S. K. Konyukhov, Prof., STI.

Apparatuses for Heating Water and Their Calculation

The use of heated water finds wide application in everyday life. We need heated water for heating; we use it in baths, bathtubs, laundries, in the kitchen, and in domestic life in general. On a large scale such water is needed in railway depots for washing boilers.

Such water finds still greater application in the textile industry. Thus, for example, to obtain fiber from flax straw, the latter is soaked for 3–4 days in retting pits at a temperature of 22°–32° C.

Finally, heated water is widely used in the chemical industry.

Thus, leaching raw soda melt in order to obtain a soda solution of high concentration is carried out with water heated to 40°C, which makes it possible to obtain a solution of 24°–25° Bé.

Leaching beet cossettes in diffusers is successfully carried out with hot water at a temperature not above 85°C. In this process, with greater or lesser speed (depending on the nature of the beet, the character of the cossettes, and other conditions), continuous extraction of sugar from the cells of the beet tissue takes place.

Very many similar examples could be cited. This is explained by the fact that an enormous number of chemical and physical processes proceed more intensively as the temperature rises. In chemical processes it is taken as a rule that the reaction rate approximately doubles when the temperature is raised by 10°C. Sometimes this rate is expressed by the formula: \(U = \sqrt{T}\), where \(T\) is the absolute temperature.

The principles on which heating apparatuses are built, and the construction of these apparatuses, are not complicated. It would be quite natural to expect that the calculation of such apparatuses also should not be intricate. The mechanical bases for such a calculation will reduce, on the one hand, to determining the thickness of the walls of the apparatus as a function of pressure, and, on the other, it is necessary to take into account the possibility of obtaining a tight and strong riveted joint. Boiler iron in this case is taken not less than 8 mm thick.

The more important aspect of the calculation of heating apparatuses is the determination of the proper heating surface, for the correct calculation of this surface guarantees the efficient operation of the installation.

All heating apparatuses for water are divided first of all into two classes: 1) periodically operating and 2) continuously operating.

In periodically operating apparatuses, the quantity of liquid being heated plays the leading role in calculating the heating surface. Professor Depp calls these apparatuses “boiler appliances.”

In continuously operating apparatuses it is always assumed that the quantity of liquid supplied for heating is equal to the quantity of liquid withdrawn. The measurement of these quantities is referred to a unit of time.

Furthermore, heating apparatuses are divided into apparatuses with one and with two currents. The latter, in turn, may be subdivided into direct-flow apparatuses, counterflow apparatuses, and vortex apparatuses.

Let the heating substance be steam.

The initial equation for determining the heating surface is the following:

\[ W = Q\,(t'_1 - t'_2), \]

where \(W\) represents the quantity of heat transferred to the water, \(Q\) is the weight of the water, \(t'_1\) is the final temperature, and \(t'_2\) is the initial temperature of the water.

Izvestiya STI, vol. 48 (II), issue 4.

According to Rietschel, the heating surface is calculated as follows:

\[ F=\frac{W}{\vartheta_m K} \]

Here \(\vartheta\) represents the mean temperature difference. Hausbrand calls the product \(\vartheta_e \vartheta_a\) the mean temperature of heat exchange, where \(\vartheta_a\) represents the maximum difference in temperatures at one end of the heating surface (\(\vartheta_e\) is the minimum difference in temperatures at the other end of the heating surface).

As for the factor \(K\), for the time being we shall call it the coefficient of proportionality.

If the values of \(\vartheta_e\) and \(K\) were known, then determining the heating surface would present no difficulty.

Let the temperature of the steam at entry into the apparatus be denoted by \(t_1\), and the same temperature of the steam at exit from the apparatus by \(t_2\).

Then the mean temperature difference may be calculated by the following approximate formula:

\[ \vartheta_m=\frac{(t_1+t_2)}{2}-\frac{(t'_1+t'_2)}{2} \]

However, in this case it is more expedient to use the more exact formula given by Grashof:

\[ \vartheta_m=\frac{\vartheta_a-\vartheta_e}{\mathrm{Lg}_n \frac{\vartheta_a}{\vartheta_e}} \]

The difference in the results of calculating \(\vartheta_m\) by these formulas may amount to 15%.

Therefore, when large installations are involved, it is better to apply the logarithmic formula at once.

The following table, compiled by Hausbrand, considerably facilitates the calculation of \(\vartheta_m\) by the logarithmic formula:

If \(\dfrac{\vartheta_e}{\vartheta_a}=\)	0.0025	0.005	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
then \(\dfrac{\vartheta_m}{\vartheta_a}=\)	0.166	0.188	0.215	0.251	0.277	0.298	0.317	0.335	0.352	0.368	0.378
If \(\dfrac{\vartheta_e}{\vartheta_a}=\)	0.10	0.11	0.12	0.13	0.14	0.15	0.16	0.17	0.18	0.19	—
then \(\dfrac{\vartheta_m}{\vartheta_a}=\)	0.391	0.405	0.418	0.430	0.440	0.451	0.461	0.466	0.478	0.489	—
If \(\dfrac{\vartheta_e}{\vartheta_a}=\)	0.20	0.21	0.22	0.23	0.24	0.25	0.30	0.35	0.40	0.45	—
then \(\dfrac{\vartheta_m}{\vartheta_a}=\)	0.500	0.509	0.518	0.526	0.535	0.544	0.583	0.624	0.658	0.693	—
If \(\dfrac{\vartheta_e}{\vartheta_a}=\)	0.50	0.55	0.60	0.65	0.70	0.75	0.80	0.85	0.90	0.95	1.00
then \(\dfrac{\vartheta_m}{\vartheta_a}=\)	0.724	0.756	0.786	0.815	0.843	0.872	0.897	0.921	0.953	0.982	1.00

It will be shown below how to use this table.

Let us now clarify the value and meaning of the proportionality coefficient \(K\). It is more properly called the coefficient of heat transfer.

If we turn to the literature data on this question, we immediately encounter a series of entirely discordant quantities.

According to Hütte, \(k = 300—600\) when heat is transferred from steam to non-boiling water.

For exactly the same conditions, other authors (Recknagel, Fischer, Rietschel) give \(K = 800—1000\).

Klinger takes \(K = 500\), while Peclet gives data according to Toma and Leranu, who assume \(K = 1720\).

Thus the difference between the extreme limits amounts almost to 500%.

If behind these numbers there did not stand the names of major authorities, one could easily ascribe the disagreement to the coarseness of the measurements. And still the question would remain unresolved: whom should one believe?

The best way out of the difficulty here would, of course, be experiment, that much-loved and much-praised means for obtaining the correct answer to the demands of life.

But one may also proceed by another route.

Let Fig. 1 represent part of the curved heating surface in section.

This surface, or wall, is made of a completely homogeneous material, is bounded by parallel sides, and around it there is absolutely no motion of air.

