Timeline for Hydropower formula:
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2016 at 9:29 | comment | added | user16324 | Easiest way is to either rewrite the formula you found to use pressure instead of "head" - or alternatively convert from your pump pressure to the equivalent "head" as in @krufra's answer and use the formula directly. If these approaches give different answers, you're doing at least one of them wrong. Keeping the water mass constant is a red herring, to calculate power you need flow rate not mass, which varies with pressure. | |
| May 4, 2016 at 23:45 | comment | added | DIYser | It seems easier to this way: If we want to keep the water mass constant & we want to increase the applied pressure, then we would need to increase the column H dimension & decrease the area of the bottom of the column proportionately (in a gravity driven system). In a pump driven system, if we wanted to keep the volume of water constant, we would need to increase the velocity of the water to get a higher pressure & higher power. Y/N? | |
| May 4, 2016 at 9:55 | comment | added | user16324 | You are keeping the mass of water constant, not the head. Thus you are increasing the column height - and that - not the area - is increasing the pressure. | |
| May 4, 2016 at 5:59 | comment | added | DIYser | Pressure appears to be dependent on area (e.g.: 1000kg * 9.81 / 1 m^2 = 9.8 kPa <verses> 1000kg * 9.81 / 10 cm^2 = 980 kPa. Since pressure varies with area, it appears If the base changes from 1 m^2 to 10 cm^2 then pressure increases 100 times. Since only base area was changed in those expressions, it seems that pressure is not "independent of area". | |
| May 3, 2016 at 19:14 | comment | added | user16324 | Because pressure is independent of area ... but proportional to the height of the column - i.e. the head ... which is where we came in. | |
| May 3, 2016 at 19:02 | comment | added | DIYser | How do you reconcile your last statement with your earlier statement "Pressure of a column of water is independent of area so you'd get the same answer for any area (because the weight of water would change proportionally." The statements seem to contradict each other. If the mass of water & the flow rate do not change--regardless of the column dimensions-then the power level would seem to remain the same, wouldn't it? | |
| May 3, 2016 at 18:53 | vote | accept | DIYser | ||
| May 3, 2016 at 18:53 | |||||
| May 3, 2016 at 10:03 | comment | added | user16324 | Not always, but 1 m^3 occupying 1m^2 area definitely is. If you distributed it over 10x10cm it would be 100x the height, so 100x the pressure, so (potentially) 100x the power at the same flow rate. | |
| May 3, 2016 at 1:16 | comment | added | DIYser | That makes sense--1 m^3 is definitely 1m in height :). Thank you! | |
| May 2, 2016 at 22:46 | comment | added | user16324 | Pressure of a column of water is independent of area so you'd get the same answer for any area (because the weight of water would change proportionally. As for the height of the column, you set that yourself, I just discovered it from your question. | |
| May 2, 2016 at 22:22 | comment | added | DIYser | Your answer checks out (9,021.52W / .92 eff / 9.8 kpa = ~1 m\$^3\$/s flow rate), however, why would you assume the value of h=1m and the area must be 1 m\$^2\$? (I would not have instinctively known to assume h=1m or use 1 m\$^2\$). | |
| May 2, 2016 at 12:35 | history | answered | user16324 | CC BY-SA 3.0 |