# Tag Info

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the outside air is colder, but the flame temperature is still much much hotter than that so the difference cold inlet air makes on the outlet temperature of the furnace will be small. The same commentary applies to the density argument: yes, but the effect is small. The fact that the outlet air is at 43 C represents the 8% efficiency loss of your furnace. ...

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Split system air conditions (heat pumps) are refrigerator based systems. The two tubes you mention transport refrigerant from the compressor in the outdoor to unit to the indoor unit & return it from the indoor unit to the compressor in the outdoor unit, in a closed loop. Water is removed from the atmosphere by condensation, by the indoor unit. This ...

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A similar graph from a another discipline is the following Basically you will need to plot for each test (which I assume had a different d/S), the values of your experiment. So for each d/S you calculated different values of the convective heat coefficient at different air speeds. So you need to plot 10 different graphs.

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Brakes primarily convert kinetic energy to heat energy. So a large area can absorb more heat lowering the peak temperatures ;of course this is strongly affected by the thickness/mass of the discs and other factors. AND the larger area can get rid or more heat . High temperatures cause deterioration of pad materials ,so lower ( not as high) temperatures ...

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I don't think it will drop the temperature at all. What it might accomplish is increase the velocity of the air in the room. That will in turn affect the convective coefficient The convective heat transfer coefficient for air flow can be approximated to (see engineering toolbox link) $$h_c = 10.45 - v + 10 v^{1/2}$$ where $h_c$ = heat transfer coefficient ($... 4 Most thermometric cooling modules (Peltier devices) are made out of Bismuth Telluride which has a thermal conductivity of 1.20 W/(m·K), similar to ordinary glass. From this wiki list of thermal conductivities: Copper, 401 W/(m·K) Aluminum, 237 W/(m·K) So for a given temperature differential, a thermometric cooling module in the off state would conduct 334 ... 3 I expect the same material as the coils would be the easiest answer ; nichrome ,Inconel ( 600, 601 or other number) , chromel. There are small differences but any one you can find would be good. That would also be practical as it would be readily available to the manufacturer. Stainless like 304 will probably work but I am uncertain of the many cycles into ... 3 IMHO the problem you observe is mainly related to limitations of your system. Notice, that up until 55 degrees C the heating element can follow the gradient of temperature wrt to time. Just after 55 degrees C, the system is unable to heat fast enough the wooden box. (probably the insulation is not sufficient). So beyond that point, the error accumulated and ... 3 Could you get it to work, sure. There are dozens of ways to turn heat into energy. Could you get it to work at scale and cost effectively? Almost certainly not. There's just not enough energy in heated air to do much, and the capital costs of the system break the effort. This is another entry into the "Why can't Stirling engines give us almost free ... 2 Generally I would recommend against heating a plastic container with a resistance heater (like a hot plate or mug warmer) unless you know for a fact it has a hardware shutoff that will kick in well below the melting point. Hardware failures do happen, and I've personally seen a runaway hot plate almost melt an unattended plastic container (full of alcohol!).... 2 If you are worried about the solar radiation (which is understandable), a much more affordable solution might be to put the whole thing in a shade if its possible, with enough clearance from the actual shading material. If you stick solar shields on the box, they are bound to get hot and, you won't be able to avoid conductive or convective heat transfer. ... 2 In general, either grating configuration should permit flow equally as well provided the slot diameters are the same, and the ratio of open area is the same. By fins in the acrylic, I assume you mean slots correct? Aluminum fins on the heat sink, slots in the acrylic housing, the acrylic material between the laser cut slots are members (slang for structural ... 2 Do this: set up the water flow plumbing with the pump OFF, turn on the water flow, and measure the flow rate by running the discharge into a bucket of known volume for a known period of time, and measure the source pressure. If the dents are restricting the flow, then 40PSI (typical for residential water supply pressure) will not result in a 0.15 gallon a ... 2 Mineral oil is not used to cool transformers because it is flammable. Transformer oil is a different compound which is not flammable and possesses high dielectric strength. In any case, the temperatures inside a welding arc are high enough to decompose almost all chemical compounds, as Blacksmith37 points out. 2 No. In water welding you make steam (> hydrogen and oxygen) and conditions need to compensate for the problem of hydrogen entry in the steel. With oil you are adding the large problem of also adding carbon to the steel. Plus the oil will be cracking and producing a whole range of smaller molecules and probably some carbon and very heavy tar-like materials.... 2 If this is a student design problem to see whether you understand the principles, make it as easy as you can, use steam and cooling water. Calculating all the pressure drops on the weird pipe systems and valves you need when using one stream to cool/heat the others will drive you insane. Not to mention what it can do to your choice of materials of ... 2 Consider that ohmic heaters are usually made of thin wire, and that the majority of the current flow going through any wire is close to the surface of the wire. Both these things mean that even if the wire itself might be a poor conductor of heat, that heat has only a very short distance to travel before it reaches the wire surface and gets carried away by ... 2 You are missing several things. You need to know the emissivity of marshmallow. That will tell you how much of the 150W of radiation is absorbed, and how much is reflected. You need to know the latent heat of fusion, which tells you how much heat is needed to change the marshmallow from "solid" to "liquid". That may not be a very well ... 2 For maximum power efficiency, you want to match the load impedance (i.e., the heating element resistance) to the source impedance (i.e., the resistance of everything else in the circuit, including the power supply). This is the maximum power transfer theorem. (Volumetric heat generation is$J^2\rho$, where$J$is the current density and$\rho$is the ... 2 Background Normally the process would go line this. Let's say the primary is hot, and the secondary is the cooler fluid. The heat transfer rate$\dot{Q} = -\dot{Q}_p = \dot{Q}_s $. I.e.: The cooler fluid (s) gains $$\dot{Q} = m_s\cdot C_{p,s}(T_{s,o}- T_{s,i})$$ The hot fluid (p) loses: $$\dot{Q} = - m_p\cdot C_{p,p}(T_{p,o}- T_{p,i})$$ Therefore the ... 2 Did you ever get a real answer to your question? I have been considering this question for years and I have no real answer. Many responses I see show me that people are a bit confused as to what you are trying to do. In a simplistic explanation you are trying to move thermal energy from a 70 degree environment into a say 75 degree environment which is very ... 2 There are a lot of reasons, but we focus more on just heat transfer. To bring a car to stop the brakes have to convert the kinetic energy of the moving car to heat on the disks and pads by friction and then transfer it to the air. The disk heat is released into the ambient air by conduction, convection, radiation. In the case of disks, direct conduction is a ... 2 Conductive The Conductive heat transfer is given by: $$\dot{Q} = \frac{k}{L} A \Delta T$$ where:$k$= is the heat conductivity of the material in this case aluminimum ($\frac{kCal}{m°C}$)$L$is the thickness of the wall A is the total exchange surface$\Delta T$the temperature difference Convective heat transfer Convective heat transfer is when a solid ... 2 Short answer , no. They are pretty much at constant temperature other than a couple minutes at start-up or change of power setting. One exception is the few highest temperature turbine blades ( in some engines ) that have a few axial holes through them for cooling air to pass through ; but again they are at constant temperature during operation. Although ... 2 How do you measure the temperature? If you use an infrared thermometer, this won‘t work because of reflection. Use a thermistor based measurement device for verification. 2 There are a lot of misconceptions in your question. Let's first start by the Carnot cycle. Given a two reservoirs at temperatures$T_H$(hot reservoir) and$T_L$(cold reservoir) where$T_H > T_L$, what is the upper limit of the mechanical work the can be generated if we added a heat engine between the two reservoirs? or in other words, what is the ... 2 An evaporative cooler works by removing the suspended layer of saturated vapor from around the filter, encouraging it to use the ambient heat from the stream of the air thus making it colder, to evaporate to replace the washed-out moisture. The latent heat for water is 2460 J/4.18 J per gram = 588.5 times more than specific heat. (Thanks To Phil Sweet for ... 1 Background Total energy lost from fluid The heat transfer rate$\dot{Q} $if you know for a pipe with fluid$f$(where f: Red, Yellow, Blue) the temperature at input and output, the mass rate, and the heat capacity of the material is given by: $$\dot{Q}_f = \dot{m}_f\cdot C_{p,f}(T_{f,o}- T_{f,i})$$ So for the Red-yellow pipe you have: $$\dot{Q}_{ry} = -\... 1 Edit to add: The OP states the applicaiton is a heat pipe, so this answer assumed the yellow fluid is a liquid and moves by free convection. It turns out the yellow fluid will undergo a phase change, which changes a lot. This is an outline for a solution that should get you going: Step one - find an equation describing the convection of the yellow medium. \... 1$$\text{Absolute maximum efficiency} = \text{Carnot Efficiency} = \frac{T_{hot}-T_{cold}}{T_{hot}}$\$ where all temperatures are expressed in degrees Kelvin (= degrees C + 273). A very well built and designed engine with high temperature differentials, high pressure and using Hydrogen (not at all recommended) or Helium working gas might achieve 10% - 50%+ of ...

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