Let us denote by \(F\) the surface giving off heat on the side \(CД\),

\(F_1\) — the surface receiving heat on the side \(AB\),

\(t_0\) and \(t\) — the temperatures of the air near \(CД\) and \(AB\),

\(\tau_0\) and \(\tau_1\) — the temperatures on the surfaces \(CД\) and \(AB\),

finally, \(\alpha_1\) and \(\alpha_2\) the coefficients of giving off and receiving heat. These coefficients denote the quantity of heat which is given off and received in one hour by one square meter when the difference between the temperatures of the heated surfaces and the air is one degree.

In the ideal case the whole quantity of heat \(W\) given off by the surface \(F\) will be received by the surface \(F_1\).

Thus one may write

\[ W = F \cdot \alpha_1(\tau_0 - t_0) \quad \text{and} \quad W = F_1 \cdot \alpha_2(t - \tau_1). \]

Whence we find:

\[ \tau_0 = t_0 + \frac{W}{F\alpha_1} \quad \text{and} \quad \tau_1 = t - \frac{W}{F_1\alpha_2}. \]

Let us single out in the wall a layer of thickness \(dx\) at a distance \(x\) from \(CД\) along the normal to the surface \(CД\) (and therefore also to \(AB\)). This layer will represent an ideally thin wall, situated concentrically to the sides \(CД\) and \(AB\) and bounded by sides with surfaces \(f\) (the difference between these surfaces will be expressed by a quantity of the second order).

Through the layer with surface \(f\), during one hour, there will pass the same quantity of heat as is given off and received by the surfaces \(F\) and \(F_1\).

Let us denote by \(\delta\) the thickness of the wall in mtr, by \(\vartheta_x\) the temperature in the layer cut out, and by \(\lambda\) the coefficient of thermal conductivity of the wall material; in other words, the quantity of heat which is transmitted in one hour through a layer one meter thick, with an area of one sq. meter, when the temperature difference is one degree.

With an established heat flow we shall have:

\[ W = \lambda \cdot f \frac{d\vartheta_x}{dx} \quad \text{or} \quad d\vartheta_x = \frac{W \cdot dx}{f \cdot \lambda}. \]

It is not hard to see that for \(x=0\), \(\vartheta_x=\tau_0\); for \(x=\delta\) the quantity \(\vartheta_x=\tau_2\).

Consequently,

\[ \tau_2-\tau_0=\frac{W}{\lambda}\int_0^\delta \frac{dx}{f}. \]

Substituting into this equation the values \(\tau_0\) and \(\tau_2\), we obtain

\[ t_1-\frac{W}{F_1\alpha_2}-t_0-\frac{W}{F\alpha_1} = W\frac{1}{\lambda}\int_0^\delta \frac{dx}{f}. \]

From this the magnitude \(W\) is found:

\[ W= \frac{t_1-t_0}{ \dfrac{1}{F_2\alpha_2} + \dfrac{1}{F\alpha_1} + \dfrac{1}{\lambda}\int_0^\delta \frac{dx}{f} }. \]

If the wall through which heat is transmitted is flat, then \(F_1=F_2=F\). In that case the expression for \(W\) takes the form

\[ W= \frac{(t_1-t_0)F}{ \dfrac{1}{\alpha_1} + \dfrac{1}{\alpha_2} + \dfrac{\delta}{\lambda} } = kF(t_1-t_0), \]

where \(K\) is determined from the equation:

\[ \frac{1}{K} = \frac{1}{\alpha_1} + \frac{1}{\alpha_2} + \frac{\delta}{\lambda} \qquad\qquad\qquad (a) \]

This dependence between \(K\) and the other quantities entering into the second part of the equation must serve as the basis for calculating the heating surface of preheating apparatuses.

According to the most recent data, for condensing water vapor, \(\alpha_1=10000\).

The quantity \(\alpha_2\) denotes the coefficient of heat transfer to the water being heated. According to Joule, it may be determined from the equation

\[ \alpha_2=300+1800\sqrt{V_f} \]

where \(V_f\) denotes the velocity of the water being heated.

If one takes into account that in equation (a), in addition to \(\alpha_1\) and \(\alpha_2\), there also appears \(\delta\)—the thickness of the wall of the heated body; that \(\lambda\) is present here, denoting the conductivity of the material; and that \(\alpha_2\) depends on the velocity of the water being heated, then the great completeness of the factors for determining the value of \(K\) becomes clear.

Sometimes formulas of a more simplified form for determining \(K\) are encountered in handbooks.

Thus, the same Joule gives the equation:

\[ K=1700\sqrt[3]{V_f}\qquad (b). \]

Mollier, on the basis of Ser’s experiments, gives the following formula:

\[ K=3300\sqrt[3]{V_f}\qquad (v). \]

In both these formulas \(V_f\) denotes the velocity of the water being heated, but, as is evident, Ser’s experiments give a value of \(K\) almost twice as large.

The divergence of the results is explained by the fact that Joule operated in his experiments with vertical tubes of small diameter, whereas Ser experimented with horizontal tubes, in which the heated particles of water can move away more quickly from the wall of the heating surface and give access to new particles of water. As for vertical tubes, in them each particle of liquid must rise along the heated surface and remain on it longer.

This explanation is justified by Nicol’s experiments, who established that the heating of a liquid occurs, with horizontal tubes, approximately one and a half times more intensively than with vertical ones.

This observation applies only to boiler fittings and open vessels. In apparatuses with two currents (parallel currents or countercurrents), however, the direction of motion is forced and is determined to a considerable extent by the choice of the velocities of the currents.

In addition to the simplified formulas of Joule and Mollier, there is also a more complex formula proposed by Hausbrand.

\[ K = 750 \sqrt{V_d}\sqrt[3]{\,0.007 + V_f\,}\cdot 0.66 \tag{г} \]

In this formula \(V_d\) denotes the velocity of the steam, and \(V_f\) the velocity of the water. The numerical factor 0.66 is taken for purely practical considerations, in order ultimately to increase the heating surface and thereby take into account circumstances disregarded in the calculation, such as, for example, the formation of scale on the walls, which considerably reduces the degree of heat transfer from the heated wall to the water. As for the coefficient 750, it represents the value of \(K\) referred to one square meter at a temperature difference of \(1^\circ\text{C}\).

The nature of the material from which the heating surface is made is taken into account as follows: in Hausbrand’s formula it is assumed that the material is copper. For brass one must introduce an additional factor of 0.95, for iron 0.75, and for lead 0.50.

The following table, taken from Wolffrum (Chem. Praktikum), gives the values of the coefficient \(K\) for copper tubes according to Hausbrand’s formula.

