Drying Paint
Most recent answer: 10/22/2007
Q:
Is there a linear relationship to water evaperatation and temperature? I can dry waterbase paint that’s been applied to a piece of wood in an oven in 1 min. The wood temp reaches on an avg. 133F. If I elect not to use as much energy and dry at a slower rate...say at room temperature I can dry on avg in 8 hrs. If the data given to you is factual and all varibles remain constant could I choose an oven mid range that will dry the wood in four hours at a tempture of 102 degrees? Please refer to the opening question.
Thanks so much,
- Greg Bowers
Bristol, IN USA
Thanks so much,
- Greg Bowers
Bristol, IN USA
A:
Whoops, this nice question fell through the cracks. Maybe it’s still worth answering.
It’s very unlikely that the linear relation you are using gives a good estimate of the drying time. In effect, you’ve asumed that the time (t) changes linearly with temperature (T). Try taking that relation just outside the limits you’ve used, say to 134°F, and you get negative time. So that can’t be right.
It might be a bit closer to guess that the rate (1/t) changes linearly with T. That would give a neagtive rate a little below room temperature, which is probably wrong but not necessarily crazy, since if the paint is water based, it could actually pick up water at low enough T.
A more likely guess would be based on the typical T-dependence of rates of chemical reactions: t= A*eB/T, where you are careful to express T as the amount above absolute zero, as in the Kelvin system, but not °F or °C, and you pick the constants A and B to fit the data points you have. (That’s two unknowns, and your two data points give two equations.) The same form works for either rates or times, just with different constants, and gives the same net results for any T.
From those numbers, I get that B is around 18,000 K, and A is around 10-22seconds. You can do the fit more carefully. My guess is that for a 4 hour time, you’d want to raise T from room temperature by about 3.4 K, or 6°F, to around 78°F or so. 102°F sounds way too high.
Mike W.
There are a great many variables in evaporation problems, such as the ambient humidity and the thermal conductivity of the drying material. Paint is a special case because it may develop a dry skin on top and the paint underneath may dry more slowly -- this effect may well be dependent on how fast the drying process is -- comparing the rate at which water diffuses through the paint and the rate at which it diffuses through the air. I’ve seen very thick skins develop on drying white school glue, for example.
Nothing beats a good experiment, of course. Mike’s parameterizations are good places to start, and are probably pretty close to reality, but try measuring them and see how far you get. This is an example of a more general problem, that of estimating function values when the function is unknown except at a few test points, and an interpolation or extrapolation is desired. A straight-line interpolation is not so bad if the test points are close together compared to, say, the local radius of curvature of the function. If you had measured the times at temperatures of 133 F and 127 F and found them to be reproducible, a linear interpolation to get the time at a temperature of 130 F probably would give a pretty close answer. As Mike says, extrapolating beyond the test points can give silly answers and a well-motivated function is needed. Still, if the answer is really important and you don’t have the luxury of making lots of test points (this happens all the time), a range of plausible model functions should be investigated and an uncertainty on the prediction can be made. In the business, this is one way we use to estimate the systematic uncertainty on a model prediction.
Tom
It’s very unlikely that the linear relation you are using gives a good estimate of the drying time. In effect, you’ve asumed that the time (t) changes linearly with temperature (T). Try taking that relation just outside the limits you’ve used, say to 134°F, and you get negative time. So that can’t be right.
It might be a bit closer to guess that the rate (1/t) changes linearly with T. That would give a neagtive rate a little below room temperature, which is probably wrong but not necessarily crazy, since if the paint is water based, it could actually pick up water at low enough T.
A more likely guess would be based on the typical T-dependence of rates of chemical reactions: t= A*eB/T, where you are careful to express T as the amount above absolute zero, as in the Kelvin system, but not °F or °C, and you pick the constants A and B to fit the data points you have. (That’s two unknowns, and your two data points give two equations.) The same form works for either rates or times, just with different constants, and gives the same net results for any T.
From those numbers, I get that B is around 18,000 K, and A is around 10-22seconds. You can do the fit more carefully. My guess is that for a 4 hour time, you’d want to raise T from room temperature by about 3.4 K, or 6°F, to around 78°F or so. 102°F sounds way too high.
Mike W.
There are a great many variables in evaporation problems, such as the ambient humidity and the thermal conductivity of the drying material. Paint is a special case because it may develop a dry skin on top and the paint underneath may dry more slowly -- this effect may well be dependent on how fast the drying process is -- comparing the rate at which water diffuses through the paint and the rate at which it diffuses through the air. I’ve seen very thick skins develop on drying white school glue, for example.
Nothing beats a good experiment, of course. Mike’s parameterizations are good places to start, and are probably pretty close to reality, but try measuring them and see how far you get. This is an example of a more general problem, that of estimating function values when the function is unknown except at a few test points, and an interpolation or extrapolation is desired. A straight-line interpolation is not so bad if the test points are close together compared to, say, the local radius of curvature of the function. If you had measured the times at temperatures of 133 F and 127 F and found them to be reproducible, a linear interpolation to get the time at a temperature of 130 F probably would give a pretty close answer. As Mike says, extrapolating beyond the test points can give silly answers and a well-motivated function is needed. Still, if the answer is really important and you don’t have the luxury of making lots of test points (this happens all the time), a range of plausible model functions should be investigated and an uncertainty on the prediction can be made. In the business, this is one way we use to estimate the systematic uncertainty on a model prediction.
Tom
(published on 10/22/2007)