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hey brother!! i love this setup!! what are you tryng to do with the water?

I am looking at ways to improve fuel economy in a semi. Could the waste heat from a semi be used to offset the workload of the alternator and significantly improve fuel mileage? A Catapillar executive said it was the fuel burned that wears a motor out not the miles driven. Some trucks now use electric APU's and these put extra demands on the alternator. Semi-trucks produce a substantial amount of waste heat in the exhaust, with exhaust gas temperatures typically ranging from 300°C to 500°C (572°F to 932°F) at the manifold. The truck’s alternator typically converts mechanical energy from the engine to electrical energy, with an efficiency of about 60-70% so lets use 65%. By offloading 5 kW of electrical demand to the ORC system, the engine’s mechanical load is reduced. I am not sure what the typical output for semi truck alternator would be for the 62 mph 8mpg example so I am describing best case. If it typically produces more then 5kw then this would be best case savings. 5 kW/0.65=7.69 kW To produce 7.69 kW of mechanical power, considering the engine’s efficiency (40%): 7.69 kW/0.40=19.22 kW of fuel energy Over an hour, this translates to: 19.22 kW×3600 seconds=69.19 MJ Diesel fuel contains 38.6 MJ per gallon, so the fuel saved per hour is: 69.19 MJ/38.6 MJ/gallon = 1.79 gallons Fuel Efficiency with ORC: To travel 62 miles at 8 mpg, the truck uses 7.75 gallons of fuel. Original fuel consumption: 7.75 gallons per hour for 62 miles. Fuel saved per hour with ORC: 1.79 gallons. New fuel consumption: 7.75−1.79 = 5.967.75−1.79 = 5.96 gallons per hour. New fuel efficiency: 62 miles/5.96 gallons≈10.4 mpg Original Scenario (Without ORC System) Fuel Efficiency: 8 miles per gallon (mpg). Distance: 10,000 miles. Fuel Consumption: Fuel Consumption=10,000 miles 8mpg = approximately 1,250 gallons Improved Scenario (With ORC System) Improved Fuel Efficiency: Approximately 10.4 mpg. Distance: 10,000 miles. Fuel Consumption: Fuel Consumption=10,000 miles 10.4mpg = approximately 961.54 gallons Fuel Savings Calculation Fuel Saved: 1,250 gallons−961.54 gallons = approximately 288.46 gallons Cost Savings Calculation To determine the cost savings, you need to know the cost of diesel fuel. Let's assume the cost of diesel fuel is $4 per gallon (this can vary depending on the region and time). A 5kw system Cost Savings: 288.46 gallons×$4/gallon = $1,153.84 A 3kw system Cost Savings 112.73*$4/gallon = $450.92 Either would be a nice monthly savings but how long would it take to recoup the costs of system and extra maintenance?

Interesting Greg, from Barcelona. Thank you for the new technology, I am interested in acquiring it.

Interesting Greg, from Barcelona. Thank you for the new technology, I am interested in acquiring it.

Interesting Greg, from Barcelona.Thank you for the new technology, I am interested in acquiring it.

how about net power achieved (Total Generator Electric output power - pump power consumption)?. voltage and RPM does not mean usable power all you know. ???

How did you built it..also is it more efficient then regular fan turbines parsons

How much are you selling a 10 kw generator?

How much does a 10KW generator cost?

That is supposed to be loose. It has a ceramic bearing and if the metal expands it would crush the bearing if it was tight.

Thank you for doing this!

Its interesting that in real life the flow is slightly different then in the simulation.

amazing content

lovely content