Engineer G. I. Fuks

On the Question of Choosing

the Most Advantageous Diameter

of a Branch Pipeline

for Compressed Gas

Tomsk—1929

Engineer G. I. Fuks.

On the Question of Choosing the Most Advantageous Diameter of a Branched Pipeline for Compressed Gas.

In designing a pipeline for compressed gas or steam, in particular an air line for a pneumatic power-distribution installation, the question of choosing the pipe diameter is fundamental. If one sets, as is usually done, the gas velocity in the pipe or some permissible pressure drop, it is easy to determine the diameter. But in doing so there is no assurance that the selected diameters are in fact the most advantageous, i.e., that they will give the lowest cost of 1 cubic meter of gas.

Blaesz*) gives a method for calculating the most advantageous velocity, from which the pipe diameter can then be calculated. This method is quite suitable for a simple pipe. But for a complex pipeline it is obviously inapplicable, since it may easily turn out that the most advantageous velocity will be different for its different parts. There remains only the path of selection: to compare several calculation variants by the cost of 1 cubic meter of compressed gas and choose the most advantageous. This path is rather long and, moreover, does not give complete confidence in the correctness of the choice of variant.

Meanwhile, the problem posed is essentially unambiguous in all its parts, i.e., it has only one possible solution under given concrete conditions.

The application of the method of equivalent lengths of the pipeline especially simplifies the practical solution of this question. As is known, this method of equivalence was first applied by Thomson in calculating an electrical network. Engineer Ya. Karchevskii**) applied this method to solve the question of choosing the diameter of a pipeline for water. The special form of the equations written by him, and the limitations, do not permit his relations to be used directly for calculating the diameters of a gas pipeline.

To avoid misunderstandings, it is necessary to stipulate that a preliminary (before design) exact solution of the question of choosing the most advantageous diameters of a branched pipeline is possible only with a certain degree of accuracy, depending in part on the experience of the designer. This becomes understandable if one takes into account that in a preliminary calculation one has to assign a number of quantities which are finally and precisely ascertained only at the end of the calculation. The method presented is characterized by the fact that, when using it:

1. The number of unknown quantities that must be assigned preliminarily is reduced to a minimum.

2. By repeated recalculation, if necessary, we obtain an increasingly exact solution of the question.

3. For any given installation (i.e., known dimensions of the pipeline) the question of whether this installation is the most advantageous is solved with complete precision.

) V. Blaesz, Die Strömung in Rohren
) Ya. Karchevskii. On the calculation of the most advantageous diameter of pipes in pumped heating. Vestnik Inzhenerov*, 1915.

Thus the work of the designer is shortened and simplified, and he is also given a reliable means of checking the results obtained. This is the justification for the indicated method.

I.

The most advantageous diameter of a pipeline for compressed gas is considered to be that diameter, or, in the case of a branched pipeline, that set of diameters, which gives the cheapest cost of one cubic meter of gas or steam at the point of consumption. The cost of 1 cubic meter is composed of a considerable number of separate components, the chief of which are:

1. Expenditure of power (energy) for compressing the gas.
2. Expenditure of water for cooling the gas being compressed.
3. Expenditure on servicing (labor, lubricating and wiping material, operating expenses, etc.).
4. Expenditure for amortization of the capital costs of the machinery installation and buildings.
5. Expenditure for amortization of pipelines.
6. Expenditures for current and capital repair of all structures for obtaining and transporting the compressed gas.

To groups 4 and 5 of expenditures one may add, if necessary, interest on the capital expended, etc.

Of these components it is quite sufficient, in a preliminary calculation, to include the most important items of expenditure. Further, there is no need to include those expenditures which, as will be seen below, will remain practically constant for different pipeline diameters. Such will be, for example, the expenditure for amortization of the machinery equipment, the expenditure for cooling water, etc. Thus we shall separate out the expenditures:

1. For the energy required to obtain compressed gas.
2. For amortization and repair of pipelines.
3. The remaining or miscellaneous expenditures.

The expenditure of energy for obtaining compressed gas naturally depends on the following conditions:

1. The quantity of gas being compressed. Let the individual consumers use compressed gas in amounts \(v_1, v_2 \ldots v_n\) cubic meters per second.

Individual consumers may use gas nonuniformly, namely:

\[ n_1,\ n_2\ .\ .\ .\ .\ .\ n_n \text{ hours per year.} \]

1. The degree of compression of the gas consumed, i.e. the ratio of the pressures of the gas consumed and that entering the compressor.

\[ \frac{p}{p_1}. \]

1. The degree of perfection of the compressor. Generally speaking, this is a variable quantity, depending on the load. But in each given case it is possible, with one or another degree of accuracy, to prescribe in advance—if only by taking into account the assumed load schedule and the test data of the installed type of compressor (the firm’s guarantees)—a certain average efficiency coefficient. We shall denote it by the letter \(\eta\), understanding by it the isothermal efficiency coefficient of a gas compressor and the adiabatic efficiency coefficient for a steam compressor.

1. The cost of energy at the compressor shaft, which we shall take as “\(a\)” rubles per horsepower-hour.

Using these notations, and also the known expression for the work expended in compressing 1 cubic meter of gas, we obtain the total annual expenditure on energy for compressing gas or steam as

\[ K'_m=\frac{a p_1 10^4 (v_1 n_1+v_2 n_2+\cdots)}{75\eta}\ln\frac{p}{p_1}\ \text{rubles}, \]

if the ideal process is taken to be an isothermal process, or

\[ K'_m=\frac{k}{k-1}\frac{a p_1 10^4 (v_1 n_1+v_2 n_2+\cdots)}{75\eta} \left[\left(\frac{p}{p_1}\right)^{\frac{k-1}{k}}-1\right]\ \text{rubles}, \]

for an ideal adiabatic process (compression of steam).

The energy expenditure for compression calculated in this way will be minimal, but not the most advantageous in the given installation, since in this case there must be no pressure losses in the pipeline. This entails infinitely large pipe diameters, i.e., the same expenditures for their amortization and repair. Therefore, in practice the compressor must compress the gas to a somewhat higher pressure than is necessary at the point of consumption. Let us call this permissible pressure drop “\(x\)” atmospheres, and then:

\[ K_1=\frac{a p_1\cdot 10^4 \sum(v\cdot n)}{75\cdot \eta}\ln\frac{p+x}{p_1}\ \text{rubles}\ ........(1) \]

\[ K_1=\frac{k}{k-1}\frac{a p_1 10^4 \sum(v n)}{75\cdot \eta} \left[\left(\frac{p+x}{p_1}\right)^{\frac{k-1}{k}}-1\right]\ \text{rubles}.(1') \]

The expenditure for amortization and repair of a pipeline consisting of individual pipes of diameters \(d_1,\ldots d_2\) meters, etc., and, correspondingly, lengths

\[ l_1,\ldots\ldots l_2,\ldots\ldots l_3,\ldots\ldots \]

and so on, in meters, will depend on the cost per unit length of each pipe in the installation and on the percentage deduction for amortization and repair allowed for the given installation. The latter quantity may be regarded as given; that is, we assume that the annual deductions for amortization and repair per ruble of the pipeline cost amount to

\[ c\ \text{rubles}. \]

The cost of 1 meter of pipe of various diameters, as calculation shows, can be expressed with a high degree of accuracy by the relation

\[ r d^f\ \text{rubles}, \]

where:

\[ r\ \text{and}\ f\ldots\ldots\ldots \text{are constants} \]

\[ d \text{ is the pipe diameter in meters.} \]

For iron pipes with couplings this dependence is perfectly accurate within the range from \(2''\) to \(8''\) pipes. A sufficiently close agreement is also obtained for other pipes.

