperature in the interval from \(X\) to \(T_0\). If by \(C\) we denote the heat capacity of the products of combustion at constant pressure, then the work delivered by all the above-mentioned ideal machines operating according to the Carnot cycle will be expressed thus:

\[ Q = \int_{T_0}^{X} C\left(1 - \frac{T_0}{T}\right)dT. \]

This thermal characteristic is a perfectly natural measure for evaluating the mechanical power of the fuel. If the quantity \(Q\) is compared with the heat-producing capacity of the fuel at constant pressure, expressed by the definite integral

\[ \int_{T_0}^{X} C\,dT, \]

then it will become clear that \(Q\) is less than the heat-producing capacity.

The analyzed process of utilization of fuel Professor Juge calls the “Peiner evolution.” This evolution thus represents the complete theoretical evolution of the steam engine.

The difference \(Q - P\) will represent the losses of the boiler.

It is now necessary to seek out these losses and to ascertain from what causes they occur.

The simplest and most illustrative way to carry out this task is with a particular example taken from life.

Example. A steam boiler with brass tubes, whose outside diameter is 50 mm and length 2500 mm, has a furnace crown at a height of 1212 mm. The draught is measured as 75 mm of water column.

The test of this boiler gave the following figures:

Gravimetric analysis of the dry fuel without ash:

Volumetric analysis of the gases in the smoke box at \(0^\circ\) and a pressure of 760 mm.

The heat balance, or the account of the work, is briefly reduced to the following:

Let us analyze these experimental data assuming a content, in 1 kg of hot fuel, of 0.012 kg of water and 0.040 kg of ash. We obtain, for 672 kg, a content of pure carbon

\[ 672(1 - 0.012 - 0.040) = 637.056 = \sim 637. \]

In order not to operate with large numbers in the calculations, let us convert our computations to grams.

Then it may be said that 672 grams of ordinary coal contain pure carbon only in the amount of 637 grams. The calorific value of one gram will be equal to 8.42 calories, and therefore 637 grams will give

\[ 8.42 \times 637 = 5363.54 = \sim 5364. \]

If it is taken into account that the process of combustion proceeds imperfectly, owing to the formation of carbon monoxide and the presence, in the slag and ash, of unburned coal, then these losses must be allowed for. It will be seen below that these losses amount to about 380 calories.

Subtracting this loss, we obtain

\[ 5364 - 380 = 4984 \text{ calories.} \]

To determine the combustion temperature one may use the equation

\[ \int_{T_0}^{X} (C' + C'' + C''')\,dT = 4984 \]

where \(C'\) represents the mean heat capacity of the products of combustion and of the steam, calculated by the formula

\[ C' = \frac{C_1 m_1 + C_2 m_2 + C_3 m_3 + \ldots} {m_1 + m_2 + m_3 + \ldots}. \]

In this formula \(C_1, C_2, C_3 \ldots\) are the heat capacities of the component parts, and \(m_1, m_2, m_3 \ldots\) the quantities by weight of these parts in kilograms.

\(C''\) represents the fines that fall through the grate.

\(C'''\) is the heat capacity of the ash.

According to Le Chatelier, \(C'' = 0.3\) and \(C''' = 0.4\). Professor Jüptner obtains

\[ C = C' + C'' + C''' = 1.96 + 0.000688T. \]

In that case we obtain

\[ \int_{T_0}^{X} (1.96 + 0.000688T)\,dT = 4984. \]

Since \(T_0 = 296^\circ\), solving the integral, we obtain

\[ 1.96X + 0.000344X^2 - 1.96 \times 296 - 0.000344 \cdot 296^2 = 4984. \]

After reduction we obtain the following quadratic equation:

\[ 0.000344X^2 + 1.96X - 5594.344 = 0. \]

Whence \(X = 2090^\circ\).

