S. K. Konyukhov, Prof., STI

Determination of the Volume of Liquid in a Cylindrical Boiler.

1. Reynolds and Catrell’s Method.

In the chemical industry, in the petroleum business, and on railways, it is sometimes necessary to determine the volume of liquid contained in a horizontal cylindrical boiler that is not filled to the top. Since the handbooks do not contain suitable formulas for solving this problem, we shall show how to solve it.

Let (Fig. 1) we have a horizontal cylindrical boiler with two end plates, described with radius \(R\). Let us denote the radius of the cylindrical part of the boiler by \(r\), and its length by \(l\). Further, let us introduce the following notation: \(ML = x\) (the depth to which the boiler is filled); \(HOL = \angle \theta\) and \(HO'L = \angle \varphi\). Let us note preliminarily that \(HKH'\) represents the surface of the level in the segment of the boiler, and that the point \(O'\), being the projection of the point \(O\) onto the line of symmetry of the level surface, is at the same time the center from which the cut is made in order to obtain the circular arc \(HKH'\). We may now write

\[ x = r - r \cos \theta = 2r \sin^2 \frac{\theta}{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (1) \]

If the point \(C\) is joined to the point \(K\), we obtain the right triangle \(CO'K\), in which \(CK = R\), \(CO' = OL = r \cos \theta\). Thus,

\[ O'K = \sqrt{R^2 - r^2 \cos^2 \theta} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (2) \]

But one may also put

\[ O'K = \lambda r \sqrt{\,1 + \frac{\sin^2 \theta}{\lambda^2}\,} = \lambda r \sqrt{\,1 + \operatorname{tg}^2 \varphi\,} \ldots \ldots (3) \]

where

\[ \operatorname{tg} \varphi = \frac{\sin \theta}{\lambda}. \]

To determine the value of the coefficient \(\lambda\), let us turn to consideration of the triangles \(OHL\) and \(O'HL\). We may now write:

\[ HL = r \sin \theta;\quad HL = O'L\, \operatorname{tg}\varphi = CO\, \operatorname{tg}\varphi = \sqrt{R^2 - r^2}\, \operatorname{tg}\varphi \ldots (4) \]

Consequently,

\[ r \sin \theta = \sqrt{R^2 - r^2}\, \operatorname{tg}\varphi \ldots \ldots \ldots \ldots \ldots (5) \]

Whence we obtain

\[ \tg \varphi=\frac{r}{\sqrt{R^{2}-r^{2}}}\Sin \theta \ldots\ldots\ldots\ldots\ldots\ldots\ldots (6) \]

i.e.

\[ \lambda=\frac{\sqrt{R^{2}-r^{2}}}{r}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (7) \]

Let us now introduce the following conventional notations

\[ X=\theta-\frac{\Sin 2\theta}{2} \]

\[ Y=-2L\theta-\frac{4}{3}\lambda^{3}\theta+\frac{2}{3}\lambda\Sin 2\theta+ \]

\[ +\frac{1}{3}\Cos\theta\left(\Cos 2\theta-6\lambda^{2}-5\right)\arctg\frac{\Sin\theta}{\lambda} +\frac{4}{3}(1+\lambda^{2})^{3/2}\arctg\left(\frac{\sqrt{1+\lambda^{2}}}{\lambda}\tg\theta\right). \]

Now the result of the integration assumes a more compact form

\[ V=Xr^{2}l+Yr^{3}\ldots\ldots\ldots\ldots\ldots\ldots\ldots (14) \]

The question is now reduced to finding the numerical value for \(X\) and \(Y\).

When it is necessary to make a more exact calculation of the volume of liquid, then in riveted boilers one must take into account the presence of rivet heads.

In the case when the rivet heads are turned with their convexity inward into the boiler, the equation for the volume takes the form

\[ V=r^{2}lX-r^{3}Y\ldots\ldots\ldots\ldots\ldots\ldots\ldots (15) \]

If one rivet is turned with its head inward and the other outward, then

\[ V=r^{2}lX\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots (16) \]

For the convenience of rapid calculations, the following tables are given of values of \(X\) and \(Y\), found by exact computations, taking into account the presence of rivet heads and boiler seams.

In these tables \(p=\dfrac{x}{2r}\). The values of \(p\) are taken within the limits of \(1/2\%\) of the diameter of the sample boiler, in which the radius of the spherical bottom is equal to the boiler diameter, or \(R=2r\).

