Single-effect Evaporators: Screencasts
Describes the operation of a single-effect evaporator that is used to concentrate a solution and shows mass balances.
We suggest that after watching this screencast, you list the important points as a way to increase retention.
Explains the energy balance for a single-effect evaporator, which is used to increase the concentration of a solute in a feed stream by evaporating some of the liquid in the feed. A feed stream of saturated steam supplies heat to evaporate the liquid.
We suggest that after watching this screencast, you list the important points as a way to increase retention.
Important Equations:
Heat transfer (assumes liquid is well mixed)
\(\dot{Q} = UA(T_s – T_{evap})\)
\(\dot{Q}\) = heat transfer rate (kJ/s)
\(U\) = overall heat transfer coefficient (kJ/(m2 s K)
\(A\) = heat transfer area (m2)
\(T_s\) = steam temperature (ºC)
\(T_{evap}\) = temperature of liquid and vapor in evaporator (ºC)
Mass balances
\(\dot{m}_f = \dot{m}_V + \dot{m}_L\)
\(\dot{m}_f\) = mass flow rate of liquid feed (kg/s)
\(\dot{m}_V\) = mass flow rate of vapor leaving evaporator (kg/s)
\(\dot{m}_L\) = mass flow rate of concentrated liquid leaving evaporator (kg/s)
\(x_f\dot{m}_f = x_L\dot{m}_L\)
\(x_f\) = mass fraction of solute in liquid feed
\(x_L\) = mass fraction of solute in concentrated liquid leaving evaporator
Energy balance (assumes saturated steam enters and saturated liquid water leaves evaporator)
\(\dot{m}_fH_f + \dot{m}_sH_s^V = \dot{m}_VH_V + \dot{m}_sH_s^L + \dot{m}_LH_L\)
\(\dot{m}_s\) = mass flow rate of steam entering evaporator (kg/s)
\(H_f\) = specific enthalpy of liquid feed (kJ/kg)
\(H_s^V\) = specific enthalpy of steam entering evaporator (kJ/kg)
\(H_V\) = specific enthalpy of vapor leaving evaporator (kJ/kg)
\(H_s^L\) = specific enthalpy of condensed steam (saturated liquid) leaving evaporator (kJ/kg)
\(H_L\) = specific enthalpy of concentrated liquid leaving evaporator (kJ/kg)
\(\dot{Q} = \dot{m}_s(H_s^V – H_s^L) = \dot{m}_s\Delta H_s^{vap}\)
\(\Delta H_s^{vap}\) = enthalpy of vaporization of steam at temperature \(T_s\)
steam economy = \(\frac{\dot{m}_V}{\dot{m}_s}\)