HYDROMECHANICS

A. ARYNOV

OUTFLOW OF GAS FROM A VESSEL WITH WALLS ENCLOSING A SMALL ANGLE \(2\theta_0\)

(Presented by Academician L. I. Sedov on 14 VI 1958)

1. The case of outflow with subcritical velocity.
In Fig. 1, \(Ox\) is the trace of the plane of symmetry; \(BB'\), \(AA'\) are the walls of the vessel; \(B'B''\), \(A'A''\) are the jet. \(\varphi\) and \(\psi\) are, respectively, the velocity potential and the stream function; \(\psi=0\) on \(Ox\), while in the region of outflow \(\psi\) varies from \(-Q/2\) to \(Q/2\), where \(\rho_0 Q\) is the discharge per 1 sec.

The problem posed can be solved analogously to Chaplygin’s problem \((^1)\).

For the stream function \(\psi\) and the velocity potential \(\varphi\), the following formulas hold:

\[ \frac{\pi}{Q}\psi = -k\theta - \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{\tau}{\tau_0}\right)^{kn} \frac{Y_{kn}}{Y_{kn,0}} x_{kn}\delta \sin 2kn\theta, \]

\[ -\frac{\pi}{Q}\varphi = c + \frac{1}{2}\int \frac{d\tau}{\tau(1-\tau)^\beta} - \frac{1}{(1-\tau)^\beta} + \frac{1}{(1-\tau)^\beta} \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{\tau}{\tau_0}\right)^{kn} \frac{y_{kn}}{y_{kn,0}} x_{kn}\cos 2kn\theta, \tag{1} \]

where \(k=\pi/2\theta_0\); \(\theta\) is the angle of inclination of the velocity; \(\tau=w^2/w_{\max}^2\); \(w\) is the modulus of the flow velocity; \(w_{\max}\) is the outflow velocity into vacuum.

Fig. 1

Using these formulas, one can find the equation of the jet boundary on which \(\tau=\tau_0\). We have

\[ \frac{\pi}{Q}w_0(1-\tau)^\beta y = -k\sin\theta - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \frac{2kn}{4k^2h^2-1} \left\{ (x_{kn}+2kn)\sin 2kn\theta \cos\theta - (1+2knx_{kn})\cos 2kn\theta \sin\theta \right\}, \tag{2} \]

where \(x_{kn}(\tau)\) is an auxiliary function introduced by Chaplygin. At the outlet \(\theta=\theta_0\), \(y=-a\). Consequently:

\[ w_0(1-\tau)^\beta \frac{\pi}{Q}a = k\sin\theta_0 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \frac{2kn}{4k^2n^2-1} (1+2knx_{kn})\sin\theta_0. \tag{3} \]

Taking into account that \(Q=2b(1-\tau_0)^\beta w_0\), we obtain

\[ \frac{\pi}{2}\frac{a}{b} = k\sin\theta_0 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \frac{2kn}{4k^2n^2-1} (1+2knx_{kn})\sin\theta_0. \tag{4} \]

Let us consider the sum in (4):

\[ I=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\,\frac{2kn}{4k^{2}n^{2}-1} \left(1+\frac{1}{4k^{2}n^{2}}+\frac{1}{16k^{4}n^{4}}+\cdots\right)(1+2knx_{kn}). \tag{5} \]

Guderley and Light-hill showed that, for Mach number \(\mathrm{M}<1\), there exists an asymptotic expansion of the form

\[ x_{kn}\sim x^{(0)}(\tau)+\frac{x^{(1)}(\tau)}{kn} +\frac{x^{(2)}(\tau)}{k^{2}n^{2}} +\frac{x^{(3)}(\tau)}{k^{3}n^{3}}+\cdots \tag{6} \]

Applying Chaplygin’s method, F. I. Frankl found the recurrence formulas:

\[ \begin{aligned} x^{(0)}&=\sqrt{1-2\beta s},\\ x^{(1)}&=-\frac{1}{2x^{(0)}}\,[s(1+s)x^{(0)\prime}+\beta sx^{(0)}],\\ x^{(2)}&=-\frac{1}{2x^{(0)}}\,[s(1+s)x^{(1)\prime}+\beta sx^{(1)}+x^{(1)^{2}}],\\ x^{(3)}&=-\frac{1}{2x^{(0)}}\,[s(1+s)x^{(2)\prime}+\beta sx^{(2)}+2x^{(1)}x^{(2)}],\\ &\qquad\qquad\ldots\ldots\ldots\ldots\ldots\ldots \end{aligned} \tag{7} \]

where \(\beta=\dfrac{1}{\kappa-1}\), \(s=\dfrac{\tau}{1-\tau}=\dfrac{1}{2\beta}\mathrm{M}^{2}\). Substituting \(s\) and \(\beta\), we obtain expressions for \(x^{(i)}\) in terms of \(\mathrm{M}\):