We shall now consider the area of the liquid level in the boiler. It consists of a central rectangle with area \(2lr\Sin\theta\) and two segments with central angle \(2\varphi\) and radius \(O'K\). The area of each segment is the difference between the areas of the sector \(O'HKH'O'\) and the triangle \(O'HH'\).

Thus we express this area as

\[ \omega=(O'K)^2\varphi-HL\sqrt{R^2-r^2}= \]

\[ =r^2(\lambda^2+\sin^2\theta)\varphi-r\sin\theta\sqrt{R^2-r^2}= \]

\[ =r^2(\lambda^2+\sin^2\theta)\varphi-r^2\lambda\sin\theta \ldots \tag{8} \]

Knowing the area of the segment, it is not difficult to find the elementary volume of the layer of water in the boiler

\[ dV=\left[2rl\sin\theta+2r^2(\lambda^2+\sin^2\theta)\varphi-2r^2\lambda\sin\theta\right]dx \ldots \tag{9} \]

But

\[ x=r(1-\operatorname{Cs}\theta), \qquad dx=r\sin\theta\,d\theta \ldots \tag{10} \]

Since in this equation the variables are the quantities \(x\), \(\varphi\), and \(\theta\), then in order to carry out the integration it is necessary to express \(x\) and \(\varphi\) in terms of \(\theta\).

Thus, we obtain

\[ V=2r^2\int_{0}^{\theta} \left[(l-r\lambda)\sin^2\theta+r(\lambda^2\sin\theta+\sin^3\theta)\operatorname{arc\,tg}\left(\frac{\sin\theta}{\lambda}\right)\right]d\theta \ldots \tag{11} \]

It is simplest to carry out the integration by parts. In this case we obtain

\[ V=r^2\left(l-2r\lambda-\frac{4}{3}r\lambda^3\right)\theta+ \frac{r^2}{6}(4r\lambda-3l)\sin2\theta+ \]

\[ +\frac{r^3}{3}\cos\theta(\cos2\theta-6\lambda^2-5)\operatorname{arc\,tg}\frac{\sin\theta}{\lambda}+ \]

\[ +\frac{4}{3}r^3(\lambda^3+1)^{1/2}\operatorname{arc\,tg}\left(\frac{\sqrt{1+\lambda^2}}{\lambda}\operatorname{tg}\theta\right)\ldots \tag{12} \]

Analytically, the quantity \(\theta\) is expressed as follows:

\[ \theta=2\arc\sin\sqrt{\frac{x}{2r}} \ldots \tag{13} \]

and varies within the limits from \(0\) to \(180^\circ\).

Finally,

\[ \operatorname{arctg}\left(\frac{\sqrt{1+\lambda^2}}{\lambda}\operatorname{tg}\theta\right) \]

varies within the limits from \(0\) to \(\pi\).

If we put \(\theta=0\) or \(\theta=\pi\), then

\[ \operatorname{arctg}\left(\frac{\sin\theta}{\lambda}\right)=0. \]

Let us show by an example how to use the compiled numerical material.

Example. Let the length of the boiler be \(e = 24\) feet, and the diameter 8 feet. The depth of the liquid in the boiler is 2 feet. Then the volume of the liquid will be

\[ V = 4^2 \cdot 24 x + 4^3 \cdot y = 384x + 64y. \]

Now it is necessary to find \(p\)

\[ p = \frac{x}{2r} = \frac{2}{8} = 0.25 \text{ or } 25\%. \]

From the table we find, for \(p = 25\%\), the values

\[ X = 0.6142 \quad \text{and} \quad y = 0.111. \]

Consequently,

\[ V = 384 \cdot 0.6142 + 64 \cdot 0.111 = 235.8528 + 7.104 = 242.96 \text{ cu. ft.} \]

Or one may resort to logarithms.

\[ \operatorname{Lg} 384X = 2.58433 + 9.78831 - 10 = 2.37264 \]

\[ \operatorname{Lg} 64Y = 1.80618 + 9.04532 - 10 = 0.85150 \]

Consequently,

\[ V = 235.8528 + 7.104 = 242.96 \text{ cu. ft.} \]

Nothing, of course, prevents calculating tables for boilers of another size.

