D. V. Spetstsi.

Studies on the Calculation

and Design

of Centrifugal Fans

Tomsk
1928

D. V. Speshin.

“Studies on the Calculation and Design of Centrifugal Fans.”

Study 1. Determination of the principal dimensions of a fan in the case where the required vacuum before the fan and the volume capacity per unit time are specified, but there is no standard characteristic.

By a centrifugal fan we shall understand an apparatus whose purpose is to move a known volume of gas while overcoming a certain resistance. If the resistance to be overcome is located before the fan, the fan is called a suction fan; if it is after the fan, a pressure fan. Finally, if the fan both draws in and delivers, it is a mixed fan.

In the most general case, the installation of a fan (Fig. 1) has a suction duct (A) and a delivery duct (B); the fan proper: the rotor (C) (the working wheel with blades) and the stator (D)—the spiral casing, the bearing of the rotor shaft (E), and a pulley or coupling (not shown in the figure).

The gas or air, drawn in and delivered by the fan, passes through the fan in the direction indicated by the arrows in Fig. 1. If we denote the static pressure loss before the fan by:

\[ p_1' - p_1 = \Delta p_s' \text{ mm. water column;} \]

the dynamic pressure before the fan by:—

\[ \frac{V_1^2}{2g}\,\gamma = \Delta p_d' \text{ mm. water column;} \]

Correspondingly, after the fan:

\[ p_2 - p_2' = \Delta p_s'' \text{ mm. w. col. and} \]

\[ \frac{V_2^2}{2g}\,\gamma = \Delta p_d'' \text{ mm. w. col., then} \]

the total pressure which the fan must overcome will be equal to:

\[ \Delta p_s' + \Delta p_s'' + \Delta p_d'' - \Delta p_d' = \Delta p \text{ mm. water col.} \tag{1} \]

The gas entering through the inlet opening of the fan into the middle of the rotor is seized by the blades of the latter at the inner ring and thrown toward the outer ring (Fig. 1).

If we denote by \(c_1\) the absolute velocity of the gas at entry into the passages formed by the blades, and by \(c_2\) the absolute velocity of the gas at exit from the passages at the outer ring, then the live force imparted by the rotor to the gas will be equal to:

\[ \frac{V \cdot \gamma \cdot c_2^2}{2g} - \frac{V \cdot \gamma \cdot c_1^2}{2g} \text{ kg-m.} \]

Here \(V\) will denote the volume of the gas being moved, independently of time. The change in volume due to compression of the gas, in view of…

the insignificance of the degree of compression in low-pressure fans, as well as the losses associated with this compression, will not be taken into account in what follows.

The gas moving through the channels formed by the blades must overcome a counter-pressure equal to:

\[ p_2-p_1=\Delta p_s' + \Delta p_s''=\Delta p_s \]

This static pressure is overcome by the centrifugal pressure of the gas developing in the channels of the rotor blades.

It is evident that the counter-pressure must be balanced by the centrifugal pressure of the gas, i.e., equality of works must hold:

\[ \frac{V}{f}(p_2-p_1)f = \frac{V}{f}\int_{r_1}^{r_2}\frac{f\cdot dr\cdot \gamma \cdot w^2 r}{g} \quad \text{kgf m.} \]

In the first part of this equality stands the work of the counter-pressure in displacing \(V\) cubic meters of gas, and in the second part—the centrifugal pressure of the gas overcoming this resistance.

In forming this expression it was assumed that the relative velocity of the gas at the entrance to the blade channels and the relative velocity of the gas at the exit from the channels formed by the blades are the same. This expression may be rewritten as:

\[ \frac{V}{f}(p_2-p_1)f = \frac{V}{f}\cdot f\cdot \gamma \cdot \frac{u_2^2-u_1^2}{2g} \quad \text{kgf m.} \]

Here the notation is as follows:

\(p\)—pressure, in kgf/m\(^2\);

\(f\)—area of the channel perpendicular to the flow axis, in square meters;

\(\gamma\)—specific weight of the gas, in kg/m\(^3\);

\(u\)—peripheral velocity, in m/sec.