\(V_f\) in mtr	\(V_d\) in mtr: 1	\(V_d\) in mtr: 2	\(V_d\) in mtr: 4	\(V_d\) in mtr: 6	\(V_d\) in mtr: 9	\(V_d\) in mtr: 12
coefficient \(K\)	coefficient \(K\)	coefficient \(K\)	coefficient \(K\)	coefficient \(K\)	coefficient \(K\)	coefficient \(K\)
0.001	150	210	300	375	450	525
0.008	187	262	375	448	562	655
0.020	225	315	450	563	675	788
0.035	262	367	524	655	786	917
0.056	300	425	600	750	900	1050
0.085	337	475	674	842	1011	1179
0.117	375	528	750	937	1125	1312
0.160	412	580	824	1030	1236	1412
0.210	450	631	900	1110	1350	1575
0.266	487	683	975	1230	1461	1704
0.500	600	846	1200	1500	1800	2100
1.000	750	1057	1550	1925	2350	2625

The same Wolff gives the relation between the length of the tubes and their diameter in the light in the following form:

\[ \frac{l}{d}=1.2\,\frac{c\cdot\gamma}{\vartheta_m}\cdot \frac{\sqrt{Vd}}{\sqrt[3]{\,0.07+V_f\,}} \]

Here \(l\) is the length of the tubes, \(d\) the internal diameter, \(c\) the heat of vaporization for 1 kg of steam, \(\gamma\) the weight of one cubic meter of steam, \(\vartheta_m\) the mean temperature difference, \(V_d\) the velocity of the incoming steam, and \(V_f\) the velocity of the water being heated.

Let us now examine how to use all the considerations given above for calculating the heating surface and other quantities, and what result is obtained thereby.

Example I. A heating apparatus operates on the counter-current principle and has a heating surface of 58 sq. meters. Under continuous and uniform operation it supplies 880,000 W. E. (heat units) per hour. The steam has a pressure of 0.1 atmosphere. The water is heated from \(55^\circ\mathrm{C}\) to \(89^\circ\mathrm{C}\). Brass tubes serve as the heating surface.

Since this example is taken from practice, let us verify the applicability of the theoretical conclusions to the calculation.

First of all, let us determine \(\vartheta_m\) by the approximate formula.

The temperature of steam at a pressure of 0.1 atmosphere, or equivalently at 1.1 absolute atmospheres, will be \(374.8-273=101.8\) (see Hütte, Handbook, vol. I, 9th ed., 1916). Without great error it may be assumed that the steam retains this temperature also at the outlet from the tubes. In that case:

\[ \vartheta_m= \frac{(t_1+t_2)}{2} - \frac{(t'_1+t'_2)}{2} = 101.8-72=29.8. \]

Here it has been assumed:

\[ t_1=t_2=101.8^\circ;\quad t'_1=89^\circ \text{ and } t'_2=55^\circ. \]

Let us find the same quantity \(\vartheta_m\) by the logarithmic formula:

\[ \vartheta_e=101.8-89=12.8;\quad \vartheta_a=101.8-55=46.8. \]

The ratio:

\[ \frac{\vartheta_e}{\vartheta_a} = \frac{12.8}{46.8} = 0.274. \]

Now we shall use Hausbrand’s table, in which we seek the number \(\frac{\vartheta_e}{\vartheta_a}\) close to 0.274. The nearest smaller number will be 0.250, and the next larger one will be 0.300. The difference between these numbers will be \(0.300-0.250=0.050\), while the difference between 0.274 and 0.250 will be 0.024.

The value \(\frac{\vartheta_m}{\vartheta_a}\) corresponding to \(\frac{\vartheta_e}{\vartheta_a}=0.250\) is 0.544, and for

\[ \frac{\vartheta_e}{\vartheta_a}=0.300 \]

the value

\[ \frac{\vartheta_m}{\vartheta_a}=0.583. \]

Thus the difference between these values of

\[ \frac{\vartheta_m}{\vartheta_a} \]

will be \(0.583-0.544=0.039\). It is now not difficult to find the correction by interpolation:

\[ x:0.039=24:50;\quad x=0.018 \]

Consequently, the true value of

\[ \frac{\vartheta_m}{\vartheta_a} \]

will be:

\[ \frac{\vartheta_m}{\vartheta_a}=0.544+0.018=0.562. \]

Hence \(\vartheta_m=0.562\,\vartheta_a=0.562 \times 46.8=26.3016\), or \(26.3\).

Let us now determine the value of \(K\).

Since we know the value \(W=880{,}000\) heat units, the heating surface is known, and, moreover, at the beginning we determined \(\vartheta_m\), we therefore have all the data for determining the actual heat-transfer coefficient.

\[ W=F\vartheta_mK;\quad K=\frac{W}{F\vartheta_m}. \]

Substituting the numerical values, we obtain:

\[ K=\frac{880{,}000}{58.26.3}=577 \]

\[ K=\frac{880{,}000}{58.29.8}=509 \]

Let us see what the formulas give us.

We shall write formula (a) in the following form:

\[ K=\frac{1}{\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{\delta}{\lambda}+\frac{\delta_1}{2}} \]

Here \(\delta_1=0.0005\) m. This is the thickness of the deposit layer.

\[ \alpha_1=10{,}000. \]

From Wolffrum’s table, by interpolation, we find that \(V_t=+0.052\).

Applying Joule’s formula, we obtain:

\[ \alpha_2=300+1800\sqrt{0.052}=300+400=710. \]

The wall thickness of the brass tubes is taken as \(0.001\) meter.

The thermal conductivity of brass (see Hütte, part I, p. 412) is \(\lambda=72\text{–}103\), i.e., on average \(90\).

Consequently,

\[ K=\frac{1}{\frac{1}{10000}+\frac{1}{710}+\frac{0.001}{90}+\frac{0.0005}{2}}=576. \]

By formula (b)

\[ K=1700\sqrt[3]{0.052}=634.5. \]

By formula (c)

\[ K=3300\sqrt[3]{0.052}=1231.7. \]

Finally, according to Hausbrand’s formula it is necessary to know in advance the steam velocity \(V_d\). Theoretically this velocity can be determined from the equation:

\[ \Delta_d=V_d\,\gamma_d\,\frac{\Pi d^2}{4}\cdot 3600, \]

where \(D\) is the weight, in kilograms, of the steam flowing through; \(\gamma_d\) is the weight, in kilograms, of one cubic meter; \(d\) is the diameter of the pipe.

Wolfrum gives the following summary table for the velocity of steam \(V_d\) in pipes of different diameter:

Diameter of the steam pipe, clear opening, mm. \((d)\)	Steam velocity in m/sec \((V_d)\)
25	8.5
30	9
35	9.5
40	10.5
45	11
50	11.5
60	13
80	14.5
100	15

Pipe length \(= 20\) meters, and steam pressure 1–6 atmospheres.

The second table gives the weight of steam passing through such pipes (in kg) at different pressures.