It is first necessary to calculate the value

\[ \lambda = \frac{\sqrt{R^2 - r^2}}{r}. \tag{17} \]

for each boiler, and then to find the values of \(X\) and \(y\) for the successive values of \(-\theta\) from the equation

\[ \theta = 2\arc \operatorname{Sin} \sqrt{\frac{x}{2r}}. \tag{18} \]

2. Kemmel’s Method.

Let the boiler be filled to more than half its height (in the direction of the vertical diameter). The volume of the liquid may conveniently be divided into three parts: 1) the central part, whose base is the sector \(SKS\) (Fig. 3), and whose height is the length of the cylinder \(E\); 2) the two lower spherical half-segments, constituting a whole spherical segment equal in volume to the bottom of the boiler; and 3) the two upper incomplete spherical segments.

The volume of the first part will be

\[ V_1 = \frac{\pi r^2 n}{360} l + OA \cdot AN \cdot l = \frac{\pi \cdot r^2 n l}{360} + c h l \ldots \tag{19} \]

Here \(n\) is the number of degrees corresponding to the arc \(NKN'\), \(OA=c\) and \(AN=b\) (Fig. 3).

The volume of the second part will be

\[ V_2=\frac{\pi\cdot BD^2(3R-BD)}{3}\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (20) \]

But

\[ BD=OD-OB=R-R\operatorname{Cos}\frac{\alpha'}{2} =R\left(1-\operatorname{Cos}\frac{\alpha'}{2}\right). \]

Consequently,

\[ V_2=\frac{\pi R^2\left(1-\operatorname{Cos}\frac{\alpha'}{2}\right)^2 \left(3R-R+R\operatorname{Cos}\frac{\alpha'}{2}\right)}{3} = \]

\[ =\frac{\pi R^3}{3} \left(2-3\operatorname{Cos}\frac{\alpha'}{2} +\operatorname{Cos}^3\frac{\alpha'}{2}\right) .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (21) \]

We now pass to the determination of the volume of the third part. If a section \(S-S'\) (Fig. 3) is drawn at a distance \(x\) from the axis of the boiler, we obtain the level of the liquid in the form of a circular segment, whose projection onto the horizontal plane passing through the axis of the boiler is represented by the segment \(CBC'E\) (Fig. 4). The area of this segment will be equal to

\[ \omega=OCHC'-OCC' =\frac{1}{2}\rho^2\operatorname{arc}\varphi -\rho\operatorname{Cos}\frac{\varphi}{2}\,\rho\operatorname{Sin}\frac{\varphi}{2} =\frac{\rho^2}{2}(\operatorname{arc}\varphi-\operatorname{Sin}\varphi) .\ .\ (22) \]

The elementary volume is expressed as

\[ dV_3=\omega\,dx=\frac{\rho^2}{2}(\operatorname{arc}\varphi-\operatorname{Sin}\varphi)\,dx .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (23) \]

\[ V_3=\int_{x=c}^{x=0}\frac{\rho^2}{2}(\operatorname{arc}\varphi-\operatorname{Sin}\varphi)\,dx .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (24) \]

In order to be able to carry out the integration, we express \(\rho\) and \(\varphi\) through \(x\) and known quantities.

\[ \rho^2=R^2-x^2;\quad \rho^2\operatorname{Sin}^2\frac{\alpha}{2}=R^2\operatorname{Sin}^2\frac{\alpha}{2}-x^2;\quad \rho\operatorname{Cos}\frac{\varphi}{2}=R\operatorname{Cos}\frac{\alpha}{2}. \]

These equations are obtained without difficulty by using Figures 2, 3, and 4.

In this case we shall have

\[ \operatorname{Cos}\frac{\varphi}{2} =\frac{R}{\rho}\operatorname{Cos}\frac{\alpha}{2};\quad \operatorname{Sin}\frac{\varphi}{2} =\frac{\sqrt{R^2\operatorname{Sin}^2\frac{\alpha}{2}-x^2}}{\rho}. \]

Hence

\[ 2\operatorname{Sin}\frac{\varphi}{2}\operatorname{Cos}\frac{\varphi}{2} =\operatorname{Sin}\varphi =\frac{2R\operatorname{Cos}\frac{\alpha}{2}\sqrt{R^2\operatorname{Sin}^2\frac{\alpha}{2}-x^2}}{\rho^2} .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (25) \]

Substituting this value of \(\operatorname{Sin}\varphi\) into integral (24), we obtain

\[ V_3= \int_{x=0}^{x=c}\frac{\rho^2}{2}\operatorname{arc}\varphi\,dx - \int_{x=0}^{x=c} R\operatorname{Cos}\frac{\alpha}{2} \sqrt{R^2\operatorname{Sin}^2\frac{\alpha}{2}-x^2}\,dx \ .\ .\ (25) \]