To determine the work obtained by the evolution of Zeuner, we use the equation

\[ Q = \int_{T_0}^{X} C\left(1 - \frac{T_0}{T}\right)\,dT + z = \int_{296}^{2090} (1.96 + 0.000688T) \left(1 - \frac{T_0}{T}\right)\,dT - z, \]

Here \(\varepsilon\) denotes the term characterizing the condensation of water vapors. Suppose that at temperature \(T\) an amount \(dm\) of water condenses; then heat \(r\,dm\) is liberated, and therefore

\[ \varepsilon=\int_{246}^{2090}\left(1-\frac{T_0}{T}\right) r\,dm. \]

It is very difficult to determine what amount of water condenses at each temperature, but in hot gases there is, in general, little moisture, and condensation in reality begins approximately at the temperature \(273^\circ+33^\circ=306^\circ\). In that case, to determine \(\varepsilon\), one may use the integral

\[ \int_{296}^{306}\left(1-\frac{T_0}{T}\right) r\,dm. \]

The value of this integral is small.

Neglecting the additional term \(\varepsilon\), we obtain

\[ \int_{296}^{2090}(1.96+0.000688T)\left(1-\frac{T_0}{T}\right)dT = \]

\[ \int_{296}^{2090}\left(1.96+0.000688T-1.96\frac{T_0}{T}-0.000688T_0\right)dT = \]

\[ = \int_{296}^{2090}\left(1.96+0.000688T-\frac{1.96\cdot296}{T}-0.000688\cdot296\right)dT \]

\[ = (1.756352\,T+0.000344\,T^2-580.16\,\mathrm{Lg}\,n\,T). \]

Substituting here the upper and lower limits for \(T\), we obtain

\[ Q=3488 \text{ calories}. \]

Let us now turn to determining the productivity of the boiler \(P\).

Let the steam feed an engine with a condenser, in which a temperature \(273^\circ+30^\circ=303^\circ\) is maintained. Thus the evolution takes place between the steam temperatures \(453^\circ\) and \(303^\circ\). Rankine’s cycle gives in this case 187 calories for 1 kg of steam, and 6.357 kg will give 1191 calories.

This will be the desired \(P\).

If by \(\gamma\) we denote the total heat capacity of 6.357 grams of water and by \(R\) the heat of vaporization at the temperature \(453^\circ\), then the quantity \(P\) may be expressed as follows:

\[ P=\int_{Tc}^{Tg}\left(1-\frac{Tc}{T}\right)\gamma\,dT+R\left(1-\frac{Tc}{Tg}\right)=1191. \]

The difference \(Q-P=3488-1191=2299\) will represent the losses in the process. These losses must be sought where, according to the Carnot–Clausius principle, irreversible phenomena occur.

1. Loss through smoke and ash.

The furnace gases and unburned particles of fuel, after heating the fire tubes, enter the smoke box, where the temperature is

\[ Tf=273^\circ+310^\circ=583^\circ. \]

Here they are cooled to the temperature

\[ 273^\circ+23^\circ=296^\circ=T_0. \]

Calling the heat capacity of the gases and unburned fuel

through \(C' + C'' = 1.952 + 0.000688T\), we shall calculate the fuel loss. Upon cooling from \(T+dT\) to \(T\) the heat loss will be

\[ (C' + C'')\,dT. \]

If, instead of being discharged into the atmosphere at temperature \(T_0\), this heat were made to work according to the Carnot cycle between the temperatures \(T\) and \(T_0\), we would obtain work, in calories,

\[ (C' + C'')\left(1-\frac{T_0}{T}\right)dT. \]

The total loss will then be equal to the work delivered by the cycle

\[ \int_{T_0}^{T}(C' + C'')\left(1-\frac{T_0}{T}\right)dT = \]

\[ = \int_{296}^{583}(1.952 + 0.000688T)\left(1-\frac{T_0}{T}\right)dT = \]

\[ = \int_{296}^{583}\left(1.952 + 0.000688T - \frac{1.952\cdot296}{T} - 0.000688\cdot276\right)dT . \]

\[ = \left(1.952T + 0.000344T^2 - 1.952\cdot296\operatorname{Lgn}T - 0.000688\cdot296T\right). \]

Substituting here the upper and lower limits, we obtain 197 calories.

As for the loss in the ash of the smoke box, it is expressed by the integral

\[ \int_{T_0}^{X} C'''\left(1-\frac{T_0}{T}\right)dT = \int_{296}^{2090} 0.008\left(1-\frac{296}{T}\right)dT. \]

The calculation gives about 12 calories.

