How to determine the vapor pressure of a glycerine and propylene glycol mixture?
Your best bet for this will be to use the Clausius-Clapeyron relation. This is an exponential function that describes the vapor pressure of a substance as a function of temperature, and uses the enthalpy of vaporization (latent heat) and a constant as parameters.

The constant is dependent on the particular substance, but there is a way to eliminate it from the equation if you can get one data point: the vapor pressure at a specific temperature. If you can do that, then you can use the two-point form to predict vapor pressure at another temperature:

$\ln\frac{p_2}{p_1} = \frac{-\Delta H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

This version only has one parameter - the enthalpy of vaporization ($\Delta H_{vap}$). $p_1$ and $T_1$ are the vapor pressure and temperature at point 1 (an arbitrary point), and $p_2$ and $T_2$ are the vapor pressure and temperature at a different arbitrary point.

For a pure substance, if you know the enthalpy of vaporization and the boiling point, you an calculate the vapor pressure at any other temperature. That is because the boiling point is the temperature at which the vapor pressure equals one atmosphere. In other words, you would set:

$p_1 = 1 \space \mathrm{atm}$

$T_1 = T_b$,

Then solve for $p_2$. The equation would be:
$p_2 = p_1\exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{T_2}-\frac{1}{T_1})) \tag{1}$

When you use this equation, make sure you first convert the temperatures to Kelvin. This only works with an absolute temperature scale. Pressure can be in whatever units you want, as long as both pressures have the same units. In other words, convert 1 atm to whatever units you want to use for vapor pressure.

The data you listed didn't have the enthalpy of vaporization for either component, but they are available from NIST:

$\Delta H_{vap} = 91.7 \space \mathrm{\frac{kJ}{mol}}$ - glycerol

$\Delta H_{vap} = 71.2 \space \mathrm{\frac{kJ}{mol}}$ - ppg

Incidentally, NIST lists different values for the boiling point than you have. I would double check those before plugging numbers in.

Now, you don't have a pure substance; you have a mixture. Depending on how non-ideal it is (which basically means how much the molecules of one component will affect the other components) you might have to account for that if you want to get very accurate results. On the other hand, if you just want an estimate and are planning to experimentally adjust the power output from there, you can get a good starting point by assuming that the mixture and the vapor are both ideal. Under those conditions, the total vapor pressure of the mixture will be equal to the sum of the partial pressures...

$p_t = p_{gly} + p_{ppg}$

... and you can use the Clausius Clapeyron equation from above for each component individually to find the respective vapor pressures. Then you can add them together if you want the total vapor pressure. However, you may be more interested in the vapor pressure of each component individually, since (from what I have heard) glycerol is more responsible for the "cloud" that forms, while propylene glycol is the better organic solvent. It would be helpful to know how much vapor of each component a given mixture would produce at a given temperature.

That's how you can predict vapor pressure from temperature. To relate this to the amount of heat you need, you will need to do another step. Temperature is a function of heat supplied and the specific heat of the substance:

$q = mC\Delta T \tag{2}$

where $q$ is the amount of heat transferred, $m$ is mass, and $C$ is the specific heat. You know the specific heat of the mixture, and the mass is an independent variable since you will control the composition and amount of the mixture. If you assume that volumes and heat capacities are additive, that mixing is fast, and that heat transfer is fast (at least locally), you can estimate the temperature (or rather, the maximum temperature) by connecting this to the power supplied by your device. Power in watts is the same as Joules/second. Therefore, you can multiply the power output by the time to get a total (maximum) amount of heat transfer (in Joules):

$q = P*t \tag{3}$

We can combine equations $(3)$ and $(2)$ and solve for the operating temperature:

$T=\frac{Pt}{mC}+T_i \tag{4}$

where $T_i$ is the starting temperature of the mixture (before you apply power).

Now we can substitute this back into $(1)$ to come up with an estimate of the vapor pressure of each component as a function of the mass of the liquid, the power applied, and the length of time that it is applied for:

$p_{vap} = (1 \space \mathrm{atm}) \exp([\frac{-\Delta H_{vap}}{R}](\frac{1}{\frac{Pt}{mC}+T_i}-\frac{1}{T_b})) \tag{5}$

The real vapor pressure will likely be lower, but I think this would get you to a good starting point.

A couple of things to remember:


$m$ refers to the mass of each component that is in close proximity to the heating element.
$Delta H_{vap}$ is in kJ/mol - make sure that the $R$ you use has the right units.
$T$ must be in Kelvin (both of them)
The units of $P$, $m$, and $C$ have to cancel out to give K.


If you want to get a (rough) upper estimate of the mass that will be vaporized, you could assume an ideal gas and use

$m=\frac{pVM}{RT}$

where $V$ is the volume of the vaporizer chamber, and $M$ is the molar mass of the component.