Associate Professor A. S. Betekhtin.

On the Most Advantageous Average Pressure in an Air-Pipeline Network and the Most Advantageous Pressure at the Compressor for Already Existing Installations

§ 1. The Lack of Elucidation of This Question.

By the most advantageous average pressure of an air-pipeline network we shall mean that pressure at which the total energy losses from leaks and from all types of pressure loss will be the smallest.

The question of the most advantageous average pressure in a network, as well as the question of the most advantageous pressure at the compressor, are questions of great theoretical and practical importance. Meanwhile, both of these questions have scarcely been elucidated in the specialized literature. The reason for this omission, so strange at first glance, lies in the great complexity of the analytical solution of the question, caused by the complexity of the interdependences of the various quantities that determine the magnitude of the various categories of losses in compressed-air networks.

In the present work we shall attempt to give a general solution of the question of the most advantageous average pressure in the network and at the compressor. The complexity of the final expressions for the minimum of losses in certain more general cases, and the possible approximate character of the solutions to a certain degree, may be justified by the complexity of the question itself and by the absence of any exhaustive solutions of it in our own and in foreign literature.

§ 2. Possible Cases in Determining the Most Advantageous Average Pressure in the Network \(P_m\) and the Most Advantageous Pressure at the Compressor.

In solving the problem posed, two principal cases may be encountered:

1. The diameters of the air pipelines are known—the case of an already existing installation.
2. The diameters of the air pipelines are unknown, i.e., the installation is in the design stage.

For both principal cases the following variants are possible:

a) The pressure at the receivers is known, identical, and remains unchanged (for example, when throttling devices are present in the network).

b) The pressure at the various groups of receivers is not identical, but the average pressure at each of the groups is known and may be regarded as unchanged; for example, in the case of an existing installation operating without throttling devices.

c) The pressure at the receivers is unknown and is a variable quantity.

In addition, in all variants one of two assumptions is possible:

1. The energy loss from cooling of the air depends little on changes of \(P_m\), so that it may be regarded as constant.

2. The changes in the magnitude of the energy losses from cooling as a function of changes in \(P_m\) cannot be neglected.

The assumption that the energy losses from air cooling remain constant when \(P_m\) changes may hold either for installations where the water consumption for cooling the air in the compressor is automatically regulated when the pressure changes, or for installations with heaters for the compressed air in various sections of the mine. In practice, one usually has to deal with a varying magnitude of energy losses from air cooling.

§ 3. Basic formulas for determining the various categories of energy losses of compressed air*)

In determining the various categories of air-energy losses, we shall use the following formulas:

For energy losses due to leakage,

\[ L_v = 2 A p_m^2 + A(5 - P_{\mathrm{cp}})p_m \quad \text{kg-m per hour} \tag{1} \]

where

\[ A = 1.4 \cdot 1900 \cdot K_c D_{\mathrm{cp}} L \cdot 10^{-4} = 0.266\, K_c D_{\mathrm{cp}} L \]

(with \(D_{\mathrm{cp}}\) in mm, and \(L\) in m).

For energy losses resulting from pressure loss due to friction:

\[ L_{\Delta pr} = \frac{10000\,T_m\,\Delta p_r}{T_0\,p_m}\,V_{\text{hour}} \quad \text{kg-m per hour} \tag{2} \]

For energy losses from air cooling:

\[ L_t = 600\,p_k\,V_{\text{hour}} \quad \text{kg-m per hour—at unchanged temperature of the intake air} \tag{3} \]

The total energy losses in the case of unchanged temperature of the intake air are:

\[ E = 2 A p_m^2 + \left[A(5 - P_{\mathrm{cp}}) + 1200\,V_{\text{hour}}\right]P + \]

\[ + \frac{10000\,T_m}{T_0}\, \Delta P_2\,\frac{V_{\text{hour}}}{P_m} - 600\,P_{\mathrm{cp}}\,V_{\text{hour}} \tag{4} \]

We shall not consider the case of varying temperature of the intake air, so as not to complicate equations that are already complex.

§ 4. The most advantageous average pressure in the network and at the compressor for an already existing installation

1. The first group of cases: energy losses from air cooling are assumed constant, i.e. \(L_t = \mathrm{const}\).

a) The simplest case—an unbranched network with identical and unchanging pressure at all receivers, equal to 5 ata.

Let the length of the network be \(L_m\), the diameter of the air pipeline \(D\) mm, the number of receivers \(n\), and the simultaneity coefficient of their operation \(K\). If all receivers are of one type, for example, pneumatic hammers, then with an average consumption by a pneumatic hammer of \(1\) cubic meter of free air per minute, the useful hourly air consumption is

\[ V_{\text{hour}} = 60\,nk \]

cubic meters per hour.

The energy losses from leakage for this case are:

\[ L_2 = 2 A p_m^2 + A(5 - P_n)P_m = 2 A p_m^2, \]

since \(p_n = 5\) ata.

The energy losses from air friction in the pipes are:

\[ L_{\Delta p} = \frac{10000\,T_m\,\Delta p_r \cdot 60\,nk}{T_0\,P_m} \quad \text{kg-m per hour, or} \quad L_{\Delta p_r} = \frac{b \cdot \Delta p_r}{P_m}, \quad \text{where } b = \]

\[ = \frac{600000\,nk \cdot T_m}{T_0} \tag{5} \]

*) See the article “On energy losses in compressed-air networks.” A. S. Betekhtin, Mining Journal, 1936, No. 1.

Since we consider the losses from air cooling to be independent of \(P_m\), it is obvious that \(T_m=\mathrm{const}\) as well.

The total energy losses for this case are:

\[ E=2Ap_m^2+\frac{b}{P_m}\Delta P_r \]

Eliminating \(\Delta p_r\):

\[ \Delta p_r=\frac{0.00129\,\beta\,P_0\,V_{\text{hour}}^2\,L}{R\,T_m\,D^5\,P_m} = \frac{0.00129\,\beta\cdot10000^2\,(60\,nk)^2\,L}{R\,T_m\,D^5\,P_m} \]

For unchanged \(n\) and \(k\), \(\beta=\mathrm{const}\); for constant \(L\), \(T_m\), and \(D\),

\[ \Delta p_r=\frac{C}{P_m}, \]

where

\[ C=\frac{0.00129\,\beta\cdot10000^2\,(60\,nk)^2\,L}{R\,T_m\,D^5}\tag{6} \]

and

\[ \frac{b\,\Delta p_r}{P_m}=\frac{bC}{P_m^2}=\frac{B}{P_m^2}, \]

where

\[ B=b\cdot C=\frac{0.00129\,\beta\,(60\,nk)^3\cdot10000^3\,L}{R\,T_0\,D^5}\tag{7} \]

Then

\[ E=2AP_m^2+\frac{B}{P_m^2} \]

Taking the first derivative of \(E\) with respect to \(P_m\):

\[ \frac{dE}{dP_m}=4AP_m-\frac{2B}{P_m^3}. \]

Taking the second derivative of \(E\) with respect to \(P_m\):

\[ \frac{d^2E}{dP_m^2}=4A+\frac{6B}{P_m^4} \]

Since the coefficients \(A\) and \(B\) are each separately greater than zero, and \(P_m\) is always greater than zero, then

\[ 4A+\frac{6B}{P_m^4}>0, \]

i.e., the function \(E\) has a minimum.

Equating \(\dfrac{dE}{dP_m}\) to zero, we obtain \(4AP_m^4=2B\), whence

\[ P_m=\sqrt[4]{\frac{B}{2A}}\tag{8} \]

This will be the most advantageous average pressure in the network for this simplest case.

The most advantageous pressure at the compressor for this case is

\[ P_k=2p_m-p_n=2\sqrt[4]{\frac{B}{2A}}-5\ \text{ata}\tag{9} \]

Numerical example. Given: \(L=1000\) m; \(nk=12\); \(D=76\) mm; \(\beta=1.06\), \(K_t=0.5\)*)

\[ \text{*) According to Fleischer’s curves, for a velocity of }7.2\ \text{m/sec and }P_m\cong6\ \text{ata.} \]

The coefficients

$$ A = 0.266 \cdot K_c D L = 0.266 \cdot 0.5 \cdot 76 \cdot 1000 = 10108 \cong 1.01 \cdot 10^4 $$

$$ B = \frac{0.00129 \cdot 1.06 \cdot (60 \cdot 12)^3 \cdot 10000^3 \cdot 1000}{29.27 \cdot 288 \cdot 76^5} = 23800000 = 2.38 \cdot 10^7. $$

$$ P_m = \sqrt[4]{\frac{2.38 \cdot 10^7}{2 \cdot 1.01 \cdot 10^4}} = \sqrt[4]{1178} \cong 5.86 \text{ ata}. $$

Then

$$ P_k = 2p_m - 5 = 11.72 - 5 = 6.72 \text{ ata}. $$

Let us turn to the expression for \(P_m\) in general form for this case.

$$ P_m = \sqrt[4]{\frac{0.00129 \, \beta \cdot P_0^3 (60V)^3 L} {RT_0 \cdot 2 \cdot 0.266 \cdot K_c D^6 L}} = \sqrt[4]{\frac{a \cdot \beta \cdot V^3}{K_c \cdot D^6}} \ldots (10), $$

where

$$ a = \frac{0.00129 \cdot 60^3 P_0^3}{RT_0 \cdot 2 \cdot 0.266} \ldots (11) $$

may be regarded as a constant quantity.

Expression (10) shows that, for the simplest case under consideration (a), the most advantageous average pressure in the network \(P_m\) does not depend on the length of the section, is directly proportional to the air consumption to the power \(3/4\), inversely proportional to the diameter of the section to the power 1.5, and depends on \(\sqrt[4]{\beta/K_c}\). This expression also shows that an error in the values of the coefficients \(\beta\) and \(K_c\) does not have a significant influence on the final result, since \(P_m\) is proportional to \(\sqrt[4]{\beta/K_c}\).

For comparatively small values of \(B\), i.e. for small air consumptions and a considerable value of the diameter \(D\), the value \(P_m\) may turn out to be less than the pressure at the receivers. This will be a case of a fictitious minimum of losses and a fictitious most advantageous pressure, since by its very nature \(P_m\) cannot be less than \(P_n\); the result \(P_m < P_n\) indicates that, for the given pressure at the receivers, with the given air-pipe diameter and the given air consumption, it is impossible to obtain a minimum of energy losses. Expression (10) indicates ways of transforming the fictitious average pressure \(P_m\) into a real one: if the pressure or the air consumption cannot be changed, the diameter of the air pipe can always be changed; reducing the diameter of the air pipe by \(a\) times increases the value of \(P_m\) by \(a^{1.5}\) times, and the diameter \(D\) can always be changed so that

$$ \left(\frac{D}{D_1}\right)^{1.5} P_{m.\mathrm{fic}} > P_n. $$

Thus, to the existing condition for the minimum of energy losses for this case,

$$ P_m = \sqrt[4]{\frac{B}{2A}}, $$

there is added a second condition, namely \(P_m > P_n\).

However, the additional condition still does not fully exhaust the question.