Subscript 1 refers to the gas inlet, subscript 2 to the outlet.

Thus the work expended by the rotor will be equal to:

\[ \vartheta = \frac{V\cdot \gamma \cdot c_2^2}{2g} - \frac{V\cdot \gamma \cdot c_1^2}{2g} + \frac{V\cdot \gamma \cdot (u_2^2-u_1^2)}{2g} \quad \text{kgf m.} \]

In the case when the relative velocity of the gas (in Fig. 1 and following denoted by \(W\)) at the beginning of the channel and at the end are not the same, it is evident that such a change in relative velocity must likewise occur at the expense of the centrifugal pressure. In this case the preceding equalities should be rewritten in the following form:

\[ \frac{V}{f}(p_2-p_1)f + \frac{V}{f} \left( \frac{w_2^2}{2g}\gamma - \frac{w_1^2}{2g}\gamma \right)f = \frac{V}{f}\int_{r_1}^{r_2}\frac{f\cdot dr\cdot \gamma \cdot w^2\cdot r}{g} \quad \text{kgf m.} \]

Whence:

\[ V(p_2-p_1) + V\left( \frac{w_2^2}{2g}\gamma - \frac{w_1^2}{2g}\gamma \right) = V\int_{r_1}^{r_2}\frac{dr\cdot \gamma \cdot w^2 r}{g} \]

\[ V(p_2-p_1) = \frac{V\cdot \gamma}{2g} \left( u_2^2-u_1^2+w_1^2-w_2^2 \right). \]

The general expression for the total work expended by the rotor will therefore be:

\[ \vartheta_{\mathrm{n}}= \frac{\gamma \cdot V \cdot \left(c_2^2-c_1^2+u_2^2-u_1^2+w_1^2-w_2^2\right)}{2g} \ \text{kg m}. \tag{2} \]

The quantity

\[ \frac{u_2^2-u_1^2}{2g}+\frac{c_2^2-c_1^2}{2g}+\frac{w_1^2-w_2^2}{2g} =H \ \text{meters of gas column} \tag{3} \]

in the theory of turbomachines is called the theoretical head (work referred to 1 kg of fluid—the Poncelet equation).

The power expended, in horsepower, will be equal to:

\[ N_{\text{theor}}=\frac{V_{\mathrm{m}}\cdot \gamma \cdot H}{75\cdot 60} \]

(where \(V_{\mathrm{m}}\) is expressed in cubic meters/min.).

Owing to aerodynamic losses, the actually produced head will be somewhat less than the theoretical one:

\[ h < H \]

The ratio of the actually produced head—\(h\)—to the theoretical head—\(H\)—is customarily called the manometric efficiency:

\[ \frac{h}{H}=\eta. \tag{4} \]

In addition to aerodynamic losses, fans also have mechanical losses, which are customarily taken into account by a certain coefficient

\[ \eta_{\mathrm{m}}. \]

Thus the total power expended on the fan is finally determined from the following expression:

\[ N_{ef}= \frac{V_{\min.}\,H\cdot\gamma}{75\cdot 60\cdot \eta_{\mathrm{m}}}, \quad \text{or,} \tag{5} \]

\[ N_{ef}= \frac{V_{\min.}\,h\cdot\gamma}{75\cdot 60\cdot \eta\cdot \eta_{\mathrm{m}}}, \quad \text{and also} \]

\[ N_{ef}= \frac{V_{\min.}\,h_{\mathrm{mm.\ water\ col.}}}{75\cdot 60\cdot \eta\cdot \eta_{\mathrm{m}}}. \tag{6} \]

