A steel pot (diameter 200 mm) is fully filled with water (2 kg of water) and the water is maintained at a constant temperature of 40°C by supplying heat from an electrical tabletop burner.

Assume the heat supplied from the burner is 0.20 kWh (kW per hour). Let us neglect the heat losses.

The pot is kept open to the surrounding (room). Hot air with a temperature of 60°C and a velocity of 1.5 m/s is blown across the water surface of the pot to assist the evaporation process.

How can the amount (mass) of water evaporated from the pot after 1 hour be calculated? What are the formulae to be considered to calculate the evaporated amount of water under forced convection?

  • $\begingroup$ If heat loses are neglected, you can link the energy given to the vaporization molar enthalpy. $\endgroup$ – user1420303 Nov 23 '18 at 19:52
  • $\begingroup$ Could you tell me how to calculate the amount of evaporated water by linking the heat supplied and the vaporization molar enthalpy? $\endgroup$ – Rabin Nov 23 '18 at 20:52
  • $\begingroup$ In 1 hour the system gets 0.20 kW, divide this value by the molar enthalpy of vaporization and multiply it with 18g/mol $\endgroup$ – user1420303 Nov 23 '18 at 21:02
  • $\begingroup$ What is your understanding of the physical mechanisms involved in determining the mass transfer rate between the liquid water and the air? $\endgroup$ – Chet Miller Nov 23 '18 at 22:12
  • $\begingroup$ The method described by @user1420303 does not guarantee that the temperature of the liquid will be maintained at 40 C. $\endgroup$ – Chet Miller Nov 23 '18 at 22:20

The physical mechanisms involved are convective heat- and mass transfer in the gas (air) phase at the interface, and, depending on whether the liquid water is stirred or unstirred, heat conduction in the liquid phase. To quantify the convective heat and mass transfer in the gas phase, you need to be able to estimate the heat transfer coefficient and mass transfer coefficient between the bulk of the air and the gas at the interface. In addition to this, you have heat of vaporization released at the interface, which comes into play in the heat transfer analysis. The mass transfer coefficient will depend on the diffusion coefficient of water vapor in air; and the mass flux at the boundary will also be determined by the difference between the equilibrium vapor pressure of water at the interface and the partial pressure of water vapor (i.e., the absolute humidity) in the bulk air. You can learn how to include all of this in a mathematical model of your system by consulting Mass Transfer Operations by Treybel. There's plenty of fundamentals for you to learn, but this quantitative treatment of the problem is doable.

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