# The "Exciting" Drilling company dug an underground canal connecting Death Valley to the Pacific. How much electricity can it generate on tidal forces?

Badwater Basin, part of Death Valley, is 282 feet (86 m) below sea level at its lowest point.

The "Exciting" drilling company, based in Nevada (they are exciting and not at all like their b**ing competitors even though they too bore), dug an underground canal connecting the Pacific to Death valley. Boom, instant saltwater lake, complete with tides.

They also installed turbines in the canal, and now are aiming to become the #1 suppliers of clean energy in the mid-west.

Is their plan viable? How much electricity will they generate per year? Assume they can convert 30% of all the tidal energy into electricity, after all efficiency issues and equipment losses.

Note: There's a sister question to this. They are NOT the same.

• Why would they bother with the tiny tides in a small basin? Generally such schemes are designed to rely directly on solar power evaporating the water in the small hot basin, for example the Red Sea to Dead Sea Water Conveyance and the Quattara Depression Project. The sun will naturally evaporate the water in the small hot basin, so that a difference of elevation can be maintained ad infinitum, allowing a hydroelectric power plant to generate power from the flow. Jul 24 at 1:25
• The point being that a small enclosed basin most usually has no tides to speak of. For example, the maximum tidal amplitude in the Black Sea or in the American Great Lakes is a few centimeters. (And in general the whole Mediterranean is known as a mostly tideless sea.) Jul 24 at 1:28
• You might make more money with a freshwater evaporation and salt collection system, with any power used to offset electrical usage. Death Valley brand salt and a blooming desert! Jul 24 at 17:02

A basic physics answer can be done, assuming a hypothetical company and a 30% figure for efficiency.

A liter of water weighs 1 kg. Gravity is 9.8 m/s2. And the water falls, say, 80 meters. Then we multiply all those to get joules (a joule is 1 kg m2/s2). So we have 784 J for every liter that falls.

In the short term, we would calculate the volume of the basin you mention. Long term we can calculate the area. We just have to figure out how fast salt water evaporates in Death Valley. I have no idea how to do that, so I went to the Web which says a pan of water evaporates at 140 inches per year in Death Valley - this seems surprisingly little, but then again, California reservoirs would otherwise go dry without being tapped. Salt water evaporates more slowly due to the energy that goes into concentrating the salt, and as the water gets brinier that will increase. Let's just call the rate 3 meters per year.

Badwater Basin is flat-looking and reportedly almost 200 square miles, though this seems hard to square with the maps I'm looking at. Say 100 km2 = 100 million m2. That's 300 million m3 of water evaporating to make room for more water you dump for profit. We multiply that by 1000 L/m3 and then by the 784 J and get 240 TJ of energy per year. Unfortunately, the awful "metric" unit of watt-hours is often heard, so we have to multiply this by (s / 3600 s) to get 65 gigawatt-hours. Then I have to remember you said 30% efficient, making it 20 GWh per year actually obtained.

• The 30% is for the impractical tidal power plants. A plain regular hydroelectric power plant is very much more efficient. Jul 24 at 1:59
• Be good to get comparison figure per year for say a large coal fired plant to get a benchmark comparison. (I tried but I couldn't find it on G/W per year.).)
– Mon
Jul 24 at 2:36
• @Mon: A typical hydroelectric power plant is about 90% efficient. A typical gas- or oil- or coal-fired thermoelectric power plant is 50% to 55% efficient. (But those numbers cannot be compared, because the two 100% out of which we compute efficiency are very different things.) A plain ordinary (and small) 1000 MW thermoelectric power plant produces about 8,750 GWh per year. Jul 24 at 16:02
• Just what the Doctor ordered! Even tripling the efficiency of the proposed hydro scheme to the industry standard (90%) and therefore tripling it's output doesn't make the proposed scheme worthwhile.
– Mon
Jul 25 at 2:00
• I have no problem with 200 square miles for Badwater Basin (I've been there multiple times), but that's not all 80m below sea level. You're not getting anything like as much power as you're calculating. Jul 25 at 2:21

Tidal power turbines are installed close to the coast and not very far away.

Badwater Basin is 282 feet (86 m) below sea level. The canal will be 250 miles long. The water flow in the canal will not be like tides. The water will flow from Pacific to Death valley (one way only).

Will the water flow stop when water level in the Death valley reaches sea level? Maybe not. In summer, the temperature is high and water will evaporate, increasing the humidity of the surroundings and lowering the water level so the water will keep flowing to the lake. In winter, the tidal energy may be used. But the amount of water that flows through the canal will depend on cross-section area of the canal.