Theory for 100ml, the practice is different and involve setting time, heat transfer and loss from oven construction.
Ideally for setup, amplified pressure sensor with 12bit, better 14bit is needed in order to detect phase3 of vapour phase.
As you can see on the numbers below, if you multiply it with 10, the power requirement is a lot and this is the reason big VPH that can load 1 Lt of galden have 3.2Kwh heaters.
As Galden is a bit hygroscopic, the first heating to 150 degree is with reduced power in order to give the alcoholic and water contens from flux and from absorbed air the possibility to
escape. This assumes a Heating power of 1200W wile monitoring the main voltage using packet modulation as usual on heaters.
Step 1: Supply energy for heater with power 514 W in 43.3 second to increasing liquid temperature from 20°C to 150°C.
Step 2: Supply energy for heater with power 1028 W in 13 second to increasing liquid temperature from 150°C to 230°C.
Step 3: Supply energy for heater with power 112 W in 15 second to keep liquid temperature at 230°C
Step 4: Cooling liquid with speed 6.1 ml/second in 18 second to cooling hot iquid from 230°C to 120°C
This is the theory and pratice is that cooling must be doubled (thermal mass of oven heat it up again, repeated later two times for 8.2 sec)
resulting in 270 sec for reflow and 530 sec for reaching 120 degree.
In theory, it is 71,3 sec for the reflow and 89,3 for the whole cycle. If you make a simulation with thermal mass, thermal loss of oven construction, .... maybe you get the same
values as the reality. This cooling is fine because it is cheap and simple to do, but not for production, only for setup.
Is that what you had wanted or not ?
Basically it start at 20 deg. and need to go up to 230 deg. this are 210K temp. diff.
Now 100ml = 182 gr and , the specific heat is 177 J/K and specific heat for producing vapour is 114.66 J/ml, heating element is 1000W as assumption.
That means 230 deg. heated 100ml fluid can be evaporated with 1kwh heater in less then 11.5 seconds or 1kg in 63 seconds.
210° * 177J = 37170J / 1000w = 37,17 sec for reaching 230 deg and then if you want evaporate 20.93ml you need 2.4 sec heating power of 1kw.