Since for the two unknowns \(P_m\) and \(\Delta p_r\), in addition to the two basic equations:

$$ 2AP_m^2 + A(5 - P_n)P_m + \frac{b \cdot \Delta p_r}{P_m} = E $$

and

$$ \Delta p_r = \frac{C}{P_m}, $$

we also have a third,

$$ P_m = \frac{2P_n + \Delta P_r}{2} $$

or

$$ \Delta p_r = 2P_m - 2P_n, $$

the possibility is not excluded—

that \(\Delta P_n\), found from the expression \(\Delta P_r=\dfrac{C}{P_m}\), at the most advantageous \(P_m\) will not correspond to the value of \(\Delta P_r\) found from the equation \(\Delta P_r=2P_m-2P_n\). Indeed, the existence of the third equation \(\Delta P_r=2P_m-2P_n\) and of the equation \(\Delta P_r=\dfrac{C}{P_m}\) essentially gives, for given network elements and a constant air flow rate, quite definite values for \(P_m\) and \(\Delta P_r\). Solving this system of equations gives:

\[ P_m=\frac{1}{2}\left(P_n+\sqrt{P_n^2+2C}\right)\ldots\ldots\ldots\ldots\ldots\ldots\ldots(12)\ \text{and} \]

\[ \Delta P_r=-P_n+\sqrt{P_n^2+2C}\ldots\ldots\ldots\ldots\ldots\ldots\ldots(13) \]

How, then, should one proceed in those cases when the value of \(P_m\), found from the condition of minimum energy losses, differs from the value of \(P_m\) found from equation (12) and representing the true actual average pressure in the network for the given conditions, in order to achieve the greatest possible agreement between the most advantageous \(P_m\) and the actual \(P_m\)?

Since the air flow rate and the length of the section can hardly be subject to change, the most accessible quantity with which one can operate is the diameter of the section. Consequently, by solving equations (10) and (12) jointly, one can find such a diameter at which the most advantageous average pressure will not differ from the actual one.

If it is possible to vary the magnitude of the air flow rate, for example, in the direction of increasing it, one can proceed in the same way, solving equations (10) and (12) with respect to \(V\).

In our numerical example we have precisely a case of discrepancy between the most advantageous \(P_m\) and the actual average pressure. The actual average pressure from equation (12):

\[ P_m=0.5\left(5+\sqrt{25+2\cdot 3.21}\right)=0.5\left(5+\sqrt{31.42}\right)=5.3\ \text{ata}, \]

whereas the most advantageous \(P_m=5.86\) ata.

The numerical value of \(C\) for our example is \(3.21\). For any new diameter

\[ C_1=\frac{3.21\cdot 76^5}{D^5} \]

Solving the equations:

\[ \sqrt[4]{\frac{a\cdot\beta\cdot V^3}{K_c\cdot D^6}} = 0.5\left(P_n+\sqrt{P_n^2+\frac{2\cdot 3.21\cdot 76^5}{D^5}}\right) \]

we obtain:

\[ \left(\frac{a\cdot\beta\cdot V^3}{K_c}\right)^{1/4}\cdot\frac{1}{D^{1.5}} -0.5P_n = 0.5\sqrt{P_n^2+\frac{6.42\cdot 76^5}{D^5}}, \]

or, denoting

\[ \left(\frac{a\cdot\beta\cdot V^3}{K_c}\right)^{1/4} \]

by \(K\) and squaring both sides, we obtain:

\[ \frac{K^2}{D^3}-\frac{P_nK}{D^{1.5}}+0.25p_n^2 = 0.25p_n^2+\frac{0.25\cdot 6.42\cdot 76^5}{D^5} \]

Whence

\[ K^2D^2-5K\cdot D^{3.5}=0.25\cdot 6.42\cdot 76^5 \quad \text{for } P_n=5\ \text{ata}. \]

In our case

\[ K=\left(\frac{0.00129\cdot 60^3\cdot 12^3\cdot 10000^3\cdot 1.06}{29.27\cdot 288\cdot 2\cdot 0.266\cdot 0.5}\right)^{1/4} =3.885\cdot 10^3; \]

\(K^2 \simeq 1.51 \cdot 10^7\) and \(0.25 \cdot 6.42 \cdot 76^5 = 4.07 \cdot 10^9\), so that, after reducing by \(10^3\), we obtain:

\[ 15100D^2 - 19.425 D^{3.5} = 4.07 \cdot 10^6, \]

whence \(D = 82.2\) mm. Thus, the nearest standard diameter \(D = 82.5\) mm, instead of the adopted 76 mm, gives a sufficiently satisfactory agreement between the most advantageous and the actual value of \(P_m\). Indeed, for this case

\[ P_m = 5.86 \sqrt[4]{\left(\frac{76}{82.5}\right)^6} = 5.19 \text{ ata}. \]

\[ P_k = 10.38 - 5 = 5.38 \text{ ata}, \quad \Delta P_r = \frac{C}{P_m} = 0.41 \text{ ata} \]

and

\[ \Delta P_{\text{act}} = 2P_m - 2P_n = 0.38 \text{ ata}. \]

The discrepancy in the value of \(\Delta P_r\) is small and in practice almost imperceptible. Thus, the most advantageous pressure at the compressor in the present case will lie within the limits \(5.38 \div 5.41\) ata. In practice, one would have to maintain a pressure \(P_k = 5.4\) ata.

From all that has been said above, the following conclusions may be drawn:

1. For a given air consumption and air-pipe diameter one can speak only of a conditional minimum of energy losses and of a conditionally most advantageous average pressure in the network.

2. If it is possible to vary some data, for example the air consumption or the diameter of the air pipe, then the conditionally most advantageous pressure may become the most advantageous actual pressure.

3. In order to obtain sufficient agreement in the value of the conditional and actual most advantageous pressure, it is necessary and sufficient:

a) In the case where the diameter of the air pipe can be changed, the new diameter, at which the conditionally most advantageous pressure becomes equal to the most advantageous actual pressure, is determined from the condition:

\[ K^2 D_1^2 - K P_n D_1^{3.5} = 0.5 \cdot C \cdot D^5 \ldots \ldots \ldots \ldots (14), \]

where

\[ K = \left[ \frac{0.00129 \cdot (60nk)^3 \cdot 10000^3 \cdot \beta^{1/4}} {R T_0 \cdot 0.266 \cdot 2 \cdot K_c D^6} \right]^{1/4}, \quad C = \frac{0.00129 \cdot \beta \cdot 10000^2 (60nk)^2 L} {R T_m D^5} \]

and \(D\) is the originally specified diameter. Then, from the new diameter, we find

\[ C_1 = \frac{C D^5}{D_1^5}, \]

from the \(C_1\) found we find

\[ P_m = 0.5\left(P_n + \sqrt{P_n^2 + 2C_1}\right), \]

from the expression

\[ \Delta P_r = \frac{C_1}{P_m} \]

we find the most advantageous \(\Delta P_r\), and, finally, from the expression

\[ P_k = P_n + \Delta P_r \]

the most advantageous pressure at the compressor.

b) If it is possible to vary the air consumption \(V\), then from the expression

\[ M^2 V_1^{3/4} - 0.5 C_2 V_1^{5/4} = M p_n \ldots \ldots \ldots \ldots (15) \]

one determines \(V_1 = 60 n_1 k_1\); from the expression

\[ P_m = 0.5 \left(P_n + \sqrt{P_n^2 + 2C\left(\frac{V_1}{V}\right)^2}\right) \]

one finds \(P_m\), and from \(P_m\), \(\Delta P_r\) and \(P_k\), as in the preceding case.

Here

\[ M = \left( \frac{0.00129 \beta \cdot 10000} {R T_0 \cdot 0.266 \cdot 2 \cdot K_c D^6} \right)^{1/4} = \frac{K}{(V^3 D^6)^{1/4}}, \]

\(V\) is the original hourly air consumption, \(C\) has its former value, and

\[ C_1 = \frac{C}{V^2}. \]

For the numerical example considered by us, the most advantageous air consumption \(V\), with the former diameter of 76 mm, would be \(580\) cubic meters per hour, or about 10 receivers of 1 cubic meter per minute each.

For this condition \(P_m = 5.265\) ata, \(\Delta P_r = 0.53\) ata, and \(P_k = 5.3\) ata \(\simeq 5.5\) ata.

It is clear that, from the standpoint of operating economy, the first variant is preferable (replacement of the pipeline \(D = 76\) mm by \(D = 82.5\) mm), since in the second variant, with a smaller number of receivers, it is necessary to maintain a higher operating pressure and to have greater total losses than in the first variant.

c) An unbranched network with identical and constant pressure at all receivers, not equal to 5 ata.

Let \(L, D, n\), and \(K\) still be given, and let the pressure at the receivers be \(P_n \gtrless 5\) ata.

In this case the useful hourly air consumption is

\[ V_{\text{hour}} = 60\,nk \,[1 + 0.2(P_n - 5)] = 12\,nkP_n \text{ cu. m per hour.} \]

The equation for the total energy losses will take the form:

\[ \Sigma = 2Ap_m^2 + A(5 - p_n)p_m + \frac{B_1}{P_m^2}, \]

where \(B_1\) will differ from \(B\) only in that, instead of \((60\,nk)^3\beta\), the expression for \(B_1\) will contain \(\beta_1(12\,nkP_n)^3\), i.e.

\[ B_1 = \left(\frac{P_n}{5}\right)^3 \frac{\beta_1}{\beta}\,B. \]

The first condition for the minimum of energy losses is:

\[ 4AP_m + A(5 - P_n) - \frac{2B_1}{P_m^3} = 0 \]

or

\[ 4AP_m^4 + A(5 - P_n)P_m^3 - 2B_1 = 0, \]

or

\[ 4AP_m^4 + A(5 - P_n)P_m^3 - 2B\left(\frac{P_n}{5}\right)^3\frac{\beta_1}{\beta} = 0 . \ . \ . \ . \ . \ . \ (16) \]

Solving this equation, we obtain a conditional value for the most advantageous \(P_m\).

The additional conditions for a real minimum of losses remain in force as before.

As is seen from equation (16), in the present case, besides \(A\) and \(B\), the magnitude of the average pressure at the receivers will also affect the value \(P_m\).

Numerical example. Let, as in the preceding example, \(L = 1000\) m, \(nk = 12\), \(D = 76\) mm, and \(P_n = 5.5\) ata. \(K_c\) and \(\beta\) vary; \(\beta_1 \simeq 1.02\), \(K_c \simeq 0.45\).

The values of the coefficients are:

\[ A = 0.266 \cdot 0.45 \cdot 76 \cdot 1000 = 9.1 \cdot 10^3, \]

\[ B = 2.38 \cdot 10^7; \qquad \frac{\beta_1}{\beta} = \frac{1.02}{1.06} = 0.962, \]

\[ \left(\frac{P_n}{5}\right)^3 = \left(\frac{5.5}{5}\right)^3 = 1.33. \]

The equation for the conditional minimum of losses will take the form:

\[ 4 \cdot 9100\,P_m^4 + 9100 \cdot P_m^3(5 - 5.5) - 2 \cdot 2.38 \cdot 10^7 \cdot 1.33 \cdot 0.962 = 0, \]

or

\[ 36400\,P_m^4 - 4550\,P_m^3 - 6.06 \cdot 10^7 = 0, \]

or

\[ P_m^4 - 0.125\,P_m^3 - 1665 = 0, \]

whence

\[ P_m = 6.42 \text{ ata} \]

and

\[ P_k = 2P_m - P_n = 12.84 - 5.5 = 7.34 \text{ ata.} \]

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Hence the conclusion that the pressure at the compressor must rise faster than the pressure at the receivers (it was 6.7 ata, became not 7.2, but 7.34 ata).