Having adopted this dependence, it is easy to calculate the cost of the entire pipeline. It is only necessary to take into account that its cost must also include the cost of connecting parts, shaped parts, fittings, laying work, etc. The indicated quantities likewise depend on the pipe diameter, but since the latter is unknown at the design stage, they also cannot be determined in advance. We shall note only that the cost of earthworks when laying pipes in the ground may here not

accounted for, since the latter practically does not depend on the pipe diameter. Consequently, this expense may be assigned to the group of other expenses. The expenses for connecting parts, etc., which depend on the diameter of the pipeline, we shall take into account by introducing, instead of the actual lengths of the separate sections

\[ l_1,\ldots l_2,\ldots l_3,\ldots \]

their fictitious lengths

\[ L'_1,\ldots L'_2,\ldots L'_n \ldots \text{ meters.} \]

We shall choose these fictitious lengths in such a way that the cost of the whole pipe, for example the first one, made with length \(L'_1\), taking into account only the cost of the pipeline, would be equal to its actual cost at length \(l_1\), with all additional expenses for connecting and shaped parts, fittings, etc. Obviously, in general:

\[ L'=\kappa_2 l \]

where

\[ \kappa_2 > 1. \]

The value “\(\kappa_2\)” in the initial design may be assigned:

a) by eye, in order, after obtaining the result, to verify whether it has been assigned correctly.

b) From the execution estimates of projects of pipelines of analogous installations, by computing the ratio of the cost of 1 meter of finished pipeline to the cost of 1 meter of pipe.

c) From an approximate preliminary calculation of the cost of 1 meter of finished pipe in the installation, assuming an average practical gas velocity.

It must be noted that the uncertainty in the choice of the value “\(\kappa_2\)” is only apparent and temporary in the preliminary calculation. From the cost summary at the end of the calculation it is at once revealed whether this value has been chosen incorrectly, and then the calculation may be repeated until sufficient agreement is obtained. Having assigned the indicated fictitious lengths of the pipeline sections, we compute the total annual cost of the pipeline

\[ K_2 = c r \left(L'_1 d_1^f + L'_2 d_2^f + \ldots \right) \text{ rubles} \tag{2} \]

The total annual costs for obtaining 1 cubic meter of compressed gas in the given installation will be expressed by the relation:

\[ K = K_1 + K_2 + K_3 \qquad \text{rubles, where} \]

\(K_1\) — the total annual expenditure for energy (equations 1 and \(1'\))

\(K_2\) — the total annual expenditure for the pipeline (equation 2)

\(K_3\) — the sum of other expenses, practically independent of the pipeline diameter.

The basic condition for the most advantageous diameters of the pipeline will obviously be

\[ K \sim \text{minimum} \ldots \tag{3} \]

Pressure drop and pipeline diameters.

Considering the quantities entering equations (1) and (\(1'\)) and (2), we see that among them are the quantities of pressure drop in the entire installation, “\(x\),” and the pipeline diameters \(d_1, d_2\), etc. The total pressure drop is composed in one way or another, depending on the geomet-

of the hydraulic relations of the pipeline and the individual pressure drops in each pipe, \(x_1\), \(x_2\), \(x_3\), etc.

In turn, the dependence between the pressure drop and the diameter of each pipe may be expressed by the well-known relation:

\[ \Delta p=\lambda \cdot \gamma' \frac{1}{d}\cdot \frac{w^2}{2g} \]

where

\[ \lambda \]
— coefficient of resistance

\[ \gamma' \]
— specific weight of the gas flowing through the pipeline, in \(\mathrm{kg}/\mathrm{m}^3\).

\[ w \]
— gas velocity, in \(\mathrm{m}/\mathrm{sec}\).

\[ g=9.81\,\frac{\mathrm{m}}{\mathrm{sec}^2}. \]

Expressing the pressure drop in atmospheres, we have:

\[ x=\frac{\lambda \gamma'}{10^4}\cdot \frac{1}{d}\cdot \frac{W^2}{2g} \]

If \(V'\) cubic meters per second of compressed gas of density \(\gamma'\) flows through the given pipe, then, denoting the volume of gas at uncompressed density \(\gamma_0\) by \(V\), we evidently have

\[ V'=V\frac{\gamma_0}{\gamma'} \]

Further, the gas velocity is:

\[ W=\frac{4V'}{\pi d^2} \]

Denoting the ratio

\[ \frac{\gamma'}{\gamma_0}=b \]

and substituting in the expression for \(x\), we have:

\[ x=\frac{B l V^2}{d^5},\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (4) \]

where:

\[ B=\frac{16\lambda \gamma_0}{10^4\cdot \pi^2\cdot 2g\,b} =816\cdot 10^{-8}\lambda \frac{\gamma_0}{b} .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (5) \]

In connection with relations (4) and (5), the following may be added: the ratio of the densities of the compressed and compressible gas \(b\) may be calculated without taking into account the pressure drop in the given pipe \((x)\), since this drop is, in practice, very small in comparison with the absolute pressure of the gas. Thus, for a gas pipeline:

\[ b=\frac{p}{p_1}\frac{273+t_0}{273+t_2} \]

where \(t_0\) and \(t_2\) are the temperatures of the uncompressed and compressed gas, respectively.

In a pipeline for compressed steam, the ratio must be taken from a table or diagram.

The coefficient of resistance \(\lambda\) is a quantity depending on many variables, including the diameter of the pipeline—

water and velocity. The best-founded dependence must be recognized as that based on the law of dynamic similarity, on the Reynolds number:

\[ R=\frac{Wd}{V} \]

where

\[ V \text{—kinematic viscosity}^{*}) \]

In what follows the dependence adopted is:

\[ \lambda = 0.3164\,R^{-0.25} \]

for \(R\) up to 200,000, and

\[ \lambda = 0.0072 + 0.6104\,R^{-0.35} \]

for \(R\) greater than 200,000. Within the practical limits of the adopted ratios, \(\lambda\) varies from 0.016 to 0.012, which makes it possible to choose this quantity in advance with sufficient reliability.

The formula (4) given for calculating the loss of pressure needs to be specified in the sense that, instead of the geometrical length \(l\), one must substitute the length

\[ L = l + l_{\mathrm{э}}, \]

where “\(l_{\mathrm{э}}\)” is the length of pipe equivalent to the local resistances of the pipeline. Denoting the coefficients of local resistances by \(\xi_1,\ \xi_2\), etc., and their sum by

\[ \Sigma \xi = \xi_1 + \xi_2 \ldots, \]

we calculate the length of pipe equivalent to the local resistances:

\[ l_{\mathrm{э}} = \Sigma \xi \frac{d}{\lambda}. \]

Thus, this quantity can be determined for a given pipeline. In general, in a preliminary calculation

\[ L = \kappa_1 l, \qquad \text{where} \qquad \kappa_1 > 1 \]

and it can also be determined in the same way as the corresponding quantity \(\kappa_2\) introduced above.