Starting from the velocity triangles (Figs. 2 and 3), the expression for

\[ H= \frac{u_2^2-u_1^2+c_2^2-c_1^2+w_1^2-w_2^2}{2g} \ \text{meters gas col.} \]

can be transformed by substituting:

\[ c_2^2=u_2^2+w_2^2+2u_2w_2\cos\alpha_2, \]

\[ c_1^2=u_1^2+w_1^2+2u_1w_1\cos\alpha_1. \]

\[ H= \frac{ u_2^2-u_1^2+u_2^2+w_2^2+2u_2w_2\cos\alpha_2 -u_1^2-w_1^2-2u_1w_1\cos\alpha_1 +w_1^2-w_2^2 }{2g} \]

\[ H= \frac{u_2^2-u_1^2}{g} + \frac{u_2w_2\cos\alpha_2-u_1w_1\cos\alpha_1}{g} \ \text{meters gas col.} \tag{7} \]

From the velocity triangles in Fig. 3 it also follows that:

\[ w_2^2 = c_2^2 + u_2^2 - 2c_2 u_2 \operatorname{cs}(c_2,u_2), \]

\[ w_1^2 = c_1^2 + u_1^2 - 2c_1 u_1 \operatorname{cs}(c_1,u_1), \]

and, for the radial direction of \(c_1\):

\[ w_1^2 = c_1^2 + u_1^2. \]

Substituting these values into expression (3) for \(H\), we obtain:

\[ H=\frac{c_2 u_2 \operatorname{cs}(c_2,u_2)-c_1 u_1 \operatorname{cs}(c_1,u_1)}{g}, \tag{8} \]

where, if the angle \(c_1,u_1=90^\circ\), \(\operatorname{cs}(c_1,u_1)=0\), then

\[ H=\frac{c_2 \cdot u_2 \cdot \operatorname{cs}(c_2,u_2)}{g}. \]

Equation (8) (Combes—Euler), after the corresponding transformation, gives an expression relating the dimensions of the fan and its operation, and the complete investigation of the fan’s operation is based on it.*)

Comparing the ideal fan (without losses) with the actual one, we must note a whole series of losses that occur in the latter.

These losses are of two kinds:

1) Losses that affect the magnitudes of the gas-flow velocities, i.e. aerodynamic losses.

The following must be assigned to such losses:

a) Shocks upon changes in the velocities of the gases and in the directions of these velocities at the inlet to and outlet from the blade passages.

b) Losses due to friction, disturbances, and as a consequence of changes in velocities as the gas passes through the rotor passages.

c) Losses due to leakage through the clearance.

2) Mechanical losses, which should include the losses due to friction of the rotor against the gas and losses due to friction in the bearings and stuffing boxes.

Losses of the first kind will subsequently be taken into account by the manometric efficiency \(\eta\).

Losses of the second kind—by the mechanical efficiency \(\eta_m\); the overall efficiency will be denoted by:

\[ \eta_e = \eta \cdot \eta_m \]

A simple calculation by the formula:

\[ h_{mp}=\frac{k\cdot L\cdot P\cdot Q^2}{S^3}\,\delta \]

shows that friction losses of the kind that we encounter when considering the motion of gases through pipelines are, in the present case, very small.

Thus, leaving aside for the time being the mechanical losses, we see that the principal causes of losses are shocks and disturbances due to changes in velocities and their directions at inlet and outlet.

) See p. 277 ff. L. Crussard. Ventilateurs et Compresseurs*. Paris, 1926.

Obviously, in eliminating these losses, the correct choice of the magnitudes and directions of the velocities is of great importance.

The velocity of gas inflow into the fan suction opening, \(V_1\), (Fig. 1) according to Pelzer*) should be selected by the following formula:

\[ V_1 = \frac{2{,}597}{1{,}94}\ \text{l.} \]

Wiesmann indicates more restricted limits for inlet velocities; he proposes the following values for \(V_1\):

\[ \begin{aligned} &\text{for low-pressure fans} && V_1 = 10\text{--}12\ \text{m/sec.},\\ &\text{for medium-pressure fans} && V_1 = 12\text{--}16\ \text{m/sec.},\\ &\text{for high-pressure fans} && V_1 = 12\text{--}18\ \text{m/sec.} \end{aligned} \]

The figure 12 m/s appears in all three cases in this table, and in most executed fan designs this is precisely the inflow velocity used.

