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I'm studying heat transfer from an electric heater to a cooking pan (Stainless steel) filled with water; specifically, my problem is related to water vaporisation, as I've already found good models to predict radiative and convection losses. The point is that vaporisation losses are easy to calculate after the water reaches the boiling point, as mass of water vaporised times enthalpy of vaporisation of water at atmospheric pressure (which is 2257 kJ/kg). But what happens before boiling? My experiments show that a considerable amount of water already evaporates starting from a temperature of 70°C. In fact, it is well known that water evaporates even at ambient temperature due to mass diffusion, and as the temperature increases the rate of evaporation is also increased. But since the water is not saturated, I guess I cannot calculate this loss by means of any enthalpy of vaporisation. How should I calculate the heat removed from the pan by the water molecules escaping from the upper surface at temperatures lower than boiling?

Thank you very much

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  • $\begingroup$ search on evaporation rate $\endgroup$
    – paparazzo
    Jan 2, 2016 at 17:30

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At the interface, the air is saturated with water vapor at the interface temperature. So you know the partial pressure of the water vapor at the interface (if you know the interface temperature). The water then diffuses into the room air above, where the bulk partial pressure of water is less than the saturation vapor pressure at the interface. There is also convective transport of the air away from the interface, and this air carries the water vapor away from the interface. So this has to be included in your model. There is a heat flux from the bulk of the water to the interface (i.e., at temperature gradient in the water below the interface), and there is a heat flux in the air above away from the interface. The difference between these two heat fluxes is equal to the heat of vaporization times the vaporization rate. There may also be natural convection currents in the water below the interface to enhance the rate of heat transfer. This is just a rough outline of what is happening, but all these things need to be considered in formulating a model of the heat and mass transfer in your system.

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  • $\begingroup$ Actually I have experimentally measured the evaporation rate of my water surface, so I know it. The problem is that it is not fitting weel my theoretical model. I tried a mass transport model both considerng or not convective effects, and in both cases the theoretically estimated rate of evaporation is much lower than what experimentally recorded. Do you have any idea why this would happen? $\endgroup$
    – Franz Eskö
    Jan 4, 2016 at 10:49
  • $\begingroup$ It's very hard to answer this question without knowing more of the details. Are you saying that the experimentally measured evaporation rate is higher than the power provided by the heater divided by the heat of vaporization? $\endgroup$
    – Chet Miller
    Jan 4, 2016 at 12:42
  • $\begingroup$ No, I'm trying to estimate the rate of evaporation by means of mass diffusion theory. It should be a function of diffusion coefficient, vapour concentration (both dependent on T), boundary layer height, surface area, and ambient relative humidity. It should not be directly affected by the heater power, which just affects the time needed to reach a certain dT, and thus the total mass evaporated. But what I obtain from this model is roughly one order of magnitude less than what emerging from experimental data $\endgroup$
    – Franz Eskö
    Jan 4, 2016 at 14:11
  • $\begingroup$ So you are looking at the transient situation and estimating the cumulative amount of evaporation up to time t? $\endgroup$
    – Chet Miller
    Jan 4, 2016 at 14:22
  • $\begingroup$ In my theoretical model yes, I'm doing so. But pay attention: I'm considering the transient situation at different temperatures (30-40-50-60-70 °C). So the "cumulative amount" from 30 to 70 degrees is estimated multiplying the rate of vaporisation of each temperature interval for the corresponding time interval, and summing all of them. As for the experimental data, I get the cumulative amount vaporised from 30 to 70 degrees simply weighing the pot before and after this temperature interval. It should be equal but is bigger by one order of magnitude. $\endgroup$
    – Franz Eskö
    Jan 4, 2016 at 14:36
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As you said, heat loss due to vaporization happens at any temperature. You can reasonably assume that the air right above the liquid is trying to remain saturated - and if you put a lid on the pot, that will indeed be the case. In that case, the heat loss due to vaporization will be quite low.

If you don't have a lid on the pot, the rate of heat loss due to vaporization becomes extremely hard to calculate, because it depends on the relative saturation of the air at the interface, which depends on the details of the air flow. Tables have been published for rates of heat loss as a function of wind velocity; see for example my earlier answer on heating swimming pools for some details. This shows that the rate of evaporation strongly depends on wind velocity: you are familiar with the concept of wind chill of which this is one manifestation.

The actual heat per unit mass removed can be determined from tables of heat of vaporization as a function of temperature - see for example https://en.wikipedia.org/wiki/Enthalpy_of_vaporization from which I reproduce this figure:

enter image description here

This shows a clear dependence of the heat of vaporization on temperature: not surprising, since you expect a molecule that escapes the liquid at a lower temperature to carry a relatively higher fraction of energy with it. I fitted (by eye) a straight line to the data between 260 and 360 K (roughly from freezing to boiling); the overlay (zoomed in) looked like this:

enter image description here

The coefficients of that straight line were

$$y = 3195 - 2.5T$$

which gives a value of 2262 J/g at 373 K - pretty close to various published values. I recommend using this relationship to estimate the heat of vaporization at temperatures below boiling - it should be close. The main problem is knowing how much liquid evaporates, and that's much harder.

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  • $\begingroup$ I assume wind speed is zero since my model refers to a controlled laboratory setting; my problem is with the heat of vaporization: I also have tables showing the correlation between heat of vaporization and temperature, but I'm guessing if using the concept of "heat of vaporization" for a liquid which is sub-cooled (lower temperature than transition phase) is correct! In fact, tables do not correlate heat of vaporization with "temperature" of water, but rather with "saturation temperature" of water [...continues] $\endgroup$
    – Franz Eskö
    Jan 4, 2016 at 9:22
  • $\begingroup$ [...follows] For a pressure of 1 bar, saturation temperature is 373,15K and heat of vap. is 2257 kj/kg. Tables show, for example, that at 300K heat of vaporization is 2438 kj/kg, but that means 300K SATURATION temperature, which requires a 0,035 bars pressure! If we are working always at 1 bar pressure, there is no heat of vaporization before transition phase, there is ony sub-cooled liquid $\endgroup$
    – Franz Eskö
    Jan 4, 2016 at 9:24
  • $\begingroup$ You are overthinking this. The vapor pressure will be less than one atmosphere but at a given temperature the air will be saturated with just a little bit of moisture. It just happens that at 373.15 K that pressure is one atmosphere so water will boil; but air can be saturated (100% relative humidity) at temperatures below the boiling point. $\endgroup$
    – Floris
    Jan 4, 2016 at 10:42
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    $\begingroup$ Ok, then is reasonable to use the heat of vaporisation. Thank you $\endgroup$
    – Franz Eskö
    Jan 4, 2016 at 10:46

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