\(d_d\)	\(V_d\)	6	5	4	3	2	1 atmosphere
25	8.5	50	42	34	26	18	—
30	9	75	63	51	39	27	—
35	9.5	107	90	73	55	38	—
40	10.5	155	130	106	81	55	—
45	11	205	173	140	107	73	38
50	11.5	265	223	181	138	95	49
60	13	431	363	294	224	153	80
80	14.5	855	720	684	446	315	159
100	15.5	1429	1204	977	746	509	275

Since the diameter of the pipes is taken as 25 mm, \(V_d = 8.5\) meters.

Substituting this value into Hausbrand’s formula, we obtain:

\[ K = 750 \sqrt{8.50}\cdot \sqrt[3]{0.007 + 0.052 \cdot 0.66 \cdot 0.45} = 534. \]

We shall now calculate the heating areas by the formula, taking \(\vartheta_m = 29.8\) and \(\vartheta_m = 26.3\); the value \(K\), however, we take from formulas (a), (b), (c), and (d).

Then we obtain:

\[ F=\frac{W}{K\cdot \vartheta_m} \]

(a)	(b)	(c)	(d)
51.27	47	24.2	55.3	\(\vartheta_m = 29.8\)
59	52.8	27.2	62.5	\(\vartheta_m = 26.3\)

From this it is evident how closely the results of calculations according to the different sources approach reality.

As can be seen, the use of the logarithmic formula for determining \(\vartheta_m\) and the determination of \(K\) by formulas (a) and (d) gives very good final conclusions, and therefore these formulas should be used in calculating the heating surface of heaters.

Example II. The apparatus has a heating surface \(F = 28\) sq. meters. It must deliver 1,025,000 W. E. (units of heat) per hour for heating water from \(70^\circ\text{C}\) to \(100^\circ\text{C}\). Steam pressure: 1.5 atm.

Let us preliminarily determine two variants of \(\vartheta_m\).

Since the temperature of steam at a pressure of 1.5 atmospheres (or 2.5 abs. atm.) is \(126.7\), then

\[ \vartheta_m=\frac{(126.7+126.7)}{2}-\frac{(100+70)}{2}=41.7. \]

By the logarithmic formula we obtain

\[ \vartheta_e=126.7-100=26.7;\quad \vartheta_a=126.7-70=56.7. \]

Hence

\[ \frac{\vartheta_e}{\vartheta_a}=\frac{26.7}{56.7}=\sim 0.471. \]

We take the approximate value \(\frac{\vartheta_e}{\vartheta_a}=0.45\) and the corresponding value \(\frac{\vartheta_m}{\vartheta_a}=0.693\). By interpolation we find the correction \(x=0.018\).

Thus,

\[ \frac{\vartheta_m}{\vartheta_a}=0.693+0.018=0.711. \]

The value

\[ \vartheta_m=0.711\cdot\vartheta_a=0.711\times56.7=40.3. \]

We shall now calculate the coefficient \(K\) by the four equations.

\[ K=\frac{1}{\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{\delta}{\lambda}+\frac{\delta_1}{2}} \]

Here, again, \(\alpha_1=10000\); \(\alpha_2=300+1800\sqrt{V_f}\). Since the tubes here are taken with a diameter of 40 mm, \(V_d=10.5\) m.

According to Fortu and Nolet, when heating with steam flowing through the tubes, the water space must be such in magnitude that the water remains in the heater no more than 12–15 minutes. The cross-sectional area of the tubes is taken as twice ...

greater than the cross-section of the pipe supplying the steam (crumpled), and the surface of the pipes must amount to 0.08 of the heating surface of the boiler. If the water moves through the tubes, then only half as much time is required, but the tube surface is doubled. (Pohlhausen Dampfkesselanlagen S. 171, Zweite Auflage).

The coefficient \(K\), on the basis of the data, assumes the values:

\[ K=\frac{1.025.000}{28\cdot 41,7}=877;\qquad K=\frac{1.025.000}{28\cdot 40,3}=987. \]

Substituting these values into Joule’s formula, we shall have:

\[ 877=1700\sqrt[3]{V_f};\qquad 987=1700\sqrt[3]{V_f}. \]

Whence we obtain:

\[ V_f=0,137 \quad \text{and} \quad V_f=0,140. \]

We therefore take from Wolfurm’s table the nearest value for \(V_f=0,160\). The value of \(a_2\) will be:

\[ a_2=300+\sqrt{0.160}\cdot 1800=1020. \]

By formula (a)

\[ K=\frac{1}{\dfrac{1}{10000}+\dfrac{1}{1020}+\dfrac{0,001}{100}+\dfrac{0,0005}{2}}=746. \]

By formula (b)

\[ K=1700\sqrt[3]{0,160}=923. \]

By formula (c)

\[ K=3300\sqrt[3]{0,160}=1792. \]

By formula (d)

\[ K=750\cdot \sqrt{10,50}\cdot \sqrt[3]{0,007+0,16\cdot 0,66\times 0,95}=838. \]

Now let us find the heating surface for \(\vartheta_m=41,7\) and \(\vartheta_m=40,3\) and for various values of \(K\)

\[ F=\frac{32,95}{34}\quad \frac{26,63}{29,55}\quad \frac{13,58}{14,05}\quad \frac{29,62}{30,34}\quad \frac{\vartheta_m=41,7}{\vartheta_m=40,3} \]

It is evident from this that Hausbrand’s formula for determining \(K\) gives the most satisfactory results; Molle’s formula gives a greatly reduced heating surface.

Example III. It is required to heat 20,150 meters of water from 19°C to 90°C. The heating surface consists of 21 copper tubes, 60 mm in diameter and 1.2 meters long. The heating surface is thus equal to 9.5 sq. meters. The steam pressure is equal to 0.15 atmosphere, or 1.15 abs. atm. From the tables we find the steam temperature to be 103°C.

With a diameter of the steam-conducting tubes of 60 mm, \(V_d=13\) m.

We determine \(\vartheta_m\):

\[ \vartheta_e=103-90=13;\qquad \vartheta_a=103-19=84 \]

\[ \frac{\vartheta_e}{\vartheta_a}=0,155. \]

To this value of \(\frac{\vartheta_e}{\vartheta_a}\) there corresponds

\[ \vartheta_m=0,456\times 84=38,30. \]

The number of calories received by the steam will be:

\[ (90-19)\,20150=1.430.650. \]

In that case \(K\) will have the value:

\[ K=\frac{W}{\vartheta_m F}=\frac{1430650}{38.3\times 9.5}=3932. \]

By Hausbrand’s formula we obtain the following value:

Since in Wolfrum’s table there are no values of \(K\) above 2625, we shall use the formula:

\[ d=\frac{1.2\cdot c\gamma}{\vartheta_m} =\frac{\sqrt[3]{V_d}}{\sqrt{0.007+V_f}}. \]

or:

\[ \frac{120}{6} = 1.2\cdot\frac{537.45\cdot 0.6618}{38.3} \cdot \frac{\sqrt[3]{13}}{\sqrt{0.007+V_f}}. \]

After multiplication we obtain:

\[ 20\sqrt[3]{0.007+V_f}=26.18. \qquad \text{Whence } V_f=2.19. \]

If \(K\) is calculated by Hausbrand’s formula, then we obtain \(K=3510\), which differs considerably from 3932.

