t(1/2)= 0,69*V/MU where: ...
Published by Apostolos Kavadias, IDRO - APOSTOLOS KAVADIAS
t(1/2)= 0,69*V/MU
where:
t(1/2)= half holding time (h). It is the time needed to lower the additive concentration at half (1/2), after stopping it's addition (dosing).
V= water system total volume (m3)
MU= Make-up water (m3/h). Equal to water looses, including blow down.