93% on the Label. 820 Kilowatt-Hours in the Cubic Meter.

820 kilowatt-hours.

Divide 56.95 kilowatts per square meter of laboratory electricity by the 69.46 liters per square meter per hour printed beside the 93.31 percent claim, and the quotient is 0.8198 kilowatt-hours per liter, 820 in a cubic meter. Nature Water published the device on October 28, 2026, with He Yang and Jianyu Zuo leading the work for Lin Jiang at Soochow University's Institute of Functional Nano and Soft Materials in Suzhou, Meiwen Peng and Yinghui Sun corresponding (Jiang's faculty page; Peng's). The abstract calls 93.31 percent an apparent energy efficiency, pairs it with that flux and with 1.64 liters in eight hours from 30.72 square centimeters, and says the yield is "sufficient to meet the daily needs of individuals," even from sediment, oil, brine, or strong acid. A same-day Nature Water index line lists the paper.

What 93 percent is counting

Apparent efficiency, in Supplementary Note 4, is evaporated-water enthalpy divided by electrical plus solar input, per square meter of projected area (Yang et al., supplementary information). Headline inputs are 100.66 kilograms evaporated per square meter per hour, an interface at 363.05 K from 294.95 K, latent heat 2283.38 kilojoules per kilogram, sensible heat 284.66, electricity 56.95 kilowatts per square meter, and sun 20, so the denominator is 76.95 and the note's arithmetic returns 93.31 percent. Supplementary Table 5 lists 52.58 at 1.5 volts and one kilowatt of light against 100.66 at 1.5 volts and 20, so about half the headline vapor precedes the twenty-sun lens, while electricity remains 74 percent of the 76.95-kilowatt input.

Collected water is the smaller pile: 69.46 divided by 100.66 is 69.0 percent of the evaporated mass, if those rates share a condition the way the abstract pairs them, and 93.31 percent times 0.690 puts the cup at 64.4 percent of the energy supplied, not 93. Supplementary Note 12 reaches 55 percent by dividing 42.32 kilowatts per square meter of condensation power, from an outdoor flux of 66.72 liters, by the laboratory 76.95, mixing an outdoor liter rate with lab power, so that 55 percent is their freshwater score and not a second meter, and the abstract does not lead with it. Check the cup.

A watt ledger that will not add

Supplementary Note 6 counts watts on a coupon of 0.000256 square meters, 2.56 square centimeters, where solar input is 5.12 watts and electrical input is 14.58 watts, and 14.58 divided by 0.000256 recovers the 56.95 kilowatts per square meter. Push 9.72 amps through a 0.048-ohm lead and 9.72 squared times 0.048 equals 4.53 watts, 31.1 percent of the electricity, lost in the lead. Optical, conduction, radiation, convection, and wire losses print as 0.09, 0.17, 0.30, 0.19, and 4.53, summing to 5.28 against an input of 19.70, and subtracting those losses is how the note prints an "actual" utilization of 73.20 percent.

Vapor on that coupon does not land on the remainder. Enthalpy is 4.18 times (363.05 minus 294.95) plus 2283.38, or 2568.04 kilojoules per kilogram, and 100.66 kilograms per square meter per hour times that figure, divided by 3600, is 71.81 kilowatts per square meter, which on 0.000256 square meters is 18.38 watts of vapor. Add the 5.28 watts of listed losses and the outputs are 23.66 watts against 19.70 in, a gap of 3.96 watts, 20.1 percent of the input. Note 5 still reports a closed balance, because the wire's 4.53 watts are divided by the wire's 0.002 square meters to give 2.27 kilowatts per square meter, then added to terms divided by the evaporator's 0.000256, whereas the same 4.53 watts on the evaporator area are 17.7 kilowatts per square meter. Radiation and convection assume 5 watts per square meter per kelvin and emissivity 1, so part of the gap can be estimated heat already counted as vapor, but two denominators in one sum are a choice, not an estimate.

820 divided by 4.13 is 198.5, and Alnajdi and colleagues put the average specific energy of 39 seawater reverse-osmosis plants at 4.13 kilowatt-hours per cubic meter in Joule on November 20, 2024. Count the sun and the total is 1,108 kilowatt-hours per cubic meter, 268 times that average. Against the separation minimum, 820 divided by 1.06 is 774. Elimelech and Phillip derive that 1.06 for 35 grams of salt per liter at 50 percent recovery and say the floor does not care whether the machine is a membrane or a still, and a 2024 roadmap restates operating plants at 3 to 5 kilowatt-hours per cubic meter (Politano and colleagues, Journal of Physics: Energy). On the boiler's own ruler, their 2260 kilojoules per kilogram at 100°C is 2260/3600 = 0.628 kilowatt-hours per kilogram, 628 per cubic meter, so 820 of electricity is 1.31 times the phase-change floor before sunlight is added. A single-effect still can only approach 628, and the vapor got close while the cup did not.

Eight hours, 1.64 liters

Twelve coupons of 2.56 square centimeters make the outdoor 30.72, and twelve times 1.5 volts is the 18 volts Supplementary Note 11 says the cells had to clear, while 1.64 liters divided by 0.003072 square meters divided by 8 hours is 66.73 liters per square meter per hour, beside the table's outdoor 66.72. Sphere's 2018 handbook sets 15 liters per person per day for drinking and domestic hygiene. WHO's emergency page asks for 15 liters as soon as possible, allows 7.5 right after impact, and treats 7.5 as enough for most people, while a quantity report puts drinking water for the highest-need group at 5.3 liters (Domestic water quantity, service level and health). The ratios are 1.64/15 = 10.9 percent, 1.64/7.5 = 21.9 percent, and 1.64/5.3 = 30.9 percent, so one published day misses that drinking floor.

Sphere's 15 liters takes 15/1.64 = 9.15 copies of the eight-hour rig, and only if the day repeats, while 15 times 0.8198 kilowatt-hours is 12.3 kilowatt-hours of lab electricity. U.S. residential customers bought 863 kilowatt-hours a month in 2024, and 863 times 12 divided by 365 is 28.4 kilowatt-hours a day (EIA Table 5.A), so 12.3 is 43 percent of that purchase, against 0.015 times 4.13 = 0.062 kilowatt-hours at the plant average. Outdoor watt-hours are unpublished, and if each cell held 14.58 watts the day used 12 times 14.58 times 8 = 1.40 kilowatt-hours, or 853 kilowatt-hours per cubic meter. Note 11 only promises a voltage above 18 volts, which is not a power reading. On August 19, 2025, in Suzhou, Supplementary Table 6 averages natural light at 0.959 kilowatts per square meter and concentrated light at 20.91, a factor of 21.8, so the lens for this pack is at least 670 square centimeters. ASTM G-173 sunlight integrates to 1,000 watts per square meter (PVEducation on ASTM G-173; NREL's spectra), and 175 watts still needs at least 0.175 square meters of perfect cell, an area the paper never states.

What You Can Do

If you referee interfacial-evaporation papers, require collected liters per kilowatt-hour beside apparent efficiency, and reject a balance that divides one loss by a different area than the input. If you buy emergency gear, do not count 1.64 liters as a person's day, since it is about a tenth of Sphere's 15, and ask for watt-hours, cell area, and a multi-day dirty-water run. If you build the next still, heat recovery across effects is what moves the 628-kilowatt-hour floor, not another point of single-boil vapor efficiency, and until watt-hours sit next to liters this is a prototype beside plants that make a cubic meter for about 4.

Limitations

Nothing here reran the coupon, outdoor electrical energy is unpublished, and the 55 percent score mixes outdoor flux with laboratory power, which is why 853 kilowatt-hours per cubic meter stays conditional. Convection and radiation use an assumed coefficient and emissivity of 1, so part of the 3.96-watt gap can be double-counted heat, Bohai ion values are plotted rather than tabulated. Feed recovery is unstated while Elimelech's 1.06 assumes 50 percent, the 4.13 average includes treatment steps the coupon skips, and a 2022 rate of 10.5 kilograms per square meter per hour under 8 kilowatts of light is taken from the institute's summary, not a reread of the paper (FUNSOM's note).

The Strongest Case Against This Analysis

A coastal plant is the wrong opponent, and 198.5-to-1 picks a fight this device never entered, because Elimelech's 1.06 kilowatt-hours describes a reversible separator on a grid, while a single-effect still, like any kettle, sits near 628 kilowatt-hours per cubic meter before losses, so 820 is 1.3 times the only floor a boiler can approach. Ninety-three percent is evidence the interlayer put most of the joules into vapor, which is the claim they made, and 1.64 outdoor liters is water somebody collected. On a day the pipes are gone and no 55-bar pump is in the bag, a still you can carry has a job even when a giant plant wins the spreadsheet. Weight, fouling, and a real drinking volume are the fair questions. Multi-effect recovery could still cut that thermal bill, and as a backpack boiler the coupon sits near the only ceiling it can hit.

The Bottom Line

Soochow built a real interlayer heater, and 93.31 percent is a real vapor score under about 57 kilowatts of electricity plus 20 of concentrated sun, while water in the cup is 69 percent of that vapor, 64 percent of the energy, and 820 kilowatt-hours of electricity per cubic meter, about 199 times a 39-plant average. One rig's 1.64 liters is a tenth of a Sphere day, so carry it after someone publishes the watt-hours, because until that page exists the label describes vapor and the cubic meter is still a power bill.

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