The choice of the velocity \(V_1\) determines one of the principal dimensions of the fan, namely:

\[ D_o \]

If, as a rule, it is assumed that the velocity of gas inflow into the suction opening is equal to 12 meters/sec. at a gas specific weight \(\gamma = 1{,}2\ \text{kg/m}^3\), then the diameter of the fan suction opening is determined as:

\[ D_o = \sqrt{\frac{4 \cdot V_{\text{sec.}}}{\pi \cdot V_1}} = \sqrt{\frac{4 \cdot V_{\text{sec.}}}{3{,}14 \cdot 12}} = 0{,}325 \sqrt{V_{\text{sec.}}}. \tag{9} \]

The same diameter may also be adopted for the cylindrical surface of the leading edges of the rotor blades. If this surface has a conical form, then this diameter will be the greatest diameter of the surface of the leading edges. Consequently,

\[ D_{\text{in}} = D_1 = 0{,}325 \sqrt{V_{\text{sec.}}}. \tag{9′} \]

For rapid orientation in choosing these diameters, Fig. 4 gives a graph (based on formula (9′)).

If this graph (Fig. 4) is used for gases of a density other than \(\gamma = 1{,}2\ \text{kg/m}^3\), then instead of the specified volume per second one should substitute the reduced volume per second of the gas.

The circumferential velocity \(u_1\) on the surface of the leading edges of the blades in completed and satisfactorily operating fans varies within considerable limits:

\[ u_1 = 15\text{--}25\text{--}40\ \text{meters/sec.} \]

For low-pressure fans this range is somewhat narrower, namely: \(u_1 = 15\text{--}25\ \text{m/sec.}\)

Starting from the expression for

\[ u_1 = \frac{\pi D_1 n}{60} \]

and from the graph (Fig. 4), one can relate \(u_1\), \(n\), and \(V_{\text{sec.}}\), which has also been done in the graph (Fig. 5).

The graphs of Figs. 4 and 5 may be used for the rapid selection of \(u_1\), \(n\), and \(D_1\) for a given \(V_{\text{sec.}}\) and \(\gamma\).

) See G. Karp. Centrifugal Fans*.

The absolute gas velocity \(c_1\) as it enters the blade passages will in all cases be taken by us as equal to the velocity in the suction opening—\(V_1\), in order thereby to avoid losses from a change in velocity.

We take the direction of the velocity \(c_1\) to be radial (see Figs. 1, 2, and 3), which, in connection with the choice of the magnitudes \(u_1\) and \(c_1\) and the direction of the latter, gives us, according to the rule of the velocity parallelogram, the direction and magnitude of the relative velocity \(w_1\) of the gas flow at the beginning of the passage. The direction of the velocity \(w_1\) determines the blade angle \(\alpha_1\) (Fig. 3) at the entry of the gas into the rotor passages.

Thus the choice of the velocity \(V_1\) of the gas entering the suction opening of the fan has enabled us to determine all dimensions relating to the suction part of the fan.

Turning to the expression for the theoretical head:

\[ H=\frac{u_2^{2}-u_1^{2}}{g}+\frac{u_2 w_2 \cos \alpha_2-u_1 w_1 \cos \alpha_1}{g}, \ldots\ldots\ldots (8) \]

we can rewrite it in the following form:

\[ H=\frac{u_2^{2}-u_1^{2}}{g}+\chi^{1}\frac{u_2^{2}-u_1^{2}}{g} =(1+\chi^{1})\frac{u_2^{2}-u_1^{2}}{g}, \]

or, if \(1+\chi^{1}=\chi\),

\[ H=\chi\frac{u_2^{2}-u_1^{2}}{g}. \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (9) \]

The value \(\chi^{1}\), depending on the type of fan, may be positive, negative, and also equal to zero, for example when \(\alpha_1=\alpha_2=90^\circ\), i.e., when the fan blades are radial.

It is evident that the negative value of \(\chi^{1}\) lies within the limits \(0>\chi^{1}>-1\), since otherwise there would be no head. It is also evident that this coefficient \((\chi^{1})\) is the smaller the longer the blades, i.e., the more \(u_2>u_1\).

In existing fan designs the coefficient \(\chi\) varies within the limits:

\[ \chi=0.65 \text{ to } 10, \]

and by fan types it has approximately the following numerical value:

1) low-pressure fans of the drum type with forward-curved blades
\[ \chi=5\text{—}6, \]

2) the same for small pressures
\[ \chi=10, \]

3) low-pressure fans with backward-curved blades
\[ \chi=0.9, \]

4) the same, medium pressure
\[ \chi=0.9, \]

5) high-pressure fan with backward-curved blades
\[ \chi=0.65. \]

The choice of the fan type determines the value of the coefficient \(\chi\) and, thereby, the outside diameter \(D_2\) of the rotor and the circumferential velocity \(u_2\) at its outer surface.

Numerical example of calculating the dimensions of a drum fan.

Given: the gas flow rate per second \(V_{\text{sec}}=20.4\ \text{m}^3/\text{sec}\);

specific weight of the gas \(\gamma=0.91\ \text{kg}/\text{m}^3\);

Resistance before the fan (in this case a smoke exhauster)

\[ \Delta p_s'=34.5\ \text{mm water column}. \]

The gas velocity in the suction opening is \(V_1 = 12\) m/sec. (for \(\gamma = 1.2\)); consequently \(V_1 = 15.8\) m/sec. (for \(\gamma = 0.91\)).

The gas velocity in the discharge opening of the fan casing according to Wiesmann:

For low-pressure fans \(V_2 = 12\text{–}16\) m/sec.
“ medium “ \(V_2 = 15\text{–}20\) “
“ high “ \(V_2 = 20\text{–}30\) “

We take \(V_2 = 16\) m/sec. (for \(\gamma = 1.2\)), which gives

\[ V_2 = 21\ \text{m/sec.}\quad (\text{for } \gamma = 0.91). \]

Thus:

\[ \Delta p'_d = \frac{V_1^2}{2g}\gamma = \frac{15.8^2}{2g}\,0.91 = 11.6\ \text{mm w.g.}; \]

\[ \Delta p''_d = \frac{V_2^2}{2g}\gamma = \frac{21^2}{2g}\cdot 0.91 = 20.6\ \text{mm w.g.} \]

\[ \Delta p''_s = 0. \]

The required pressure in mm of water column will be:

\[ h = \Delta p - \Delta p'_s - \Delta p'_d + \Delta p''_s + \Delta p_d = 34.5 - 11.6 + 0 + 20.6 = \]

\[ = 43.5\ \text{mm w.g.} \]

We take the mechanical efficiency of the fan as:

\[ \eta_m = 0.95 \]

and the manometric efficiency as

\[ \eta = 0.60. \]

Wiesmann gives the values of the overall efficiency coefficient

\[ \eta_e = \eta \cdot \eta_m \]

within the following limits:

low-pressure fans \(\eta_e = 0.5\text{–}0.6\)
“ medium “ \(\eta_e = 0.5\text{–}0.7\)
“ high “ \(\eta_e = 0.7\text{–}0.8\)

With the adopted efficiencies, we determine the total power consumed by the fan:

\[ \mathrm{Nef} = \frac{V_{\text{sec}}\cdot h}{75\cdot \eta \cdot \eta_m} = \frac{V_{\text{sec}}\cdot h}{75\cdot \eta_e} = \frac{20.4\cdot 43.5}{75\cdot 0.60\cdot 0.95} = 20.8\ \text{h.p.} \]

From the graph (Fig. 4), by interpolation between the curves \(\gamma = 0.8\) and \(\gamma = 1\), we find that the inner diameter \(D_1 = D_0 = 1.26\) m.

If, however, this fan is designed with double suction

\[ \left( V_{\text{sec}} = \frac{20.4\ \text{m}^3/\text{s}}{2}\cdot 2 \right), \]

then \(D_1 = D_0 = 0.9\) m.

We adopt double suction; then, according to the graph (Fig. 5), the number of revolutions is determined as:

for \(u_1 = 25\) m/sec. \(\qquad n = 530\)
“ \(u_1 = 15\) m/sec. \(\qquad n = 320\)

When determining the number of revolutions for gases of density other than \(1.2\ \mathrm{kg/m^3}\), one may use the special scales plotted in Fig. 5, or else the curve \(\gamma = 1.2\) from Fig. 4, if both graphs (Figs. 4 and 5) are placed one above the other and have the same abscissa scale.

The theoretical head in meters of gas column is

\[ H=\frac{h}{\eta\cdot\gamma}=\frac{43.5}{0.60\cdot 0.91}=79.5\ \mathrm{m}. \]

We determine the peripheral velocity \(u_2\) at the outer surface of the rotor from the expression:

\[ H=x\cdot\frac{u_2^2-u_1^2}{g}, \]

where \(x=5.5\) is adopted.

\[ 79.5=5.5\,\frac{u_2^2-u_1^2}{g}, \]

whence

\[ u_2=\sqrt{142+u_1^2}. \]

Let us examine both cases:

\[ \begin{aligned} 1)\quad &u_1=25\ \mathrm{m/s},\qquad u_2=\sqrt{142+625}=27.7\ \mathrm{m/sec},\\[4pt] 2)\quad &u_1=15\ \mathrm{m/s},\qquad u_2=\sqrt{142+225}=19.2\ \mathrm{m/sec}. \end{aligned} \]

Outer diameter of the rotor:

\[ u_1=25\ \mathrm{m/s},\qquad D_2=\frac{D_1 u_2}{u_1}=\frac{0.9\cdot 27.7}{25}=0.98\approx 1\ \mathrm{meter}. \]

\[ u_1=15\ \mathrm{m/s},\qquad D_2=\frac{D_1 u_2}{u_1}=\frac{0.9\cdot 19.2}{15}=1.15\ \mathrm{m}. \]

Radial height of the blade:

\[ \text{1st case}\qquad \kappa=\frac{D_2-D_1}{2}=\frac{1-0.9}{2}=0.05\ \mathrm{m}=50\ \mathrm{mm}. \]

\[ \text{2nd case}\qquad \kappa=\frac{D_2-D_1}{2}=\frac{1.15-0.9}{2}=0.125\ \mathrm{m}=125\ \mathrm{mm}. \]

Distance between blades (pitch) at the middle of their radial height for a fan of any type and pressure may be taken within the limits \(t=(0.75-0.9)\kappa\); then:

\[ \begin{aligned} \text{for the 1st case}\qquad &t=0.75\cdot\kappa=0.75\cdot 50=37.5\ \mathrm{mm}=0.0375\ \mathrm{m},\\ \text{for the 2nd case}\qquad &t=0.75\cdot 125=94\ \mathrm{mm}=0.094. \end{aligned} \]

Number of blades:

\[ \text{1st case}\quad n=\frac{\pi\cdot(D_1+D_2)}{2\cdot t} =\frac{3.14\,(0.9+1.0)}{2\cdot 0.0375} \approx 80\ \text{blades}. \]

\[ \text{2nd case}\quad n=\frac{\pi\cdot(D_1+D_2)}{2\cdot t} =\frac{3.14\,(0.9+1.15)}{2\cdot 0.094} \approx 34\ \text{blades}. \]

The relative velocity \(w_1\) at entry (see Figs. 2 and 3):

\[ \begin{aligned} \text{1st case}\quad w_1&=\sqrt{c_1^2+u_1^2}=\sqrt{15.8^2+25^2}=\sqrt{875}=29.6\ \text{m/sec},\\ \text{2nd case}\quad w_1&=\sqrt{c_1^2+u_1^2}=\sqrt{15.8^2+225}=\sqrt{475}=21.8\ \text{m/sec}. \end{aligned} \]

The angle of inclination of the blades at entry, \(\alpha_1\), is determined as follows:

\[ \text{for the 1st case}\quad \tg(180-\alpha_1)=\frac{c_1}{u_1}=\frac{15.8}{25}=0.63 \]

\[ \text{for the 2nd case}\quad \tg(180-\alpha_1)=\frac{15.8}{15}=1.05. \]

\[ \begin{aligned} \text{1st case}\quad &180-\alpha_1=32^\circ 10', \quad \alpha_1=147^\circ 50',\\ \text{2nd case}\quad &180-\alpha_1=46^\circ 20', \quad \alpha_1=133^\circ 40'. \end{aligned} \]

The axial width of the rotor (the useful axial length of the blade) \(b_1\) at entry and \(b_2\) at exit is determined from the following considerations.

The surface of the inlet edges of the rotor blades, just like the surface of the outlet edges, in the case where both of these surfaces are cylindrical, represents rectangles obstructed by the blades.

For each of these surfaces one may write, on the basis of continuity of motion (see Fig. 6):

\[ \pi D_1 \Sn(180-\alpha_1) w_1 \cdot b_1 = \pi D_2 \Sn \alpha_2 \cdot w_2 \cdot b_2 \]

In the general case \(b_1 \ne b_2\), but for drum fans the equality \(b_1=b_2\) greatly simplifies the construction of the entire rotor.

Taking into account, on the one hand, the obstruction of the channels between the blades by the thickness of the latter, and, on the other hand, the non-filling of the rectangular channels between the blades by the gas flow (Wiesmann accounts for this non-filling by a coefficient \(\mu=0.7\text{–}0.85\), the smaller the greater the number of blades), we may write:

\[ V_{\text{sec.}}=\frac{20.4}{2}\ \text{m}^3/\text{sec.} = [\pi D_1 \Sn(180-\alpha_1)-n\cdot s]\,w_1\cdot b_1\cdot \mu \]

For the 1st case:

\[ 10.2\ \text{m}^3/\text{sec.} = [\pi\cdot0.9\,\Sn 32^\circ 10' - 80\cdot0.003]\cdot29.6\cdot0.7\cdot b_1 \]

\[ b_1= \frac{10.2}{0.7\cdot29.6(1.5-0.24)} = \frac{10.2}{26} \simeq 0.4\ \text{m}. \]

For the 2nd case:

\[ 10.2 = [\pi\cdot0.9\,\Sn 46^\circ 20' - 34\cdot0.003]\,21.8\cdot0.75\cdot b_1 \]

\[ b_1= \frac{10.2}{0.75\cdot21.8(2.04-0.104)} = \frac{10.2}{31.6} = 0.325\ \text{m}. \]

In determining \(b_1\), for the first case \(\mu=0.7\) was adopted, and for the second \(\mu=0.75\).

The blade thickness for each case was taken as \(S=3\) mm.

For the outlet surfaces we shall have:

\[ (\pi D_2 \Sn \alpha_2 - n\cdot s)\,w_2\cdot b_2\cdot\mu = V_{\text{sec.}}, \]

whence:

\[ \pi D_2 \Sn\alpha_2\cdot w_2\cdot b_2\cdot \mu - n\cdot s\cdot w_2 b_2\,\mu = \]

\[ \pi D_1\cdot \Sn(180-\alpha_1) w_1\cdot b_1\cdot \mu - n\cdot s\cdot w_1 b_1\,\mu \]

neglecting the quantities \(n \cdot s \cdot w_2 \cdot b_2 \cdot \mu\) and \(n \cdot s w_1 b_1 \mu\), since they are almost equal to each other and small in comparison with the other two terms, we obtain:

\[ w_2 \operatorname{Sn}\alpha_2 = \frac{D_1 \operatorname{Sn}(180-\alpha_1)\, w_1 b_1}{D_2 \cdot b_2}. \]

For the example under consideration, if we assume that \(b_1=b_2\):

\[ \text{1st case.}\qquad w_2 \cdot \operatorname{Sn}\alpha_2 = \frac{0.9 \cdot 0.532 \cdot 29.6}{1} = 14.18, \]

\[ \text{2nd case.}\qquad w_2 \cdot \operatorname{Sn}\alpha_2 = \frac{0.9 \cdot 0.723 \cdot 21.8}{1.15} = 12.34. \]

From the equation of the theoretical head (8)

\[ w_2 \operatorname{cs}\alpha_2 = \frac{Hg-u_2^2+u_1^2+u_1 w_1 \operatorname{cs}\alpha_1}{u_2}. \]