If we had neglected the term \(P_m^3\), \(P_m = 6.4\) ata, i.e., the error would have been less than \(1/3\%\) and would have lain beyond the accuracy limits of the pressure gauge.

As in the preceding example, here too there will be a discrepancy in magnitude between the conventional most advantageous \(P_m\) and the actual mean pressure in the air pipeline. The actual mean pressure

\[ P_m' = 0.5\left(5.5 + \sqrt{5.5^2 + 2 \cdot 3.21 \cdot \left(\frac{5.5}{5}\right)^2}\right) = 0.5(5.5 + 6.165) = 5.83 \text{ ata} \]

instead of 6.42 ata. Replacing the 76 mm diameter by a diameter of 82.5 mm will reduce \(P_m\) to

\[ P_m = 6.42 \sqrt[4]{\left(\frac{76}{82.5}\right)^6} \]

ata (without taking into account the insignificant change in \(K_c\)) or to 5.68 ata.

At the same time \(P_m^1\) will be

\[ 0.5\left(5.5 + \sqrt{5.5^2 + 6.42 \cdot 1.21 \cdot \left(\frac{76}{82.5}\right)^2}\right) = 5.765. \]

As can be seen, the actual mean pressure decreased with the change in diameter by only 0.065 at, or by 1.1%, whereas the conventional most advantageous pressure \(P_m\) decreased by 0.84 ata, or by 13%. Obviously, by leaving part of the air pipeline at the former diameter, we can raise the value of \(P_m\) from 5.68 to \(5.765 \div 5.8\), i.e., achieve complete coincidence of the actual and conventional most advantageous pressure also for this case. The part of the length of the air pipeline to be replaced by a pipeline of diameter 82.5 mm will be found from the equation

\[ 5.8 = 6.42 \sqrt[4]{\left(\frac{76}{82.5}\right)^6 \cdot \frac{1000}{X + (1000 - X)\left(\frac{76}{82.5}\right)^5}} \]

whence \(X = 740\) m.

Since the mean pressure at the receivers in practice rarely exceeds 6 ata, the term with \(P_m\), as well as \(\beta_1/\beta\), may be neglected, and the condition for the real minimum of energy losses for this case may be represented in the form:

\[ P_m = \sqrt[4]{\frac{B \cdot p_n^3}{250A} \cdot \left(\frac{D}{D_1}\right)^{1.5}} = 0.5\left(p_n + \sqrt{p_n^2 + 2C\left(\frac{D}{D_1}\right)^5}\right) \ldots \tag{17} \]

or in the form

\[ M V_1^{3/4} - 0.5 C_2 V_1^{5/4} = M p_n \ldots \ldots \tag{18} \]

depending on which of the factors can be subjected to change—the diameter of the air pipeline or the quantity of air.

§ 5. Determination of the most advantageous mean pressure of a branched network with constant losses from cooling of the air.

In view of the fact that in this case, although constancy of pressure at the various groups of receivers is attainable, equality of pressures at all receivers in the network is very difficult to attain, we shall first consider a more general case, namely: with equality of pressures at the receivers of one and the same group, the pressures at different groups of receivers are different and, moreover, depend on the useful flow of air in the given section of the network. Constancy of pressure in the sections and, finally, equality of pressure at all groups of receivers may subsequently be considered as simplified variants of this more complex case.

In the general case of a branched network considered by us, the coefficients \(A\) and \(B\) will have a somewhat different mathematical expression, namely: \(A = 0.266\,K_c \cdot D_{\mathrm{av}}\,L \cdot 10^{-4}\).

Here \(D_{\mathrm{av}}\) is the average diameter of the air-pipeline network, determined from the expression

\[ D_{\mathrm{av}}=\frac{D_1 l_1+D_2 l_2+\cdots+D_n l_n}{l_1+l_2+\cdots l_n}, \]

\(L=\sum l\), i.e., the total length of the network, and \(K_c\), the coefficient of imperfection of the network, is determined from the average air velocity, found analogously to the average diameter of the network.

The coefficient \(B=b\cdot C\) will in this case have a more complicated form:

\[ \text{Indeed}\quad b=\frac{10000\cdot 60\,nk\,T_m}{T_0} =\frac{10000\cdot 12\cdot nk\,P_{\mathrm{av}}\cdot T_m}{T_0} \]

\[ \text{and}\quad C=\frac{0.00129\,\beta\cdot 10000^2\,(12\,nk\,P_{\mathrm{av}})^2\,L_{\mathrm{e}}}{R\,T_m\,D^5}. \]

In the expressions for \(b\) and \(C\), the complex quantity will be \(P_{\mathrm{av}}\), since, when the pressures differ for different groups of receivers, \(P_{\mathrm{av}}\) will be a function of the pressure loss from the compressor to the receivers of each section. In addition, \(B\) will have different values for different sections, as a result of which the expressions for \(C\) become still more complicated, so that \(C\) can be taken as a constant coefficient only under special conditions. Let us find these conditions.

The nonuniformity of the pressure drop to different working sections makes it necessary to introduce some average pressure drop due to friction \(\Delta P_{r,\mathrm{av}}\). Since, in the general case, the quantity of air per hour \(V\), the diameter of the air pipeline, and its length are different for different sections, the determination of \(\Delta P_{r,\mathrm{av}}\) presents considerable difficulties. In a first approximation,

\[ \Delta P_{r,\mathrm{av}} = \frac{ \Delta P_{r,\max} + \frac{l_1}{l_{\max}}\Delta P_{r,\max} +\cdots+ \frac{l_n}{l_{\max}}\Delta P_{r,\max} }{n_y} = \]

\[ = \frac{ \Delta P_{r,\max}\,(l_{\max}+l_1+l_2+l_3+\cdots l_n) }{ l_{\max}\cdot n_y } \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (19) \]

where \(\Delta P_{r,\max}\) is the pressure drop on the most distant section of length (measured from the compressor station) \(l_{\max}\); \(l_1, l_2, l_3,\ldots l_n\) are the distances from the compressor station to the other sections; and \(n_y\) is the number of all sections, or the number of groups of receivers. However, expression (19) is valid only in the case where the pressure drop per unit length is the same both for the main line and for the branches, which usually is not the case.

Another method of determining \(\Delta P_{r,\mathrm{av}}\) consists in replacing the lengths of all branches and secondary mains by certain “lengths reduced to the cross-section of the main line,” i.e., by such fictitious lengths of sections with diameters everywhere equal to the diameter of the principal main line on which, with the quantity of air flowing through them equal to the full hourly or minute air consumption \(V_u\), the same pressure is lost as on the real sections. Let us explain this by an example. Suppose there is a comparatively simple network shown in Fig. 1. The lengths and diameters of the air pipelines on the individual sections, as well as the air consumption in them, are presented in Table 1.

Table 1.

No. in order	Section designation	Section length, m	Average air flow, cu. m/min	Average air flow, kg/hour	\(\beta\)	Air-pipe diameter, mm	Air velocity, m/sec	Note
Without No.	\(KA\)	500	50	3400	0.87	106.5	13.4	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).
1	\(AB\)	1500	23	1560	0.98	94.5	7.81	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).
2	\(Bb_1\)	600	10	678	1.09	70	6.2	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).
3	\(Bb_2\)	400	13	881	1.05	76	6.8	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).
4	\(AF\)	1000	27	1830	0.96	94.5	9.27	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).
5	\(Ff\)	800	15	1015	1.03	82.5	6.68	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).
6	\(EF\)	500	12	814	1.07	76	6.30	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).
7	\(Ee_1\)	400	4	271	1.25	50	4.85	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).
8	\(Ee_2\)	300	8	542	1.14	70	5.14	Mean temperature of the air in the air main \(25^\circ\) C. Mean temperature of the intake air \(15^\circ\) C and \(\gamma_0 = 1.186\).

Let us now suppose that we would like to replace the length of section \(Ee_2\) by a length reduced to the cross-section of the main line. Let us call this length \(l_{э,8}\). It is determined from the following considerations:

The real pressure drop on section \(Ee_2\) is

\[ \Delta P_{r,8}=\frac{0.00129\,\beta_8\,P_0^2\,V_8^2\,l_8}{R\,T\,d_8^5\,P_m} \]

or, if the quantity

\[ \frac{0.00129\,P_0^2}{R\,T}, \]

which is constant for all sections, is denoted by \(C_0\),

\[ \Delta P_{r,8}=\frac{C_0\,\beta_8\,V_8^2\,l_8}{d_8^5\,P_m}\ . . . . . . . . . . . . . . . . . . . . . . . . . . . (20) \]

The same pressure drop \(\Delta P_{r,8}\), obtained under the condition that the entire quantity of air passing through the main line, i.e. \(V_u\), flows through the section \(l_8\), and that the diameter of the branch \(l_8\) is equal to the diameter of the main line \(D\), will be expressed as:

\[ \Delta P_{r,8}=\frac{C_0\,\beta_u\,V_u^2\,l_{э,8}}{D^5\,P_m}\ . . . . . . . . . . . . . . . . . . . . . . . . . (21), \]

where \(\beta_u\) is the coefficient of friction corresponding to the hourly air flow \(V_u\).

Equating the right-hand sides of equations (20) and (21) and solving the resulting expression for \(l_{э,8}\), we have:

\[ l_{э,8}=l_8\left(\frac{D}{d_8}\right)^5\left(\frac{V_8}{V_u}\right)^2\frac{\beta_8}{\beta_u}\ \text{m}.\ . . . . . . . . . . . . . . . . . . . . . . (22) \]

In particular, for the data of Table No. 1,

\[ l_{э,8}=300\left(\frac{106.5}{70}\right)^5\left(\frac{8}{50}\right)^2\frac{1.14}{0.87}\simeq 84\ \text{m}. \]

In a similar way, the reduced lengths will also be found for the other sections.

Replacing the real lengths by lengths reduced to the cross-section of the main line makes it possible to consider any of the sections \(KAFEe_1\), \(KAFEe_2\), \(KAFf\), \(KABb_1\), and \(KABb_2\) under identical operating conditions, since the quantities of air flowing through them and the cross-sections of the air pipes are the same. We thus obtain a number of sections that are, as it were, independent of one another, differing only in length. Therefore the pressure drops in these sections will be related as the corresponding reduced lengths. Calling the pressure drop for the section with the greatest reduced length \(\Delta P_{\max}\) and designating the sections in abbreviated form as \(Ke_1\) (section \(KAFEe_1\)),

\(Kb_2\) (section \(KABb_2\)), etc., we obtain an expression for the mean pressure drop due to friction:

\[ \Delta P_{r,\mathrm{cp}}= \frac{ \Delta P_{r\max} +\Delta P_{r\max}\frac{l_{ke_2}}{l_{\max}} +\Delta P_{r\max}\frac{l_{kf}}{l_{\max}} +\cdots }{n_y} \quad \text{or} \]

\[ \Delta P_{r,\mathrm{cp}}= \frac{ \Delta P_{r\max}\left(l_{\max}+l_{ke_2}+l_{kf}+\cdots+l_{bb_2}\right) }{ l_{\max}\cdot n_y } \ .\ .\ .\ .\ .\ .\ .\ (23) \]

where \(l_{\max}\) is the reduced length corresponding to the length of the section with the greatest pressure drop (i.e., the maximum of the reduced lengths), and \(n_y\) is the number of sections.