\[ \begin{aligned} x^{(0)}&=\sqrt{1-\mathrm{M}^{2}},\\ x^{(1)}&=\frac{\kappa+1}{8}\,\frac{\mathrm{M}^{4}}{1-\mathrm{M}^{2}},\\ x^{(2)}&=-\frac{\kappa+1}{32}\, \frac{\mathrm{M}^{4}}{(1-\mathrm{M}^{2})^{5/2}} \left[4-(3-2\kappa)\mathrm{M}^{2}-\frac{3\kappa-1}{4}\mathrm{M}^{4}\right], \end{aligned} \tag{8} \]

\[ \begin{aligned} x^{(3)}&=-\frac{[2+(\kappa-1)\mathrm{M}^{2}]\,\mathrm{M}^{4}} {64(1-\mathrm{M}^{2})^{4}} \Bigl\{[2+(\kappa-1)\mathrm{M}^{2}] [\kappa+6(1-\kappa)\mathrm{M}^{2}+(5\kappa-1)\mathrm{M}^{4}\\ &\qquad-\kappa\mathrm{M}^{6}](1-\mathrm{M}^{2})^{1/2} +\frac{2(\kappa-1)(1-\mathrm{M}^{2})^{7/2}}{\mathrm{M}^{2}} [2+5\mathrm{M}^{2}(1-\mathrm{M}^{2})]\\ &\qquad-(\kappa+1)\mathrm{M}^{2} \left[4-(3-2\kappa)\mathrm{M}^{2}-\frac{3\kappa-1}{4}\mathrm{M}^{4}\right]\Bigr\}. \end{aligned} \]

Substituting the value of \(x_{kn}\) into equation (5), we obtain

\[ \begin{aligned} I&=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} \left(\frac{1}{2kn}+\frac{1}{8k^{2}n^{2}}+\frac{1}{32k^{4}n^{4}}+\cdots\right) \left(1+2knx^{(0)}+2x^{(1)}+\right.\\ &\qquad\left.+\frac{2x^{(3)}}{kn}+\frac{2x^{(3)}}{k^{2}n^{2}}+\cdots\right)\\ &=0.693x^{(0)}+0.822\,\frac{1/2+x^{(1)}}{k} +0.901\,\frac{x^{(2)}+\tfrac14 x^{(0)}}{k^{2}} +0.951\,\frac{1/8+\tfrac14 x^{(2)}+x^{(3)}}{k^{3}}+\cdots . \end{aligned} \]

Substituting this sum into (4), we obtain

\[ \frac{\pi}{2}\frac{a}{b}\sim \frac{\pi}{2}\frac{\sin\theta_{0}}{\theta_{0}} +\sin\theta_{0} \left[ 0.693x^{(0)} +0.822\,\frac{1/2+x^{(1)}}{\pi}\,2\theta_{0} +0.901\,\frac{x^{(2)}+\tfrac14 x^{(0)}}{\pi^{2}}\,4\theta_{0}^{2} +0.951\,\frac{1/8+\tfrac14 x^{(2)}+x^{(3)}}{\pi^{3}}\,8\theta_{0}^{3} +\cdots \right]. \]

or

\[ \frac{\pi}{2}\frac{a}{b}\sim \frac{\pi}{2}+0.693\chi^{(0)}\theta_0- \left[\frac{\pi}{2}-1.644\,\frac{1/2+\chi^{(1)}}{\pi}\right]\theta_0^2+ \]

\[ +\left[\frac{3.604}{\pi^2}\left(\chi^{(2)}+\frac14\chi^{(0)}\right) -\frac{0.693}{6}\chi^{(0)}\right]\theta_0^3+ \]

\[ +\left[\frac{\pi}{240}+\frac{7.608}{\pi^3}\left(\frac18+\frac14\chi^{(2)}+\chi^{(3)}\right) -\frac{0.822}{3\pi}\left(\frac12+\chi^{(1)}\right)\right]\theta_0^4+\cdots \]

Finally we have

\[ \frac{b}{a}\sim 1-0.441\chi^{(0)}\theta_0+ \left(0.194{\chi^{(0)}}^2+0.332\chi^{(1)}\right)\theta_0^2- \]

\[ -\left(0.292\chi^{(0)}\chi^{(1)}-0.232\chi^{(2)}+ 0.015\chi^{(0)}+0.085{\chi^{(0)}}^3\right)\theta_0^3+ \]

\[ +\left[0.017+0.055\chi^{(1)}-0.003\chi^{(3)} -0.882\chi^{(0)}\left(0.232\chi^{(2)}-0.015\chi^{(0)}\right)+ \right. \]

\[ \left. +0.11{\chi^{(1)}}^2+0.128{\chi^{(0)}}^2\chi^{(1)} \right]\theta_0^4+\cdots \tag{9} \]

In order to determine \(b/a\) for a diatomic ideal gas, we substitute its value \(1.4\) for \(\chi\) in formulas (8) and (9).