In order to come closer to reality, it is necessary to increase the velocity \(V_d\). Let us take the following number, 14.5. Then by Hausbrand’s formula we obtain 3715.

Engineer K. Schubart gives the following values for \(K\):

\[ K_{\text{observ.}}=3870;\qquad K_{\text{calc.}}=3600. \]

The difference in the results arises because K. Schubart takes \(V_f=1.98\) and \(\vartheta_m=38.38\). However, I did not obtain these numbers.

For the calculation of \(W\) we have so far used the equation

\[ W= \frac{t_1-t_0}{ \frac{1}{F_1\alpha_1} + \frac{1}{F\alpha_2} + \frac{1}{\lambda}\int_0^\delta \frac{dx}{f} }. \]

For a flat wall this equation becomes:

\[ W= \frac{(t_1-t_0)F}{ \frac{1}{\alpha_1} + \frac{1}{\alpha_2} + \frac{\delta}{\lambda} }. \]

In heating apparatuses with two streams, the question of which principle should be preferred is a very important one.

If the technical task is reduced to bringing the heated water to as high a temperature as possible, then apparatuses with counterflow must always be chosen. If this is not the case, then apparatuses with parallel flows are chosen, because in them the heating surface, for a large temperature difference, comes out smaller.

In practice, however, even in this case counterflow apparatuses are often chosen in order to attain the highest possible water temperature, by virtue of which, as for example in feeding steam boilers, one may obtain a small quantity of water receiving heat. In turn, this reduction in the quantity of heated water affects the size of the feed pumps. Naturally—

Naturally, at the same time the expenditures for the drive are reduced, which compensates for the depreciation costs of the expensive counterflow apparatuses.

Let us consider an example. Suppose we have a water-to-water heating surface. Let the temperature of the heating water be \(+90^\circ\mathrm{C}\), and that of the return water \(+70^\circ\mathrm{C}\). The water to be heated must be brought from \(+10^\circ\mathrm{C}\) to a temperature of \(+80^\circ\mathrm{C}\). In this case it is necessary to use an apparatus operating on the counterflow principle, because the water being heated must flow toward the point at which the heating water enters.

When using the principle of parallel currents, one can count on heating only up to \(+60^\circ\mathrm{C}\).

The mean temperature of heat exchange is found for these cases as follows:

A. Counterflow.

\[ \vartheta_e = 80 - 70 = 10 \qquad \vartheta_a = 90 - 10 = 80 \qquad \frac{\vartheta_e}{\vartheta_a} = 0.125. \]

From Hausbrand’s table we find:

\[ \frac{\vartheta_m}{\vartheta_a} = 0.418 + 0.006 = 0.424. \]

Consequently,

\[ \vartheta_m = 0.424 \times 80 = 33.92 = \sim 34. \]

B. Parallel currents

\[ \vartheta_e = 90 - 60 = 30 \qquad \vartheta_a = 70 - 10 = 60 \qquad \frac{\vartheta_e}{\vartheta_a} = 0.5. \]

From Hausbrand’s table we find:

\[ \frac{\vartheta_m}{\vartheta_a} = 0.724; \qquad \vartheta_m = 0.724 \times 60 = 43.44. \]

It is clear from this that with parallel currents the heating surface will be smaller.

If it is required to deliver, by counterflow, 3,000,000 heat units to water heated from \(10^\circ\mathrm{C}\) to \(80^\circ\mathrm{C}\), then the amount of heating water required is:

\[ \frac{3{,}000{,}000}{(80 - 10)\cdot 1000} = 42.85 \text{ cubic meters}. \]

When heating from \(+10^\circ\mathrm{C}\) to \(60^\circ\mathrm{C}\) by parallel current, the amount required is:

\[ \frac{3{,}000{,}000}{(60 - 10)\cdot 1000} = 60 \text{ cubic meters}. \]

These quantities must also be supplied by the corresponding heater.

For apparatuses with parallel currents, more energy must be expended on supplying the water than for counterflow apparatuses. For this reason, apparatuses of the latter type, despite their greater cost, are preferred to apparatuses with parallel currents.

A few more words should be said concerning heating apparatuses operating in one current and combined single-current vortex apparatuses, such as boiler devices and open vessels with a coil.

Let us again take, for heating water, steam and the low velocity of water that is observed in these apparatuses.

This insignificant velocity of the water has a strong effect in lowering the coefficient of heat transfer and causes boiling of the water. In most cases

these apparatuses operate periodically and, thus, the quantity of liquid intended for heating is a constant value. If one observes the heating surface in such apparatuses, one may note a more considerable circulation of water at the inlet than at the outlet. This phenomenon is especially noticeable when the water is boiling. Hence it is clear that the elasticity and temperature of the steam in different places of the device are different. The elasticity of the steam at the end is very small (close to zero); it gradually increases with increasing heating and finally passes into an unchanging form. It follows from this that the diameter and length of the pipes must be in a definite ratio.

Taking this consideration into account, Rietschel and Brabbée calculate coils with allowance for the fall in elasticity.

The same idea is also pursued by Hausbrand, expressing the proposition that the heat given off by the heating tube must be equal to the heat of the steam supplied, i.e.,

\[ F \cdot K \cdot \vartheta_m = \frac{\Pi d^2}{4} V_d \cdot 3600 \cdot c \cdot \gamma . \]

Here \(C\) denotes the heat of evaporation, \(\gamma\)—the weight of a cubic meter of steam. The velocity of the water in apparatuses is taken as comparatively not large, \(V_f = 0.02\text{—}0.3\) meters, with the smaller numbers referring to large vessels and low liquid temperatures, approximately \(\leq 60^\circ C\).

For small vessels and higher temperatures of \(60^\circ\)—\(100^\circ C\), larger numbers must be taken.

