37

Trees increase the turbulence of the air that reaches the turbines. That creates all sorts of uneven, rapidly-shifting loads on the blades and structure. That increases the maintenance costs, decreases availability, decreases the capacity factor, and decreases the life expectancy of the turbine. So, higher costs, lower revenue. One of the ways we measure ...


11

If you can replicate what you report above under controlled conditions then you can retire for life next year. Really. The overwhelming opinion by all reputable test houses or suitably competent careful and honest individuals is that for combustion improvement by treating the **fuel* with magnets you'd be just as well off rubbing on snake oil - or drinking ...


11

There are a lot of questions in your question, and it should probably be broken up into multiple different questions. I don't want to wait until that happens though so I'll address the ones that I know the answers to. How is heat converted to an electrical signal (current or voltage)? A microbolometer is just a special case of a bolometer which ...


10

In a similar vein to my answer about calculating the lever force in a continuous situation; you need to use integration. You start by taking the standard heat law that you are familiar with $$ \Delta Q=c\ m\ \Delta T $$ and replacing the $\Delta$s with differentials: $$ dQ=c(T)\ m\ dT. $$ This new equation reads: For an infinitesimal (very tiny) change in ...


10

Yes, there are indeed combined PV-T (photovoltaic-thermal) hybrid panels that turn some of the incident light into electricity, and have a circulating fluid and heat exchanger to put some of the heat into a heat store. There's very little take-up of them, because they're just not economic in most circumstances. The theory looks great: PV panels are more ...


9

I am guessing, you are solving the following system: $$\partial_t T = \alpha_i \partial_x^2 T$$ with $\alpha_i=\kappa_i/\rho_i c_{p,i}$ for material $i$, subject to the conditions: $$T\left(x,0\right)=T_\infty\quad T\left(0,t\right) = T_h \quad -\kappa\partial_xT\left(L,t\right)=h\left[T\left(L,t\right)-T_\infty\right]$$ Now you can approach simulating ...


8

Illuminate the room indirectly via dichroic reflectors at 45 degrees to each transparent wall. These will reflect light at 90 degrees (i.e. into the room) but allow the heat to pass straight through. (NB you may want watercooled heatsinks behind the reflectors, or some other adequate arrangement to dispose of the heat.)


7

Considering the three methods of moving heat: Convection - To keep the IP66 rating of the enclosure, you can't add any holes for exhaust fans. Radiation - At the temperatures that you are talking about, radiation will not be removing much heat. Conduction - This is a viable alternative since it could work through the walls of the enclosure. You can add ...


7

A simple approach would be to just consider the heat added. The specific heat equation gives the temperature change due to heating as $$Q=mc\Delta T$$ where $Q$ is heat added, $m$ is mass. $c$ is the specific heat of the material in question (~1.005kJ/kgK for air at room temperature) and $\Delta T$ is the temperature change. A 5x5x5 cm box will have an ...


7

If a process is both isothermal and adiabatic, it is implied that the work done on the system is being stored somewhere other than the internal energy of the working fluid. (Or conversely, if the system is doing work, the energy is coming from somewhere other than internal energy.) The classic example of such a process is free expansion of an ideal gas (...


7

The efficiency of any radiator (heat exchanger) is a function of the temperature difference between the two fluids in question. All else being equal, a heat exchanger with a greater temperature differential will transfer more heat. Each radiator will have a temperature gradient across it. (Here I'm talking about how much the temperature of each fluid ...


7

Mainly cost difference. And also, a fan is sufficient for the job. The problem being solved is thermal stratification. That's caused by a lack of vertical mixing; the warm air rises to the top, the cold air sinks. Normally, there's quite a bit of movement in the first two metres or so from the ground - thermal stratification starts when the ceiling gets to ...


7

Because they're two completely different systems that have, in some circles, accidentally been given the same names. The things that are sometimes (particularly in the USA) called home geothermal systems are ground-source heat pumps. The energy for these comes from sunlight on the ground over the year. Electricity is used to move heat from a source ...