Not differing in outward form from formula (19), formula (23) is considerably more accurate than it, since it is valid for any law of pressure loss per unit length in any of the sections of the air pipeline.

The mean pressure at the receivers, also necessary for determining the amount of losses, is determined on the basis of the following considerations: let \(P\) be the pressure at the most distant group of receivers in ata, and \(\Delta P_t\) the pressure drop due to cooling of the air, which, at a distance from the compressor of 1.5 km and more, is in fact the same for all groups of receivers. Then the pressure at the compressor may be expressed, on the one hand, as \(P_k=P+\Delta P_t+\Delta P_{r,\max}^{*}\), and on the other hand as \(P_k=P_{\mathrm{cp}}+\Delta P_t+\Delta P_{\mathrm{cp}}\).

Equating the right-hand sides and solving the resulting equation with respect to \(P_{\mathrm{cp}}\), we find:

\[ P_{\mathrm{cp}} = P+\Delta P_{r,\max}-\Delta P_{\mathrm{cp}} = P+\Delta P_{r,\mathrm{cp}} \frac{l_{\max}\cdot n_y}{l_{\max}+l_1+l_2+\cdots+l_n} - \Delta P_{r,\mathrm{cp}} = P+\Delta P_{r\mathrm{cp}} \left( \frac{l_{\max}\cdot n_y}{\sum l_\mathrm{э}}-1 \right), \]

where \(l_1,l_2,l_3,\ldots,l_n\) are the reduced lengths of the sections \(Kb_1\), \(Kb_2\), etc.; or, denoting

\[ \left( \frac{l_{\max}\cdot n_y}{\sum l_\mathrm{э}}-1 \right) \]

by \(\alpha\), we obtain:

\[ P_{\mathrm{cp}}=P+\alpha\Delta P_{r,\mathrm{cp}} \ .\ .\ .\ .\ .\ (24) \quad \text{and} \quad \Delta P_{r,\max}=(\alpha+1)\Delta P_{r,\mathrm{cp}}. \]

The coefficient \(\alpha\) may be regarded as the degree of nonuniformity of the pressure loss up to the receivers.

The hourly air consumption by the receivers will be expressed as:

\[ V_{\text{hour}}=12\,nk p_{\mathrm{cp}} \quad \text{or} \quad 12\,nk\left(P+\Delta P_{\mathrm{cp}}\right) \ .\ .\ .\ .\ .\ .\ .\ .\ (25), \]

if the air consumption by an individual receiver at 5 ata is \(v_0=1\) cubic m per min., and

\[ V_{\text{hour}}=12\,nk\cdot v_{\mathrm{cp}}\left(P+\alpha\Delta P_{r,\mathrm{cp}}\right), \]

if \(v_{\mathrm{cp}}\) is the average minute consumption of air per receiver when the receivers are of different types.

The energy loss due to pressure loss will be expressed as:

\[ L_{\Delta p_2} = \frac{ 10000\cdot 12\,nk\left(P+\alpha\Delta P_{r,\mathrm{cp}}\right)T_m\cdot \Delta P_{r,\mathrm{cp}} }{ T_0 P_m } \]

But

\[ \Delta P_{r,\mathrm{cp}} = \frac{ 0.00129\cdot 10000^2 \beta_{\mathrm{ш}} \left[12\,nk\left(P+\alpha\Delta P_{r,\mathrm{cp}}\right)\right]^2 \sum l_\mathrm{э} }{ R T_m\cdot D^5 P_m n_y } \ .\ .\ .\ .\ .\ (26) \]

* Although the energy losses from cooling of the air will be constant, the pressure drop due to cooling \(P_t\) must nevertheless be taken into account, since it enters as a term in the expression \(P_k\).

so that, in the final analysis,

\[ L_{\Delta p_r}= \frac{0.00129\,\beta_u\cdot 10000^3\cdot (12nk)^3\,(P+\alpha\Delta P_{r,\mathrm{cp}})^3\,T_m\cdot \sum l_{\mathrm{э}}} {T_0\cdot R\cdot T_m\cdot D^5 P_m^2 n_y} \ldots \tag{27} \]

Hence the coefficient

\[ B= \frac{0.00129\,\beta_u\,10000^3\cdot 12^3\cdot (nk)^3\sum l_{\mathrm{э}}} {T_0\,R\,D^5 n_y} = \frac{7.62\cdot 10^{10}\,\beta_u\,(nk)^3\sum l_{\mathrm{э}}} {T_0\,D^5 n_y} \ldots \tag{28} \]

\[ \Delta P_{r,\mathrm{cp}} = 6.35\cdot 10^5\,\beta_u\, \frac{(nk)^2\,(P+\alpha\Delta P_{r,\mathrm{cp}})^2\sum l_{\mathrm{э}}} {T_m\cdot D^5\cdot P_m n_y} \ldots \tag{29} \]

and

\[ \Delta P_{r,\max}=(\alpha+1)\Delta P_{r,\mathrm{cp}} \ldots \tag{30} \]

We replace \(\alpha\Delta P_{r,\mathrm{cp}}\) by \(P_m\).

On the one hand,

\[ P_m=\frac{P_k+P_{\mathrm{cp}}}{2} = \frac{P_k+p+\alpha\Delta P_{r,\mathrm{cp}}}{2}. \]

On the other hand, \(P_k=P+\Delta P_t^*)+\Delta P_{r,\max}\). Substituting \(P_k\) into the preceding expression, we obtain:

\[ P_m= \frac{P+\Delta P_t+\Delta P_{r,\max}+P+\alpha\Delta P_{r,\mathrm{cp}}}{2} = \frac{2p+\Delta P_t+\Delta P_{r,\mathrm{cp}}(1+2\alpha)}{2} \tag{31} \]

whence

\[ \Delta P_{r,\mathrm{cp}}= \frac{2p_m-2p-\Delta P_t}{(1+2\alpha)} \ldots \tag{32} \]

Then

\[ P+\alpha\Delta P_{r,\mathrm{cp}} = \frac{P+2\alpha p+(2P_m-2p-\Delta P_t)\alpha}{1+2\alpha} = \frac{2\alpha P_m}{1+2\alpha} + \frac{P-\alpha\Delta P_t}{1+2\alpha}. \]

Finally, denoting \(P-\alpha\Delta P_t\) by \(F\), we obtain:

\[ P_{\mathrm{cp}}= P+\alpha\Delta P_{r,\mathrm{cp}} = \frac{2\alpha P_m}{1+2\alpha} + \frac{F}{1+2\alpha} \ldots \tag{33} \]

The total energy losses will be:

\[ E= 2Ap_m^2 + A\left[ 5-\left( \frac{2\alpha P_m}{1+2\alpha} + \frac{F}{1+2\alpha} \right) \right]P_m + B\left[ \frac{2\alpha P_m}{1+2\alpha} + \frac{F}{1+2\alpha} \right]^3 \frac{1}{P_m^2} \tag{34} \]

The condition for minimum energy losses:

\[ 4AP_m - A\left(5+\frac{F}{2\alpha+1}\right) - \frac{4A\alpha P_m}{2\alpha+1} + \frac{B}{(2\alpha+1)^3} \left[ 3(2\alpha P_m+F)^2\,2\alpha P_m^2 - 2P_m(2\alpha P_m+F)^3 \right] \frac{1}{P_m^4} =0, \]

or, after transformations,

\[ 2A(2\alpha+1)^2(\alpha+1)P_m^4 + \left[ \frac{A(2\alpha+1)^3}{2} \left(5-\frac{F}{2\alpha+1}\right) + 4\alpha^3 B \right]P_m^3 \]

\[ -3\alpha BF^2P_m-BF^3=0. \]

Dividing all terms of this equation by \(B\), we obtain:

\[ \frac{2A}{B}(2\alpha+1)^2(\alpha+1)P_m^4 + \left[ 4\alpha^3+ \frac{A(2\alpha+1)^2}{2B} \bigl(5(2\alpha+1)-F\bigr) \right]P_m^3 - \]

\[ -3\alpha F^2P_m-F^3=0 \ldots \tag{35} \]

In most cases the term \(4\alpha^3\) may quite well be neglected. Indeed, in networks with properly selected diameters \(\alpha \ll 0.1\), \(P_m\) is hardly more than \(7\) ata and \(F\) is of the order of 5. Neglecting the term \(4\alpha^3\) will give an error of less than \(1\%\), i.e. less than the accuracy of reading from the manometer. In particular cases, for small values of \(\alpha\) and \(\Delta P_t\), the entire term containing \(P_m^3\) may be neglected, and then the equation for the conditional minimum of losses is considerably simplified.

In what follows, to determine the most advantageous average pressure in the network, we shall use the expression

\[ \frac{2A}{B}(2\alpha+1)^2(\alpha+1)F_m^4 + \frac{A(2\alpha+1)^2}{2B}\bigl(5(2\alpha+1)-F\bigr)P_m^3 - \]

\[ -3\alpha F^2P_m-F^3=0 \ldots\ldots\ldots\ldots\ldots\ldots (36) \]

except in those cases where \(\alpha>0.1\).

As is evident from expression (36), the dependence of \(P_m\) on the various factors characterizing a branched network proves to be rather complex. Only for networks for which \(\alpha<0.1\) and \(B<30A\) can the most advantageous value of the conditional \(P_m\) be determined with a sufficient degree of accuracy from the equation

\[ \frac{2A}{B}(2\alpha+1)^2(\alpha+1)P_m^4 - 3\alpha F^2p_m - F^3 = 0 \ldots\ldots\ldots (37) \]

Numerical example. Let us find the most advantageous value \(P_m\) for a particular case—namely, for the network shown in Fig. 1. The characteristic quantities for this network are given in Table 1. We shall take the pressure at the most distant receiver as \(P=5.0\) ata. The receivers are jackhammers; the useful consumption of free air per hammer is \(1\) cu. m per min at 5 ata, allowing for wear of the hammer.

We shall take the pressure drop due to cooling \(\Delta P_t\) as \(1.2\) ata. *)

Calculation procedure:

1. We find the total length of the network \(\Sigma l = 6000\ \text{m}\).

2. We find the average diameter of the air-pipeline network.

\[ D_{\mathrm{av}} = \frac{ 106.5\cdot500 + 94.5(1500+1000) + 82.5\cdot800 + 76(400+500) + }{6000} \]

\[ + \frac{ 70(600+300) + 50\cdot400 }{6000} = \frac{506900}{6000} = 84.48\ \text{mm} \simeq 85\ \text{mm} \]

1. We find the average air velocity in the network:

\[ C_{\mathrm{av}} = \frac{ 13.4\cdot500 + 7.8\cdot1500 + 6.2\cdot600 + 6.8\cdot400 + 9.27\cdot1000 + 6.68\cdot800 + }{6000} \]

\[ + \frac{ 6.3\cdot500 + 4.85\cdot400 + 5.14\cdot300 }{6000} = \frac{46001}{6000} = 7.58 \simeq 7.6\ \text{m/sec}. \]

1. From the curves of Fig. 2 we find the coefficient \(K_F\) for the least favorable case—namely, for \(P_m=7\) atm. abs., \(K_F=0.60\). We take the type of installation to be ordinary, i.e. \(K_n=1\); there is no pipe welding, i.e. \(K_s=1\).

