I have a cylindrical tank with an 'm' amount of water in it. The tank is sealed except through an exhaust with a fan that keeps a lower than atmospheric pressure (but not vacuum) inside the tank. The fan also works to expel forming gases. The vessel is kept at a constant temperature 'T'. This is a description of an actual device being built in which the lower pressure inside the vessel is used to lower the boiling point of water (and to remove it faster). Is there a formula anyone can recommend to calculate the rate at which water evaporates and exits the vessel?

Question

Answer 1

E_{m a s s} = \frac{m R_{n} + ρ_{a} c_{p} (δ e) g_{a}}{λ_{v} (m + γ)}

Where

E_{m a s s}

= mass of the evaporated water
m = Slope of the saturation vapor pressure curve

(P a K - 1)

R_{n}

= Net irradiance

(W m - 2)

ρ_{a}

= density of air

(k g m - 3)

c_{p}

= heat capacity of air

(J k g - 1 K - 1)

g_{a}

= momentum surface aerodynamic conductance

(m s - 1)

δ e

= vapor pressure deficit

(P a)

λ_{v}

= latent heat of vaporization

(J k g - 1)

γ

= psychrometric constant

(P a K - 1)

Answers (1)

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