I What will happen to the heat input of the two-phase system?

  • Thread starter nesca
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If I have a tank half-filled with low boiling point liquid at ambient temperature, then this tank is placed inside dry oven whose temperature is kept to be around 5℃ above the liquid boiling temperature. A fan is placed inside the dry oven as an effort to make temperature distribution even or at least minimize large gap to happen. Because of this the liquid temperature is increased, yet there is no bubble developed, or in other words there is no liquid bulk motion inside the tank.

After boiling temperature is reached, the tank is tightly sealed, thus the pressure is increased due to liquid evaporation.

In this case, to calculate heat input, is it okay to ignore heat convection both in liquid and gas and focus with heat conduction only?

Is the equation below can be used for heat input calculation?


What if the density and liquid specific capacity vary with temperature? How to determine the mass and specific capacity?

If the gas phase is equal to liquid vapor and there is no information about vapor properties, what is the best way to at least obtain the density, thermal conductivity, and heat capacity for the gas phase?


Science Advisor
What about the heat of evaporation?
What about the heat of evaporation?
If I'm not mistaken, to determine the heat of evaporation, we have to obtain internal energy of the system, and to do it ones should know the heat input of the system, shouldn't they?


Science Advisor
No, if the system reaches equilibrium, the liquid will evaporate either until it is all vapour, or until the vapour pressure equals the equilibrium vapour pressure at the oven temperature (depending on how much liquid there is and the volume of the oven). You can work out how much liquid evaporates, and knowing the molar (or specific) heat of vaporisation, you can work out the heat input for vaporisation.
If you have no information on vapour properties, you can as a first approximation assume it's an ideal gas and calculate what its properties would be.

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