Thus, finally,

\[ x=\frac{BLV^2}{d^5}\ldots\ldots\ldots\ldots\ldots\ldots\ldots (4') \]

The fundamental equation of the most advantageous diameter of the pipeline, written above in the general form (3), is expressed mathematically, as is known, by the fact that the derivative of \(K\) must be equal to zero. The choice of the independent variable with respect to which the derivative is taken plays no role. In the present case the quantity \(x\)—the pressure drop from the compressor to the consumer—is taken as the independent variable. The equation of the minimum is:

\[ \frac{dK}{dx}=0 \ldots\ldots\ldots\ldots\ldots (6) \]

From this equation one can find \(x\), and then, by relation (5), find the corresponding pipe diameter.

\[ \begin{aligned} ^{*})\ &\text{Blausius, Das Aehnlichkeitsgesetz bei Reibungsvorgängen in Flüssigkeiten. Mitt. u.}\\ &\text{Vorsch. VDI H 131.}\\ &\text{Ombek, Druckverlust strömendes Flüssigkeit in geraden zylindri Röhrleitungen. M. u.}\\ &\text{VDJ H 158—159.} \end{aligned} \]

II.

Special cases.

1. The installation consists of a single pipe through which compressed gas is delivered to the consumer.

Diagram: a single pipeline labeled \(x, d, B, L, L'\), with length \(l\).

Depending on the type of installation, the annual cost of the energy for compression will be:

\[ K_1=\frac{a p_1 10^4 (v\cdot n)}{75\,\eta}\ln\frac{p+x}{p_1} \]

or

\[ K_1=\frac{k}{k-1}\cdot \frac{a p_1 10^4\cdot (v\cdot n)}{75\cdot \eta} \left[\left(\frac{p+x}{p_1}\right)^{\frac{k-1}{k}}-1\right]. \]

The annual cost of the pipeline expenses is

\[ K_2=crd^f L'. \]

The total cost is

\[ K=K_1+K_2+K_3. \]

The condition for a minimum is:

\[ \frac{dK}{dx}=\frac{dK_1}{dx}+\frac{dK_2}{dx}=0. \]

Taking into account the relation \(x=\dfrac{B L V^2}{d^5}\) and carrying out the necessary calculations, we have:

\[ x^{1+0.2f}=\frac{crf}{5A}\cdot (p+x)^{\frac{1}{k}} L' (B L V^2)^{0.2f}\ldots\ldots (7) \]

where:

\[ A=\frac{a p_1 (v n) 10^4}{75\,\eta_{is}}\ldots\ldots (8') \]

for the basic isothermal compression, and

\[ A=\frac{a p_1^{\frac{1}{k}}10^4\cdot (v\cdot n)}{75\cdot \eta_{ad}}\ldots\ldots (8'') \]

for the basic \((8'')\) adiabatic process.

Both of these formulas can conveniently be reduced to one:

\[ A=\frac{a p_1^{\frac{1}{k}}10^4(v n)}{75\cdot \eta}\ldots\ldots (8) \]

which passes into formula \((8')\) when the ideal process is isothermal \((k=1)\). At the same time, equation \((7)\) also becomes simpler when \(k=1\). We shall not dwell on the simple proof that equation \((7)\) indeed gives the minimum value of the function \(K\).

An exact solution of equation (7) is possible graphically, as the point of intersection of the curve:

\[ y=x^{(1+0.2f)\kappa} \]

and the straight line:

\[ y=\left[\frac{crf}{5A}\,L'(BLV^2)^{0.2f}\right]^{\kappa}(p+x). \]

In ordinary practical cases, in view of the smallness of the quantity \(x\) in comparison with \(p\), one may take:

\[ x\sim \left[ \frac{crf}{5A}\,L'(BLV^2)^{0.2f} \right]^{ \frac{1}{1+0.2f}\,\kappa(1+0.2f) } p \tag{9} \]

From the value thus found for \(x\), the pipe diameter \(d\) is determined, and the problem is thereby solved.

The direct determination of the desired pipe diameter may be carried out from this same equation (9). Substituting for \(x\) the diameter from the relation:

\[ x=\frac{BLV^2}{d^5} \]

and taking into account the value of \(B\) according to equation (5), we have:

\[ d^{m/m}=1000\left(\frac{D}{N}\right)^{\frac{0.2}{1+0.2f}} \tag{10} \]

where:

\[ D=\frac{54.5\,a(vn)\gamma_0}{\gamma_1crf\cdot b}\cdot 10^{-4} \tag{10'} \]

and

\[ N=\frac{L'}{L}\cdot\frac{1}{\lambda V^2} \tag{10''} \]

The solution of equations (9) and (10) is naturally to be carried out by trial, since beforehand, before determining the diameter, we cannot know the values \(L'\), \(L\), and \(\lambda\). For a single pipe the ratio \(\frac{L'}{L}\) should not deviate appreciably from unity, which value may be adopted in the initial calculation.

2nd special case.

The installation consists of two pipes connected in parallel.

[Diagram: two branches from a common node. The upper inclined branch is labeled \(x\, d_1\, B_1\, L_1\, L'_1\, V_1\); the lower horizontal branch is labeled \(x\, d_2\, B_2\, L_2\, L'_2\, V_2\).]

As before:

\[ K_1=\frac{\kappa}{\kappa-1}\, \frac{a p_1\cdot 10^4 \sum(vn)}{75\cdot\eta} \left[ \left(\frac{p+x}{p_1}\right)^{\frac{\kappa-1}{\kappa}}-1 \right] \]

The annual cost of the pipeline

\[ K_1 = cr\,[L_1' d_1^f + L_2' d_2^f] \]

Between the pipe diameters and the pressure drop in them there must exist the relation

\[ x=\frac{B_1 L_1 V_1^2}{d_1^5}=\frac{B_2 L_2 V_2^2}{d_2^5} \]

Thus, mathematically, \(d_1\) and \(d_2\) are functions of \(x\), i.e.

\[ \frac{dK_2}{dx} = \frac{\partial K_1}{\partial d_2}\cdot \frac{dd_1}{d.x} + \frac{\partial K_2}{\partial d_2}\cdot \frac{dd_2}{dx} \]

Finding, analogously to the 1st case, the derivative \(\dfrac{dK}{dx}=0\), we have:

\[ x^{1+0.2f} = \frac{crf}{5A}(p+x)^{\frac{1}{k}} \left[ L_1'(B_1L_1V_1^2)^{0.2f} + L_2'(B_2L_2V_2^2)^{0.2f} \right] \ldots\ldots(11) \]

where the value \(A\), according to (8):

\[ A=\frac{ap_1^{\frac{1}{k}}\sum v\eta)10^4}{75\cdot \eta} \]

The very course of calculating relation (11) shows that it can be applied to any number of parallel pipes. In the general case we have:

\[ x^{1+0.2f} = \frac{crf}{5A}(p+x)^{\frac{1}{k}} \sum_1^n \left[ L'(BLV^2)^{0.2f} \right] \ldots\ldots(11') \]

Let us note that in this case the diameters of the individual pipes of the system are connected by the relation:

\[ \frac{d_1}{d_2} = \left( \frac{B_1L_1V_1^2}{B_2L_2V_2^2} \right)^{0.2} \]

\[ \frac{d_1}{d_3} = \left( \frac{B_1L_1V_1^2}{B_3L_3V_3^2} \right)^{0.2} \quad \text{and so on} \ldots\ldots(12) \]

3rd special case.

The installation consists of two pipelines connected in series. In the general case \(V_1 \ne V_2\). The quantities \(K_1\) and \(K_2\) are computed in the general form, as in case 2.