Leaving the first integral aside, let us note that the second integral is evaluated without difficulty, since the quantities \(R\Cos\dfrac{\alpha}{2}\) and \(R\Sin\dfrac{\alpha}{2}\) are constants. Write this integral as

\[ R\Cos\dfrac{\alpha}{2}\int_{x=0}^{x=C}\sqrt{R^2\Sin^2\dfrac{\alpha}{2}-x^2}\,dx= \]

\[ =R\Cos\dfrac{\alpha}{2}\left[ \dfrac{x}{2}\sqrt{R^2\Sin^2\dfrac{\alpha}{2}-x^2} +\dfrac{R^2\Sin^2\dfrac{\alpha}{2}}{2}\arc\Sin\left(\dfrac{x}{R\Sin\dfrac{\alpha}{2}}\right) \right]= \]

\[ =R\Cos\dfrac{\alpha}{2}\left( \dfrac{C}{2}\sqrt{R^2\Sin^2\dfrac{\alpha}{2}-C^2} +\dfrac{R^2\Sin^2\dfrac{\alpha}{2}}{2}\arc\Sin\dfrac{C}{R\Sin\dfrac{\alpha}{2}} \right). \tag{26} \]

Let us now undertake the finding of the integral

\[ \int_{x=0}^{x=C}\dfrac{\rho^2}{2}\arc\varphi\,dx = \int_{x=0}^{x=C}\rho^2\arc\left(\dfrac{\varphi}{2}\right)\,dx = \]

\[ =\int_{x=0}^{x=C}\rho^2\arc\Cos\left(\dfrac{R\Cos\dfrac{\alpha}{2}}{\rho}\right)\,dx. \tag{27} \]

To carry out the integration it is necessary to replace \(\rho\), expressing it as a function of \(x\). Then we obtain

\[ \int_{x=0}^{x=C}(R^2-x^2)\arc\Cos \dfrac{R\Cos\dfrac{\alpha}{2}}{\sqrt{R^2-x^2}}\,dx. \tag{28} \]

The integration is most conveniently performed by parts, according to the well-known formula

\[ \int V\,dU=UV-\int U\,dV. \tag{29} \]

putting

\[ dV=(R^2-x^2)\,dx \quad\text{and}\quad U=\arc\Cos\dfrac{R\Cos\dfrac{\alpha}{2}}{\sqrt{R^2-x^2}}. \tag{30} \]

Hence we have

\[ V=R^2x-\dfrac{x^3}{3}=\dfrac{x}{3}(3R^2-x^2). \tag{31} \]

Therefore,

\[ \int_0^C (R^2-x^2)\arc\Cos \dfrac{R\Cos\dfrac{\alpha}{2}}{\sqrt{R^2-x^2}}\,dx = \dfrac{x}{3}(3R^2-x^2)\arc\Cos \dfrac{R\Cos\dfrac{\alpha}{2}}{\sqrt{R^2-x^2}} = \]

\[ + \int_0^c \frac{3R^2\chi-\chi^3}{3}\cdot \frac{R\chi\cos\frac{\alpha}{2}\,d\chi} {(R^2-\chi^2)\sqrt{R^2\sin^2\frac{\alpha}{2}-\chi^2}} \ldots \ldots \ldots \ldots (32) \]

It is not difficult to see that the solution of the problem of integration has now been reduced to finding the integral

\[ \int_0^c \frac{(3R^2x^2-x^4)\,dx} {(R^2-x^2)\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2}} \ldots \ldots \ldots \ldots (33) \]

This integral may be decomposed, using the consideration that

\[ \frac{3R^2x^2-x^4}{R^2-x^2} = x^2-2R^2+\frac{2R^4}{R^2-x^2} = \frac{R^2x^2-2R^4-x^4+2R^2x^2-2R^4}{R^2-x^2}. \tag{34} \]

In turn, one may write

\[ \frac{2}{R^2-x^2} = \frac{1}{R(R+x)} + \frac{R(R-x)}{1} = \frac{R^2+Rx+R^2-Rx}{R^2(R^2-x^2)}. \ldots (35) \]

\[ \int_0^c \frac{(3R^2x^2-x^4)\,dx} {(R^2-x^2)\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2}} = \int_0^c \frac{x^2\,dx} {\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2}} -2R^2 \]

\[ \int_0^c \frac{dx} {\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2}} + R^4 \int_0^c \frac{2\,dx} {(R^2-x^2)\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2}} \ldots (36) \]

