Your Questions About Energy Of The Wind Density

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More bicycle paths along side our roadways and tax reductions for motorcycle and scooters to start.

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C

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Hey Z, there is an effect for all of this, quite obviously. Take the old mills from the turn of the last century, they had 10 or 15 large, flat, wide metal blades, so almost all the swept area had some blade area exposed. These were actually quite inefficient it turns out, a long, slender blade with a high aspect ratio operating at a higher speed works much better. This is why we have the type turbine blades you see today, usually 2 or 3 long skinny blades and the rest of the area is open.

What is missing from your formula is turbine efficiency. The formula you are working with is used to calculate the amount of power available in a column of moving air. What the turbine is able to extract from this moving column of air is a great deal less than this.

Modern turbines today have efficiencies around 20 to 35 %. So let’s say you do your calculation for swept area, speed and air density and determine that 1500 watts are available. A good quality turbine running near its design rpm will only extract about 450 watts. There is a theoretical limit on turbine efficiency, around 59%, called the Betz limit, named after the French engineer that discovered it. You mentioned a VAWT, or vertical axis wind turbine. These type turbines, previously called, “Darrieus,” Rotors have horrible efficiency ratings, usually around 5 ot 8 %. The reason for this is that most of the energy absorbed by the downwind blade is being used to force the upwind blade back into position for the next cycle, so very little is left to produce energy. They are really neat, look more attractive than horizontal turbines, and might be easier to mount, but you can never get any real power out of one. The turbine we have at our home is about the size of a large ceiling fan, 7 foot or so in diameter, yet on a windy night it can produce over 1000 watts. You’ll never get production like that from a similar sized Darrieus rotor, but if you’re building one as an example, or maybe to run a backyard pond pump, it might be fine, just do the math first.

There is some real technical information at the American Wind Energy Associations website you can look into. Specifically, I would look or articles written by Mick Sagrillo, he is considered today to be the modern day guru on wind turbines. He has one article that explains all the misinformation surrounding Darrieus rotors, or VAWT’s, as well as a better explanation of wind power equations and rotor efficiencies. Take care Z, Rudydoo

adminsta answers:

1 watt is about 0.10197 kg-m/s (where kg is kg-force). See reference link at end.

You stated that a 150 kg weight would travel through 1 meter lift 60 times per minute, or 1 time per second. The power required to perform this work is calculated as:
P = 150kg x 1m/rev x 1rev/s / (0.10197kg-m/s-w)
P = 1471W = 1500W (rounded)

The blades produce sufficient power to raise the weight.

I don’t see where the “20 kg power” comes from. The entire blade (except hub) extracts energy from the air stream. Not just the tip.

The blade is shaped with a large angle of attack the closer toward the hub it is because the blade is moving through the air at a slower speed. For slower speeds, a larger surface area is helpful.

Does surface area matter? Yes. The more blades there are, the less efficient the wind turbine. Two are most efficient, but they have trouble starting rotation in slow breezes, so three blades are the typical number used.

I think the efficiencies given for wind turbines based on number of blades, assume that the designer has optimized the shape of the blade to approach maximum theoretical efficiency. When I was a kid, I made a windmill using a rectangular board for each blade. It didn’t work to well…

adminsta answers:

Pumped-storage hydro is a commonly used method. There are hundreds of working installations around the world.

Http://en.wikipedia.org/wiki/Pumped-storage_hydroelectricity

In Europe, about 5% of their total capacity is pumped storage. In the USA about 2.5%.

There is a basic, often-overlooked problem with compressed air storage (as well as hydrogen and CNG on vehicles). When you compress air (or a gas), it gets hot. Decompress air, it gets cold.

That leaves you with two options, both unpleasant:

1. Intercool the compressed air, and throw away all that heat energy, which you paid for. This slaughters your efficiency. And gives you a second problem: when that air expands, it will turn cold. If it is highly compressed, it will turn very cold. In the machine shop using air at 8 bar, it works in your favor because air tools need cooling. But it could be a serious problem in a large, powerplant-sized facility.