The total pressure drop

\[ x=x_1+x_2 \ldots\ldots(13) \]

\(x_1\) and \(x_2\) are the pressure drops in the 1st and 2nd pipe, respectively. The relation between the pressure drop in each pipe and the diameter of the latter is

\[ x_1=\frac{B_1L_1V_1^2}{d_1^5}; \qquad x_2=\frac{B_2L_2V_2^2}{d_2^5} \ldots\ldots(14) \]

The problem of finding the minimum of the function

\[ K = K_1 + K_2 (+K_3) \ldots \ldots \ldots \ldots \ldots \ldots \ldots (15) \]

is reduced to determining the minimum of a function of 4 variables

\[ x_1,\ x_2,\ d_1\ \text{and}\ d_2, \]

connected with one another by the additional relations (12) and (13).
Carrying out the calculations, we have:

\[ x_1^{1+0.2f} = \frac{crf}{5A}(p+x)^{\frac{1}{k}}L'_1(B_1L_1V_1^2)^{0.2f} \]

\[ x_2^{1+0.2f} = \frac{crf}{5A}(p+x)^{\frac{1}{k}}L'_2(B_2L_2V_2^2)^{0.2f} \ldots \ldots \ldots (16) \]

Whence it is easy to obtain:

\[ x^{1+0.2f} = \frac{crf}{5A}(p+x)^{\frac{1}{k}} \left\{ \left[L'_1(B_1L_1V_1^2)^{0.2f}\right]^{\frac{1}{1+0.2f}} + \left[L'_2(B_2L_2V_2^2)^{0.2f}\right]^{\frac{1}{1+0.2f}} \right\}^{1+0.2f} \ldots \ldots \ldots (17) \]

In practice, when seeking the most advantageous diameter (pressure drop) of a pipeline composed of pipes connected in series, one may also use equations (16) and (17). The total pressure drop \(x\) is usually small in comparison with \(p\), and formulas (14) and (16) may be written analogously to (9). From this one may draw the conclusion that the magnitude of the diameter of each pipe of a pipeline connected in series is only weakly influenced by the diameters of the remaining pipes, and that they may be computed, at least in a first approximation, independently of one another.

The course of reasoning given here will evidently not change if the number of pipes is greater than two.

The number of equations (16) will be equal to the number of individual pipes, while the general equation (17), in the case of a large number of them, will be written as

\[ x^{1+0.2f} = \frac{crf}{5A}(p+x)^{\frac{1}{k}} \left\{ \sum_{1}^{n} \left[ L'(BLV^2)^{0.2f} \right]^{\frac{1}{1+0.2f}} \right\}^{1+0.2f} \ldots \ldots \ldots (17') \]

Let us note that in this case the diameters of the individual pipes of the system are also connected by a definite relation, which is obtained by eliminating \(x_1\) and \(x_2\) from systems (14) and 16). Carrying out the necessary calculations, we have:

\[ \frac{d_1}{d_2} = \left[ \frac{L'_2}{L'_1} \frac{(B_1L_1V_1^2)}{(B_2L_2V_2^2)} \right]^{\frac{0.2}{1+0.2f}} \]

\[ \frac{d_1}{d_3} = \left[ \frac{L'_3}{L'} \frac{(B_1L_1V_1^2)}{(B_3L_3V_3^2)} \right]^{\frac{0.2}{1+0.2f}} \ldots \ldots \ldots (18) \]

It may be considered that, in general form, the problem has been finally solved for both of the last particular cases. Having found the total pressure drop \(x\) (or the partial drops \(x_1, x_2\), etc.), we determine the required diameters. Of course, in practice, in these cases also, the problem is solved not

at once, in view of the indeterminacy in solving for the values \(L\), \(L'\), and \(B\) \((\lambda)\). But by successive trial the problem is practically solved rather quickly.

III.

Equivalent Pipe Length with Respect to Resistance and Cost

The method given above for determining the most advantageous pressure drop in a pipeline, i.e. the most advantageous diameter of its individual sections, is a general method.

With its help one can solve the posed problem for any branched pipeline, since it always consists of a system of parallel and series pipelines.

But as the number of individual branches of the system increases, the number of unknown quantities grows and, most importantly, the equations themselves become so complicated that in practice they are inconvenient to use. Therefore this method may be considered convenient only for simple schemes made up of parallel or series pipes alone. In the remaining, more complex cases, it is more convenient to use the method of equivalent lengths with respect to resistance and cost, in order to simplify the writing and solution of the minimum equations.

By a pipe equivalent in resistance to some given pipe of diameter \(\delta\), we shall mean such a pipe of diameter \(d\) as will have the same pressure drop as the given one. It is obvious that, in such a definition, not everything is yet unambiguous. In each case it is necessary to indicate for what gas flow the equivalent pipe is being computed. If the length of the pipe of diameter \(\delta\), determining its resistance, is denoted by the letter \(L\), and the length of the pipe of diameter \(d\) equivalent to it in resistance by the letter \(L_e\), then evidently, in general,

\[ L_e \ne L. \]

It goes without saying that one can also define a pipe equivalent in resistance to 2, 3, or more separate pipes connected with one another in any way. The diameter of the equivalent pipe may be chosen arbitrarily equal to the diameter of one of the pipes of the system.

In what follows, in the case of parallel pipes, we shall agree to compute the equivalent length, taking its diameter equal to the diameter of one of them, for example the 1st, and computing it for a gas flow \(V\) equal to the sum of the flows in the parallel branches, i.e.:

\[ V = V_1 + V_2 + V_3 + \ldots . \]

where \(V_1, V_2 \ldots\) are the gas flows through the individual parallel pipes.

The resistance coefficient of this equivalent pipe, since it depends on the gas velocity in the pipe, will not, generally speaking, be equal to the resistance coefficient of the first pipe \(\lambda_1\). Nevertheless, it may be taken equal to \(\lambda_1\), since this, in turn, will cause only a corresponding change in the length of the equivalent pipe, to which we may ascribe an exclusively mathematical, and not a real, meaning.

Starting from the definition, for a system of parallel pipes one may write the relation

\[ x = \frac{B_1 L_1 V_1^2}{d_1^5} = \frac{B_2 L_2 V_2^2}{d_2^5} = \ldots = \frac{B_1 L_{er} (V_1 + V_2 + \ldots)^2}{d_1^5}. \]

Whence

\[ \frac{L_{\mathrm{er}}}{L_1} = \frac{V_1^2}{(V_1+V_1\ldots)_1} \quad\text{or} \]

\[ L_{\mathrm{er}} V^2 = L_1 V_1^2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (19) \]

Relation (19) finally and unambiguously determines the equivalent length of a pipe, in resistance, for a given system of parallel pipes \((L_{\mathrm{er}})\).

In determining the equivalent length for a system of pipes connected in series, we shall agree to take the diameter of the equivalent pipe as equal to the diameter of the pipe that comes first in the direction of gas flow. The gas flow through the equivalent pipe shall be taken as equal to the gas flow through the main pipe \(V_1\); likewise, we shall regard the coefficients of resistance of the equivalent pipe and of the 1st pipe as identical.