If the velocity of the water is unknown, then for determining \(K\) one may use the following formula:

\[ K = 0.65 \cdot 225 \sqrt{V_d} \sim 0.65 \cdot 450 \sqrt{V_d}. \]

If we substitute the values of \(F\) and \(K\) into the previously written formula, we obtain:

\[ \Pi d l \cdot 0.65 \cdot 225 \sqrt{V_d} \cdot \vartheta_m = \frac{\Pi d^2}{4} V_d \cdot 3600 \cdot c \cdot \gamma . \]

Hence we obtain:

\[ \frac{l}{d} = \frac{5.95 \sqrt{V_d} \cdot c \cdot \gamma}{\vartheta_m} \quad \text{and} \]

\[ \frac{l}{d} = \frac{3.08 \sqrt{V_d} \cdot c \cdot \gamma}{\vartheta_m}. \]

The following ratio is often encountered:

\[ \frac{l}{d} = \frac{5.5 \sqrt{V_d} \cdot c \cdot \gamma}{\vartheta_m}. \]

Example 3. Suppose it is necessary to select a heater for 2,000 liters of water, which must be heated from \(10^\circ\) to \(90^\circ C\) within one and a half hours. The heating surface is a copper coil, whose outside tube diameter is 76 mm, and inside diameter 70 mm. The total length of the coil tube is 14 m. The heating surface will be:

\[ \Pi d l = 3.14 \cdot 0.076 \cdot 14 = 3.34 \ \text{m}^2 \]

\[ \Pi d l = 3.14 \cdot 0.07 \cdot 14 = 3.077 = \sim 3.1 \ \text{sq. m}. \]

Steam pressure is 0.4 atm. or 1.4 abs. atmospheres. We take \(V_d = 17\) and \(V_f = 0.166\). The velocity \(V_f = 0.166\) represents a number close to the mean value of normal velocities (range \(0.02\)—\(0.3\)). As for \(V_d = 17\), its choice must be justified.

Let us first determine \(\vartheta_m\). The temperature of the steam at an absolute pressure of 1.4 atmospheres will be 108.7.

Consequently:

\[ \begin{gathered} \vartheta_e = 108.7 - 90 = 18.7, \\ \vartheta_a = 108.7 - 10 = 98.7, \end{gathered} \qquad \frac{\vartheta_e}{\vartheta_a}=\frac{187}{987}=0.189. \]

From Hausbrand’s table we find:

\[ \frac{\vartheta_m}{\vartheta_a}=0.489;\qquad \vartheta_m=0.489 \times 98.7 = 48.26. \]

Further, according to Hütte (Part I, p. 444, 9th edition, 1916), we find:

\[ \text{Heat of vaporization—}533.7. \]

\[ \text{Weight of 1 cubic meter of steam—}0.7955. \]

Substituting these values into the formula, we obtain:

\[ \frac{1}{d} = \frac{5.5 \times 533.7 \times 0.7955}{47.16} \cdot \sqrt{V_d} = \frac{1400}{7} = 200 \]

or

\[ 9652 = 2335\sqrt{V_d};\qquad V_d = 17.05. \]

The quantity of heat transmitted to the water by the steam when heating from \(10^\circ\mathrm{C}\) to \(90^\circ\mathrm{C}\) during one hour will be equal to

\[ W= \frac{2000(90-10)\cdot 2}{3} = \sim 107.000\ \text{W. E.} \]

Thus, the heat-transfer coefficient will be:

\[ K=\frac{170\cdot 000}{48.26\times 3.1}=\sim 715. \]

According to Hütte’s formula,

\[ K= \frac{1}{ \frac{1}{10000} + \frac{1}{1032} + \frac{0.003}{335} + \frac{0.0005}{2} } =\sim 709. \]

According to Hausbrand’s formula:

\[ K = 0.65\cdot 225\sqrt{V_d}\ \text{to}\ 0.65\cdot 450\sqrt{V_d} \]

i.e.

\[ K=0.65\cdot 225\sqrt{17}=603. \]

\[ K=0.65\cdot 450\sqrt{17}=1206. \]

The following Hausbrand formula is often used:

\[ K=0.65\,\Pi\,250\sqrt{V_d}=670. \]

Let us now turn to the question of which diameter of tubes conducting steam or water for heating the water should be preferred: large or small.

If it is necessary to transmit a definite number of heat units per hour, then the basic equation for calculating this problem is the following:

\[ \text{W. E./hour}=\Pi d l (T-t)K. \tag{a} \]

Here: \(d\) — diameter of the tube in meters
\(\quad l\) — length of the tube in meters
\(\quad T\) — mean temperature inside the tube
\(\quad t\) — mean temperature outside the tube
\(\quad K\) — coefficient of heat transfer per hour through 1 sq. meter of heating surface with a temperature difference of \(1^\circ C\).

The value of the coefficient \(K\) was taken as follows:

\[ K=\frac{1}{\frac{1}{\alpha_1}+\frac{\delta}{\lambda}+\frac{1}{\alpha_2}}. \]

Equation (a) is nothing other than the simplified Fourier equation.

A more exact equation will be the following:

\[ \text{W. E./hour}= \pi d l (T-t)\cdot \frac{1}{ \frac{1}{d_1\alpha_1} + \frac{1}{2\lambda}\,\mathrm{Lg}_n\frac{d_a}{d_i} + \frac{1}{d_2\alpha_2} }. \]

This equation applies to cylindrical tubes, which are most often encountered in practice. In the more exact equation the following are denoted: \(\alpha_1\) and \(\alpha_2\) are the coefficients of heat transfer per hour through 1 sq. meter with a temperature difference of \(1^\circ C\); \(\lambda\) is the coefficient of thermal conductivity per hour through one meter with a temperature difference of \(1^\circ C\); \(\delta\) is the thickness of the tube wall in meters. The numbers \(\alpha_1\), \(\alpha_2\), and \(\lambda\) are expressed in heat units.

The exact formula passes into the simplified one if we put \(d_i=d_a\) and

\[ \mathrm{Lg}_n\frac{d_a}{d_i}=2\delta/d. \]

For investigation, the simplified formula has a more compact and easily surveyed form, while the essence of the conclusions does not change. We shall therefore take up the study of the simple formula. From the structure of this formula it is clear that, for a definite tube length \(l\), a definite temperature difference \(T-t\), and coefficient \(K\), the heat capacity of the tube is directly proportional to the diameter of the tube. Thus, by increasing or decreasing \(d\), we shall increase or decrease the heat capacity of the tube.

For a unit length of tube we obtain:

\[ \frac{\text{W. E.}}{l}=K\cdot \pi d (T-t) \]

and for a unit area of tube,

\[ \frac{\text{W. E.}}{\pi d l}=K(T-t), \]

Consequently, if \(\pi d l=\mathrm{Const}\), then the heat capacity remains unchanged.

If it is necessary to transfer 100,000 W. E. from water to water, when \(K=400\), with a temperature difference of \(50^\circ C\), then it would be possible to take tubes 100 meters long and \(0.0127\) m (\(1/2''\)) in diameter, or tubes 50 meters long and \(0.025\) m in diameter, or 25 meters long and \(0.0508\) m (\(2''\)) in diameter. Such arbitrariness in the choice of tubes for the heating surface would have its good side, giving the designer the possibility of settling on that grade of tubes which is cheaper, which is at hand, etc.

This conclusion found confirmation in the experiments of Stanton, Sera, Josse, Tann, Joule. In determining the magnitude \(K\), these investigators give material for determining \(\alpha_2\), where the velocity \(V_f\) plays a large role. It is enough only to take the formulas of Mollier (Sera) and Joule in order to confirm this idea.

