Aeromechanical Calculation of the Cooling Channel of the Magnetizing Winding of a Betatron

V. V. Ivanov, G. F. Shilin

(Presented by Prof. Dr. G. I. Fuks)

In designing the air-cooling system of a betatron, it is necessary to know the aeromechanical resistances of its individual parts. One element of the betatron that releases a considerable amount of Joule heat \(q_v\) and therefore requires intensive air cooling is the magnetizing winding.

This article presents the results of a study of the aeromechanical resistance of the spiral magnetizing winding of a betatron, made of a rectangular row of insulated bars (Fig. 1). The equivalent coefficient of thermal conductivity of the winding \(\lambda\) and the coefficient of heat transfer from the surface to the cooling air \(\alpha\) are constant quantities.

If we assume that the temperature of the cooling air is constant and equal to \(T_{\mathrm{в}}\), then the temperature field in a rectangular row of bars is described by the following system of equations:

\[ \frac{\partial^2 T(x,y)}{\partial x^2} + \frac{\partial^2 T(x,y)}{\partial y^2} + \frac{q_v}{\lambda} =0, \]

\[ -a \leq x \leq a,\qquad -b \leq y \leq b, \]

\[ -\lambda \frac{\partial T(a,y)}{\partial x} = \alpha \left[T(a,y)-T_{\mathrm{в}}\right], \]

\[ -\lambda \frac{\partial T(x,b)}{\partial y} = \alpha \left[T(x,b)-T_{\mathrm{в}}\right], \]

Fig. 1. Cooling channel of the magnetizing winding of a betatron.

\[ \frac{\partial T(0,y)}{\partial x}=0,\qquad \frac{\partial T(x,0)}{\partial y}=0 . \]

The solution of this system is given in [1] and has the form

\[ T(x,y)=T_{\mathrm{v}}+\frac{q_v a^2}{\lambda} \left[ \frac{1}{2}\left(1-\frac{x^2}{a^2}\right)+\frac{\lambda}{a\alpha} -\frac{4a\alpha}{\lambda} \sum_{n=1}^{\infty} \frac{ \sin \psi_n \cdot \operatorname{ch}\psi_n \frac{y}{a}\cdot \cos \psi_n \frac{x}{a} }{ \psi_n^3(2\psi_n+\sin 2\psi_n) \left(\operatorname{th}\psi_n \frac{b}{a}+\frac{a\alpha}{\lambda\psi_n}\right) \operatorname{ch}\psi_n \frac{b}{a} } \right]. \]

The heat flux coming from the surface of the magnetizing winding is found on the basis of Fourier’s law:

\[ q(x)=-\lambda \frac{\partial T(x,b)}{\partial y} =4q_v a^2 \frac{\alpha}{\lambda}\sum_{n=1}^{\infty} A_n\cdot \cos \psi_n \frac{x}{a}, \]

where

\[ A_n= \frac{ \sin\psi_n \cdot \operatorname{th}\psi_n \frac{b}{a} }{ \psi_n^2(2\psi_n+\sin 2\psi_n) \left(\operatorname{th}\psi_n \frac{b}{a}+\frac{a\alpha}{\lambda\psi_n}\right) }. \]

For the case of motion of the cooling air in vertical channels, the heat-balance equation over the segment \(dx\) (Fig. 1) is written as follows:

\[ G\cdot C\cdot dT_{\mathrm{v}}=q(x)\cdot U\cdot dx, \tag{1} \]

where \(G\) is the air flow rate \([\mathrm{kg/sec}]\),
\(C\) is the heat capacity of air \([\mathrm{J}/\mathrm{kg}\cdot{}^\circ\mathrm{K}]\);
\(U\) is the perimeter of the flow cross section \([\mathrm{m}]\).

Integrating (1), we obtain

\[ T_{\mathrm{v}}(x)= \frac{U}{G\cdot C}\cdot 4q_v\cdot a^3\frac{\alpha}{\lambda} \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\psi_n}\cdot \sin \psi_n \frac{x}{a} +B . \]

Since for \(x=a\), \(T=T_{\mathrm{in}}\), it follows that

\[ B=T_{\mathrm{in}}+ \frac{U}{G\cdot C}\cdot 4q_v\cdot a^3\frac{\alpha}{\lambda} \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\psi_n}\cdot \sin \psi_n . \]

Then the temperature distribution in the flow of cooling air along the length of the channel of the magnetizing winding takes the form

\[ T_{\mathrm{v}}(x)=T_{\mathrm{in}}+ \frac{U}{G\cdot C}\cdot 4q_v a^3\frac{\alpha}{\lambda} \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\psi_n} \left(\sin \psi_n \frac{x}{a}+\sin\psi_n\right). \tag{2} \]

The temperature of the cooling air at the outlet is determined as

\[ T_{\mathrm{out}}=T_{\mathrm{v}}(a)=T_{\mathrm{in}}+ \frac{U}{G\cdot C}\cdot 8q_v\cdot a^3\frac{\alpha}{\lambda} \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\psi_n}\cdot \sin\psi_n . \tag{3} \]

Combining (2) and (3), we obtain (Fig. 2)

\[ T_{\mathrm{в}}(x)=T_{\text{in}}+\frac{T_{\text{out}}-T_{\text{in}}}{2}\cdot \frac{\displaystyle \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\psi_n}\cdot \left(\sin \psi_n \frac{x}{a}+\sin \psi_n\right)} {\displaystyle \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\psi_n}\cdot \sin \psi_n}. \tag{4} \]