By condition, for pipes in series:

\[ x = x_1 + x_2 + x_3 + \ldots \]

\[ \frac{B_1 L_{\mathrm{ec}} V_1^2}{d_1^5} = \frac{B_1 L_1 V_2^2}{d_1^5} + \frac{B_2 L_2 V_2^2}{d_2^5} + \frac{B_3 L_3 V_3^2}{d_3^5} +\ldots \]

Whence:

\[ \frac{L_{\mathrm{ec}}}{L_1} = 1 + \frac{B_2 L_2 V_2^2}{B_1 L_1 V_1^2} \left(\frac{d_1}{d_2}\right)^5 + \frac{B_3 L_3 V_3^2}{B_1 L_1 V_1^2} \left(\frac{d_1}{d_3}\right)^5 +\ldots \]

Using relation (18), we have:

\[ \frac{L_{\mathrm{ec}}}{L_1} = 1 + \left(\frac{L_2'}{L_1'}\right)^{\frac{1}{1+0.2f}} \left(\frac{B_2 L_2 V_2^2}{B_1 L_1 V_1^2}\right)^{\frac{0.2f}{1+0.2f}} + \]

\[ + \left(\frac{L_3'}{L'}\right)^{\frac{1}{1+0.2f}} \left(\frac{B_3 L_3 V_3^2}{B_1 L_1 V_1^2}\right)^{\frac{0.2f}{1+0.2f}} +\ldots \ldots \ldots \ldots (20) \]

Relation (20) finally and unambiguously determines the length of a pipe of diameter \(d_1\), equivalent in resistance to a system of pipes in series \((L_{\mathrm{ec}})\).

By a pipe equivalent in cost to some given pipe of diameter \(\delta\), we shall mean such a pipe of diameter \(d\) that will have the same cost as the given one. Generalizing this definition, we arrive at a pipe equivalent in cost to a given system of pipes.

If a system of pipes is composed of separate pipes whose diameters and lengths are respectively

\[ d_1,\ d_2,\ d_3 \ldots \text{ etc.} \]

\[ L_1',\ L_2',\ L_3' \ldots \text{ etc.} \]

then the total cost of these pipes is:

\[ K_2 = r\left(L_1' d_1^f + L_2' d_2^f + \ldots\right). \]

By the definition of a pipe equivalent in cost, taking its diameter equal to the diameter of the 1st pipe:

\[ r L_{\mathrm{e}}' d_1^f = r\left(L_1' d_1^f + L_2' d_2^f + \ldots\right). \]

Whence:

\[ \frac{L_{\mathrm{e}}'}{L_1'} = 1 + \left(\frac{d_2}{d_1}\right)^f \left(\frac{L_2'}{L'}\right) +\ldots \]

When replacing a system of parallel pipes by an equivalent pipe, we have, according to the relation between the diameters of the pipes (12):

\[ \frac{L'_{\mathrm{ep}}}{L'_1} = 1+ \left(\frac{B_2 L_2 V_2^{2}}{B_1 L_1 V_1^{2}}\right)^{0.2 f} \frac{L'_2}{L'_1} + \left(\frac{B_3 L_3 V_3^{2}}{B_1 L_1 V_1^{2}}\right)^{0.2 f} \frac{L'_3}{L'_1} +\cdots \tag{22} \]

This equality makes it possible to calculate the cost-equivalent length of the pipe for the given system of pipes connected in parallel \((L'_{\mathrm{ep}})\).

The corresponding equality for a system of pipes in series is written, using (18),

\[ \frac{L'_{\mathrm{es}}}{L'_1} = 1+ \left(\frac{L'_2}{L'_1}\right)^{\frac{1}{1+0.2 f}} \left(\frac{B_2 L_2 V_2^{2}}{B_1 L_1 V_1^{2}}\right)^{\frac{0.2 f}{1+0.2 f}} + \]

\[ + \left(\frac{L'_3}{L'_1}\right)^{\frac{1}{1+0.2 f}} \left(\frac{B_3 L_3 V_3^{2}}{B_1 L_1 V_1^{2}}\right)^{\frac{0.2 f}{1+0.2 f}} +\cdots \tag{23} \]

Let us note that from (20) and (23) it follows directly:

\[ \frac{L_{\mathrm{es}}}{L_1} = \frac{L'_{\mathrm{es}}}{L'_1}, \quad \text{which may also be written as:} \quad \frac{L'_{\mathrm{es}}}{L_{\mathrm{es}}} = \frac{L'_1}{L_1}. \tag{24} \]

When a system of pipes in series is reduced to an equivalent pipe, the ratio of the equivalent lengths by cost and by resistance is equal to the ratio of the length by cost and by resistance for the pipe to which the reduction is made. Of course, this relation is correct only if all the indicated rules of reduction are observed. It is important because it facilitates calculations when reducing pipes to equivalent ones.

IV.

Reduction of a system of pipes to an equivalent pipe

Applying the method of equivalent lengths by resistance and cost, it is easy to reduce the last two special cases analyzed above to the first, i.e. to an equivalent pipe for which we have ready solutions (9) and (10).

Indeed, a system of parallel pipes, according to (19) and (22), can be reduced to a single pipe characterized by the following properties.

1. Its length by resistance (19)

\[ L_{\mathrm{ep}} = \frac{L_1 V_1^{2}}{(V_1+V_2+\cdots)^2} \]

1. Its length by cost (22)

\[ L'_{\mathrm{ep}} = L'_1 + L'_2 \left(\frac{B_2 L_2 V_2^{2}}{B_1 L_1 V_1^{2}}\right)^{0.2 f} + L'_3 \left(\frac{B_3 L_3 V_3^{2}}{B_1 L_1 V_1^{2}}\right)^{0.2 f} +\cdots \]

1. The resistance coefficient and the condition of the gas in this pipe are equal to the analogous quantities of the first pipe.

2. The gas flow through this pipe is equal to the total flow of the system.

\[ V = V_1 + V_2 + V_3^{2} + \cdots \]

Writing the equation of the most advantageous pressure drop for this pipe, according to equation (7), we have:

\[ x^{1+0.2 f} = \frac{c r f}{5 A} (p+x)^{\frac{1}{k}} L'_{\mathrm{ep}} (B_1 L_{\mathrm{ep}} V^2)^{0.2} \cdots \tag{25} \]

Substituting the corresponding values of \(L'_{\mathrm{er}}\) and \(L_{\mathrm{er}}\) from the conditions written above, and carrying out the corresponding transformations, we have:

\[ x^{1+0.2f} = \frac{crf}{5A}(p+x)^{\frac{1}{k}} \left( L_1'(B_1L_1V_1^2)^{0.2f} + L_2'(B_2L_2V_2^2)^{0.2f} +\ldots \right), \]

which identically coincides with equation (11).

Let us note that the result will not change if we first reduce pipe 2 and pipe 3 to an equivalent one with diameter \(d_2\), and then reduce the resulting pipe and pipe 1 to an equivalent one with diameter \(d_1\).

