Cooling of a surface due to fluid passing over it
I am working on a project that requires me to measure the cooling effect of a liquid flowing through a surface. In order for me to effectively calculate the cooling effect, the solution of the below problem will be very helpful to me.
The surface that is being cooled by a liquid is rectangular in shape, with a length $a$ and a breadth $b$. It is being cooled by a continuously flowing mass of liquid with a height $l$ flowing across the surface with a constant velocity $v$. A lot of heat is being generated below the surface, so the surface can be assumed to be at a constant temperature $T_h$ all along the rectangle. The liquid that is being pumped in to cool the surface is initially at a temperature $T_0$, and as it picks up heat from the surface, its temperature slowly increases as it flows out. 707
The thermal conductivity of the liquid is $\kappa$ and has a heat capacity per unit mass of $C$ and density of $\rho$.
I have tried solving the problem as follows:
Since the problem involves the flowing of the liquid at different time instants, I would like to instead simplify the calculations involved by considering a stationary pool of liquid with a height $l$ on the surface with temperature $T_h$ and area $A$. The liquid has a temperature of $T_0$ at $t=0$ and if I am able to solve for $T(x,t)$, where $x$ is the vertical height above the surface and $t$ is the time. I am using this as a simplification of the original problem, where the time $t$ corresponds to the position of the liquid as it flows along the rectangle. Thus, $t=\frac{a}{v}$ corresponds to the liquid that is flowing out at the end of the surface. I know that this is an approximation (as this calculation will neglect the heat conduction through the liquid, along the length of the rectangle), but since the liquid in my case moves with a good speed $v$, and has a small thickness $l$, so I feel that I am justified in making the approximation.
Now, for my solution:
(Note: I am taking $x$ to be the distance along the vertical, ie., along the height of the liquid layer, $t$ to denote the time since the initial condition, and Q(x,t) to be the rate of flow of heat through a cross section of the liquid)
The conduction equation: \begin{equation} \kappa A \frac{\partial T}{\partial x}= Q(x,t) \end{equation}
Now, for accounting for the absorption of heat by the differential layers of the liquid: \begin{equation} dQ=-\rho AC\frac{\partial T}{\partial t}dx \end{equation}
Since all the heat current passing through a given cross section is used up in heating the layers above it, \begin{equation} \int_{x}^{l}\rho A C \frac{\partial T}{\partial t}= \kappa A \frac{\partial T}{\partial x} \end{equation}
Taking the partial derivative of the above equation wrt $x$, \begin{equation} -\rho C \frac{\partial T}{\partial t} = \kappa \frac{\partial ^2 T}{\partial x^2} \end{equation}
After this, I assume that the temperature is of the form $T_1 (x)T_2(t)$ (Is this assumption wrong? If so, can someone please help me solve the differential equation?)
Using the above, I get: \begin{equation} -\rho C \frac{1}{T_2 (t)} \frac{\partial T_2 (t)}{\partial t} = \kappa \frac{1}{T_1 (x)} \frac{\partial^2 T_1 (x)}{\partial x^2} \end{equation}
Since they are functions of independent variables being equal to each other, they both must be a constant ($=\lambda$).
Solving the above, I get: \begin{equation} T(x,t)=Ae^{\sqrt{\frac{\lambda}{\kappa}} x - \frac{\lambda t}{\rho c}} \end{equation} ($A$ is just a constant)
and the boundary condition is that $T(x,0)=T_0$ for all $x$. However, from the form of the solution I have obtained, clearly such a boundary condition is not possible. 707