Re: Power - how much is too much
- --- In Distillers@yahoogroups.com, "rye_junkie1" <rye_junkie@...> wrote:
>You could do all the heat loss calculations but it gets messy because you have losses due to convection, radiation and conduction if the boiler is in contact with the floor, a much simpler way is to distill some water with zero reflux, time the length of run and measure the volume of water collected
> --- In Distillers@yahoogroups.com, "rosnekcaj" <jensor4@> wrote:
> > Thanks to your suggestions. I read Mike Nixon's treatise on vapor speed, However, his analysis omits any heat lost. That is, heat generated by the heating element that escapes to the surrounding environment. A submersion heater would impart 100% of its wattage input to the creation of vapor, whereas an external heat source will lose some of its input wattage to the surrounding environment.
> > I assume that there should be a way of testing a given column to find out what heat must be supplied for it to operate efficiently so if using an external heat source such as a hot plate, it could then be adjusted to make up for heat losses.
> > I created my own controller to provide continous heat. I setup a 555 timer circuit to run at a frequency of 1 Hz. I think a frequency of 1 hertz should be sufficiently fast enough to be damped out by the heat coupling to the liquid. The circuit includes a potentiometer to vary the duty factor over a range of less than 1% to greater than 99%. Its output then switches a SSR (solid state relay with zero crossing switching) that switches the power to the heating element. It works very well.
> Even with internal elements there are losses in heat through the boiler walls and lid. These losses vary with the amount,type and placement of insulation put on the boiler, lid and column. With that in mind I do not know of an accurate way to account for these variables. I believe someone said once that an uninsulated boiler would lose some 500 watts through the walls. The best test that i know of is to find what the maximum product output is for a given heat input and adjust the reflux ratio/takeoff to that. I dont know the math but if you know the max output for a given heat input you should be able to calculate that into vapor volume, speed and then back into heat input. For me, if the column is shaking back and forth like a leaf in a breeze then its time to turn the heat down.
1000 watts will produce 26.5ml water per minute near as dammit
if using internal element subtract output from input to give total losses, it will probably be quicker than doing all the heat loss
it should also be noted that mains voltage is quoted +/- 10% so a 1000 watt rated element could in fact be running anything between 810 and 1210 watts so you need to be able to measure the input power also