Since equation (25) is quite analogous to (7), from it it is easy to obtain approximate equations analogous to (9) and (10),

\[ x \sim \left[ \frac{crf}{5A}\, L'_{\mathrm{er}}(B_1L_{\mathrm{er}}V^2)^{0.2f} \right]^{\frac{1}{1+0.2f}} p^{\frac{1}{k(1+0.2)f}} \ldots \tag{25} \]

and

\[ d_1^{m}/m = 1000\, \frac{ \left[ \dfrac{54.5\cdot 10^{-4}\,a\sum(vn)\gamma_o}{\eta crfb_1} \right]^{\frac{0.2}{1+0.2f}} }{ \left[ \dfrac{L'_{\mathrm{er}}}{L_{\mathrm{er}}} \dfrac{1}{\lambda_1V^2} \right]^{\frac{0.2}{1+0.2f}} } \ldots \tag{26} \]

The latter equation is more conveniently used in a somewhat modified form, since by definition \(L_{\mathrm{er}}V^2=L_1V_1^2\):

\[ d^m/m = 1000 \left( \frac{D}{N} \right)^{\frac{0.2}{1+0.2f}} \ldots \tag{26} \]

where:

\[ D= \frac{54.5\cdot 10^{-4}\cdot a\sum(vn)\gamma_o}{\gamma_1crfb_1} \quad\text{and}\quad \ldots \tag{26} \]

\[ N= \frac{L'_{\mathrm{er}}}{\lambda_1L_1V_1^2} \ldots \tag{26} \]

For a system of pipes connected in series, reduced to one equivalent pipe, the minimum equation (7) will be written as follows:

\[ x^{1+0.2f} = \frac{crf}{5A} (p+x)^{\frac{1}{k}} L'_{\mathrm{es}}(B_1L'_{\mathrm{es}}V_1^2)^{0.2f} \ldots \tag{27} \]

And since for this system:

\[ L'_{\mathrm{es}} = L_1' + L_1' \left( \frac{L_2'}{L_1'} \right)^{\frac{1}{1+0.2f}} \left( \frac{B_2L_2V_2^2}{B_1L_1V_1^2} \right)^{\frac{0.2f}{1+0.2f}} +\ldots \]

\[ L_{\mathrm{es}} = L_1 + L_1 \left( \frac{L_2}{L_1} \right)^{\frac{1}{1+0.2f}} \left( \frac{B_2L_2V_2^2}{B_1L_1V_1^2} \right)^{\frac{0.2f}{1+0.2f}} +\ldots \]

then, by substitution and the corresponding transformations, we obtain:

\[ x^{1+0.2f} = \frac{crf}{5A} (p+x)^{\frac{1}{k}} \left\{ \left[ L_1'(B_1L_1V_1^2)^{0.2f} \right]^{\frac{1}{1+0.2f}} + \right. \]

\[ \left. + \left[ L_2'(B_2L_2V_2^2)^{0.2f} \right]^{\frac{1}{1+0.2f}} +\ldots \right\}^{1+0.2f} \]

which coincides with equation (17).

And here, as is not difficult to verify directly, the result will not change if one first finds the pipe equivalent to the 2nd and 3rd with diameter \(d_2\), and then the pipe equivalent to this found one and the first with diameter \(d_1\).

Starting from the exact equation (27), one may write the approximate ones:

\[ x \sim \left[ \frac{c r f}{5 A}\, L'_{\mathrm{ec}} \left(B_1 L_{\mathrm{ec}} V_1^2\right)^{0.2 f} \right]^{\frac{1}{1+0.2 f}}\, p^{\frac{1}{\kappa(1+0.2 f)}} . \tag{28} \]

\[ d_1 = 1000 \left(\frac{D}{N}\right)^{\frac{0.2}{1+0.2 f}} . \tag{29} \]

where:

\[ D = \frac{54.5 \cdot a \sum (v n)^\gamma \cdot 10^{-4}} {\gamma c r f b_1} \tag{29'} \]

and:

\[ N = \frac{L'_{\mathrm{ec}}}{L_{\mathrm{ec}}}\, \frac{1}{\lambda_1 V_1^2} \tag{29''} \]

and finally, making use of (24), with the same accuracy:

\[ d_1 = 1000 \left(\frac{D}{N}\right)^{\frac{0.2 f}{1+0.2 f}}, \tag{30} \]

where the quantity \(D\) is as above (equation 29), and:

\[ N = \left( \frac{L_1'}{L_1}\, \frac{1}{\lambda_1 V_1^2} \right), \tag{30'} \]

Every pipeline, however complex its scheme may be, can be regarded as consisting of a series of parallel and successive pipes.

Applying, beginning with the pipes most distant from the beginning, the reduction to a single equivalent pipe, for each pipeline we arrive at one pipe equivalent in resistance and cost, whose diameter is computed from equation (26) or (30).

The determination of the remaining diameters is reduced to calculating them by formulas (12) or (18), depending on the manner in which they are connected to the preceding pipe.

Obviously, in the general case as well, we are essentially dealing with the selection of the most advantageous diameters. The quantities \(L'\) and \(L\), as well as the coefficient of resistance \(\lambda\) \((B)\), may, so long as the diameters are unknown, be chosen only by eye, more or less successfully. Having assigned them, we determine the diameters, compute from the latter \(L'\), \(L\), and \(\lambda\) more accurately for each pipe, and carry out the calculation again. Usually the second recalculation, taking into account the necessity of rounding the obtained diameters to the sizes of the assortment, gives results so accurate that there is no need for further calculations.

V

Summary and simplification of the calculation formulas.

For parallel pipes, the equivalent length by resistance is computed from the relation

\[ \frac{L_{\mathrm{er}}}{L_1} = \frac{V_1^2}{(V_1 + V_2)^2} \tag{19} \]

Equivalent length by cost:

\[ \frac{L'_{\mathrm{er}}}{L'_1} = 1+ \left(\frac{L'_2}{L'_1}\right) \left( \frac{B_2 L_2 V_2^{\,2}}{B_1 L_1 V_1^{\,2}} \right)^{0.2 f} \tag{22} \]

Using (5), we have:

\[ \frac{L'_{\mathrm{er}}}{L'_1} = 1+ \frac{\left(\dfrac{L'_2}{L'_1}\right)} {\left( \dfrac{b_2 \lambda_1 L_1 V_1^{\,2}} {b_1 \lambda_2 L_2 V_2^{\,2}} \right)^{0.2 f}} \tag{31} \]

Having determined the diameter \(d_1\), we determine the diameter of the 2nd pipe by the formula:

\[ \frac{d_1}{d_2} = \left( \frac{B_1 L_1 V_1^{\,2}}{B_2 L_2 V_2^{\,2}} \right)^{0.2} \tag{12} \]

the simplification of which gives:

\[ \frac{d_1}{d_2} = \left( \frac{b_2 \lambda_1 L_1 V_1^{\,2}} {b_1 \lambda_2 L_2 V_2^{\,2}} \right)^{0.2} \tag{32} \]

For successive pipes, the equivalent lengths by resistance and by cost are computed from the relation:

\[ \frac{L'_{\mathrm{es}}}{L'_1} = \frac{L_{\mathrm{es}}}{L_1} = 1+ \left(\frac{L'_2}{L'_1}\right)^{\frac{1}{1+0.2 f}} \left( \frac{B_2 L_2 V_2^{\,2}}{B_1 L_1 V_1^{\,2}} \right)^{\frac{0.2 f}{1+0.2 f}} \tag{20 and 23} \]

Using (5), we have:

\[ \frac{L'_{\mathrm{es}}}{L'_1} = \frac{L_{\mathrm{es}}}{L_1} = 1+ \frac{ \left(\dfrac{L'_2}{L'_1}\right)^{\frac{1}{1+0.2 f}} }{ \left( \dfrac{b_2 \lambda_1 L_1 V_1^{\,2}} {b_1 \lambda_2 L_2 V_2^{\,2}} \right)^{\frac{0.2 f}{1+0.2 f}} } \tag{33} \]

Having determined the diameter of the first pipe, we determine the diameter of the second \((d_2)\) by the formula:

\[ \frac{d_1}{d_2} = \left( \frac{L'_2}{L'_1} \frac{B_1 L_1 V_1^{\,2}}{B_2 L_2 V_2^{\,2}} \right)^{\frac{0.2}{1+0.2 f}} \tag{18} \]

the simplification of which gives:

\[ \frac{d_1}{d_2} = \left( \frac{L'_2}{L'_1} \frac{b_2}{b_1} \frac{\lambda_1 L_1 V_1^{\,2}}{\lambda_2 L_2 V_2^{\,2}} \right)^{\frac{0.2}{1+0.2 f}} \tag{34} \]

VI.

