Hi,
chilling out on a sunday afternoon, vaping away on my Novice mod; the juice is Primoroso from CCvapes, very delicious aromatic tobacco! GF's out of town so I got bored and thought about the heat-up time of my coils. Well, yeah...
So, obviously, a longer coil would take longer to heat up. But I wanted to know how much longer; for example, does it increase in linear proportion to the length of the wire? Also, how does it depend on the thickness of the wire?
Here's what I came up with. The time it takes to heat the wire by some amount ΔT (temperature difference in Kelvin or Celsius; that's 5/9 times the temperature difference in Fahrenheit) is
(1) t = (L/U)2 d r c ΔT
with L the length of the wire in cm and U the voltage. Specific properties of Kanthal A-1 are listed here: density d=7.1 g/cm3, resistivity
r = 1.45E-4 Ω cm, specific heat capacity c ~ 0.55 J/(g K); it depends a bit on temperature though. With these units, t is in seconds.
The point is that if you're vaping at a given voltage the time does not depend on the thickness (or cross section) of the wire. It is proportional to the square of the length of the wire and inversely proportional to the square of the voltage. So, if the voltage of the battery drops from 4.2V to 3.5V, the wire takes (4.2/3.5)2 ~ 1.44 times as long to heat by the same ΔT. Also, if one increases the length of the coil by 40% the heat-up time doubles (√2 ~ 1.4).
If you're vaping at given power P then formula (1) from above turns into
(2) t = (L A / P) d c ΔT ,
where A is the cross section of the wire, i.e. pi/4 times the square of its diameter in cm. So now the time t does indeed increase in proportion to L and A and is inversely proportional to P.
So, the way you'd think about this issue depends on VV vs VW. For a mech mod like my Novice I simply have t proportional to L2, nothing else I have an influence on.
Ok, now back to vaping my Primoroso. Hope you're enjoying a vape, too!
FYI here's how I got to those formulas (1) and (2). The heat (in Joule) supplied to the metal wire is Q = P t, assuming that one can neglect losses to the wick & juice during the short heat-up time. Also, Q = m c ΔT with m the mass of the metal wire (in g), which in turn we can write in terms of its density as m = d L A. Then, t = Q/P = (L A / P) d c ΔT, which is formula (2).
To get (1) replace P by U2/R and finally write the resistance in terms of the resistivity as R = r L / A.
chilling out on a sunday afternoon, vaping away on my Novice mod; the juice is Primoroso from CCvapes, very delicious aromatic tobacco! GF's out of town so I got bored and thought about the heat-up time of my coils. Well, yeah...
So, obviously, a longer coil would take longer to heat up. But I wanted to know how much longer; for example, does it increase in linear proportion to the length of the wire? Also, how does it depend on the thickness of the wire?
Here's what I came up with. The time it takes to heat the wire by some amount ΔT (temperature difference in Kelvin or Celsius; that's 5/9 times the temperature difference in Fahrenheit) is
(1) t = (L/U)2 d r c ΔT
with L the length of the wire in cm and U the voltage. Specific properties of Kanthal A-1 are listed here: density d=7.1 g/cm3, resistivity
r = 1.45E-4 Ω cm, specific heat capacity c ~ 0.55 J/(g K); it depends a bit on temperature though. With these units, t is in seconds.
The point is that if you're vaping at a given voltage the time does not depend on the thickness (or cross section) of the wire. It is proportional to the square of the length of the wire and inversely proportional to the square of the voltage. So, if the voltage of the battery drops from 4.2V to 3.5V, the wire takes (4.2/3.5)2 ~ 1.44 times as long to heat by the same ΔT. Also, if one increases the length of the coil by 40% the heat-up time doubles (√2 ~ 1.4).
If you're vaping at given power P then formula (1) from above turns into
(2) t = (L A / P) d c ΔT ,
where A is the cross section of the wire, i.e. pi/4 times the square of its diameter in cm. So now the time t does indeed increase in proportion to L and A and is inversely proportional to P.
So, the way you'd think about this issue depends on VV vs VW. For a mech mod like my Novice I simply have t proportional to L2, nothing else I have an influence on.
Ok, now back to vaping my Primoroso. Hope you're enjoying a vape, too!
FYI here's how I got to those formulas (1) and (2). The heat (in Joule) supplied to the metal wire is Q = P t, assuming that one can neglect losses to the wick & juice during the short heat-up time. Also, Q = m c ΔT with m the mass of the metal wire (in g), which in turn we can write in terms of its density as m = d L A. Then, t = Q/P = (L A / P) d c ΔT, which is formula (2).
To get (1) replace P by U2/R and finally write the resistance in terms of the resistivity as R = r L / A.
But isn't it easier to calculate the temperature as a delta of mass, then you only need to calculate the mass of the particular wire diameters or lengths and apply the variable?
