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The Richardson number $\mathrm{Ri} = \frac{\mathrm{Gr}}{\mathrm{Re}^2} = \frac{\text{buoyancy term}}{\text{flow gradient term}}$ (1) can be used to determine if forced or natural convenction dominate in thermal convection. $\mathrm{Re}$ = Reynolds number $\mathrm{Gr}$ = Grashof number You can neglect natural convection for $\mathrm{Ri} < 0.1$ and ...

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As alephzero mentioned the difference is huge. I will explain mathematically and then physically. Mathematically, For any given partial differential equation of the form, $$A \frac {\partial^2\phi}{ \partial x^2} + B \frac {\partial^2\phi}{ \partial x \partial y} + C \frac {\partial^2\phi}{ \partial y^2} + \rm first\ derivative \ terms\ =0$$ We get, $... 3 You have the right general approach, and the numbers look reasonable. Your airspeed isn't very fast, so I'm not surprised that the convection coefficient you calculate is less than 10 W m-2 K-1. Note, however, that you're applying the heat flux you calculated at a plate temperature of 288 K for all other temperatures as well. In reality, the convective flux ... 3 From "Heat Transfer, 2nd Edition" by A.F. Mills, equation 4.51 gives a formula for the average Nusselt number for flow between two parallel plates. This might be appropriate for your duct because you can probably ignore the sides and just treat it like two parallel plates: $$\overline{Nu} = 7.54 + \frac{0.03\frac{D_H}{L}Re\cdot Pr}{1+0.016 \left[ \frac{... 2 If your temperature sensor is far away from the heat source, the detected temperature may increase even after the heat gun is turned off due to a preexisting temperature gradient evening itself out. 2 You certainly need to take account of the shape of the object somehow. As a simple example, consider a sphere of radius r in space where everything except solar radiation can be ignored. The cross section area of the sphere which is absorbing solar radiation is \pi r^2 but the surface area which is emitting it is 4 \pi r^2. So the radiative heat ... 1 Nothing will change beyond L3. As you increase the length of the heat exchanger, the hot will reach the cool temperature, and beyond that point there will be no change. So beyond L3, the temperature of the cool flow and the hot flow will be equal to the cool flow inlet temperature t_{c,i}. 1 You can buy an incense stick, which burns slowly and produces perfumed smoke. you then position the incense stick at different points in the sauna and make a video of the resulting smoke trails. 1 This NASA article discusses some of the common wind tunnel flow visualization methods. A fog machine generally uses a heated mixture of polypropylene glycol and water. You can test different mixture concentrations in an electric kettle, and use a plastic tube to plumb the fog stream to the locations you are interested in visualizing the flow. Tufts or pieces ... 1 The flux of a quantity is by definition the rate per time per area of a given quantity normal to the area in question. If you use the original direction of the particles and this direction is not normal to the surface in question, then you are not evaluating the flux through the surface. To use the original particle direction, you would need to use a surface ... 1 Stand at any center point along the axis of the pipe where the temperature is T. Assume that you can obtain h_i, the internal convection heat transfer coefficient, k the thermal conductivity of the pipe with thickness w, and h_o the external convection heat transfer coefficient. Assume an external temperature of T_\infty. The heat transfer rate \... 1 For modelling the rate of air flow through the opening, empirical relationships in the ASHRAE Fundamentals Handbook seem relevant here. Volumetric flowrate Equation 37 on page 25.13 of the 1997 version, section "Flow Caused by Thermal Forces" may be useful for calculating flow rate:$$Q=C_D \cdot A \cdot \sqrt{2 \cdot g \cdot \Delta H_{NPL} \cdot (T_i - ... 1 There are several issues going on that I believe are making the model not very accurate. They all in some way relate to your question about the convection: Breaking up the flow into 5 discrete chunks is a good start, but probably not a sufficient number. Keep increasing the number of nodes in your model until the answer stops changing. The next thing to ... 1 Assuming incompressibility you can work with volume-flow to calculate (a first guess) the velocity,$u$with the following equations. Probably you have some volume or mass$\dot{m}$flow entring at (A). This mass-flow is entering the box through a opeing with a certain cross-section$A_\mathrm{A}$. The relation is:$u_\mathrm{A} = \frac{\dot{m}}{\rho\, A_\...

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Some solar water heating systems use a tube with a fluid that evaporates / condenses which moves the heat from end to end as you describe. They are designed to work preferentially in one direction as one end has a bulb or larger diameter to better transfer the heat. Will post an image when on the laptop, but, for the moment here is a link : http://www....

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Okay, I'm not 100% sure if I got the question right but, to combine natural and forced convection for flow over a cylinder (I assumed horizontal) you have the following formula: $$Nu_{comb}=(Nu_{free}^n+Nu_{forced}^n)^{1/n}\tag{1}$$ where $n$ depends on your cylinder type (horizontal or vertical). Next you need to calculate the Nu for each regime (free and ...

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You should be fine to neglect kinetic energy from a thermal perspective. The kinetic energy per unit volume will be $1/2 \rho u^2$, which is about 2800 $J/m^3$ for your conditions. Given that the volumetric heat capacity is ~ 1200 $J/m^3.K$ The most you'd get out of your KE is a few degrees K (if you're lucky), due to viscosity, rather than compression. ...

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