Example of finding the most advantageous diameters of a branched pipeline.

It is required to find the most advantageous diameters of the air pipeline of a power-distribution pneumatic installation. The air consumption is assumed as follows:

During 7 hours per day, 55 cubic meters of air per minute, reduced to atmospheric conditions, are consumed. In addition, during 12 hours per day, devices operate with a consumption of 25 cubic meters of air per minute. The number of working days per year is 220. The air pressure at the consumer is 5 atmospheres by gauge.

Pipeline layout of the installation

The lengths of the individual pipes, the air flow rate, the fittings and local resistances are given in the following table.

Table No. 1.

Pipe No.	Length, meters	Flow rate in cubic meters of air	Valve	Bend	Tee	Moisture separator
1	300	30	1	2	—	1
2	80	10	2	—	—	1
3	450	40	1	—	—	2
4	120	20	1	1	1	—
5	120	20	1	1	1	—
6	60	8	1	—	—	—
7	80	35	—	—	1	—
8	40	15	1	—	—	—
9	50	25	1	1	1	—

Piston compressors with isothermal efficiency \(\eta_{is} = 0.65\) are proposed for the installation. The drive is by electric motors, whose efficiency \(\eta_i = 0.96\). The cost of current is 4 kopecks per kilowatt-hour.

The state of the air is to be taken as follows: atmospheric—1 absolute atmosphere, temperature \(15^\circ\) C. The temperature of the compressed air in pipes 1 and 3 is \(25^\circ\) C, in pipe 2—\(35^\circ\) C, and in the remaining pipes—\(15^\circ\) C. For the installation, black iron pipes are proposed, joined by threaded couplings, laid externally. Their cost is to be taken from the “Handbook of Prices for Labor, etc., as of May 1, 1928, for the city of Moscow,” No. 8, with an addition of 15%. The cost of the work of assembling and laying the pipes is to be calculated according to the “Valuation Schedules” for construction work for the city of Tomsk of May 26, 1928, with an addition of 20%. Annual deductions for depreciation and repair of the pipes are to be taken as 18%.

Using the price data of the Handbook, a graph of pipe prices with a 15% addition as a function of diameter was constructed in logarithmic coordinates. From the graph we have:

\[ r = 229.5 \qquad f = 1.456. \]

The specific weight of air at 1 atm. abs. and \(15^\circ C\) is \(\gamma_1 = 1.183\).

Further,

\[ b=\frac{p}{p_1}\frac{273+t_0}{273+t_2} \]

for pipes 1 and 3 . . . . . . . . . \(b=5.8\)

for pipe 2 . . . . . . . . . \(b=5.61\)

for the other pipes . . . . . . . . . \(b=6.0\)

For the preliminary calculation, \(\lambda=0.015\) was adopted.
For orientation as to the magnitudes \(k_1\) and \(k_2\), the values \(L\) and \(L'\) were calculated for pipe No. 4, assuming a velocity \(W \sim 5\ \frac{\mathrm{m}}{\mathrm{sec}}\). This calculation showed that one may take

\[ k_1=k_2=1.5. \]

According to the assignment:

\[ \Sigma vn=\left(\frac{55}{60}\cdot 7+\frac{25}{60}\cdot 12\right)220=2510 \]

The cost of 1 horsepower-hour is

\[ a=0.04\cdot 0.736=0.0295\ \text{rubles}. \]

For convenience of calculation, \(V\) and \(V^2\) were computed for each pipe.

No.	1	2	3	4	5	6	7	8	9
\(V\) . . .	0.5	0.167	0.667	0.333	0.333	0.134	0.583	0.25	0.417
\(V^2\) . . .	0.25	0.0278	0.440	0.111	0.111	0.018	0.340	0.062	0.174

Next, we replace pipes by equivalent ones, beginning with pipes 4 and 5.

a) Reduction of pipes 4 and 5 to pipe IV.*

The pipes are parallel (formula 31)

\[ \frac{L_{IV}'}{L_4'}= 1+ \frac{ \left(\frac{L_5}{L_4'}\right) }{ \left( \frac{b_5\lambda L_4 V_4^2}{b_4\lambda L_5 V_5^2} \right)^{0.2f} } = 1+ \frac{ \left(\frac{120}{120}\right) }{ \left( \frac{6.0}{6.0}\frac{0.015}{0.015}\cdot\frac{120}{120}\cdot\frac{0.111}{0.111} \right)^{0.791} } =2. \]

b) Reduction of pipes IV and 1 to pipe I.

The pipes are in series (formula 32).

\[ \frac{L_I'}{L_1'}=\frac{L_I}{L_1} = 1+ \frac{ \left(\frac{L_{IV}'}{L_1'}\right)^{\frac{1}{1+0.2f}} }{ \left( \frac{b_4\lambda L_1 V_1^2}{b_1\lambda L_4 V_4^2} \right)^{\frac{0.2f}{1+0.2f}} } \qquad \frac{1}{1+0.2f}=0.775 \]

\[ \frac{0.2f}{1+0.2f}=0.2255 \]

\[ \frac{L_I'}{L_1'}=\frac{L_I}{L_1} = 1+ \frac{ \left(\frac{2\cdot 120}{300}\right) }{ \left( \frac{6.0}{5.8}\frac{300\cdot 0.25}{120\cdot 0.111} \right)^{0.225} } =1.568 \]

* The numerical computations were carried out using a logarithmic slide rule.

Omitting the intermediate calculations, we shall give only the final results for the remaining pipes.

Reduction of pipes 8 and 9 to the equivalent VIII pipe gives

\[ \frac{L_{\mathrm{VIII}}'}{L_8'} = 2.81 \]

Reduction of pipes VIII and 7 to the equivalent VII pipe gives

\[ \frac{L_{\mathrm{VII}}'}{L_7'}=\frac{L_{\mathrm{VII}}}{L_7}=1.76 \]

Reduction of pipes VII and 6 to the equivalent VI pipe gives

\[ \frac{L_{\mathrm{VI}}'}{L_6'}=8.13 \]

Reduction of pipes VI and 3 to the equivalent III pipe gives

\[ \frac{L_{\mathrm{III}}'}{L_3'}=\frac{L_{\mathrm{III}}}{L_3}=1.32 \]

Reduction of all pipes to one equivalent pipe.

