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stockzahn

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Dear all,

I would need mathematical help to solve for the temperature field in an annular geometry (you find a picture attached below the text):

A copper pipe containing a boiling two-phase flow (in the stratified regime) is immersed in a liquid bath, which temperature ##T_{IY}## is assumed to be constant in the entire cross section. The heat transfer coefficient between the bath and the outer surface of the copper pipe(##h_{out}##) is constant.

The liquid phase and the gas phase of the inside flow are in thermodynamic equilibrium (##T_{liq}=T_{gas}=T_{BHX}##). The heat transfer coefficient between the copper pipe and the gas ##h_{Dry}## is much lower than the heat transfer coefficient between the copper pipe and the liquid ##h_{Wet}##.

The geometry is symmetric with respect to the vertical axis, the liquid level can be expressed as the angle ##\varphi_{LL}##. ##R_{in}## and ##R_{out}## denote the inner and the outer radii of the copper pipe respectively.

2-dimensional Laplace equation for polar coordinates:

$$\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial^2 T}{\partial \varphi^2} = 0 \quad for\;R_{in}\leq r\leq R_{out},\;0\leq \varphi\leq 2\pi$$

With separation of the variables I obtained two correlations:

$$ g''\left(\varphi\right) + \lambda g\left(\varphi\right) =0$$

$$ r^2 f''\left(r\right) + r f'\left(r\right)-\lambda f\left(r\right) =0$$

Each linear combination of the solutions of the previous equations can be a solution, which generally reads

$$ T\left(r,\varphi\right)=c_0 + d_0 ln\left(r\right)+\sum\left(c_n r^{-n}+d_n r^{n}\right)\left(a_n cos\left(n\varphi\right)+b_n sin\left(n\varphi\right)\right) $$

Now, there are several boundary conditions for the first derivative in radial direction (##k## denotes the (constant) heat conductivity of the copper):

$$\frac{\partial T}{\partial r}\bigg{|}_{r=R_{out}}=\frac{h_{out}}{k} \left(T_{IY}-T\right)\quad for\;0\leq\varphi\leq

\varphi_{LL}$$

$$\frac{\partial T}{\partial r}\bigg{|}_{r=R_{out}}=\frac{h_{out}}{k} \left(T_{IY}-T\right)\quad for\;\varphi_{LL}\leq\varphi\leq \pi$$

$$\frac{\partial T}{\partial r}\bigg{|}_{r=R_{in}}=\frac{h_{Dry}}{k}\ \left(T-T_{BHX}\right)\quad for\;0\leq\varphi\leq \varphi_{LL} $$

$$\frac{\partial T}{\partial r}\bigg{|}_{r=R_{in}}=\frac{h_{Wet}}{k} \left(T-T_{BHX}\right)\quad for\;\varphi_{LL}\leq\varphi\leq \pi $$

Due to the symmetry, also for the first derivative in tangential direction two boundary conditions can be found:

$$ \frac{\partial T}{\partial \varphi}\bigg{|}_{\varphi=0}=0 $$

$$ \frac{\partial T}{\partial \varphi}\bigg{|}_{\varphi=\pi}=0 $$

Finally, at the position ##\varphi=\varphi_{LL}##, two transitional conditions can be set up:

$$T_{Dry}{\left(\varphi= \varphi_{LL}\right)}=T_{Wet}{\left(\varphi= \varphi_{LL}\right)}$$

$$ \frac{\partial T_{Dry}}{\partial \varphi}\bigg{|}_{\varphi=\varphi_{LL}}=\frac{\partial T_{Wet}}{\partial \varphi}\bigg{|}_{\varphi=\varphi_{LL}}$$

All together I have eight boundary conditions, but twelve unknown constants:

$$\underbrace{c_{0},d_{0},a_{n},b_{n},c_{n},d_{n}}_{\cdot 2\;for\;the\;wetted\;perimeter\;and\;the\;dry\;perimeter}$$

Unfortunately I'm not even sure, if this could be a way to solve the problem or if I made elementary mistakes. I hope that I used the mathematical symbols and notations correctly and gave any necessary information. Any help is appreciated.