7

They are very nearly equal for typical four-stroke non-turbo diesels under load. A turbo diesel under load should have slightly more radiator loss than exhaust loss. At the bottom is a link to the technical spec sheet for a Cat 3412 powered genset. It's a probably a bit bigger than what you had in mind. It is a turbo with aftercooler (A/C in the doc below). ...


6

This will depend very much on the heat flow around the enclosure. There is a thing called the H factor which describes the heat transport properties between a surface and a fluid. The value of H varies with surface properties and flow. So to do your calculation you need a few different assumptions. Let's simplify. 1) the air in the box is uniformly heated ...


6

Neither. In this sort of situation, there's no "simple" linear solution; you need to use integral calculus to add up the incremental heat absorbed at each temperature along the way. The only time that this calculation becomes a simple multiplication is when the quantity being integrated (the specific heat) is a constant over the range of the integration.


6

Those two holes by themselves will have little effect. The pressure difference between the two ends of the holes will be so small that only a little air will flow. If you want to go thru all this trouble, then put a hole at the bottom too, with a fan that actively blows cold air from the bottom of the living room into the bedroom. Hot air will then find ...


6

Heat pipes are very useful when you have limited space to work with or you want an entirely sealed and self contained unit. An interesting example of this is that heat pipes are used to cool the legs of oil pipelines which run through permafrost areas, the cooling is required to stop the heat generate by friction in the pipe and solar heating from ...


6

Orientation does often matter. As Carl's answer mentions, the liquid can get from the condenser to the hot interface via capillary action, but most common heat pipes are designed assuming gravity will do the job. Capillary action is much more effective in space where there is no gravity, but produces very little flow when it has to work against gravity. ...


5

If you get complete combustion of any hydrocarbon fuel, your products will be primarily $CO_2$ and $H_2O$. Other products will be present in quantities too small to be relevant to the heat losses that you're looking at. These two gases will change the heat capacity, conductivity and density of the flame and your exhaust gases. So heat heat transfer ...


5

Assumptions: The copper side with the traces is modeled as a sheet of copper rather than traces. The body is thin enough that thermal conductivity within the body is unimportant, and the entire device is considered to be at a uniform temperature. Only the two broad surfaces contribute to the heat loss, the sides are neglected. The surroundings, including ...


5

First, to counter what some people are saying about printed metals, selective laser sintering can produce parts that are as thermally conductive as their base metals. They are limited only by porosity, and state of the art machines can produce fully-dense (zero porosity) metal parts. Second, you probably don't need to take radiation into account because ...


5

Neither. As has already been pointed out, this is not trivial to do, but here is a suggested method: accurately measure out a certain quantity of fuel, then burn that fuel and use a material with a very constant or otherwise well known specific heat capacity to determine how much energy your test piece is receiving through time by recording it's temperature....


5

What you want is called a heat exchanger. Imagine two long air tubes with a thin wall between them. The air exiting the house travels in one tube, and entering air in the other, but in opposite directions. Over the length of the tubes, heat transfers thru the thin wall. Ideally, by the time the house air gets to the far end where the outside air comes in,...


5

If you are interested in knowing why your tea is cooling in different times, you already answered the question yourself. It is due to the different thermal conductivities of the different items you use it to store. $\dot{Q} = A U \Delta T \tag{1}$ is the heat flux you want to know. This gives you a relationship between the heat that is transfered to the ...


5

There is a non-engineering reason as well: the people most open to placing turbines on their land are farmers who are interested in the extra income. Turbines are not (yet) placed near built-up areas, which generally leaves large farms, undeveloped lots and nature reserves. Getting a building permit for a commercial enterprise in a nature reserve is close ...


5

I have done a lot of work with small sterling heat engines. While decreasing the size generally does have a favorable effect on your heat transfer, your energy density doesn't increase substantially over the feasible range. Other issues come into play, like being able to fabricate an engine that small while keeping internal friction low. Also your cylinder ...


5

Direct sunlight increases power usage and wear of the unit. The cooling efficiency depends directly on temperature gradient - difference between temperature of the coolant and ambient air pushed through it. With direct sunlight the chassis heats up, the protective mesh on the fan heats up, surrounding walls and floors heat up from sunlight, the air passes, ...


5

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. ...


5

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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