Vs is the relative humidity of the medium. Hence the change in volume of water in the container due to the evaporation is. A ÏL. â H. â t. = -A. 1. L - H. w PS.
��������������������������
�������������������������������� ������������� Here we consider a liquid in a closed container. We will consider a moving control volume V(t) that encompasses the liquid in the container. The interface height is given by H(t) which is changing with time. Thus the volume of liquid at any time t is � (�) = � � (�)
(1)
where A is the cross-sectional area of the container. We will consider first the isothermal quasi-steady evaporation of the water. The macroscopic mass balance over the liquid is ⅆ ⅆ�
ρ� ⅆ� + ρ� (� - �) · � ⅆ� = �
� (�)
(2)
�
where w=dH/dt is the velocity of the liquid gas interface and v is the velocity of the liquid. This can be expressed as ⅆ� ⅆ�
= - ρ� (� - �) · � ⅆ� = -���� �
(3)
�
where jrel kgm2 ·s is the mass flux leaving the moving control volume due to evaporation, and the mass of liquid in the container at any time t is �=�
ⅆ(ρ� �) ⅆ�
Since the evaporation is slow, we can assume that the concentration of water vapor in the gas phase is quasi-static ( inertial effects are not important). The quasi-static concentration field in the ambient gas that constitutes the surrounding gas phase ignoring (convective effects) must satisfy ⅆ� � ⅆ��
=�
(4)
Let the concentration of the water in the gas phase at the liquid surface be equal to its equilibrium concentration cs and sufficiently far away from the interface let it be c∞ . Then the concentration field for the water species can be found be solving Eq. (4) to give � (�) =
(�∞ - �� ) �-�
(� - �) + �∞
(5)
At the surface of the evaporating liquid (y=H) the mass flux is ���� = -
ⅆ� = - ⅆ�
(�∞ - �� ) �-�
=
� �-�
(�� - �∞ )
(6)
Thus the total mass leaving the surface is ���� � = �
� �-�
(�� - �∞ )
(7)
2 ���
BGHNotesEvaporatingLiquidInterface.nb
Thus the change in mass of the liquid in the container is given by �
ⅆ� = -�
�-�
ⅆ�
(�� - �∞ )
(8)
If we assume the vapor phase is an ideal gas then we can write (�� - �∞ ) =
� ℛ�
��� - ��∞
(9)
where PVi is the vapor pressure of the evaporating liquid at station i. Here w is the molecular weight of the liquid and ℛ is the universal gas constant. It is customary to take PVs to be the equal to the saturation vapor pressure of the evaporating species PSVs at temperature T, i.e., ��� = ��� �
(10)
Then in the case when water is evaporating in air, we can write (�� - �∞ ) =
� ℛ�
�� ��� � - �∞
(11)
=
� ��� � ℛ�
(� - ��)
Vs where RH = PVs ∞ PS is the relative humidity of the medium. Hence the change in volume of water in the
container due to the evaporation is � ρ�
�
� ��� �
�-�
ℛ�
ⅆ� = -� ⅆ�
(� - ��)
(12)
and simplifying gives ⅆ�
=-
ⅆ�
� ��� �
�
ℛ�
ρ� � - �
(� - ��)
(13)
If the temperature is constant during evaporation ( or more precisely quasi-steady), then we have ��� - �� + � � (�� - �) = -β ��
����
β=�
� ��� � ℛ�
(� - ��) (14)
��� - �� + � � (�� - �) = -�
� ��� � ℛ�
(� - ��)
The solution to this problem is ��������
��� = ������� �[�] ⩵ -
� (�) = � -
β � - �[�]
� �[�] ⩵ ��� �[�]� �
�� � - � � � � + � � + � � β
(15)
Thus all the water has evaporated when �=
�� (� � - �� ) �β
Note that temperature effects enter through β. In reality T will be changing albeit slowly. To account for
BGHNotesEvaporatingLiquidInterface.nb
���3
the change in temperature we examine the energy balance.
������������������� Temperature effects can best be understood by examining the macroscopic energy balance for the evaporating system. But first we start with the thermal energy point equation for the liquid which is written as �� = ∇ · (� ∇ �) ρ� �� (16) �� In the point equation we are ignoring viscous dissipation effects and pressure work related terms which is a reasonable assumption for slow evaporation. At the liquid/gas interface of the evaporating liquid we have the following jump boundary conditions ( mass and energy): ρ� (�� - �) · � = ρ� (�� - �) · � (17)
ρ� �� (�� - �) · � + �� · � = ρ� �� (�� - �) · � + �� · � where Hi is the enthalpy per unit mass of phase i. These two expressions can be combined to give (�� - �� ) · � = λ ρ� (�� - �) · ��
λ = �� - ��
(18)
where λ is the latent heat of evaporation per unit mass. Integrating point energy equation (16) over the liquid volume V(t) gives ⅆ ⅆ�
ρ� �� � ⅆ� +
� (�)
ρ� �� � (�� - �) · � ⅆ� = � (�)
�� · � ⅆ�
(19)
� (�)
Substituting Eq. (18) into Eq.(19) to eliminate qL gives ⅆ ⅆ�
The LHS of Eq. (21) can be simplified by making the following assumptions: For a small liquid volumes, we can safely assume that T, CP and ρL are approximately uniform within the liquid so that ⅆ ⅆ�
ρ� �� � ⅆ� + ρ� �� � (�� - �) · � ⅆ� ≈
� (�)
�
ⅆ ⅆ�
(�� � �) - �� �
ⅆ� ⅆ�
= � ��
ⅆ� ⅆ�
(22)
Thus the energy balance becomes � ��
ⅆ� ⅆ�
= � �� · � - λ
ⅆ�
(23)
ⅆ�
In the absence of radiation and convective effects the heat transferred from the gas phase to the liquid is ⅆ� = -� �� (�∞ - �� ) � ≡ � �� · � = -� �� ⅆ�
�/�
(24)
where Ts is the surface temperature of the liquid. Thus we can write the energy balance as � ��
ⅆ� ⅆ�
= -� �� (�∞ - �� ) - λ
ⅆ� ⅆ�
(25)
4 ���
BGHNotesEvaporatingLiquidInterface.nb
Then for a quasi-static process we can assume that ⅆ T /ⅆ t ≈ 0, so that (�∞ - �� ) = -
�
ⅆ� (26)
λ
� �� ⅆ�
But we have already shown that �
� ��� �
�-�
ℛ�
ⅆ� = -� ⅆ�
(� - ��)
(27)
Thus the temperature difference is (�∞ - �� ) =
λ
� ��� �
(� - �) ��
ℛ ���
(� - ��)
(28)
and we have already found that � - � (�) =
(29)
�� � - � � � � + � � + � � β
so that � ��� �
λ
(�∞ - �� ) = �
�
� � - � � � � + � + � � β ��
(� - ��)
ℛ ���
(30)
�������� The approach taken here is based on the fact that temperature and concentration changes are relatively slow so that the complicated effects to fluid inertia, and time dependent effects due to convective transport whether from temperature or mass can be neglected. We have not shown how to compute the surface temperature of the water. This requires an energy balance over the liquid phase and then one must account for the energy input ( heating) to the liquid. The simplest approach would be to consider only conduction effects so that the energy balance reduces to ∇ · (�� ∇ �)
(31)
Integrating this over the volume of the liquid and assuming that the side walls of the container are insulated, we would get ∇ · (� ∇ �) ⅆ� = � · (�� ∇ �) ⅆ� = ��� - ��� �
�
ⅆ� ⅆ�
=� �
Thus the heat input Qin to the bottom of the container is related to ��� = � ��
