Tank Drain Time & Flow Rate Calculator — Fill Time, Gravity Drain & Orifice Flow

Calculate Flow In Or Out Of A Tank: One Formula, Six Situations

Almost every water-moving question about a tank reduces to one relationship: flow equals a coefficient times an opening area times a velocity, and that velocity comes from the height of water (or pressure) pushing on the opening. The six modes of the calculator above — orifice, gravity drain, pump, two-tank transfer, pressurized outflow, and net accumulation — are all rearrangements of that single idea. If you need the tank's capacity first, the tank volume calculator gives you gallons from geometry, and the all tank calculators hub lists every specialized tool on the site.

💧Gravity does the pumping — head sets the speed

Gravity does the pumping — head sets the speedPhoto: Unsplash

Calculate Flow Out Of A Tank: Torricelli At The Drain Hole

Torricelli's law says water escaping a hole under gravity leaves at the same speed a dropped object would gain falling that same distance: v = √(2g·h), with g = 32.174 ft/s² and h measured in feet. Thirty inches of head — 2.5 ft — produces a jet at √(2 × 32.174 × 2.5) = 12.7 ft/s. Real holes never achieve the theoretical speed across the full opening because the flow contracts, so we multiply by a discharge coefficient Cd: 0.62 for a sharp-edged hole punched in plate, 0.82 for a short pipe stub, and 0.97 for a properly rounded entrance.

Calculate Flow Rate Out Of A Tank: Worked Numbers

Take the default case: a 1 inch hole sitting under 30 inches of water. The hole area is π·(0.5)²/144 = 0.00545 ft². The jet moves at 12.68 ft/s, so the theoretical flow is 0.00545 × 12.68 = 0.0691 ft³/s; apply Cd = 0.62 and the real flow is 0.0429 ft³/s. Multiply by 448.831 to convert to gallons per minute and you get 19.3 GPM, or 1.21 L/s. Scale the opening down to a 1/8 in nail hole and the area — and the flow — falls 64-fold to roughly 0.3 GPM, which is why a punctured rain barrel weeps for days while a 1 in drain fitting can empty one in minutes.

How To Calculate Flow Rate From A Tank Drain Hole, Step By Step

1. Measure the head h from the water surface down to the center of the hole, in inches, and divide by 12.
2. Compute the hole area A = π·(d/2)²/144 in square feet.
3. Get the jet velocity v = √(2·32.174·h_ft) in feet per second.
4. Multiply Q = Cd·A·v in ft³/s, then × 448.831 for GPM or ÷ 15.850 for liters per second.

Calculate How Long It Takes To Fill A Tank With A Pump

Filling is the friendly sibling of draining: a pump delivers a constant rate, so there is no square root and no calculus. To calculate time to fill a tank or empty one at constant rate, the whole formula is minutes = gallons ÷ GPM. The only trap is unit honesty — a pump rated in GPM but a tank planned in cubic feet will be off by a factor of 7.48, and a "gallon per hour" dosing pump is 60 times slower than the same number in GPM.

🚰Constant-rate fill: volume ÷ flow, no calculus needed

Constant-rate fill: volume ÷ flow, no calculus neededPhoto: Unsplash

Calculate Time To Fill A Tank: Two Worked Examples

Fill-time math also anchors aquarium planning: working out how long a partial water change takes is the step that decides how patient you must be before running a fish tank stocking calculator, because new livestock should only meet stable, fully conditioned water.

How To Calculate Flow Of Pressurized Tank: PSI As Virtual Head

Pressure is just head measured a different way. For water, every 1 psi of gauge pressure equals 2.3077 ft of water column, so every 10 psi ≈ 23.1 ft of extra head — head the tank geometry never has to provide. The pressurized mode adds that virtual head to the real water head and runs the same orifice math: total head H = 2.3077·P + h/12, then v = √(2g·H) and Q = Cd·A·v.

Worked default: 10 psi on a tank with the hole at water level, 1 in hole, Cd 0.62. Total head is 23.08 ft, jet velocity is √(2 × 32.174 × 23.08) = 38.5 ft/s, and flow is 0.62 × 0.00545 × 38.5 × 448.83 ≈ 58.5 GPM. Because velocity scales with the square root of head, tripling the pressure to 30 psi raises flow only to about 101 GPM — not 175. Pressure is powerful but it hits diminishing returns, which is why high-throughput systems enlarge the outlet instead. (The static side of this — pressure at the bottom outlet from depth alone, about 1.08 psi under 30 in of water — is the domain of the tank pressure calculator.)

Calculate Speed Of Draining Tank: Velocity At The Hole

Speed and flow are easy to confuse. The jet speed depends only on head — 12.7 ft/s at 30 in, 25.4 ft/s at 10 ft, 38.5 ft/s at 23 ft of pressurized head. The volumetric flow is that speed times hole area times Cd, so a tank can drain fast (high jet speed) through a tiny hole yet still move few GPM. When someone asks how fast a tank is draining, ask whether they mean feet-per-second at the outlet or gallons-per-minute past the meter — the answer differs by a factor of the hole area.

Calculate Liquid Accumulation Rate In A Tank: Net Inflow Minus Outflow

When a tank has both a supply and a demand, the question stops being "when does it empty?" and becomes "how fast is the level moving right now?" Net accumulation is inflow minus outflow in GPM, converted to a level speed by the tank's surface area: inches per minute = (net GPM × 231) ÷ area in in², since each gallon occupies 231 cubic inches.

The defaults show the arithmetic: 10 GPM in, 7.5 GPM out, net 2.5 GPM. A 24 in diameter tank presents π·12² = 452 in² of surface, so the level climbs at 2.5 × 231 ÷ 452 = 1.28 in per minute — 76.6 in per hour. From a 30 inch working level, an unattended tank overflows in under half an hour, which is why accumulation rate belongs next to every high-level alarm setpoint. Reverse the flows (7.5 in, 10 out) and the same math returns a falling level at 76.6 in/hour — the calculator flags the direction for you.

Corner Cases: Tips, Warnings And Where The Model Breaks

Calculating Time A Tank Takes To Drain By 3 Ft

Depth intervals are where the square root earns its keep. Take a 24 in diameter tank (At = 3.1416 ft²) draining from 48 in down to 12 in through a 1 in hole with Cd 0.62. The constant from the derivation above is 231.7 s per √ft, and the head term is √4 − √1 = 2 − 1 = 1, so t = 231.7 × 1 = 232 seconds, about 3.9 minutes for the first 3 ft. Draining the next 3 ft — from 12 in to 0 — costs another 231.7 × (√1 − √0) = 232 seconds. Equal depth drops, equal times: that additivity of √h differences is a quick sanity check on any hand calculation.

Extreme Values And The Last Inch

The formula's edges behave sensibly. At h = 0 the drain time contribution is zero — no head, no flow. As h grows the time grows only as √h, so quadrupling the head merely doubles the drain time. At the far end, though, reality diverges from the model: as the tank empties, a vortex forms over the outlet and air entrainment chokes the flow, meaning the real last inch drains slower than the formula predicts. If the final minutes matter — tank turnover schedules, batch release times — either switch to a measured Cd from a bucket test or enlarge the outlet and add an anti-vortex plate so the last inch behaves like the rest.