2. Or, store the air “hot”, keep the energy, but raise daunting problems in the storage of this very hot air. Aside from finding a pressure vessel that wouldn’t be melted by the heat, you’d also need to insulate the vessel to keep the heat contained, and then you’d need a much larger tank to store the same mass of compressed air, since hot air has proportionately more volume than cool.

I would guess #2 is so difficult as to be impracticable. In many cases it is definitely not an option, such as a vehicle which runs on compressed air. Hydrogen cars also have considerable losses in this area because they also discard the potential energy of compression; they only want the chemical energy available as a fuel.

Battery electric drive is about 90% efficient at converting joules in battery to rubber meets the road. This still ignores the “regenerative braking” capability of electric-transmission cars, which could push it over 100%.

Http://en.wikipedia.org/wiki/Electric_car#Energy_efficiency

Tesla claims a “Well to wheels” energy efficiency nearly 10 times that of an average automobile. Their page here compares it to the most efficient alternative fuel vehicles.

Http://www.teslamotors.com/efficiency/well_to_wheel.php

Gasoline engine drive is less than 20% effective at turning “joules in the gas tank” into joules of zoom. Fuel cells are about 50% but only if you ignore the considerable energy losses of compressing the hydrogen.

If you want to maximize the efficiency of “substation to tire”, battery electric is vastly superior to other modes.

You claim that battery electric cars will use primarily coal. I dispute that assertion. If electric cars were commonplace today, that would be true. But they are not, and by the time they are, the shape of electric generation is likely to change. Remember, the same social/economic/environmental forces which push us toward alternative cars, also push us toward alternative power generation as well.

In California, our power is only 1/3 coal and about 1/3 carbon-free. (most of that being large hydro, we hate nukes.) http://www.energy.ca.gov/sb1305/power_content_label.html

You are exactly correct that measuring the number of joules of “potential” energy loaded onboard a vehicle is not a particularly useful metric, and wrongly casts battery-electric in a terrible light. Once on a mailing list, someone claimed that batteries couldn’t possibly have enough useful range. He said “There will never be a battery that has the same energy content (joules) as a 12 gallon tank of gas.” I ran the math: if he had such a battery, using well-known “miles per KWH” figures from real electrics, he could drive from NYC to Colorado on one charge. The question is, could he stay awake that long?

Fred makes an interesting point that nighttime charging of electric cars is a very useful form of energy storage. More than that, if your car has a 120V inverter on it (as the weak 2005 Silverado hybrid did http://www.edmunds.com/insideline/do/Drives/FirstDrives/articleId=104820 ), you can power your house from your car during an outage.

adminsta answers:

Hey Music Man, the basic formula for power of a wind turbine is P=.5 X A X rho X V^3 X %eff. In this equation, P is power output in watts, A is the swept area in square meters, rho is the air density in kg/cubic meter, V^3 is the velocity in meters per second cubed, and % is the efficiency of the turbine. Most turbines run in the 25 to 30 % range today, there is a theoretical limit, called the Betz Limit, but I don’t recall what it is. In your case, you are looking for the amount of energy of the air moving through a 60 m diameter blade area, you can omit the % since you are not looking for the turbines output, but rather the energy available in the air movement. See if you can work out your answer from this. The formula can be found at the American Wind Energy Association website, or wikipedia, or some other places. If you google, “Wind Turbine Power Formula,” you’ll find lots of sites to check out. Good luck, and take care, Rudydoo

adminsta answers:

First, you need to find the mass flow rate of air through the windmill.

Mass flow rate = volume flow rate x density
= Area of windmill x airspeed x density

Available Power from wind = 0.5 x mass flow rate x airspeed^2

Elec pwr o/p = 0.6 x above

Remember to be consistent in your units.