The total loss will be

\[ A_1 = 197 + 12 = 209 \text{ calories}. \]

2. Losses due to radiation and conduction.

Cooling from \(2090^\circ\) to \(583^\circ\), the furnace gases and the unburned fuel give up heat

\[ \int_{Tf}^{X}(C' + C'')\,dT = \int_{583}^{2090}(1.952 + 0.000688T)dT. \]

The calculation of this integral gives 4317 calories.

Only part of this heat is transmitted to the boiler water, while the remainder is lost to radiation and conduction.

Since the feed water for the boiler is supplied at a temperature of \(294.3^\circ\), the boiler takes

\[ \int_{T_0^1}^{Tg} \gamma\,dT + R = (453 - 294.3) + 482.95 = 641.65 \]

\[ 641.65 \times 6.357 = 4078.97 = \sim 4079. \]

Thus, the difference

\[ 4317 - 4079 = 238 \text{ calories} \]

will be the loss due to radiation and conduction.

Let us find the corresponding loss of work. It will, of course, depend on the temperature at which the various parts of the 238 calories are given off. Let \(C^{IV}\,dT\) be the heat lost at temperature \(T\). Then one may write

\[ \int_{Tf}^{X} C^{IV}\,dT = 238. \]

Hence it is easy to find that \(C^{IV} = 0.158\).

The corresponding loss due to heat radiation and conduction will be

\[ A_2 = \int_{Tf}^{X} C^{IV}\left(1-\frac{T_0}{T}\right)dT = \int_{583}^{2090} 0.158\left(1-\frac{T_0}{T}\right)dT = 163. \]

3. Loss during heating.

Heat in the amount

\[ \int_{Tf}^{X} (C' + C'' - C^{IV})\,dT = \int_{T_0}^{Tg} \gamma\,dT + R = 4074 \]

is brought by the gases and transferred to the walls of the boiler tubes. This process of heat transfer is an irreversible phenomenon. Indeed, the temperature of the steam in the boiler is \(273^\circ + 180^\circ = 453^\circ\), while the temperature of the gases varies within the limits from \(X = 2090^\circ\) to \(Tf = 583\).

Here the greatest loss is obtained.

Heat in the amount

\[ (C' + C'' - C^{IV})\,dT \]

is given off by the gases at temperature \(T\).

Let us make this heat perform work according to the Carnot cycle between \(T\) and \(Tg\). Then we obtain the work

\[ (C' + C'' - C^{IV})\left(1-\frac{T_0}{T}\right)dT. \]

This expression represents the mechanical power of the heat.

This heat goes to steam generation, giving a thermal effect in the form of

\[ (C^1 + C'' - C^{IV})\left(1-\frac{T_0}{Tg}\right)dT \]

The difference between these quantities of heat will represent the loss

\[ T_0(C^1 + C'' - C^{IV})\left(\frac{1}{Tg}-\frac{1}{T}\right)dT \]

The total loss will be expressed by the definite integral

\[ A_3 = T_0 \int_{Tf}^{X} (C^1 + C'' - C^{IV})\left(\frac{1}{Tg}-\frac{1}{T}\right)dT = \]

\[ = 296 \int_{583}^{2090} (1.952 + 0.000688\,T - 0.158) \left(\frac{1}{453}-\frac{1}{T}\right)dT \]

\[ = 296 \int_{583}^{2090} (1.794 + 0.000688\,T) \left(\frac{1}{453}-\frac{1}{T}\right)dT = \]

\[ \ldots = \frac{296}{453}\int_{583}^{2090} (1.704 + 0.000688T)\left(\frac{T - 453}{T}\right)dT = 1703\ \text{calories} \]

Thus, \(A_3 = 1703\) calories.

4. Loss during feed.

Among the phenomena that cannot be neglected is the transfer of boiler heat to the feed water.

The water is taken from the condenser. In this case its temperature rises gradually from \(303^\circ\) to \(453^\circ\). When the temperature is raised from \(T - dT\) to \(T\), heat \(\gamma\,dT\) is taken from the boiler.

