+259 Hey there,
I wanna say in advance this question is not about general math but towards physic and tidal energy.
So I tried to calculate the potential Energy of a Tidal barrage. I took the formula E = 0.5 * A * p * h^2 *g.
A as the size of the tidal barrage in m^2, p as the density of water 1025 kg/m^3, h as the tidal range and g as the gravity.
I took the tidal barrage in sihwa-ho as an example and put in all the values. I took the values from wikipedia. A was 56.5km^2 and h 8m.
0.5*64*1025*5.65e+8*9.8
It got me a total value of 181613600000000 in short, 50.448 Gw/h.
Now it does say, that this tidal barrage gets around 550 Gw/h per year, so where is my error?
I know that 50.4Gw/h only includes one maximum per day, while there are 2 (every 12hours, since potential energy on ebb is 0).
Any help is appreciated :)
+36916 .5 x 64 x 1025 x 5.65 x10^8 x 8^2 x 9.8 = 116232704 x 10^8
kg/m^3 m^2 m^2 m/s^2 = kg m^2/s^2 (correct units?)
where did 64 come from? 8 should be squared according to your equation......
Can you answer these questions for me..... and where did you GET this eqution?
I'll see if I can help further....
+36916 E=1/2 Apgh^2 from
http://hydropower-tidalpower.blogspot.com/2009/07/energy-calculations_07.html
Which has a sample calc.....which will step you through your calculations to see where you might have erred
your '8' for tidal height....is that in meters? (should be...not in feet)
Did you divide by the seconds in a day (86400) / Did you multiply by 2 (tides per day) as in the example????
Let me know ! G'Day !!
+259 Jup, its 8 meters. I squared 8, so it gave me the 64 in my calculation. I also transfered every unit into meter. I used the sample calculation as a refference. I didnt divided it by 86400 seconds, as i didnt wanted mw/day, but gw/h to get a better refference.
I got the equation from many reliable resources, also the one you linked :)
+259 Im still editing too much, welp cant edit my first post thou since its a reply.
I didnt multpiplied it by two, but i mentioned it in my main post in the last sentence. But still with multiplying it doesnt get close to the 550 Gw/h stated on wiki. Also where did u get the 64 and 8^2 in ur first reply? Shouldnt there be only one squared tidal high?
+36916 Yes....I saw after I posted that the 64 was 8^2 and the other 8 was an exponent.....
Perhaps Wikipedia has the wrong answer !! Perhaps look somewhere else...they may have made a typo ......
Your formula is : E = 0.5 * A * p * h^2 *g. All the numbers you have plugged in appear to be OK except one. A, the size of tidal barrage is 56.5Km^2 is way too big. It could be 5.65 Km^2 =5.65e6. So that you total will be:1,817,989,200,000J. Now you have 2 high tides per day.So, 1,817,989,200,000J x 2 =
3,635,978,400,000J. Therefore, the mean power generation potential = Energy generation potential / time in 1 day =3,635,978,400,000J /86,400s =42,083,083 / 10^6 =42 MW - per day.
Sorry, that is the best I can do. The only Math error in your calculations is 56.5Km^2 =5.65e+7 NOT 8.
+259 Oh jea, your right. I only converted 5.6 km^2 while it is 56 km^2. Mhmm that makes my calculation even more irritating having only 5.4Gw/h with one high or 10.8 Gw/h with two highes.
+36916 A was 56.5km^2 and h 8m.
Here is the conversion: 56.5 km^2 = 56.5 x 10^6 m^2 = 5.65 x 10^7 m^2
This will reduce your answer by a factor of 10.....
The remaining question is Is A really 56.5 km^2 ?
+259 Jup it is 56.5 km^2. At least referring to wikipedia. I actually dont need any correct answer, just a value that gets near the 550 Gw/h. I mean, the formula 0.5*A*p*h^2*g describes the maximal potential Energy in one high tide, meaning it cant get over this value. Either theres a mistake in my logic or calculation or the values given by wiki are wrong. only thing that would come to mind are the difference in high tides during the year between springtide and nipptide and difference in position between moon, sun and earth...