The same formulas with \(\chi=2\) apply to the flow of water in a shallow channel \((^2)\), and also to an ultrarelativistic gas and to a photon gas \((^3)\). In Table 1 and in Fig. 2 the values of \(b/a\) are given.

2. Case of outflow of a liquid with critical velocity

In this case Chaplygin’s auxiliary function is given by the asymptotic formula of F. I. Frankl \((^4)\)

\[ x_{kn}\simeq x^{*}_{kn}\sim \frac{C_0}{n^{1/3}k^{1/3}}+\frac{C_1}{kn} +\frac{C_2}{n^{5/3}k^{5/3}}+\cdots \tag{10} \]

Fig. 2

Table 1

Substituting \(x^{*}_{kn}\) into formula (4), we obtain

\[ \frac{\pi}{a}\frac{a}{b}\sim k\sin\theta_0+\sin\theta_0 \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\, \frac{1}{2kn} \left(1+\frac{1}{4k^2n^2}+\frac{1}{16n^4k^4}+\cdots\right)\times \]

\[ \times \left[ 1+2kn\left( \frac{C_0}{n^{1/3}k^{1/3}} +\frac{1/2+C_1}{nk} +\frac{C_2}{n^{5/3}k^{5/3}}+\cdots \right) \right]= \]

\[ = k\sin\theta_0+\sin\theta_0 \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} \left( \frac{C_0}{n^{1/3}k^{1/3}} +\frac{1/2+C_1}{nk} +\frac{C_2}{n^{5/3}k^{5/3}}+\cdots \right). \]

After summation we obtain

\[ \frac{\pi}{a}\frac{a}{b} = k\sin\theta_0 + \sin\theta_0\left[0.6C_0\theta_0^{2/3} + 0.523(0.5+C_1)\theta_0 + 0.649C_2\theta_0^{5/3}\right]. \tag{11} \]

By means of transformations analogous to those carried out in the preceding paragraph, we obtain

\[ \frac{a}{b}\sim 1+\frac{1.2}{\pi}C_0\theta_0^{4/3} +\left[\frac{1.04(1.5+C_1)}{\pi}-0.1666\right]\theta_0^2 +\frac{1.29}{\pi}C_2\theta_0^{8/3}. \tag{12} \]

In the case of an ideal diatomic gas \((\varkappa=1.4)\), \(C_0=0.7745\), \(C_1=-0.3901\), \(C_2=0.2053\) \((^4)\). Substituting these values of \(C_i\) into formula (12), we obtain

\[ \frac{a}{b}\sim 1+0.295\theta_0^{4/3} -0.13\theta_0^2 +0.084\theta_0^{8/3}+\cdots, \tag{13} \]

whence

\[ \frac{b}{a}\sim 1-0.295\theta_0^{4/3} +0.13\theta_0^2 -0.171\theta_0^{8/3}+\cdots \tag{14} \]

In the case of the outflow of water and a photon gas \((\varkappa=2)\) we have \(C_0=0.834\), \(C_1=0.450\), so that, according to (12):

\[ \frac{a}{b}\sim 1+0.318\theta_0^{4/3} -0.15\theta_0^2+\cdots, \]

whence

\[ \frac{b}{a}\sim 1-0.518\theta_0^{4/3} +0.15\theta_0^2+\cdots, \]

Consequently, for \(M=1\), \(\varkappa=1.4\),

\[ \left.\frac{b}{a}\right|_{5^\circ}=0.9880,\qquad \left.\frac{b}{a}\right|_{10^\circ}=0.9737,\qquad \left.\frac{b}{a}\right|_{15^\circ}=0.9547. \]

If \(\varkappa=2\), then

\[ \left.\frac{b}{a}\right|_{5^\circ}=0.9882,\qquad \left.\frac{b}{a}\right|_{10^\circ}=0.9739,\qquad \left.\frac{b}{a}\right|_{15^\circ}=0.9555. \]

Kabardino-Balkarian
State University

Received
14 VI 1958

REFERENCES CITED

1. S. A. Chaplygin, On Gas Jets, Collected Works, 2, 1948.
2. F. I. Frankl, Transactions of the Physico-Mathematical Faculty, Kirghiz State University, 2, 47 (1953).
3. F. I. Frankl, DAN, 123, No. 1 (1958).
4. F. I. Frankl, DAN, 58, No. 3 (1947).