# Tag Info

34

Electricity production optimization is a very complex subject. It is also affected by many parameters , which I will try to outline below. TL;DR: Demand is constantly monitored and supply is constantly adjusted TL;DR 2: Lshaver's excellent post is a suggested reading after reading this, because it expands and explains what happens at timescales ranging from ...

18

Timescales on the grid Power demand fluctations can be broken into timescales from micro-seconds to decades. On the "decades" end of the scale, the power industry and utility regulators work together to plan and fund construction of power plants and the associated transmission and distribution infrastructure. When you turn your air conditioner off ...

15

Your question sort of has two parts: How to supply heat, and how to keep it in. Large open rooms with a high ceilings are most efficiently warmed with radiant ceiling heat. Warm air rises, which renders forced-air systems inefficient because the pumped heat ends up at the ceiling and the coldest part of the room is near the floor where you actually want ...

14

According to Wikipedia: It [Sabatier reaction ] involves the reaction of hydrogen with carbon dioxide at elevated temperatures (optimally 300–400 °C) and pressures in the presence of a nickel catalyst to produce methane and water. $$CO_2 + 4H_2 \rightarrow CH_4 + H_2O + \text{Energy}$$ as you can see from the previous chemical reaction, to obtain 1 ...

11

I think you're overcomplicating things. To push 85 kg up a 15% slope against gravity of 9.8 m/s2 requires a force of $$\sin (\arctan (0.15) ) = 0.1483 \approx 0.15$$ $$F = 85 kg \cdot 9.8 \frac{m}{s^2} \cdot 0.15 = 125 N$$ With an 80 mm wheel, this requires a torque of $$T = 125 N \cdot 0.04 m = 5 Nm$$ To do this at a forward speed of 11.11 m/s ...

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, they monitor demand and due to differences in how different power stations are controlled they can increase or reduce output to match demand. Some power plants like nuclear run at full output as they are slow to change, but others like Dinorwig (in the UK) can go from standby to max output in 12 seconds and reduce to half quickly. Danish researchers did ...

9

It's not a simple relationship. First let's deal with the kinetic energy of the wind passing through the rotor. The mass of air passing through it in one second, $m$, is equal to the the density of the air ($\rho$), times the surface area of the rotor (${\pi}r^2$), times the velocity of the air ($v$). i.e. ${\rho}{\pi}r^2v$. The kinetic energy of that air ...

8

Another option to the radiant solution above would be under floor air distribution systems (UFAD) using displacement ventilation. These systems require a raised floor plenum which works well in theatre style auditoriums, where you often have this anyway. The plenum essentially becomes your supply duct. The systems are sized to condition the occupied zone ...

7

One of the problems that plagued older rechargeable batteries (e.g. Nickel Cadmium ($\text{NiCad}$) and Nickel Metal Hydride ($\text{NiMH}$)) was the memory effect. The memory effect occurs when a rechargeable battery is not fully discharged. It then "forgets" that it has a greater capacity than it thinks it has, and so in the future it discharges less. A ...

7

Flywheels are used in space, but not for storage of energy. They are used for attitude adjustment, as the spinning of a large mass has significant implications in an object which is free to move. Unlike on Earth, where a flywheel or gyro can be spun up against the force holding it in place (bolted to foundations), and change in spin speed will cause the ...

6

There is another reason for this in multi-cylinder engines. While one cylinder is compressing the air, another is decompressing it. The net result is that the torque as a function of position gets smoother as the number of cylinders is increased.

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

Peak Demand One of the biggest issues in the electricity business is managing peak demand. This is the maximum amount of electricity needed at any one time. Typically this will be the middle of the day on the hottest days in the summer, as that combines household usage (everyone is awake), office usage (everyone is at work, or at least before the pandemic...)...

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

It isn't completely infeasible. Just to get a ROM (rough order of magnitude), let's assume that a typical household that isn't using electricity for heating uses about 1 kW on average, and that you'd like to be able to store a half day's energy, or 12 kWh, which is roughly 45 MJ. Commercial air compressors can easily achieve 15 bar or so (over 200 PSI). The ...

5

Because this question has proven a little more derisive than I had anticipated (where's that engineering sense of fun/humour?)- I thought I'd attempt an answer as well, if only to prove it's an engineering question which can be answered using the engineering method; although hopefully someone using something more than Google will be able to chip in with a ...

5

Nuclear power plants are always working, except circa one month/year for maintenance. It's mainly because stopping the nuclear reaction and cooling the reactor down is a very long and difficult process (a few weeks). So "turning on" and "turning off" a Nuclear Power Plant (NPP) is not as simple as pushing a button. The primary circuit (containing the water ...

5

There are some people who cannot be told. They think they know everything and that they are never wrong. My advice is do not try to convince your friend that your friend is wrong. You will always lose the argument. State what you think and if your friend rejects that, leave it at that and talk about something else. The other thing you can do is to tell him ...

5

Get a small DC motor from a cheap toy or something and put a gear or a flywheel or something on the shaft to make it easy to spin with your fingers. You should find these in the toy as well. Ask your friend to spin the motor while nothing is attached to the terminals. He will find that it spins easily and some momentum keeps it going for a short while. ...

5

To convert a plot of J versus s to a plot of J/s vs s, take the derivative. Here is my quick-and-dirty answer after digitizing the data.

5

TL;DR: the MM in MMtoe is probably related to the Latin Numeral M (for 1000). In that context MM means one thousand thousands (i.e. one million) Although this is very confusing, the bottom line is that to (my understanding) $$1 \text{ MMtoe} = 1 \text{ Mtoe} = 10^6\text{ toe}$$ Where: 1 toe = 11.63 megawatt-hours (MWh) 1 Mtoe = 1 MMtoe= 11.63 terawatt-...

4

Here are some ways to keep the warm air from escaping: Revolving doors prevent a constant stream of air exchange between the inside and outside. Air lock areas - a double-set of sliding doors can also prevent continuous air-exchange. Positive pressure by the entrance, usually supplied by a fan heater above the inner entrance door ensures the main bulk of ...

4

I think Failmond's first post does a solid job with the analysis. You have a voltage potential across a resistor(the air). You can not remove air from the equation as it is what is sourcing and sinking the voltage potential in question. The maximum energy you can harness is what it is already sinking. The amount of power harnessable is indeed very low, but ...

4

In theory, it should perform as you say. This is how prop starting an aircraft engine works. An internal combustion engine, unlike your bike pump, is designed for high speed, high power, and high temperatures. These design considerations give it higher internal friction and leakage (valves and cylinder seals) when doing a slow and cold propping operation. ...

4

I'd try and fit the material to a model. The Debye model is the "standard". (sorry the wiki article is a bit over the top.) In the Debye model the material can be fit with one "Debye temperature". Edit upon request. (though, I would trust the wiki article over my answer.) At high temperatures, (but not too high) materials have a heat ...

4

The force on each car is purely longitudinal because the restraining force of necessity runs thru the central point (origin). For example, the East and West forces cancel so far as the origin is concerned, so no lateral force exists from the viewpoint of the North or South car. The restraining force on each car, then, is the same as if it were tied to an ...

4

It's a very poor analogy, especially in presence of ion engines used for space travel, where the propellant is ionized inside the engine, then accelerated outwards. Besides, the electric cars are based on standard electric motors - all the ion transmission happens in batteries, which are not a part of the engine. The engine is based upon rotating (or ...

4

Batteries can't deliver unlimited current. Internal resistance, and limitations of the chemical reaction within the battery make that it can only deliver so much current. And the more current is drawn, the bigger the voltage drop will be because of above mentioned reasons. Specific power (W/kg) and capacity, specific charge or specific energy(Wh/kg or J/kg) ...

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