The loss in heating in this case will be

\[ A_4 = T_0 \int_{t_c}^{T_g} \left(\frac{1}{T} - \frac{1}{T_g}\right)\gamma\,dT = \]

\[ = 296 \int_{303}^{453}\left(\frac{1}{T} - \frac{1}{453}\right)6.357\,dT = \sim 131. \]

Since the feed water is taken not at the condenser temperature \(303^\circ\), but at the temperature \(294.3^\circ\), this circumstance also must be taken into account:

\[ T_0 \int_{T_0}^{T_c}\left(\frac{1}{T_0} - \frac{1}{T_g}\right)\gamma\,dT. \]

But this is still not all. The water from the condenser, being in contact with the atmosphere, is brought to the temperature \(T_0\).

The corresponding loss of work will then be the following:

\[ \int_{T_0}^{T_c}\left(1 - \frac{T_0}{T}\right)\gamma\,dT. \]

Since \(T_0 > T_0'\), the water would have to be cooled, yielding a negative loss equal to

\[ -\int_{T_0'}^{T_0}\left(\frac{T_0}{T} - 1\right)\gamma\,dT. \]

In final form, the additional loss in feeding the boiler therefore takes the following form:

\[ A_5 = \left(1 - \frac{T_0}{T_g}\right)\int_{T_0'}^{T_0}\gamma\,dT = \left(1 - \frac{296}{453}\right)\int_{294}^{303} 6.357\,dT = 19.8\ \text{calories}. \]

5. Loss in the condenser.

The condenser quite naturally must be connected to the boiler, and therefore, when the quantity \(P\) was calculated, it was assumed that the evolution of steam occurred between the temperatures \(T_g = 453\) and \(T_c = 303^\circ\), and did not fall to \(T_0 = 296^\circ\).

When, however, the quantity \(Q\) was calculated, it was assumed that all machines operating according to the Carnot cycle have the final temperature \(T_0\). For this reason, in \(Q - P\) the corresponding loss must necessarily appear.

And indeed, the water condensing in the condenser gives up heat to the water of the cooler; and the water of the cooler gives up heat to the atmosphere. This process is irreversible, since it takes place between bodies at different temperatures.

Thus, the Rankine cycle takes from the boiler the heat

\[ \int_{T_c}^{T_g} \gamma\, dT + R \]

and gives the work

\[ \int_{T_c}^{T_g} \gamma\left(1-\frac{T_c}{T}\right)dT + R\left(1-\frac{T_c}{T_g}\right), \]

and gives up to the condenser

\[ \int_{T_c}^{T_g}\frac{T_c}{T}\,\gamma\,dT + R\frac{T_c}{T_g}. \]

If this heat is made to work according to the Carnot cycle, it will yield the work

\[ A_6= \left(1-\frac{T_0}{T_c}\right) \left[ \int_{T_c}^{T_g}\frac{T_c}{T}\,\gamma\,dT + R\frac{T_c}{T_g} \right] = \left(1-\frac{296}{303}\right) \left[ \int_{303}^{453}\frac{303}{T}\,6.357\,dT + R\frac{303}{453} \right] =66. \]

This, then, represents the loss in the condenser.

There are no further irreversible processes in the boiler.

The correctness of the conclusions is verified by the fact that

\[ Q=P+A_1+A_2+A_3+A_4+A_5+A_6;\quad 3488=1191+209+163+1709+ \]

\[ +131+19.8+66=\sim 3482.8. \]

As is seen, the difference obtained is insignificant, only about \(0.15\%\).

It remains for us to calculate the losses due to incomplete combustion of the coal \(A_7\), and due to the fact that not exactly as much air is supplied as is necessary for complete combustion, but with a certain excess. This loss will be \(A_8\).

To calculate \(A_7\), we shall first make use of the equation by which the temperature in the firebox can be determined.

\[ \int_{296}^{X} (1.96+0.000688\,T)\,dT=5364. \]

Evaluating this integral, we obtain

\[ 1.96X+0.000344X^2-1.96\times296-0.000344\cdot296^2=5364. \]

Thus, we obtain a quadratic equation, from which it is determined that

\[ X=\sim 2200^\circ. \]

To determine the work for these conditions, we shall use the equation

\[ (1.752T+0.000344T^2-580.16\,\mathrm{Lgn}\,T)=Q. \]

Substituting here the limits \(2200\) and \(296\), \(Q=3840\).