*) It depends on the temperature in the air receiver and on the average temperature at the receivers. For an existing mine both are known, and therefore \(\Delta P_t\) is also known.

1. We find the coefficient of imperfection of the network:

\[ K_c = K_F \cdot K_n \cdot K_c = 0.6 \cdot 1 \cdot 1 = 0.6. \]

1. We find the coefficient \(A\):

\[ A = 0.266 K_c D_{\mathrm{av}} \Sigma l = 0.266 \cdot 0.60 \cdot 85 \cdot 6000 = 8.14 \cdot 10^4. \]

1. We determine the lengths of the individual sections, reduced to the cross-section of the main line:

\[ l_{21} = l_{AB}\left(\frac{D}{d_1}\right)^5 \left(\frac{V_1}{V_u}\right)^2 \frac{\beta_1}{\beta_u} = 1500 \left(\frac{106.5}{94.5}\right)^5 \left(\frac{23}{50}\right)^2 \frac{0.98}{0.87} = 666\ \mathrm{m} \]

\[ l_{32} = 600 \left(\frac{106.5}{70}\right)^5 \left(\frac{10}{50}\right)^2 \cdot \frac{1.09}{0.87} = 240\ \mathrm{m} \]

\[ l_{33} = 400 \left(\frac{106.5}{76}\right)^5 \left(\frac{13}{50}\right)^2 \cdot \frac{1.05}{0.87} = 180\ \mathrm{m} \]

\[ l_{34} = 1000 \left(\frac{106.5}{94.5}\right)^5 \left(\frac{27}{50}\right)^2 \cdot \frac{0.96}{0.87} = 600\ \mathrm{m} \]

\[ l_{35} = 800 \left(\frac{106.5}{82.5}\right)^5 \left(\frac{15}{50}\right)^2 \cdot \frac{1.03}{0.87} = 314\ \mathrm{m} \]

\[ l_{36} = 500 \left(\frac{106.5}{76}\right)^5 \left(\frac{12}{50}\right)^2 \cdot \frac{1.07}{0.87} = 196\ \mathrm{m} \]

\[ l_{37} = 400 \left(\frac{106.5}{50}\right)^5 \left(\frac{4}{50}\right)^2 \cdot \frac{1.25}{0.87} = 158\ \mathrm{m} \]

\[ l_{38} = 300 \left(\frac{106.5}{70}\right)^5 \left(\frac{8}{50}\right)^2 \cdot \frac{1.14}{0.87} = 84\ \mathrm{m}. \]

1. We find the reduced lengths from the compressor to each group of receivers:

\[ l_1 = l_{kb_1} = 500 + 666 + 240 = 1406\ \mathrm{m} \]

\[ l_2 = l_{kb_2} = 500 + 666 + 180 = 1346\ \mathrm{m} \]

\[ l_3 = l_{kf} = 500 + 600 + 314 = 1414\ \mathrm{m} \]

\[ l_4 = l_{ke_1} = 500 + 600 + 196 + 158 = 1454\ \mathrm{m} \]

\[ l_5 = l_{ke_2} = 500 + 600 + 196 + 84 = 1380\ \mathrm{m}. \]

1. We choose the section with the maximum reduced length \(l_{\max}\). Obviously, this will be the section \(l_{ke_1}\) with length \(1454\ \mathrm{m}\).

2. We determine the average pressure drop to the receivers as a function of \(\Delta P_{\max}\):

\[ \Delta P_{r,\mathrm{av}} = \Delta P_{\max} \left( \frac{1454 + 1406 + 1346 + 1414 + 1380}{1454 \cdot 5} \right) = \]

\[ = \frac{7000}{7270} = 0.963\,\Delta P_{\max} \]

1. We determine the coefficient \(\alpha\):

\[ \alpha = \frac{l_{\max} n_y}{\Sigma l_z} - 1 = \frac{7270}{7000} - 1 = 0.0386 \]

•

1. We determine the coefficient \(B\):

\[ B=\frac{7.62\cdot 10^{10}\beta_n(nk)^3\Sigma l_{\mathrm{э}}}{T_0 n_y D^5},\quad \text{or} \]

\[ B=\frac{7.62\cdot 10^{10}\cdot 0.87\cdot 50^3\cdot 7000}{288.5\cdot 106.5^5} =2.94\cdot 10^6 \quad \text{for } T_0=288^\circ K \]

1. We determine the value of \(F\) for the given case, with the prescribed values of \(P\) and \(\Delta P_t\):

\[ F=P-\alpha\Delta P_t=5-0.0386\cdot 1.2=5-0.0463=4.9537 \]

1. We find the condition for the minimum losses for this particular case:

\[ \frac{2A}{B}(2\alpha+1)^2(\alpha+1)P_m^4+ \frac{A}{2B}(2\alpha+1)^2(5(2\alpha+1)-F)P_m^3 -3\alpha F^2P_m=F^3,\quad \text{or} \]

\[ \frac{2\cdot 8.14\cdot 10^4\cdot 1.077^2\cdot 1.0386}{2.94\cdot 10^6}P_m^4+ \frac{8.14\cdot 10^4\cdot 1.077^2}{2\cdot 2.94\cdot 10^6}(5\cdot 1.077-4.954)P_m^3- \]

\[ -3\cdot 0.0386\cdot 4.954^2P_m=4.954^3,\quad \text{or} \]

after multiplying both sides of the equation by 100:

\[ 6.675P_m^4+0.673P_m^3-293.0P_m\simeq 12150 \]

1. Solving the last equation with respect to \(P_m\), we obtain:

\[ P_m=6.75\ \text{ata}. \]

1. We find \(\Delta P_{r,\mathrm{cp}}\):

\[ \Delta P_{r,\mathrm{cp}}= \frac{2P_m-(2P+\Delta P_t)}{2\alpha+1} = \frac{13.5-10-1.2}{1.0772} \simeq 2.135\ \text{ata}. \]

1. We find

\[ \Delta P_{r,\max}=(\alpha+1)\Delta P_{r,\mathrm{cp}} =1.0386\cdot 2.135=2.22\ \text{ata}. \]

1. We find the most advantageous pressure at the compressor:

\[ P_k=P+\Delta P_t+\Delta P_{r,\max}=5+1.2+2.22=8.42\ \text{ata}. \]

The average pressure drop \(\Delta P_{r,\mathrm{cp}}\), found from equation (29), will be

\[ 1.887\ \text{ata} \]

and the actual

\[ P_m=0.25[2P+\Delta p_t+\sqrt{(2p+\Delta p_t)^2+8c\left(\frac{P_{\mathrm{cp}}}{P}\right)^2}] = \]

\[ =6.6\ \text{ata},\quad \text{since } P_{\mathrm{cp}}=5+0.0386\cdot 1.887=5.081 \]

and \(c\) for \(\Delta P_{r,\mathrm{cp}}\) is 12.27, and for \(\Delta P_{r,\max}\), \(c=12.74\).

Thus, in this case as well, there is a discrepancy between the conditional and actual value of \(P_m\). True, we have nowhere taken into account the pressure drop in the fittings and shaped parts, which is on average equal to 10 to 20% of the actual drop. But even taking these additional losses into account, coincidence of the actual and the most advantageous value of \(P_m\) for an already existing network may be regarded as a happy exception; usually there is always a more or less substantial discrepancy between the value of the most advantageous average pressure in the network and its actual value. Elimination of this discrepancy is possible, although it presents certain difficulties. It must proceed by changing the equivalent lengths of one or several sections, i.e.

along the path of changing the diameters of these sections. This causes a simultaneous change in \(A\), \(B\), \(\alpha\), and \(F\).

Then the conditionally most advantageous average pressure for the new conditions is found and compared with the actual value \(P_m\) for this case. Sometimes the changes in the magnitude of \(\alpha\) and \(F\) are so insignificant that they may be neglected, and then the solution of the question is considerably simplified.

For our numerical example, replacing the diameter in section \(AB\) from \(94.5\) mm to \(100.5\) mm gives the following results:

\[ \Sigma l_э = 6696,\quad \alpha = 0.086,\quad F = 4.90 \]

The equation for the conditional minimum of losses will take the form:

\[ 8.26 P_m^4 + 1825 P_m^3 - 620 P_m = 11780 \]

This equation is satisfied by the value \(P_m = 6.60\) ata:

\[ \Delta P_{r,\mathrm{cp}} = \frac{13.2 - 11.2}{1.172} = 1.7 \text{ ata.} \]

\[ \Delta P_{r,\max} = 1.086 \cdot 1.7 = 1.846 \text{ ata,} \quad \text{the actual } \Delta P_{r,\max} = \frac{12.74}{25}\frac{(5.18)^2}{6.6} = \]

\[ = 2.03,\quad \text{the actual } P_m\ (\text{with } P_{\mathrm{cp}} = 5 + 0.086 \cdot 2.03 = 5.18)\ 6.63 \text{ ata.} \]

The pressure at the compressor, found from the conditional value \(P_m\):

\[ P_k = 5 + 1.2 + 1.85 = 8.05 \text{ ata,} \]
and the actual pressure,
\[ P'_k = 5 + 1.2 + 2.03 = 8.23 \text{ ata.} \]

As can be seen, in this case we already have smaller discrepancies between the conditional and actual value of \(P_m\). Thus, in the first case the discrepancy in the value of \(P_m\) was \(6.75 - 6.13 = 0.62\) ata, or \(10\%\), whereas now it is \(6.63 - 6.60 = 0.03\) ata, or about \(0.5\%\). Since the losses are ultimately expressed as a function of \(P_m\), when the conditional and actual \(P_m\) agree sufficiently, in order to obtain the minimum losses it is sufficient to maintain at the compressor the pressure determined by the actual magnitude of the pressure losses and the lowest pressure at the receivers, i.e., \(P + \Delta P_t + \Delta P_{\max}\), or \(P_{\mathrm{cp}} + \Delta P_t + \Delta P_{r,\mathrm{cp}}\), as the mean of the pressures \(P_k\) and \(P'_k\). In our example, about \(8.15\) ata.

In the case where the pressure at the receivers differs, but does not depend on \(\Delta P_{r,\mathrm{cp}}\), the equation for the conditional minimum of energy losses is considerably simplified, namely:

\[ 2AP_m^2 + AP_m(5 - P_{\mathrm{cp}}) + B\frac{P_{\mathrm{cp}}}{P_m^2} = E, \]

or

\[ 4AP_m + A(5 - P_{\mathrm{cp}}) - \frac{2BP_{\mathrm{cp}}}{P_m^3} = 0, \]

whence

\[ 4AP_m^4 + A(5 - P_{\mathrm{cp}})P_m^3 = 2BP_{\mathrm{cp}}, \]

or

\[ P_m^4 + \frac{A}{4A}(5 - P_{\mathrm{cp}})P_m^3 = \frac{2BP_{\mathrm{cp}}}{4A} \qquad \cdots\cdots (38) \]

\(P_{\mathrm{cp}}\) may be regarded in this case as

\[ \frac{P_1 n_1 + P_2 n_2 + \cdots + P_n n_n}{\Sigma n}, \]

where \(P_1, P_2, P_3\) are the averages of the measured pressures at groups of receivers with numbers \(n_1, n_2, n_3\).