Thanks in advance,

stockzahn

I would need mathematical help to solve for the temperature field in an annular geometry (you find a picture attached below the text):

A copper pipe containing a boiling two-phase flow (in the stratified regime) is immersed in a liquid bath, which temperature ##T_{IY}## is assumed to be constant in the entire cross section. The heat transfer coefficient between the bath and the outer surface of the copper pipe(##h_{out}##) is constant.

The liquid phase and the gas phase of the inside flow are in thermodynamic equilibrium (##T_{liq}=T_{gas}=T_{BHX}##). The heat transfer coefficient between the copper pipe and the gas ##h_{Dry}## is much lower than the heat transfer coefficient between the copper pipe and the liquid ##h_{Wet}##.

The geometry is symmetric with respect to the vertical axis, the liquid level can be expressed as the angle ##\varphi_{LL}##. ##R_{in}## and ##R_{out}## denote the inner and the outer radii of the copper pipe respectively.

2-dimensional Laplace equation for polar coordinates:

$$\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial^2 T}{\partial \varphi^2} = 0 \quad for\;R_{in}\leq r\leq R_{out},\;0\leq \varphi\leq 2\pi$$

With separation of the variables I obtained two correlations:

$$ g''\left(\varphi\right) + \lambda g\left(\varphi\right) =0$$

$$ r^2 f''\left(r\right) + r f'\left(r\right)-\lambda f\left(r\right) =0$$

Each linear combination of the solutions of the previous equations can be a solution, which generally reads

$$ T\left(r,\varphi\right)=c_0 + d_0 ln\left(r\right)+\sum\left(c_n r^{-n}+d_n r^{n}\right)\left(a_n cos\left(n\varphi\right)+b_n sin\left(n\varphi\right)\right) $$

Now, there are several boundary conditions for the first derivative in radial direction (##k## denotes the (constant) heat conductivity of the copper):

$$\frac{\partial T}{\partial r}\bigg{|}_{r=R_{out}}=\frac{h_{out}}{k} \left(T_{IY}-T\right)\quad for\;0\leq\varphi\leq

\varphi_{LL}$$

$$\frac{\partial T}{\partial r}\bigg{|}_{r=R_{out}}=\frac{h_{out}}{k} \left(T_{IY}-T\right)\quad for\;\varphi_{LL}\leq\varphi\leq \pi$$

$$\frac{\partial T}{\partial r}\bigg{|}_{r=R_{in}}=\frac{h_{Dry}}{k}\ \left(T-T_{BHX}\right)\quad for\;0\leq\varphi\leq \varphi_{LL} $$

$$\frac{\partial T}{\partial r}\bigg{|}_{r=R_{in}}=\frac{h_{Wet}}{k} \left(T-T_{BHX}\right)\quad for\;\varphi_{LL}\leq\varphi\leq \pi $$

Due to the symmetry, also for the first derivative in tangential direction two boundary conditions can be found:

$$ \frac{\partial T}{\partial \varphi}\bigg{|}_{\varphi=0}=0 $$

$$ \frac{\partial T}{\partial \varphi}\bigg{|}_{\varphi=\pi}=0 $$

Finally, at the position ##\varphi=\varphi_{LL}##, two transitional conditions can be set up:

$$T_{Dry}{\left(\varphi= \varphi_{LL}\right)}=T_{Wet}{\left(\varphi= \varphi_{LL}\right)}$$

$$ \frac{\partial T_{Dry}}{\partial \varphi}\bigg{|}_{\varphi=\varphi_{LL}}=\frac{\partial T_{Wet}}{\partial \varphi}\bigg{|}_{\varphi=\varphi_{LL}}$$

All together I have eight boundary conditions, but twelve unknown constants:

$$\underbrace{c_{0},d_{0},a_{n},b_{n},c_{n},d_{n}}_{\cdot 2\;for\;the\;wetted\;perimeter\;and\;the\;dry\;perimeter}$$

Unfortunately I'm not even sure, if this could be a way to solve the problem or if I made elementary mistakes. I hope that I used the mathematical symbols and notations correctly and gave any necessary information. Any help is appreciated.

Thanks in advance,

stockzahn

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