I don't have any pics of my set-up, although I'll take a few tomorrow and post them. Either my pond is a lot smaller than I thought (could be, it's shrunk a lot in the past few months), or my well pumps out a lot more than I thought. I'm electrically challenged, and math isn't my forte, so I contacted a friend who is better than I am, who has a friend that is better than him, and what he sent me is posted below.
Is that clear as mud?
My pond is very irregular shaped, and I'm at a loss how to figure out the area until Google updates their satellite map. I was guessing that the pond when at a level that is 2' higher than it is now, was around an acre and a half.
Well pump is 1 hp, rated at 28 gpm. I didn't pump for 24 hr, and the level dropped 1". I pumped for 12 hr. and the level rose 3/8". The pond slope varies around the pond from 2:1 - 4:1. Ambient temp that day was 91* during the day, down to 70* at night. Daytime Rh was 60% I have a yardstick on a stake, so I can measure the water level pretty accurately.
Answer part 1:
Pretty sure there's enough info to solve this, which I'll attempt later this evening. We know the inflow rate (Qi) is 28 gpm and we know that after 24 hours we can raise the pond level 3/4 inch if we pump continuously.
Since the pond drains 1" in 24 hours absent our pumping, we therefore know that the inflow rate is 1.75 times the outflow rate. Now we have all flows in terms of a known flowrate; so far so good.
If we neglect evaporation (probably pretty reasonable over a single day) and we assume that the slope of the pond bank is also negligible over an inch of level change (seems pretty reasonable), we can approximate the shape of the pond by an easily computed volume, say a slice of a cylinder. Now we've got a formula for Volume in terms of area and level and it's off to the races.
From here, it's a matter of writing the equation for water balance, dV/dt=Qin-Qout, substituting pond height for level, rearranging to solve for area and Bob's your uncle!
I think.
Part 2
Data:
Pump discharge = 28 gpm
Pond level subsidence rate in absence of pumping = 1"/day Pond level increase rate with pumping = 3/8" per 12 hrs = 3/4" per day
Assumptions:
1. Near the surface, the sides of the pond have insignificant slope (makes it easier to solve, although still amenable to solution if we assume some arbitrary slope).
2. The shape of the pond's surface can be approximated as a circle.
3. Water loss due to evaporation is negligible (once we estimate the surface area, we can backtest this assumption by choosing a range of evaporation rates per unit area based on temperature and wind speeds).
3. Water temperature during the time period in question is constant, so volume per unit mass of water is constant.
4. Inflow due to rain or other natural causes is negligible.
Solution:
1) Water balance for pond
Rate of change of pond volume equals the inflow (pump discharge) minus the outflow (leakage through the bottom and sides of the pond), in symbols:
dV/dt = Qi - Qo
where
V = pond volume, gallons
Qi = rate of water flow in = pump discharge rate = 28 gpm Qo = rate of water leakage
Qo is not given explicitly, but we know that in the absence of pumping, the pond level subsides 1" per day. When the pump is running, the pond level increases 3/8" in 12 hours, or 3/4" per day. Therefore, the pump provides not only the volume lost to leakage, but an additional 3/4" per day. From this information we can say that the pump discharge is 1-3/4 (or 7/4) times as great as the leakage rate, or:
Qi = Qo / (7/4)
Qo = 4/7*28 gpm = 16 gpm
so:
dV/dt = 28 gpm - 16 gpm = 12 gpm
Our model of the pond near its surface is a cylinder (circular surface with straight sides), so the volume of a section is:
V = Area * Height
so:
dV/dt = d(Area * Height)/dt
Since the Area is constant, remove it from the differential and solve for it:
(dV/dt)/(dH/dt) = Area
we know dH/dt = 3/4"/day, so, solving and using unit conversions as necessary, we have:
12 gallons | 1440 min | day | 1 acre-ft | 12 in | = 0.85 acre
min | day | 0.75 in | 325,851 gallons | ft |