If the pressure at all receivers is the same, one may use the preceding equation, taking \(P_n\) instead of \(P_{\mathrm{av}}\). In the simplest case, when

\[ P_n=5, \]

the equation becomes

\[ P_m=\sqrt[4]{\frac{B\cdot P_n}{2A}}=\sqrt[4]{2.5\,\frac{B}{A}}. \]

§ 6. Branched Network with Variable Losses from Cooling and Unequal Pressure at the Receivers

The total energy losses for this case, according to the equations, can be expressed as*):

\[ E=2AP_m^2+\left[A(5-P_{\mathrm{av}})+14400\,nk\,P_{\mathrm{av}}\right]P_m +B^1P_{\mathrm{av}}\frac{\Delta P_{r,\mathrm{av}}}{P_m} -7200\,nkP_{\mathrm{av}} . \tag{39} \]

Let us first consider the general case—\(P_{\mathrm{av}}\) depends on the pressure drop, on friction, and on the cooling of the air; no throttling devices are installed.

For the majority of modern air-pipeline networks, the pressure drop due to air cooling \(\Delta P_t\) may be considered practically identical for all groups of receivers, owing to the considerable length of the networks. The quantity \(\Delta P_t \simeq 0.165P_k\), according to our assumptions*).

To express \(P_{\mathrm{av}}\) and \(\Delta P_{r,\mathrm{av}}\) as functions of \(P_m\), we have the following equations:

\[ a)\; P_{\mathrm{av}}=P+\alpha\Delta P_{r,\mathrm{av}};\quad b)\; \Delta P_{r,\mathrm{av}}=\frac{C^1P_{\mathrm{av}}^2}{P_m};\quad c)\; P_m=\frac{2P_{\mathrm{av}}+\Delta P_{r,\mathrm{av}}+\Delta P_t}{2}. \]

Replacing \(\Delta P_t\) by \(0.165P_k=0.165(2P_m-P_{\mathrm{av}})\) and solving equations \((a)\) and \((c)\) with respect to \(P_{\mathrm{av}}\), we obtain:

\[ P_{\mathrm{av}}=\frac{1.67\,\alpha P_m}{1.84\alpha+1} +\frac{P}{1.84\alpha+1}\;\ldots\ldots\ldots\ldots\ldots \tag{40} \]

Then

\[ \Delta P_{r,\mathrm{av}}= \frac{C^1}{P_m}\left[ \frac{1.67\alpha P_m}{1.84\alpha+1} +\frac{P}{1.84\alpha+1} \right]^2 \;\ldots\ldots\ldots\ldots \tag{41} \]

where \(P\) is the pressure at the most distant group of receivers, in ata.

Since \(\alpha\) is usually not greater than 0.2, we round \(1.835\alpha\) to \(1.84\alpha\). Substituting \(P_{\mathrm{av}}\) and \(\Delta P_{r,\mathrm{av}}\) into the basic equation for the total losses, we obtain

\[ \begin{aligned} E={}&2AP_m^2+ \left[ A\left( \frac{9.2\alpha+5-P}{1.84\alpha+1} -\frac{1.67\alpha P_m}{1.84\alpha+1} \right) +14400\,nk\left( \frac{1.67\alpha P_m}{1.84\alpha+1} \right.\right.\\ &\left.\left. +\frac{P}{1.84\alpha+1} \right) \right]P_m +\frac{B^1C^1}{P_m^2} \left( \frac{1.67\alpha P_m+p}{1.84\alpha+1} \right)^3 -7200\,nk\left( \frac{1.67\alpha P_m+p}{1.84\alpha+1} \right) \;\ldots \tag{42} \end{aligned} \]

We set the first derivative of \(E\) with respect to \(P_m\) equal to zero:

\[ 4AP_m+ \frac{A}{1.84\alpha+1} \left(9.2\alpha+5-P-3.34\alpha P_m\right) +\frac{14400\,nk}{1.84\alpha+1} \left(3.34\alpha P_m+P\right) + \]

\[ +\frac{B^1C^1}{(1.84\alpha+1)^3} \left( \frac{1.67^3\alpha^3P_m^3-3\cdot1.67\alpha P^2P_m-2P^3}{P_m^3} \right) -\frac{7200\cdot1.67\alpha nk}{1.84\alpha+1}=0. \]

*) See the above-mentioned work by A. S. Betekhtin, “On Energy Losses in Compressed-Air Networks.”

After collecting like terms and clearing the denominator, we obtain:

\[ (1.84a+1)^2[4A(a+1)+14400\cdot 3.34a\, nk]P_m^4+ \{[A(9.2a+5-P)+ \]

\[ +7200(2P-1.67a)](1.84a+1)^2+1.67^3a^3B^1C^1\}P_m^3-3\cdot 1.67aB^1C^1P^2P_m- \]

\[ -2B^1C^1P^3=0 \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (43) \]

It is not difficult to see that the very complicated expression (43), when the pressure at all receivers is equal, i.e. when \(a=0\), takes the simpler form:

\[ 4AP_m^4+[A(5-P)+14400\, nkP]P_m^3-2B^1C^1P^3=0 \]

Since \(B^1C^1\) is our former coefficient \(B\), then, dividing all terms of the final equation (43) by \(2B\) and denoting:

\[ \frac{4A(a+1)+14400\cdot 3.34a\, nk}{2B}=M \ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (44) \]

\[ \frac{[A(9.2a+5-P)+7200\, nk(2P-1.67a)](1.84a+1)^2+1.67^3a^3B^1C^1}{2B}=N . (45), \]

we obtain:

\[ (1.84a+1)^2MP_m^4+NP_m^3-2.5aP^2P_m-P^3=0 \ . . . . . . . . . . . . . . . . . . . . (46), \]

i.e. an equation similar in form to equation (37), with the same powers of \(P_m\), but only with other, more complicated coefficients.

Equation (46), as well as the complexity of the expressions for the coefficients \(M\) and \(N\), involuntarily suggest that, if any mining administration decided to organize constant monitoring of energy losses in compressor installations, its first measure should be to equalize the pressures at all groups of receivers (i.e. to achieve \(a=0\)), at least by installing throttling devices at the branch lines. Otherwise, with almost daily changes in the value of \(a\), monitoring losses would be practically very difficult to carry out, since solving equation (46) with the subsequent coordination of the values of the nominal and real \(P_m\) is a rather difficult matter.

Let us find the most advantageous pressure in the network and at the compressor for the numerical example considered by us in § 4 of the present article.

We had \(a=0.0386\); \(A=8.14\cdot 10^4\); \(B=2.94\cdot 10^6\); \(nk=50\) and \(P=5\) ata. Hence:

\[ (a+1)=1.0386;\quad (1.84a+1)^2=1.148;\quad 1.67a=0.0645;\quad 9.2a=0.355;\quad M=0.0301; \]

\[ (1.84a+1)^2M=0.03534;\quad N=0.692;\quad 2.5aP^2=2.412\ \text{and}\ P^3=125. \]

Equation (46) will take the form (after multiplying both sides by 100):

\[ 3.534P_m^4+69.2P_m^3-241.2P_m=12500 \]

The value of \(P_m\) satisfying this equation will be \(P_m=5.40\) ata.

The value obtained for \(P_m\) shows that, under the given conditions, in particular with the specified diameters of the air pipelines, only a fictitious minimum of losses is possible, since \(P_m\) cannot be less than

\[ \frac{1.84P_{\mathrm{cp}}+\Delta P_{r,\mathrm{cp}}}{1.67}, \]

as follows from the equations:

\[ P_m=\frac{2P_{\mathrm{cp}}+\Delta P_{r,\mathrm{cp}}\Delta P_t}{2} \]

and

\[ \Delta P_t=0.165(2P_m-P_{\mathrm{cp}}) \]

Even if \(P_{\mathrm{av}} = 5\) (and in fact it is somewhat greater) and \(\Delta P_{r,\mathrm{av}}\) is taken (according to § 5) as 1.85 ata, then \(P_m\) should be on the order of 6.6 ata. To obtain a real minimum of losses, two paths are possible: a) reducing the working pressure at the receivers and b) reducing the diameters of the air pipelines. The first path is quite acceptable if the initially specified pressure \(P_n\) exceeded 5 ata. The second path, although it reduces the quantitative losses, leads to a strong increase in pressure losses from friction. The first path reduces the coefficient at \(P_m^3\) in the equation for the minimum of losses almost directly proportionally to the decrease in \(P\); the second path leads to a decrease in both coefficients \(M\) and \(N\).

It is obvious that, in the present state of the mine compressed-air economy of the USSR, the second path is the most natural, i.e. the replacement of pipes in certain sections by pipes of another diameter, with subsequent systematic verification of the most advantageous average pressure in the network \(P_m\), and hence also of the most advantageous pressure at the compressor.

§ 7. Influence of the average diameter of the network on the magnitude of the fictitious and real most advantageous pressure at the compressor.

The basic equation for the most advantageous average theoretical pressure in the network,

\[ (1.84\alpha + 1)^2 M P_m^4 + N P_m^3 - 2.5\alpha P^2 P_m - P^3 = 0 \]

shows that, for a specified pressure at the most separate group of receivers,*) the value \(P_m\) depends on \(\alpha\), \(M\), and \(N\). Each of these coefficients is a function of the pipe diameters in the individual sections of the network, and hence also a function of its average diameter. At the same time, the dependence of the quantities \(\alpha\), \(M\), and \(N\) on \(D_{\mathrm{av}}\) is rather complex. In particular, for example, a simultaneous increase or decrease in the cross sections of all pipelines by one and the same percent does not change the values of the equivalent lengths (as follows from equation 22), and consequently the value of \(\alpha\), whereas the coefficients \(M\) and \(N\) change their values.

The value of \(M\) changes especially sharply in this case. Indeed,

\[ M = \frac{4A(\alpha + 1) + 14400 \cdot 3.34\alpha\, nk}{2B}; \]

but the quantity \(A\) is directly proportional to the value of the average diameter of the network, and with an increase of the diameters by some \(\%\) it increases by the same \(\%\); while the quantity \(B = B^1 \cdot C^1\), where the quantity \(C^1\) is inversely proportional to the fifth power of the diameter of the main line, and with an increase of \(D\) by \(X\%\) decreases by

\[ \left(1 + \frac{X}{100}\right)^5 \]

times. Hence, for \(\alpha = 0\), the quantity \(M\), when the diameters of all pipelines are changed by \(X\%\), will increase by

\[ \left(1 + \frac{X}{100}\right)^6 \]

times; when \(\alpha \ne 0\), the change in the quantity \(M\) will be the sharper, the smaller \(\alpha\) and \(nk\) are, i.e. the smaller the share of the quantity \(14400 \cdot 3.34\alpha\, nk\) in the total quantity \(4A(\alpha + 1) + 14400 \cdot 3.34\alpha\, nk\). For the coefficient \(N\), the change in its value with a change in the diameters of the pipelines, including the main line, will be less sharp, since in this case the prevailing term is \(7200\,nk(2P - 1.67\alpha)\); the term \(A(9.2\alpha + 5 - P)\), with increasing \(D\), increases, while the term \(1.67^3 \alpha^3 \cdot B^1 C^1\) decreases, so that even when \(\alpha = 0\) an increase in the diameter of the main line by \(X\%\) can reduce the value of \(N\) at most

*) More precisely, at the group operating under the most difficult conditions, in particular at a group which, although not the most distant from the compressor, has abnormally small branch cross sections.

by \(\left(1+\dfrac{X}{100}\right)^5\) times. If, however, as most often happens in reality, only the magnitudes of the diameters of the branches and secondary mains are changed, then the quantity \(\alpha\) will change comparatively little, as will the quantities \(M\) and \(N\), since in this case \(C^1\), which is one of the factors in the expression \(B=C^1 B_1\), will be proportional only to \(\Sigma l_{\mathrm{e}}\), and moreover the quantity \(\Sigma l_{\mathrm{e}}\) will change not in the ratio \(\left(1+\dfrac{X}{100}\right)^5\), but considerably less, and the less so the greater the length of the main line and the smaller the number of sections to be re-equipped. The coefficient \(M\) will also change less sharply. In any case, one thing is beyond doubt: when the diameters of the network are increased on its separate sections, the coefficients \(M\) and \(N\) will increase, and when the diameters are decreased they will decrease, with comparatively small fluctuations in the value of \(\alpha\). Since an increase in \(M\) and \(N\), according to equation (46), causes a decrease in \(P_m\), the most advantageous theoretical average pressure in the network \(P_m\), with an increase in the diameters of the air ducts at least on individual sections, decreases. From this point of view, increasing the diameters of air ducts is desirable.