Parallel system (f. 31):

\[ \frac{L_{\mathrm{e}}'}{L_1'} = 1+ \frac{\left(\dfrac{L_2'}{L_1'}\right)} {\left(\dfrac{b_2}{b_1}\,\dfrac{\lambda L_1 V_1^2}{\lambda L_2 V_2^2}\right)^{0.291}} + \frac{\left(\dfrac{L_{\mathrm{III}}'}{L_1'}\right)} {\left(\dfrac{b_3}{b_1}\,\dfrac{\lambda L_1 V_1^2}{\lambda L_{\mathrm{III}} V_3^2}\right)^{0.291}} = \]

\[ = 1+ \frac{\dfrac{80}{1.568\cdot 300}} {\left(\dfrac{5.61}{5.80}\cdot \dfrac{1.568\cdot 300\cdot 0.25}{80\cdot 0.0278} \right)^{0.291}} + \]

\[ + \frac{\dfrac{1.32\cdot 450}{1.568\cdot 300}} {\left(\dfrac{1.568\cdot 300\cdot 0.250}{1.32\cdot 450\cdot 0.440}\right)^{0.291}} = 2.654. \]

Next we have (f. 26″):

\[ D= \frac{54.5\cdot 10^{-4}\cdot a\sum(vn)} {\eta c r f b_1}\,\gamma_0, \]

which, on substitution, gives:

\[ D= \frac{54.5\cdot 10^{-4}\cdot 0.0295\cdot 2510\cdot 1.185} {0.65\cdot 0.96\cdot 0.18\cdot 229.5\cdot 1.456\cdot 5.8} = 21.95\cdot 10^{-4} \]

\[ N= \frac{L_{\mathrm{er}}'}{\lambda_1 L_1 V_1^2} = \frac{2.654\cdot 300\cdot 1.5\cdot 1.568} {0.015\cdot 1.568\cdot 300\cdot 0.25\cdot 1.5} = 758 \]

\[ \frac{D}{N}=2.900\cdot 10^{-6}. \]

According to (26′)

\[ d_1 = 1000\left(\frac{D}{N}\right)^{\frac{0.2}{1+0.2f}} = 132\ \mathrm{m/m}. \]

Determination of the remaining diameters:

Pipe 2 (formula 32)

\[ \frac{d_1}{d_2} = \left( \frac{b_2}{b_1}\, \frac{\lambda L_1 V_1^2}{\lambda L_2 V_2^2} \right)^{0.2} = \left( \frac{5.61}{5.80}\, \frac{1.583 \cdot 300 \cdot 0.25}{80 \cdot 0.0278} \right)^{0.2} = 2.2 \]

\[ d_2 = 60 \text{ m/m}. \]

Omitting the further analogous calculations, we give only their final results

\[ \frac{d_3}{d_5}=1.16 \qquad d_3=153 \text{ m/m} \]

\[ \frac{d_1}{d_4}=1.28 \qquad d_4=103 \]

\[ d_4=d_5 \qquad d_5=103 \]

\[ \frac{d_5}{d_6}=2.29 \qquad d_6=66.8 \]

\[ \frac{d_7}{d_6}=2.14 \qquad d_7=143 \]

\[ \frac{d_7}{d_8}=1.53 \qquad d_8=93.5 \]

\[ \frac{d_9}{d_8}=1.28 \qquad d_9=119.5 \]

Rounding the obtained figures to the standard sizes of the assortment, we obtain the sizes placed in the following Table No. 2. In the same table are brought together the data necessary for further calculations. Using the data of these tables, we calculate, by the same formulas (31) and (33), the equivalent lengths of the pipes more accurately.

Reducing pipes 4 and 5 to the equivalent pipe IV gives

\[ \frac{L'_{IV}}{L'_4}=2 \]

“ ” IV and 1 “ ” I

\[ \frac{L'_I}{L'_1}=\frac{L_I}{L_1}=1.525 \]

Reducing pipes 8 and 9 to the equivalent pipe VIII

\[ \frac{L'_{VIII}}{L'_8}=3.1 \]

“ ” 7 and VIII “ ” VII

\[ \frac{L'_{VII}}{L'_7}=\frac{L_{VII}}{L_7}=1.94 \]

“ ” 6 and VII “ ” VI

\[ \frac{L'_{VI}}{L'_6}=8.22 \]

“ ” VI and 3 “ ” III

\[ \frac{L'_{III}}{L'_1}=\frac{L_{III}}{L_1}=1.31 \]

Pipes 1 and 2 and 3 are reduced to one pipe (1):

\[ \frac{L'_3}{L_1}=2.58 \]

Table No. 2

No. of pipe	Diameter in mm \(d\)	Area in cm² \(f\)	Quantity of gas, m³/sec \(V\)	\(b\)	Velocity in m/sec \(w\)	Kinematic viscosity \(\nu\)	Reynolds number \(R\)	Coeff. of resistance \(\lambda\)	Sum of local resistances \(\sum \xi\)	\(\dfrac{d}{\lambda}\)	\(\sum \xi \dfrac{d}{\lambda}\)	Length \(B\), m	\(L\)	\(k_1\)
1	125	122,5	0,5	5,8	7,04	2,315·10⁻⁶	380000	0,0140	21,5	8,93	192	308	492	1,64
2	63	31,2	0,167	5,61	9,55	2,458 ·	245000	0,0151	21,0	4,17	88	80	168	2,1
3	150	176,5	0,667	5,8	6,51	2,315 ·	422000	0,0137	21	10,95	230	450	680	1,51
4	100	78,5	0,333	6,0	7,09	2,215 ·	320000	0,0144	8,5	6,95	59	120	179	1,49
5	100	78,5	0,333	6,0	7,09	—	320000	0,0144	—	—	—	—	—	—
6	63	31,2	0,134	—	7,16	—	201500	0,0157	7	4,01	28	60	88	1,47
7	150	176,5	0,583	—	5,50	—	372000	0,0140	1	10,08	11	80	91	1,14
8	89	62	0,260	—	6,83	—	273000	0,0149	7	5,96	42	40	82	2,05
9	125	122,5	0,417	—	5,67	—	320000	0,0144	8,5	8,7	74	50	124	2,48

Determination of \(L'\) was made in the following table:

No. of pipe	\(d\) m/m	Cost of 1 meter	Cost of branch	Sum	Cost of pipe + work	Cost of fittings and flanged parts	Full cost	\(L'\)	\(k_2\)
1	125	10,62	1,64	12,26	3780	1134	4914	462	1,50
2	63	4,20	1,00	5,20	416	164	580	138	1,72
3	150	14,80	1,97	16,77	7540	2168	9708	655	1,46
4	100	7,73	1,31	9,04	1087	238	1325	172	1,43
5	100	7,73	1,31	9,04	1087	238	1328	172	1,43
6	63	4,20	1,00	5,20	312	55	367	88	1,46
7	150	14,80	1,97	16,77	1340	384	1724	117	1,45
8	89	6,38	1,31	7,69	308	62	370	58	1,45
9	125	10,62	1,64	12,26	614	234	848	80	1,6

Whence a more precise value is obtained

\[ N = 690 \qquad \text{and} \qquad d_1 = 134\ \mathrm{m}/\mathrm{m}. \]

Determining, by formulas (32) and (34), the values of the remaining diameters, we have their values, collected in the following table. The third line gives the diameter values closest to the standard assortment.

Pipe No.	1	2	3	4	5	6	7	8	9
\(d\)	134	65	155	105	105	69	140	99	\(132\ \mathrm{m}/\mathrm{m}\)
\(d\)	125	63	150	100	100	63	150	100	\(125\ \mathrm{m}/\mathrm{m}\)

The difference in the pipe diameters according to the first variant is so insignificant that, in practice, the selection of pipes could be considered completed after the first calculation.