The mean temperature of the air in the channel is found from the relation

\[ T_{\text{avg}}= \frac{\displaystyle \int_{-a}^{a} T_{\mathrm{в}}(x)\cdot dx}{2a} = \]

\[ = \frac{ \displaystyle \int_{-a}^{a} \left[ T_{\text{in}}+ \frac{T_{\text{out}}-T_{\text{in}}}{2}\cdot \frac{\displaystyle \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\psi_n} \left(\sin \psi_n \frac{x}{a}+\sin \psi_n\right)} {\displaystyle \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\psi_n}\sin \psi_n} \right]dx }{2a} = \]

\[ =\frac{T_{\text{in}}+T_{\text{out}}}{2}. \tag{5} \]

Fig. 2. Distribution of the cooling-air temperature \(T_{\mathrm{в}}\) along the length of the cooling channel.

To determine the pressure loss in the cooling channel, we write the differential equation of motion on the selected segment \(dx\) in the form

\[ -dP=\xi \frac{\rho W^{2}}{2}\cdot \frac{dx}{d_{\text{eqv}}}+\rho W\cdot dW. \tag{6} \]

Here \(\rho\) is the air density \([\mathrm{kg}/\mathrm{m}^{3}]\);
\(W\) is the air velocity \([\mathrm{m}/\mathrm{s}]\);
\(d_{\text{eqv}}\) is the equivalent diameter \([\mathrm{m}]\).

Multiplying (6) by \(P\) and using the continuity equation,

\[ \frac{G}{f}=W\cdot \rho=\mathrm{const}, \]

where \(f\) is the area of the flow cross section \([m^2]\), and the gas equation of state

\[ P=\rho RT, \]

we obtain

\[ -P\cdot dP=\xi\left(\frac{G}{f}\right)^2\cdot \frac{1}{2d_{\mathrm{eq}}}\,RT_{\mathrm{b}}(x)\,dx+ \left(\frac{G}{f}\right)^2 P\cdot d\left(\frac{1}{\rho}\right). \tag{7} \]

Integrating (7), taking into account (5), (4),

\[ \begin{aligned} \frac{P_{\mathrm{in}}^2-P_{\mathrm{out}}^2}{2} &=\xi\left(\frac{G}{f}\right)^2\cdot \frac{1}{2d_{\mathrm{eq}}}\,R \int_{-a}^{a} \left[ T_{\mathrm{in}}+\frac{T_{\mathrm{out}}-T_{\mathrm{in}}}{2}\times \right.\\ &\quad \left. \times \frac{ \displaystyle \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\nu_n} \left(\sin \mu_n\cdot \frac{x}{a}-\sin \mu_n\right) }{ \displaystyle \sum_{n=1}^{\infty} A_n\cdot \frac{1}{\nu_n}\cdot \sin \nu_n } \right] dx +\left(\frac{G}{f}\right)^2 \left[ RT_{\mathrm{b}}(x)\right]_{T_{\mathrm{in}}}^{T_{\mathrm{out}}} -\int_{P_{\mathrm{in}}}^{P_{\mathrm{out}}} RT_{\mathrm{avg}}\frac{dP}{P} \\ &= \left(\frac{G}{f}\right)^2 R \left[ \xi\cdot \frac{a}{d_{\mathrm{eq}}}\cdot T_{\mathrm{avg}} +(T_{\mathrm{out}}-T_{\mathrm{in}}) -T_{\mathrm{avg}}\ln\frac{P_{\mathrm{in}}}{P_{\mathrm{out}}} \right]. \end{aligned} \]

Taking into account that \(\ln \dfrac{P_{\mathrm{in}}}{P_{\mathrm{out}}}\approx 0\), we finally find

\[ P_{\mathrm{in}}-P_{\mathrm{out}}=\Delta P\simeq \left(\frac{G}{f}\right)^2\frac{1}{\rho_{\mathrm{avg}}} \left[ \xi\frac{a}{d_{\mathrm{eq}}} + \frac{T_{\mathrm{out}}-T_{\mathrm{in}}}{T_{\mathrm{avg}}} \right]. \tag{8} \]

The total aeromechanical resistance of the cooling channel of the magnetizing winding of the betatron is determined by the relation

\[ \Delta P_n=\Delta P+\Delta P_{\mathrm{in}}+\Delta P_{\mathrm{out}}, \]

where \(\Delta P_{\mathrm{in}}\) and \(\Delta P_{\mathrm{out}}\) are the local aeromechanical resistances at the inlet to and outlet from the channel. Recommendations for determining \(\Delta P_{\mathrm{in}}\) and \(\Delta P_{\mathrm{out}}\) are given in [2].

Conclusions

Equation (8) has been constructed for calculating the aeromechanical resistance of the cooling channel of the magnetizing winding of a betatron, taking into account the nonisothermal motion of air.

LITERATURE

1. N. N. Lebedev, I. P. Skalskaya, Ya. S. Uflyand. Collection of Problems in Mathematical Physics. GITTL, 1955.
2. A. A. Uginchus. Hydraulics and Hydraulic Machines. Gosenergoizdat, 1954.