If we turn to the real value of \(P_m\), then, solving jointly the equations

\[ P_m=\frac{2P_{\mathrm{cp}}+\Delta P_{r,\mathrm{cp}}+\Delta P_t}{2},\quad P_{\mathrm{cp}}=\frac{1{,}67\alpha P_m+P}{1{,}84\alpha+1},\quad \Delta P_{r,\mathrm{cp}}=\frac{C^1 P_{\mathrm{cp}}^2}{P_m} \]

and

\[ \Delta P_t=0{,}165(2P_m-P_{\mathrm{cp}}) \]

with respect to \(P_m\), we obtain that in the general case the quantity \(P_m\) must satisfy the equation:

\[ 1{,}67[(1{,}84\alpha+1)^2-(1{,}84\alpha+1)1{,}84\alpha-1{,}67\alpha^2 C^1]P_m^2 -[2\cdot 1{,}67\alpha P C^1+ \]

\[ +1{,}84\cdot(1{,}84\alpha+1)P]P_m-C^1P^2=0, \]

whence \(P_m\), after collecting like terms in \(P_m^2\), is expressed as:

\[ P_m= \frac{3{,}34\alpha P C^1+1{,}84(1{,}84\alpha+1)P+}{3{,}34(1{,}84\alpha+1-1{,}67\alpha^2 C^1)} \]

\[ +\frac{\sqrt{P^2[3{,}34\alpha C^1+1{,}84(1{,}84\alpha+1)]^2+4\cdot1{,}67(1{,}84\alpha+1-1{,}67\alpha^2 C^1)C^1P^2}}{3{,}34(1{,}84\alpha+1-1{,}67\alpha^2 C^1)}. \tag{47} \]

As is evident from the last expression, with an increase in the diameters of individual sections of the network, \(P_m\) will decrease, although slowly. Indeed, \(C^1=\varphi(\Sigma l_{\mathrm{e}})\); but the equivalent length of any section, with an increase in its diameter, decreases by \(\left(\dfrac{d_1}{d_0}\right)^5\) times, where \(d_0\) is the original diameter, and \(d_1\) the increased diameter of the air duct on the given section. However, \(\Sigma l_{\mathrm{e}}\) will decrease more slowly than the individual equivalent lengths decrease, and the more slowly, the longer the main line in comparison with \(\Sigma l_{\mathrm{e}}-l_{\mathrm{mag}}\). It is necessary here to note that in the expression under the radical \(C^1\) enters both with a plus and with a minus sign, and the resulting term with \(C^1\) will have the form:

\[ 3{,}34C^1[3{,}34\alpha^2 C^1+(1{,}84\alpha+1)(3{,}67\alpha+2)-3{,}34\alpha^2 C^1]= \]

\[ =3{,}34C^1(1{,}84\alpha+1)(3{,}67\alpha+2)=6{,}68C^1(1{,}84\alpha+1)^2 \]

The influence of \(C^1\) on the magnitude of the denominator in the expression for \(P_m\) will be negligible, since in properly designed networks \(\alpha\) is of the order of \(0{,}1\) and even less;

and \(C^1\) will not exceed \(0.5\)*), so that the quantity \(1.67 a^2 C^1\) may be neglected. Thus, in final form:

\[ P_m = \frac{ 3.34aPC^1 + 1.84(1.84a+1)P + P(1.84a+1)\sqrt{6.68C^1+1.84^2} }{ 3.34(1.84a+1) } \]

or

\[ P_m = P\left[ \frac{aC^1}{1.84a+1} +0.55 +0.3\sqrt{6.68C^1+1.84^2} \right] \qquad \ldots\ldots (48) \]

Proceeding from the last expression, one may assert that, with an increase in the diameters of the air pipelines, \(P_m\) decreases comparatively slowly; moreover, in networks with unequal pressure at the receivers, the decrease of \(P_m\) with an increase in the diameter of the sections (other conditions being equal) proceeds more slowly than in networks with equal pressure at the receivers. In the latter case

\[ P_m = P\frac{1.84+\sqrt{1.84^2+6.68C^1}}{3.34} \]

or

\[ P_m = P\left(0.55+0.3\sqrt{6.68C^1+1.84^2}\right) \qquad \ldots\ldots (49) \]

Analysis of the expressions for the fictitious value \(P_m\), at which the minimum loss of compressed-air energy occurs, and of the real value \(P_m\), as functions of pressure losses and pressure at the receivers, makes it possible to indicate a way of eliminating the discrepancy that usually exists between these quantities. Since changing the diameters of air pipelines in individual sections of the network causes, for both \(P_m\), a change in their magnitude in the same direction, but with different intensity, it is obvious that such a change in the diameters of air pipelines in individual sections is always possible in which the real value of \(P_m\) coincides with the fictitious one. The greater the initial discrepancy between both values of \(P_m\), the sharper the changes in the diameters of individual sections must be, and the larger the number of sections that must be re-equipped. For the first time, solving this question is associated with a considerable expenditure of time, since it will probably be necessary to try several combinations in order to obtain a satisfactory agreement between the values of \(P_m\). However, subsequently, when over time the configuration of the network and the number of operating receivers change, it will be considerably easier to maintain agreement between both values of \(P_m\), since for this purpose a change of section on one or two secondary mains will be required. In any case, the economy that will inevitably be obtained when operating with a minimum loss of energy will always more than pay for the labor spent in finding the most advantageous real \(P_m\).

Thus, in installations of medium capacity, on the order of \(100\) cubic m per min., with an average network length of \(15\)—\(20\) km, air losses due to leakage are taken as about \(20\%\) of the useful air consumption, which in the present case will amount to \(20\) cubic m per min. If, for each cubic m of usefully employed air, only \(1000\) kg/m is saved in energy losses—and such saving is quite possible—then, at a cost of one kg/m

*) From the expression

\[ \Delta P_{r,\mathrm{avg}}=\frac{C^1 P_{\mathrm{avg}}}{P_m} \]

for \(\Delta P_{r,\mathrm{avg}}=2\) ata, \(P_{\mathrm{avg}}=5\) ata and \(P_m=6\) ata, we obtain

\[ C^1=\frac{12}{25}=0.48. \]

With an increase in \(P_{\mathrm{avg}}\) and a decrease in \(\Delta P_{r,\mathrm{avg}}\), \(C^1\) will decrease still further.

at 2.4·10\(^{-5}\) kopecks, the annual saving due to the reduction of qualitative losses will be

\[ \frac{2.4\cdot 10^{-5}\cdot(100-20)\cdot60\cdot5.5\cdot3\cdot345\cdot1000}{100} = 6557\ \text{rub.}\ 76\ \text{k.} \]

with three-shift operation and an installation operating time of 5.5 hours per shift.

Quantitative losses can also be reduced, at a minimum, by 15%, or to 17 cubic meters per minute instead of 20, which, at a cost of 1 cubic meter of air of 0.6 kopecks, will give during the year another

\[ \frac{0.6\cdot20\cdot15\cdot60\cdot5.5\cdot3\cdot345}{100} = \]

\[ = 6147\ \text{rub.}\ 90\ \text{kop.} \]

Thus, the total saving in losses in value terms will amount, for the present case, to about 6557 rub. 76 k. \(+\) 6147 rub. 90 k. \(=\) 12705 rub. 66 kop., or, in round figures, 12700 rubles per year.

In reality, however, the saving will be considerably greater, since practice shows that the actual cost of 1 cubic meter of free air is considerably higher than the design cost: thus, for the Prokop’evsk Mine Administration of Kuzbassugol’, the actual cost of 1 cubic meter of air proves to be 1.07 kopecks instead of the design 0.5 kopecks.

The costs of maintaining the additional staff in charge of loss control will be on the order of 7000–8000 rubles per year, so that organizing an air-loss control service is economically advisable even for an installation of medium capacity. As for larger compressed-air power installations, such as, for example, at Anzherka, at mine 9—15, at the Prokop’evsk Coking Plant, at the “STO” mine of the Krasnogvardeisk Mine Administration, at the Temir-Tau mine of the Stalin Metallurgical Plant, and many others, here the profitability of loss control admits of no doubt whatever.

§ 8. Influence of Individual Categories of Compressed-Air Energy Losses on the Magnitude of the Average Most Advantageous Pressure in the Network

Analysis of the coefficients of equation (46), expressing the conditions for the minimum of energy losses for the most general case, permits the following conclusions to be drawn:

1. With an increase in energy losses from leaks and from cooling of the air, the magnitude of the most advantageous pressure in the network \(P_m\) tends to decrease. Indeed, with an increase in the coefficient \(A\), characterizing the energy losses from air leaks, and in the magnitude \(nk\), which enters as a component part into the expression for losses from air cooling, the coefficients \(M\) and \(N\) in equation (46) increase, and consequently \(P_m\) decreases.

2. With an increase in energy losses from friction of the air in the pipes, the magnitude \(P_m\) also tends to increase. Indeed, in the expression for the coefficients \(M\) and \(N\), the coefficient \(B\), characterizing the magnitude of energy losses from friction of the air in the pipes, enters in the denominator; with an increase in \(B\), the coefficients \(M\) and \(N\) decrease, and consequently \(P_m\) tends to increase.

3. Welding of pipes reduces the magnitude \(A\), characterizing the magnitude of energy losses from leaks, and thereby promotes an increase in \(P_m\) and a reduction of the gap between the fictitious most advantageous and the real value of \(P_m\).

4. The magnitude of the diameters of air conduits, as well as the method of joining the air-conduit pipes, do not exert a substantial influence on the magnitude of energy losses from cooling of the air. Under the existing operating conditions of mine compressed-air power installations, this category of energy losses is least amenable to our influence.