Agreement in the observations of a large number of investigators (with the exception of Peclet) would seem to have provided grounds for the confident application of the formulas in practice. However, practice showed for the first time that something in the theory was not altogether satisfactory. In some cases the results obtained were better, in others worse, and since these results are translated in real life into the language of economy, it became necessary again to take up the study of the settled question by deepening the experiments. Thus arose the investigations of Holborn, Dittenberger, Austin, Engels, Donkin.

In Stanton’s formula the diameter of the tube appears

\[ WE=\frac{V^{m-1}}{d^{2-m}}(T-t)(1+\lambda T)(1+\beta t)\frac{k p^{2-m}}{4}. \]

(see Z. d. V. d. Ing. Forschungsheft 108–109)

In this case it follows that

\[ \alpha_2=C\frac{p^{2-m}}{d^{2-m}}V^{m-1}. \]

The quantity \(m\) has the following value

\[ m=1.84. \]

Nusselt, who constructed his theory of heat transfer on the Stokes–Kirchhoff law, came to exactly the same conclusion.

According to Nusselt:

\[ \alpha_2=1\lambda \cdot \frac{V^n}{d^{1-n}}\left(\frac{S}{\eta}\right)^n\left(\frac{C p \eta}{\lambda}\right)^m \]

Further, Zeneken, by exact experiments carried out in the laboratory of Professor Knoblauch in Munich, established the following relationships:

\[ \alpha_2=735\frac{V^{0.7}}{d^{0.3}}(1+0.014T_i) \]

for rough iron tubes in the transfer of heat from metal to water;

\[ \alpha=1020\frac{V^{0.69}}{d^{0.31}}(1+0.006T_i) \]

for rough iron tubes in the transfer of heat from water to metal.

Here \(T_i\) is the temperature of the tube wall.

The exponents of the velocity and of the diameter complement each other to unity.

Other investigators have different exponents for \(d\).

Thus, for example, Rietschel has \(d^{0.16}\); Holborn has \(d^{0.13}\), \(d^{0.18}\); Pensen has \(d^{0.164}\). On the contrary, Vamsler gives the same result as Zeneken.

Vamsler’s empirical formula has the following form:

\[ K=\frac{p(t_1-t_2)^{0.233}}{d^{0.3}}. \]

Here \(p\) represents a constant quantity, equal to 0.97 for cast iron, 0.91 for wrought iron, and 1.04 for copper. The quantity \(t_1\) denotes the temperature of the tube surface, and \(t_2\) the temperature of the surrounding air.

If we have the case of heating water by means of hot water, then Zeneken’s formula takes the following form:

\[ W \cdot E = \Pi d l (T-t)\cdot \frac{1}{ 1020 \frac{V^{0.69}}{d^{0.31}(1+0.006T)}+\frac{\delta}{\lambda} +735 \frac{V^{0.7}}{d^{0.3}(1+0.14t)} }. \]

This formula can be considerably simplified if one takes into account its application to the heating of water in heating devices or apparatuses. To begin with, the greatest permissible velocity is \(0.1—0.2\) mtr/sec. (of the heating water), while the velocity of the water being heated is not higher than \(1—1.5\) mtr/sec. Further, with respect to the temperatures of the tube walls, one may assume from purely practical considerations that \(T=70^\circ C\) (roundly), and \(t=40^\circ C\).

With these limitations we obtain:

\[ \frac{1}{\alpha_1}=\frac{d^{0.3}}{500} \quad \text{and} \quad \frac{1}{\alpha_2}=\frac{d^{0.3}}{1500}. \]

If these values are introduced into the previously written formula, it acquires the very simple form:

\[ W \cdot E = \sim \Pi d l (T-t)\cdot \frac{1}{\frac{d^{0.3}}{400}+\frac{\delta}{\lambda}}. \]

The simplification can be carried still further, since \(\frac{\delta}{\lambda}=\frac{0.003}{50}\) is a very small fraction, so that we finally obtain:

\[ W.E = \sim 400 \Pi l (T-t)d^{0.3}. \]

The influence of the diameter of the tube on the magnitude of the heat given off when in contact with air (at various temperatures) can be seen clearly from the diagram (Fig. 2). This dependence is given by Dr. Wamsler.

In the diagram (Fig. 3) the relation is given between the magnitude of the coefficient of heat transfer \(\alpha\) and the diameter of the tube according to Ten Bosch.

Finally, the diagram (Fig. 4) gives a clear representation of the number of heat units transmitted by one square meter of heating surface per hour, at a temperature difference of \(1^\circ C\), as a function of the tube diameter. This dependence is given by Dr. Zeneken, and it is based on the conditions: \(V_1=0.1\), \(V_2=1\) m/sa, \(T=70^\circ C\), \(t=40^\circ C\). The transfer of heat takes place from water to water.

In all the diagrams one and the same tendency is readily observed: the amount of heat transmitted by the heating surface decreases as the diameter of the tubes increases.

Consequently, when tubes of smaller diameter are used, a smaller heating surface is required in order to obtain the same heat effect.

In Dr. Zeneken’s diagram there is also a curve of the purchase cost of one running meter of tubes of various diameters (in peacetime). This dependence was given by Dr. Pfeiderer in 1914. It turns out that the costs of installing tubes, beginning with a diameter of 100 mm, increase much more noticeably than the thermal capacity of one running meter of tube; i.e., here too it proves more advantageous to employ tubes of small diameter.

Finally, from this same diagram it is seen that, for example, when the tube diameter is increased fourfold (from 50 mm to 200 mm), the thermal capacity of one running meter at a temperature difference \((T. D)\) of \(1^\circ C\) increases only twofold or a little more (from 100 \(W.E\) to 240 \(W.E\)), whereas

the specific thermal capacity, or heat output, of one square meter of tubular heating surface falls from 535 W. E. to 356 W. E. per sq. meter, i.e., by 33%.

Sometimes thermal capacity is expressed not in heat units (W. E.), but in kilowatts. The conversion from heat units to kilowatts is made according to the equation:

\[ 1\ \text{kilowatt-hour} = 864\ \text{W. E.} \]

\[ 1\ \text{kilowatt} = \frac{864\ \text{W. E.}}{\text{hour}}. \]

Therefore the following relation, written in Kalender für Gesundheitstechniker H Recknagel (28th edition, 1924), is quite incorrect:

\[ 1\ \text{kilowatt-hour} = 864\ \text{W. E. hours}!! \]

One may also use the following relation:

\[ 1\ \text{W. E.} = 4184\ \text{watt seconds}. \]

In Figs. 5–18 are shown various embodiments of water heaters built according to the counterflow principle. The designs are so clear that they require absolutely no explanation or description.

However, these designs do not exhaust all the variety of heaters, both of the more complex kind and of the simpler kind, which are encountered in practice.

In the next article we shall try, as far as possible, to gather everything known to us on this question, without, however, making any claim to exhaustive completeness.

Let us now turn to Fig. 19, which represents an apparatus invented by the Belgian engineer A. Heintz and bearing the name “saturator without pressure” (Satourateur sans pression).

