With thanks to Mr. Olson for his kind words allow me to add some hopefully useful information...
(Flow x Pressure)/ 1460= Electric Brake HP (this is a legitimate formula based on experience, not an engineering based theoretical value from charts showing relative component efficiency)
Brake HP x 1.5 = gasoline engine HP (again based on experience in life)
Water weighs 8.4#/US gallon or 1kg/litre (fact)
Impact force is (GPM x square root of PSI)/18.92 (this is theoretical but close enough to use)
Therefore at 7.7 GPM the operator will be dealing with 64.7# per minute coming from the nozzle.
7.7 GPM @ 3000 PSI would require a 15.8 HP electric motor or a 23.7 HP gasoline engine, minimum.
A 525 gallon tank will have a useful 473 gallons (90% utilization) or 61 1/2 minutes of run time. There is also the same amount of discharge that needs to be dealt with. Most municipal water supply is less than 6 GPM, few rural pumps offer much more.
Impact force would be 22.3# at 7.7 GPM, 5.6 GPM would give 16.2#. So there would be more cleaning available and would take ~27% less time to do the same job, with one gun or two.
A rule of thumb is that you will need 100,000 Btu to raise 1 gallon of water 140 degrees Farenheit. Therefore conventional wisdom says that you will need an 800,000 Btu burner. For LP this would burn about ~37# per hour; oil would be ~4.5 GPH.
Hope this helps? Cheers.