Hence the loss due to incomplete combustion is obtained as

\[ A_7=3840-3488=352. \]

The loss \(A_8\) for rarefaction of the gases, owing to a certain excess of supplied air, is not large, and it is calculated at only 40 calories.

\[ A_8 = 40 \text{ calories}. \]

Let us now draw up a summary of productivity and losses and express these quantities as a percentage of the calorific value.

If we compare this heat balance of a steam boiler with the balance that was given at the beginning of the article, it at once becomes clear wherein lies the error of the generally accepted method of heat balancing a boiler installation.

Thus, for example, the loss through gases and ash is determined in the old method as 12.1%, whereas in the new method these losses amount to only 5.4%.

The productivity of the boiler in the former method is 75.9%, while here it reaches only 31%.

Finally, the loss for heating is entirely left out of account, and this loss constitutes one of the most important items in the heat balance of a steam boiler. The most important defect of boilers, both of erroneous and of inescapable character, consists in the difference in temperatures that constantly exists between the furnace gases and the water of the generator.

Thus, in order to reduce this defect as far as possible, one must strive to increase the temperature of the steam generator, i.e. of the boiler, and this increase in temperature is closely connected with raising the steam pressure.

Until recent times this tendency toward increasing pressure always encountered a limit: the fear of raising the pressure too much because of a possible boiler explosion and the grave consequences connected with it, both for people and for the surroundings.

But once people have already moved forward, risking a break with the methodical increase of steam pressure; once installations exist that operate for long periods at pressures such as 60–100 atmospheres and even higher, this path should not be abandoned. It is necessary to study the conditions of using such high pressures, to study the properties of steam, and to strive to develop such a structural material for boilers as would be capable of withstanding both great pressure and high temperature.

Routine metallurgy, with its traditional methods, cannot satisfy the needs of machine building. It is necessary to create a new metallurgy, one that in its progress does not lag behind the demands of life.

If metallurgical minds are not capable of coping with this problem, then science in the broad sense of the word must be called upon for this task.

Such is the urgent task of the times, such are the demands of life, requiring an answer. Somehow a stereotyped practice has become established of praising internal-combustion engines and disparaging steam technology. As though internal-combustion engines have no shortcomings of their own, or these shortcomings are so insignificant that they are not worth mentioning, let alone constantly emphasizing.

Fairness requires noting that the vulnerable point of internal-combustion engines is the loss of heat through the exhaust gases. The temperature of these gases is always very high.

If in steam engines the defect is the comparatively low temperature of live steam, then in internal-combustion engines this defect is shifted toward the high temperature of the spent gases. In one case the head aches, and in the other—the feet.

We have already mentioned that the use of high-pressure steam brought steam installations, in terms of efficiency, closer to internal-combustion engines. Professor Juge foresaw this long ago, and therefore it is a special pleasure to note a new triumph of scientific research. Only insufficient awareness of scientific research pushes people onto the long road of achievements through practice. Even a well-conducted practical example needs further improvement in its theoretical foundation.

Medicine men, too, had great practice; but it is better not to mention what they gave humanity.

It is true that the path of science is strewn with thorns, against which an unfitted head is painfully pricked.

The tasks of science in the broad sense now come down to introducing organization into scientific work. On this subject a good article has been written by S. F. Oldenburg: “The Question of the Organization of Scientific Work” (Scientific Chemical-Technical Publishing House. Scientific-Technical Department of the V.S.N.Kh. Petrograd, 1923. Collection Creativity).

By this one must understand “that organization which we expect, and which we have the right to demand, so that the most precious thing created by humanity through personal creativity should not perish without revealing itself sufficiently or without bearing true fruit, so that great and valuable thoughts should not remain buried in the pages of manuscripts and printed books as material for future historians of science.”

When this organization of scientific work arises, then practice will be assigned its proper place, and there will be no need to call attention to and prove the necessity, for the tasks of life, of studying science.