The operation of this apparatus is as follows: steam enters, after the steam-distributing valve, into pipe H, box K, and, through valve S, into the annular space and then into the pipe. In this process the contents of box K are sucked in. Here we have simply an injector, through which steam passes, carrying air into the heating system. Having passed through this system, the exhaust steam is drawn in by the same injector J, together with air and condensate water, into pipe N. The mixture of air and steam rushes upward into box K, while the heavier water flows down into the discharge pipe L. Since pipe L communicates with the atmosphere, the whole system is called open, so that the pressure in it does not exceed atmospheric pressure. If it happens that the pressure rises above atmospheric pressure, the excess steam will escape through the same pipe to the outside. Conversely, if the pressure in the system falls below atmospheric pressure, atmospheric air will rush into pipe L and will produce an increase in pressure. By means of valve S, located in box K, it is possible to vary the relative quantities of steam and air and to obtain the proper temperature in the network within the range from 10° to 95°C. Automatic regulation of the inflow of steam into this system is achieved by the use of a “thermal spring” (T. R.). Between valve S and handle C, which can move the valve by means of screw B, there is inserted a hollow steel bent tube filled with an easily boiling liquid. When valve S is opened to a certain inflow of steam into the system, a certain temperature is established in the network. If the temperature of the steam were to rise (for example, with an increase in the pressure of steam in the boiler room), this would have to affect the temperature in the network. The most sensitive part of the network—the thermal spring—then expands and moves the valve to the left, whereby the gap for the passage of steam is narrowed. And conversely.

From this it is clear that with Geinde’s open saturator the following are achieved: 1) circulation of steam in the system, 2) removal of condensed water, and 3) regulation of the temperature in the network.

Steam of any pressure may be used for heating needs, beginning with \(1/3\) atmosphere. No jolts, shocks, or settling of water in the system are observed.

In Fig. 19 a diagram is given of a saturator with pressure.

The operation of this saturator proceeds in the following manner. The steam, after passing through the distributing valve A, goes to valve B and through pipe MN to injector R. Having passed through the injector, the steam enters the system heated by it (drums, plates, heaters, dryers, double bottoms of apparatuses, etc.), and from the other end of this system, together with the condensed water, it goes into vessel S through fitting E. Here the condensed water flows to the bottom and is discharged outward through fitting F by the trap, while the steam is sucked by the action of injector R through pipe KL and again goes into operation.

Here, in exactly the same way, there is a thermal spring, which moves valve B, located on the cover of vessel S. The thermal spring regulates both the temperature in the network and the pressure by moving valve B. Thus the pressure regulator performs the following functions: 1) it gives a rapid circulation of steam in the heating system, 2) owing to the high velocity of the circulating steam, the condensed water is quickly driven off the walls that transmit heat, and 3) the quantity of steam entering the system is strictly regulated. All this causes an increase in heat transfer and, consequently, accelerates the process of heating and drying, and increases the speed of movement of the material being dried, as, for example, in drying drums, plate dryers, sizing and other machines.

In the proceedings of the Ivanovo-Voznesensk Branch of the Russian Technical Society (1911, No. 1), V. Temkin gives the following table:

Name of factory	Duration of tests	Type of drying machine	Increase in production	Saving in steam consumption
Weaving factory of A. V. Smirnov	5 hrs.	Sizing drum	17.7%	16.94%
” ” Iv. Garelin	6 hrs.	—	7%	20.6%
” ” N. Garelin	6 hrs.	—	—	15.6%
” ” Gorbunov	8 hrs.	—	18%	15.5%
Calico-printing, Ryabov Manufactory in Moscow	8 hrs.	Plate dryer	—	33%
Calico-printing, Society of N. Derbenev	6 hrs.	Horizontal drum	30%	20%
” ” Neburnilov	3 hrs.	Plate dryer	14.2%	12.1%
Danilov Society, sugar-refining factory	50 min.	Heating chamber	24%	25.6%
Confectionery factory Ding	4 hrs.	Pastille dryer	—	20.3%
” ” Einem	4 hrs.	—	—	13.1%
Calico-printing factory of Katsevich	6 hrs.	Vertical drums	14%	15%

For the article by Prof. S. K. Konyukhov

Apparatuses for Heating Water

Bittner system.

Drawing 6.

Outlet

Steam inlet

Condensate outlet

Drawing 5.

steam

steam

water

sludge

water

Drawing 1.

…water.

Drawing 7.

Drawing 8.

Drawing 1.

Graph labels:

according to Weisbach

• 3000
• 2000
• 1000
• 0
• 20
• 40
• 60
• 80
• 100
• 120
• diameter
• temperature difference: 150°C, 100°C, 50°C

according to Ten Bosch

• 2
• 1.5
• 1
• 0.5
• 0
• 50
• 100
• 150
• 200
• 250
• Pipe diameter in mm.

Fig. 8.

Labels visible in the drawing:

• steam admission
• water
• [[unclear: fragment ending in “... temperature”]]
• steam

Visible dimensions: 570; 1500; 275; 700.

Fig. 9.

Labels visible in the drawing:

• water
• water
• steam
• steam
• water

Labels in the drawing: steam; condensate; water; water; water; dirt.

Labels in the drawing: water; steam.

steam

Fig. 5.

Labels in the drawing:

• steam
• water
• sludge
• water

Fig. 1.

Visible labels in the drawing:

• \(T\)
• \(T_0\)
• \(T_{\mathrm{c}}\)
• \(z\)
• \(\alpha\)
• \(\delta\)

Fig. 19.

Visible labels in the drawing:

• steam
• Steam, air, condensate
• water
• \(TR\)
• \(R\)
• \(S\)
• \(H\)
• \(K\)
• \(L\)
• \(B\)
• \(C\)

Visible graph label:

• \(W\,/\,m^2 /{}^\circ C\)
• \(100\), \(200\), \(300\), \(400\)

according to Wansler

Temperature difference: 150°C, 100°C, 50°C

Diameter

according to Ten-Bosch

Diameter of pipe in mm.

according to Zelonek

\(Wc/\mathrm{m}^2 / 1^\circ C / 20\ \mathrm{s}.\)

\(K/\) linear meter

Operating

\(WE\) per linear meter of pipe \(1^\circ C\) \((D)\)

Diameter of pipe in mm.

[[unclear: fragment ending with “... temperature”]]

p

Fig. 19-a.

Fig. 18.

steam inlet

Fig. 19-b.

Heating system

live steam

return steam

live steam

condensate

250 300 350 mm

mm.

[[unclear: small label on graph]]

Fig. 13.

water outlet

steam inlet

steam

condens.

steam

water

condens.

cleaning

Fig. 17.

Fig. 16.

Fig. 15.

Cartographic Establishment at the Tomsk Railway Administration. Steam.

Fig. 13.

Fig. 15.

Fig. 14.

Tomsk Colonization-Resettlement Party, boiler No. 8.