# Air fan problem - fluid dynamic

For preparing an exam me and my classmates are unable to solve this question. Can anybody help us. Thanks.

At a conference my colleagues had a room without air conditioning. At the reception they could get a fan to cool down the warm room.

(a) Assume that the open part of the window is a vertical opening not much wider than the fan, but much higher in height. Where would it be better, for my colleagues to put the fan, at the bottom of the opening, in the middle or at the top? Why? What physical limitation do you have to make on the blowing speed of the fan to get to physically relevant results?

(b) The flow rate of the inflow caused by the fan is Q and the inflow air has a temperature Ti , and the volume of the chamber is Vk. Assume (as opposed to question a) that the outflow from the window at any time is the average temperature of the chamber T(t) (and thus there is a very efficient mixing of the air). Set up a differential equation for T(t) and calculate how long it takes for the room to cool from a high initial temperature $$T_0$$ to $$T = (T_0 + T_i)/2$$\

(c) The cooling effect of a fan is caused by the layer of air around the skin (heated by thermal conduction) being replaced by cool air of the passing air flow of the fan. We can write the influence of the thermal conduction with the help of the P´eclet number, which takes the ratio of the advection transport of thermal energy and the conductive transport. Calculate this dimensionless number (analog of the Reynolds number) using the equation for temperature transport: $$\frac{dT}{dt}=k \nabla^2T$$ where κ is the thermal conduction coefficient

• I think this is a question for physics.stackexchange.com Jan 15, 2021 at 21:36
• There is a typo in part (c) : Peclet number is mistakenly written P' eclet number. This makes the problem harder to understand for non-physicists. Jan 18, 2021 at 12:30

We are assuming air flow to be incompressible in this example. Therefore, the flow rates of inlet and exit are equal ($$Q$$). In the rest of this post we assume that $$Q$$ is volume flow rate (and not mass flow rate). In an infinitesimally short duration, $$dt$$ , volume of in-coming air is $$Qdt$$ . This air, with temperature $$T_i$$ mixes with the air in the rest of the room (with volume $$V-Qdt$$) at temperature $$T(t)$$ . As a result, the temperature of the room changes by $$dT$$. The temperature of the mixture is a weighted average of the temperatures of the two volumes of air, with weights being proportional to their respective volumes. $$T(t)+dT = \frac{Qdt}{V}T_i + \frac{V-Qdt}{V}T(t)$$ Rearranging the terms, we will have: $$\frac{dT}{dt} + \frac{Q}{V}T(t) = \frac{Q}{V}T_i \qquad(1)$$ This is a first-order linear ordinary differential equation (ODE) that is not homogeneous. From the text of the problem I assume you are familiar with ODEs, so I will suffice to a hint for solving equation (1). You must find the general solution of (1) and then apply the initial condition $$T(0) = T_0$$ . The general solution is the sum of the solution of the homogeneous equation (in which the right hand side is zero) and a particular solution of the non-homogenous equation. The solution of the homogeneous equation $$y'+ry=0$$ is $$y_h=e^{-rx}$$. To find a particular solution to the non-homogeneous equation $$y'+ry=A$$ , try $$y_p=y_h+B$$ and use it in the equation to find $$B$$. The general solution will be $$y_g=cy_h+y_p$$ , wherein $$c$$ is a constant that should satisfy the initial condition.
• how do you become (1)? Because: $T(t)+dT=(Qdt T_{i}/V)+(V-Qdt)T(t)/V)$ gives $dT/dt+1=QT_{i}/V+T/dt-QT/V$ Jan 16, 2021 at 13:05