5. In order to reduce the gap between the most advantageous fictitious and the real average pressure in the network, it is useful and necessary to vary from time to time

from time to time, the diameters of the air pipelines in individual sections. Preliminary calculations show that, for a whole series of installations already in operation, such a change should be made in the direction of reducing the existing pipe diameters. This conclusion contradicts the established view, according to which one must strive for the smallest possible pressure loss due to friction (not more than 0.1 ata), which can be achieved only by increasing pipe diameters. Such a divergence of views on so important a question has arisen because we have considered the economy of the mine compressed-air installation as a whole, whereas advocates of increasing pipe diameters proceed from the condition that the air pressure at the receivers should not fall below the value permissible for them. We have already seen above that increasing the diameters of air pipelines increases leakage losses in proportion to

\[ \frac{D_1}{D_0}, \]

whereas the energy losses from friction decrease by

\[ \left(\frac{D_1}{D_0}\right)^5 \]

times. It would seem that increasing pipe diameters is always advantageous, since one of the components of the total losses grows comparatively slowly, while the other falls very rapidly. However, this is true only when the initial energy losses from friction are large and the energy losses from leakage are expressed by a comparatively modest figure. If, however, as is most often the case in networks of great extent, the losses from leakage are considerably greater than the energy losses from friction, then increasing pipe diameters will not only fail to reduce the total losses, but will even increase them. Let us explain this by an example: suppose the energy losses from leakage are \(2 \cdot 10^6\) kgm per hour, and those from pressure loss are 40,000 kgm per hour; the total losses will be 2,400,000 kgm per hour. With an increase of diameters by 10%, the losses will respectively be 2,200,000 and

\[ \frac{400000}{1.1^5}=250000, \]

and the total losses 2,450,000 kgm, i.e. 50,000 kgm more than in the first case. This simple example shows that, from the point of view of reducing total energy losses, it is sometimes advantageous not to increase pipe diameters, but, on the contrary, to reduce them. It is clear that, in recommending a reduction of pipe diameters in individual sections of the network, we at the same time leave in full force the condition that the pressure at the receivers must not fall below the value permissible for them.

§ 9. On the influence of fittings and valves on the magnitude of the most advantageous pressure in the network and at the compressor.

The discrepancy usually observed between the theoretically most advantageous pressures in the network and at the compressor and the actual values of these quantities is sometimes reduced very considerably if the influence of fittings and valves is taken into account. Indeed, the presence of fittings and valves in an air-pipeline network is equivalent to increasing the length of the network by 10–15%. The increase in network length, owing to the addition of fictitious lengths compensating for the presence of valves and fittings, will have almost no effect, or only a very small effect, on the magnitude of the theoretically most advantageous \(P_m\). Indeed, the value of \(P_m\) in this case is determined chiefly by the ratio of the coefficients A and B, other conditions being equal. Therefore, in those cases where losses from cooling of the air may be considered constant, introducing fictitious network lengths into the calculation instead of the real ones will hardly change the value of \(P_m\), since both A and B will increase almost in the same ratio, i.e.

\[ \frac{A}{B} \]

will remain almost unchanged.

It is a different matter with the pressure drop due to friction. As the length of the network increases, it naturally increases by the same percentage, if all the other quantities (the diameters of the air ducts, the air consumption) remain unchanged. Since in most cases the theoretically most advantageous pressure \(P_m\), which depends, among other things, on the pressure drop due to friction, is obtained as smaller for the given specific conditions, introducing into the calculation the fictitious lengths of the network instead of its actual lengths is not only more correct in principle, but also gives a picture closer to reality and at the same time reduces the discrepancy between the theoretical and the actual values of \(P_m\).

The matter will be somewhat more complicated in the case where the energy losses from cooling cannot be regarded as constant. In this case, taking account of the influence of shaped pieces and fittings on the value of \(P_m\) will have a smaller effect, although it will still act in a direction favorable to reducing the above-mentioned discrepancy. However, since we ourselves are inclined to regard the solution of the question of the most advantageous pressure in the network and at the compressor for this case as a very approximate solution, not claiming accuracy, we consider it superfluous to examine here in greater detail the influence of shaped pieces and fittings in the case of varying losses from air cooling.

§ 10. Graphical-analytical method for determining the most advantageous average pressure in the network.

The complexity of the final expressions for the most advantageous average pressure in an air-duct network and the usually inevitable discrepancy between the value of the fictitious most advantageous pressure and the real pressure in the network under given conditions create great difficulties in determining \(P_m\) in each specific case. However, these difficulties can be considerably alleviated by applying a graphical method both for finding the quantities whose knowledge is necessary for determining \(P_m\), and for solving analytically complex equations.

First of all, the equivalent lengths for individual groups of receivers can be found graphically. Indeed, transforming the expression \(l_{\mathrm{e}} = l \left(\dfrac{D}{d_x}\right)^5 \left(\dfrac{V_x}{V}\right)^2 \dfrac{\beta_x}{\beta}\) into the equivalent expression \(l_{\mathrm{e}} = l \left(\dfrac{D}{d_x}\right)^5 \left(\dfrac{V_x}{V}\right)^{1.852}\), i.e., freeing ourselves from two variables, we can determine \(l_{\mathrm{e}}\) for each specific case by using the nomogram shown in Fig. 3. In doing so, for \(l\) it is more correct to take the fictitious lengths, i.e., the actual lengths of the sections multiplied by a coefficient of \(1.1\)–\(1.2\), and on average \(1.15\).

With the aid of nomograms the following can also be found:

a) The air velocities in individual sections, needed for determining the average air velocity in the network, from which, using Fleischer’s curves, Fig. 2, the coefficient of the dynamic regime is determined.

Nomograms for determining velocities in individual pipelines are presented in Figs. 4 and 5.

The nomograms are constructed from the formula

\[ C = \frac{4V}{60\pi d^2 P_m}. \tag{50} \]

where \(V\) is the quantity of free air flowing through the cross-section of the air duct, in cubic meters per minute (including losses), \(d\) is the diameter of the air duct in m, and \(P_m\) is the average air pressure in the given section. Without great error \(P_m\) may be taken as \(6.75\) atm; under this assumption the air velocities in some sections turn out to be underestimated (at branches), and in others overestimated, so that possible errors

partly compensate one another. The formula for the velocity will take a simpler form if \(d\) is in mm:

\[ C=\frac{4V1000^2}{60\pi\cdot 6.75\cdot d^2} \quad \text{or} \quad C=\frac{3200V}{d^2}\ \text{m per sec.} \ . . . (51) \]

b) Coefficient \(A\). The nomogram for determining it is presented in Fig. 6.

The nomogram is constructed on the basis of the equation: \(A=0.266K_cD_{\mathrm{av}}\Sigma l\), and for \(\Sigma l\) it will be most correct to take the sum of the fictitious lengths, i.e. the sum of the actual lengths increased by \(10 \div 15\%\).

The method of using the nomogram is very simple. From the point on the abscissa axis corresponding to the value of the given average diameter, a perpendicular is raised until it intersects the ray \(\Sigma l\). From the point of intersection a straight line is drawn parallel to the abscissa axis. This straight line intersects the \(\Sigma l\) scale at some point. Drawing a ray from this point, one raises a perpendicular from the point on the abscissa axis corresponding to the value \(K_c\) until it intersects this ray. From the point of intersection a straight line is drawn parallel to the abscissa axis, which cuts off on the \(A\) axis the sought value of it. The nomogram shows the case of determining \(A\) for the following data:

\[ D_{\mathrm{av}}=110.0\ \text{mm}, \quad \Sigma l=16000\ \text{m and } K_c=0.7. \]

c) Coefficient \(B\). From the expressions:

\[ B=\frac{7.62\cdot 10^{10}\cdot \beta_u(nk)^3\Sigma l_e}{T_0D^5n_y} \quad \text{and} \quad \beta_u=\frac{2.86}{G^{0.148}} \]

after eliminating \(\beta_u\) we obtain:

\[ B=\frac{4.05\cdot 10^9\cdot (nk)^{2.852}\cdot l_{e.\,\mathrm{av}}}{D^5} \ . . . . . . . . . (52) \]

The nomogram for determining \(B\) (in the form of logarithmic scales) is presented in Fig. 7. The method of using it is clear from the nomogram.

The coefficient \(C_1\), needed for determining the real pressure drop in the network, can be readily found from the expression

\[ C'=\frac{B}{1.3\cdot 10^5(nk)} \ . . . . . . . . . (53) \]

d) Finally, in those cases where the losses in the network from cooling of the air may be regarded as unchanged and the term \(P_m^3\) may be neglected, the equation
\(2A(2\alpha+1)^2(\alpha+1)P_m^4-3\alpha BF^2P_m-BF^3=0\) may be solved graphically.

At our request, Associate Professor of the Department of Mathematics of the Siberian Mining Institute A. E. Milovidov developed a simple and elegant method for the graphical solution of this equation. The nomogram of the equation is presented in Fig. 8. It consists of four scales: three fixed ones—the scale \(E\), the scale \(D\), and the scale \(X\) (or \(P_m\))—and one movable scale \(X\) (or \(P_m\)). To solve the equation it is necessary first to determine the quantities \(E\) and \(D\) for the given specific case. The expressions from which they are determined are indicated on the nomogram. Then the movable scale is set so that its zero coincides with the obtained value \(E\). Next, a ruler is taken and, around the point corresponding to the value \(D\) for the given case as a center, it is rotated until the values \(X\) on the movable and fixed scales coincide. Thus, for example, for the case \(E=1600\) and

\(D = 0.91\) (\(D\) is generally close to 1 and will most often lie within the limits \(0.9—0.95\)), \(X = P_m = 7.1\) ata.

Conclusion.

The present work, of course, does not claim to provide an exhaustive solution to the question. The complexity of the interdependences of the various quantities determining the value of \(P_m\) involuntarily led to replacing these dependences with others, mathematically simpler ones, which give identical results but do not express the true essence of the processes taking place. Certain assumptions, especially for the case of varying losses from air cooling, compel us to regard the solution of the question for this case only as a first approximation to reality. However, since the questions we have raised have been scarcely treated in either our domestic or foreign literature, we consider their solution, even in this form, to be a certain step forward.

Fig. 2

Nomogram for determining \(\ell_3\)

\[ \ell_3=\ell\left(\frac{D}{d_x}\right)^5\left(\frac{V_x}{V}\right)^{2.852} \]

Legend:

\[ \frac{V}{V_x} \]

\[ \frac{D}{d_x} \]

\[ 0 \]

\[ \ell \]

Fig. 3.

\[ C=\frac{3200V}{d^{2}} \]

Fig. 4. Nomogram for determining the air velocity in an air pipeline for \(V\) from 1 to 2.5 cubic meters per minute.

Fig. 5

NOMOGRAM FOR DETERMINING \(B\)

\[ B=\frac{4.05\cdot 10^{9}(nK)^{2.852}\,\ell_{\mathrm{av}}}{D^{5}} \]

— \(nK\) —
— \(D\) —
— \(Q\) —
— \(\ell_{\mathrm{av}}\) —

Fig. 7

Nomogram of the equation

\[ 2A(2\alpha+1)^2(\alpha+1)x^4 - 3\alpha B F^2 x - B F^3 = 0 \]

\[ 4 < x = P_m < 8 \]

\[ \varepsilon = \frac{B F^2}{2A(2\alpha+1)^2(\alpha+1)} \]

\[ \frac{l}{D} = 1+ \frac{3B\alpha F^2}{2000A(2\alpha+1)^2(\alpha+1)} \]

Labels and notes visible on the nomogram:

• \(D\)
• \(\varepsilon\)
• \(x\)
• Scale markings on the inclined line: \(0.9,\ 0.8,\ 0.7,\ 0.6,\ 0.5\)
• “The movable scale is lowered downward by a length \(=\varepsilon\)!”
• “Movable scale \(x\) (Fig. 8) (for the left \(\varepsilon\) scale)”

Fig. 8