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user81881
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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/s^2s2. And the water falls, say, 80 meters. Then we multiply all those to get joules (a joule is 1 kg m^2m2/s^2s2). 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 km^2km2 = 100 million m^2m2. That's 300 million m^3m3 of water evaporating to make room for more water you dump for profit. We multiply that by 1000 L/m^3m3 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.

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/s^2. And the water falls, say, 80 meters. Then we multiply all those to get joules (a joule is 1 kg m^2/s^2). 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 km^2 = 100 million m^2. That's 300 million m^3 of water evaporating to make room for more water you dump for profit. We multiply that by 1000 L/m^3 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.

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.

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Mike Serfas
Mike Serfas
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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/s^2. And the water falls, say, 80 meters. Then we multiply all those to get joules (a joule is 1 kg m^2/s^2). 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 km^2 = 100 million m^2. That's 300 million m^3 of water evaporating to make room for more water you dump for profit. We multiply that by 1000 L/m^3 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.