On Wikipedia it states that the main benefit of a water jet cutter is that there is no heat-affected zone.

I am wondering about the physics behind this. I have found experiments supporting this, but I cannot produce a theoretical argument. Am I wrong in assuming that by the time it would take to heat up the material, a new influx of cold water keeps the temperature down?

  • $\begingroup$ What material are you cutting? Compare to a laser or plasma cutter... $\endgroup$
    – Solar Mike
    Commented Dec 7, 2017 at 16:09
  • 1
    $\begingroup$ "No heat-affected zone" is not the same as "no heat". $\endgroup$ Commented Dec 7, 2017 at 17:06

2 Answers 2


The item to be cut is completely submerged in a large water filled tank. Some as large as 10x10 feet, or more. What little heat that is created by the abrasion, and the pump, is immediately carried away from the kerf. I've had slots cut in Kydex® plastic with no apparent heat increase in the material.


I can't speak to the mathematics of the heat generated during water jet cutting, but the following points may clarify the situation for you.

Heat affected zone is generally with metals where the cutting process will generate enough heat to change the mechanical properties of the metal, essentially locally heat treating the material at and nearby the cutting area.

Waterjet does not get near hot enough to create a heat affected zone. Remember, It's water at ambient pressure so it would become steam if it were any hotter than boiling temperature. That would make it an ineffective medium to carry the abrasive to blast through the metal.

Also, let's assume that the machine's water reservoir/supply is large enough where its temperature is stable. The pumping/pressurizing will create heat and the abrasive cutting action will create heat, but there is a "constant" supply of water at a stable temperature being applied to the material which would regulate the materials temperature.

That being said, a waterjet part could still become warm to the touch after being cut.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.