Substituting into this expression the quantities already found earlier, we obtain:

\[ \text{1st case.}\qquad w_2 \operatorname{cs}\alpha_2 = \frac{79.5 \cdot 9.81 - 27.7^2 + 25^2 - 25 \cdot 29.6 \cdot 0.846}{27.7} = 0.47 \]

\[ \text{2nd case.}\qquad w_2 \operatorname{cs}\alpha_2 = \frac{79.5 \cdot 9.81 - 19.2^2 + 15^2 - 25 \cdot 21.8 \cdot 0.69}{19.2} = 18.3 \]

\[ \text{1st case.}\qquad \frac{w_2 \operatorname{sn}\alpha_2}{w_2 \operatorname{cs}\alpha_2} = \operatorname{tg}\alpha_2 = \frac{14.18}{0.47} = 30,\quad \alpha_2=88^\circ \quad\text{and} \]

\[ w_2 = \frac{14.18}{\operatorname{sn}\alpha_2} = \frac{14.18}{0.999} = 14.2\ \text{m/sec.} \]

\[ \text{2nd case.}\qquad \operatorname{tg}\alpha_2 = \frac{12.34}{18.3} = 0.674,\quad \alpha_2=34^\circ \]

\[ w_2 = \frac{12.34}{\operatorname{sn}\alpha_2} = \frac{12.34}{0.559} = 22.1\ \text{m/sec.} \]

With this we have finished determining all the dimensions of the rotor and the angles of inclination of its blades. With respect to the two cases considered by us, it should be pointed out that the second case is more advantageous for us, since in it we have \(w_2 \cong w_1\), whereby the known loss due to the change in relative velocity is avoided. With respect to \(w_1\) and \(w_2\), it should be noted that in almost all examples analyzed by Wiesmann \(w_1 > w_2\).

Karp recommends taking \(w_2 \gg w_1\), which is more rational in view of the considerations indicated above, but is not obligatory for the designer if there is the possibility of obtaining a good efficiency even with variable \(w\).

In Fig. 6 the blade is drawn, the dimensions and angles of inclination of which we determined for the 2nd case.

In Fig. 7 a method is given for constructing the spiral casing from four centers—the corners of the construction square.

To determine the side of the construction square we determine the dimensions of the outlet section of the casing at the velocity adopted by us

\(V_2 = 21\) m/sec. We take the coefficient of incomplete filling of the rectangular outlet opening as \(\mu = 0.9\). For one half, we take the width of the casing to be

\[ b = 1.7 b_2 = 1.7 \cdot b_1 = 1.7 \cdot 0.325 \simeq 0.550 \]

(the full width is \(2 \cdot 0.55 = 1.1\) m.)

\[ a \cdot b \cdot V_2 \cdot 0.9 = 10.2 \ \text{m}^3/\text{sec.}, \]

whence:

\[ a = \frac{10.2}{0.55 \cdot 21 \cdot 0.9} = 0.98 \ \text{m.} \]

The side of the construction square for our case is:

\[ o = \frac{d - k}{4} = \frac{0.98 - 0.125}{4} = \frac{0.855}{4} = 0.216 \ \text{m.} \]

(Taking \(k\) into account, it is necessary to increase \(b\) somewhat so that the adopted velocity \(V_2\) does not change.)

Fig. 1.

Visible labels in the diagram include: $A$, $B$, $D$, $D_0$, $D_1$, $D_2$, $\rho_1$, $\rho_2$, $u_1$, $u_2$, $c$, $c_2$, $w_1$, $w_2$, $a$, $b$, $\delta$, and the ratios $\dfrac{u_2^2}{c_2^2}=2$, $\dfrac{c_2}{u_2}=1$.

Velocity triangles in the blades.

Fig. 2.

Velocity triangles at the inlet (a)
and at the outlet (b).

Fig. 3.

Fig. 6.

\[ \frac{u D_{1}\operatorname{Sn}(180-\alpha_{1})}{n} \]

Fig. 7.