Let us now take up the evaluation of these integrals separately

\[ \int_0^c \frac{x^2\,dx} {\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2}} = -\frac{x}{2}\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2} + \frac{R^2\sin^2\frac{\alpha}{2}}{2} \]

\[ \int_0^c \frac{dx} {\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2}} \ldots \ldots \ldots \ldots (37) \]

(see L. Kiepert, II Teil, Integral-Rechnung, p. 95)

\[ -2R^2\int_0^c \frac{dx} {\sqrt{R^2\sin^2\frac{\alpha}{2}-x^2}} = -2R^2\arc\sin \frac{x}{R\sin\frac{\alpha}{2}} \ldots \ldots \ldots (38) \]

Thus, the first two integrals give

\[ -\frac{x}{2}\sqrt{R^{2}\operatorname{Sin}^{2}\frac{\alpha}{2}-x^{2}} + \frac{R^{2}\left(\operatorname{Sin}^{2}\frac{\alpha}{2}-4\right)}{2} \operatorname{arc\,Sin}\frac{x}{R\operatorname{Sin}\frac{\alpha}{2}} . \tag{39} \]

We shall find the third integral as follows:

\[ R^{4}\int_{0}^{c} \frac{2\,dx}{(R^{2}-x^{2})\sqrt{R^{2}\operatorname{Sin}^{2}\frac{\alpha}{2}-x^{2}}} = \]

\[ = R^{3}\int_{0}^{c} \frac{1}{R-x}\, \frac{dx}{\sqrt{R^{2}\operatorname{Sin}^{2}\frac{\alpha}{2}-x^{2}}} + R^{3}\int_{0}^{c} \frac{1}{R+x}\, \frac{dx}{\sqrt{R^{2}\operatorname{Sin}^{2}\frac{\alpha}{2}-x^{2}}}. \tag{40} \]

Let us find these integrals separately. Introduce a new variable, putting \(R-x=\)

\[ =\frac{1}{u}. \]
In this case
\[ 1)\; U=\frac{1}{R-x};\quad 2)\; x^{2}=R^{2}-\frac{2R}{u}+\frac{1}{u^{2}} \]

and

\[ 3)\; dx=\frac{du}{u^{2}} \ldots \tag{41} \]

Substituting these values into the first integral, we obtain

\[ R^{3}\int_{0}^{c} \frac{1}{(R-x)} \frac{dx}{\sqrt{R^{2}\operatorname{Sin}^{2}\frac{\alpha}{2}-x^{2}}} = R^{3}\int \frac{U\,dU}{U^{2}\sqrt{R^{2}\operatorname{Sin}^{2}\frac{\alpha}{2}-R^{2}-\frac{1}{u^{2}}+\frac{2R}{U}}} \]

\[ = \int \frac{dU}{\sqrt{U^{2}R^{2}\operatorname{Sin}^{2}\frac{\alpha}{2}-U^{2}R^{2}-1+2RU}} = \int \frac{dU}{\sqrt{2Ru-1-\operatorname{Cos}^{2}\frac{\alpha}{2}\,R^{2}U^{2}}}. \]

We evaluate this integral (see Kiepert, II Teil Integrale Rechnung, p. 703):

\[ \int \frac{dU}{\sqrt{2RU-1-R^{2}\operatorname{Cos}\frac{\alpha}{2}}} = -\frac{1}{R\operatorname{Cos}\frac{\alpha}{2}} \operatorname{arc\,Sin} \frac{R\cdot \operatorname{Sin}^{2}\frac{\alpha}{2}-x} {(R-x)\operatorname{Sin}\frac{\alpha}{2}} + C_{1}. \tag{44} \]

In the same way we shall find the second integral, putting

\[ 1)\; R+x=\frac{1}{s};\quad 2)\; x^{2}=\frac{1}{s^{2}}+R^{2}-\frac{2R}{s};\quad 3)\; dx=-\frac{ds}{s^{2}}. \]

Then we shall have

\[ \int \frac{dx}{(R+x)\sqrt{R^{2}\operatorname{Sin}^{2}\frac{\alpha}{2}-x^{2}}} = -\frac{1}{R\operatorname{Cos}\frac{\alpha}{2}} \operatorname{arc\,Sin} \frac{R\operatorname{Sin}^{2}\frac{\alpha}{2}+x} {(R+x)\operatorname{Sin}\frac{\alpha}{2}} + C_{2}. \tag{45} \]

