PV Gap

Gwendolyn Said:

Will this solar USB charger charge my iPod Touch 2nd gen?

We Answered:

Sorry, no.

An iPod Touch needs 500 mA of charging current, even a 300 mA charger won't work (I tried it myself, it doesn't).

Assuming you could actually get 0.4 W out of that panel at 5 volts, that's only 80 mA of current. But small panels tend to have optimistic specifications, so the actual current you get could be just 20 mA.

You would look for a panel that says 5 volt output, 500 mA or more. Such a panel would likely be 1 foot square, or 1 foot by 2 feet.

Judy Said:

Could these institutions help reverse climate change, shift off of non-renewable energy, help the environment,

We Answered:

I think so. The US government represents about 20% of the US economy and so should be able to contribute that much to the US portion of climate change. The US produces about 25% of the total output of greenhouse gases and so the US government is responsible for about 5% of the entire world production of greenhouse gases. This does not count state governments which I would guess are around another 10% of the US economy.

You might be interested to know that the United States Air Force has begun construction on the largest U.S. solar photovoltaic system at Nellis Air Force Base. This system when completed will produce over 25 million kilowatt-hours of clean electricity annually.

Jack Said:

Question about Piezoelectricity?

We Answered:

There's much better ways to generate power from rain. Catch it and run it through a turbine, for instance. It wouldn't be difficult at all to make it with a piezoelectric crystal - you just need to find a way to make it cause vibration.

It's estimated that the sum of the noise energy released by a stadium of cheering fans in a football game is enough to heat one cup of tea.

Gabriel Said:

two quantitative literacy questions?

We Answered:

Q1) The first paragraph tells you that solar cells, when exposed to direct sunlight, could produce 1000 watts of power for each square metre of surface if they were 100% efficient. For lower efficiencies eg Q%, the output will be reduced proportionately, by the factor Q/100.

So if the efficiency of an array of solar cells with a surface area of 1 square metre is
60 %, the power output in direct sunlight is (60/100)*1000 = 600 watts.

You should also note the relationship between energy and power - the power is the rate at which energy is produced, so 1 watt is defined as the production of 1 Joule each second. So the power output of 600 watts is equal to 600 Joules per second. In an average day, the cells receive 6 hours or 6 times 3600 seconds of direct sunlight (or its equivalent), so the total energy produced is 600*6*3600 = 12960000 J, which can be written as 12.96 MJ (megaJoules) per day.

The cells produce 600 watts when illuminated ie for 6 hours each day, and nothing when it's dark, so the average power output is 600*(6/24) watts/day. Note that it is important to include the period over which the average is taken in the answer - you should not say simply '150 watts', you should include 'per day' at some point or other.

Q2) The difficult part of this problem is how you allow for the spiral wrapping of the ribbon around the column, so that it reaches the top of the column in 6 turns. If you were to flatten out the surface of the column required for one turn, it would form a rectangle of width 2 feet - the circumference - and height 10/6 feet. So if the ribbon runs along the diagonal of this rectangle, it would have a length, from Pythagoras' theorem, of ?[2² + (10/6)²] = 2.603 feet. This is a reasonable approximation, but not exact.

That gives you the length of ribbon for one turn. You have 6 turns on each column, and a total of 18 columns. So the total length required is 6*18*2.603 = 281.169 ft, which rounds up to 281.2 ft for one decimal place.

HTH