Collecting together the results of the integration, we obtain the following equation for the expression of the volume

\[ \begin{aligned} V_{3}={}& \int_{0}^{c}\rho^{2}\arc\varphi\,dx -\int_{0}^{c}R\Cos\frac{\alpha}{2} \sqrt{R^{2}\Sin^{2}\frac{\alpha}{2}-x^{2}}\,dx \\[4pt] ={}& \frac{c}{3}\,(3R^{2}-c^{2})\arc\Cos \frac{R\Cos\frac{\alpha}{2}}{\sqrt{R^{2}-c^{2}}} -\frac{2Rc}{3}\Cos\frac{\alpha}{2} \sqrt{R^{2}\Sin^{2}\frac{\alpha}{2}-c^{2}} \\[4pt] &+\frac{R^{3}}{3}\left( \arc\Sin \frac{R\Sin^{2}\frac{\alpha}{2}+c}{(R+c)\Sin\frac{\alpha}{2}} - \arc\Sin \frac{R\Sin^{2}\frac{\alpha}{2}-c}{(R-c)\Sin\frac{\alpha}{2}} \right) \\[4pt] &+\frac{R^{3}}{2}\Cos\frac{\alpha}{2}\, \arc\Sin \frac{c}{R\Sin\frac{\alpha}{2}} \left[ \frac{1}{3}\left(\Sin^{2}\frac{\alpha}{2}-4\right) -\Sin^{2}\frac{\alpha}{2} \right]. \tag{46} \end{aligned} \]

After a few transformations we shall have

\[ \begin{aligned} V^{3}={}& \frac{c}{3}(3R^{2}-c^{2})\arc\Cos \frac{R\Cos\frac{\alpha}{2}}{\sqrt{R^{2}-c^{2}}} -\frac{R^{3}}{3} \left(3\Cos\frac{\alpha}{2}-\Cos^{3}\frac{\alpha}{2}\right) \arc\Sin\frac{c}{R\Sin\frac{\alpha}{2}} \\[4pt] &-\frac{2}{3}Rc\Cos\frac{\alpha}{2} \sqrt{R^{2}\Sin^{2}\frac{\alpha}{2}-c^{2}} +\frac{R^{2}}{3} \left[ \arc\Sin \frac{R\Sin^{2}\frac{\alpha}{2}+c}{(R+c)\Sin\frac{\alpha}{2}} \right. \\[4pt] &\left. \qquad\qquad -\arc\Sin \frac{R\Sin^{2}\frac{\alpha}{2}-c}{(R-c)\Sin\frac{\alpha}{2}} \right]. \tag{47} \end{aligned} \]

This, then, is the final formula for determining \(V_{3}\). Let us check it.
Suppose the boiler is completely filled. Then \(C=R\Sin\frac{\alpha}{2}\). In equation (47) the first term and the third will vanish, since

\[ 1)\quad \arc\Cos \frac{R\Cos\frac{\alpha}{2}} {\sqrt{R^{2}-R^{2}\Sin^{2}\frac{\alpha}{2}}} =0, \qquad 2)\quad \sqrt{R^{2}\Sin^{2}\frac{\alpha}{2}-c^{2}}=0 . \]

The equation itself will take the simplified form

\[ V=\frac{\pi R^{3}}{3}-\frac{\pi R^{3}}{6} \left(3\cos\frac{\alpha}{2}-\cos^{3}\frac{\alpha}{2}\right). \tag{48} \]

\[ =\frac{\pi R^{3}}{6} \left(2+\cos^{3}\frac{\alpha}{2}-3\cos\frac{\alpha}{2}\right). \tag{49} \]

i.e., precisely one half of the volume of a spherical segment.

Fig. 1.

\[ M\mathcal{L}=x \]

\[ \angle HO\mathcal{L}=\varphi \]

\[ \angle HOx=\theta \]

\[ MO=M'O=z \]

\[ CM'=R \]

Fig. 2.

Fig. 3.

\[ \rho \sin \frac{\alpha}{2} \]

\[ \rho \sin \frac{\varphi}{2} \]

Fig. 4.

\[ R\sin \frac{\alpha}{2} \]

\[ \rho \sin \frac{\varphi}{2} \]

For the article by Prof. S. K. Konyukhov,
“Determination of the Volume of Liquid in a Cylindrical Boiler.”