Mathematic modeling of evaporators

Evaporation is a energy intensive operation using thermal energy to make steam, To do this on an industrial scale evaporators are widely used. The main use of evaporators is in air conditioning. It can also be used to remove water or other liquids from the mixtures The basic mechanism of all these processes is evaporation. The evaporator is a main component for heat transfer in an air-conditioner system, its characteristics have effects on cooling capacity and COP directly. A double-systems air conditioner is composed of a compressor, a condenser, and two evaporators which are connected in parallel and controlled by two electric expansion valves separately. This configuration results in the same suction pressure of the two paralleled evaporators. Because the two evaporators are controlled separately, and their conditions for heat transfer are different, thetwo evaporators are independent to some extent. The valves not only have effects on controlling the superheat degrees, but also have effects on allocating the flow amount. So the double-evaporators have different characteristics compared with the single evaporator

Evaporators are classified in these categories

On the basis of type of film

• Falling film evaporator
• Rising film (Long Tube Vertical) evaporator
• Climbing film and falling film plate evaporator.

On the basis of process

• Single effect evaporators.
• Multiple effect evaporators.

Modeling of single effect evaporators

The mathematical model Describing the single effect evaporator is given by material balance,component balance,Energy balance on single effect evaporators.

Component material balance on solute and solvent are

$m_{f}*x_{f}=m*y_{v}$ ...1

$m_{f}(1-x_{f})=m_{v}+m(1-y_{v})$ ...2

Where, $m_{f}$ = feed flow rate( kg/sec)

$m$ =liquid flow rate (kg/sec)

$m_{v}$= Vapour flow rate (kg/sec)

$m_{f}$=feed flow rate(kg/sec)

$m_{s,i}$=liquid flow rate initially before heating(kg/sec)

$m_{s,o}$= liquid flow rate after heating(kg/sec)

$x_{f}$ = mass fraction of solute in feed

$y_{v}$ =mass fraction of solute in the product or vapour

$x$= mass fraction of bottom liquid product

Total material balance is given by

$m_{f}=m_{v}+m$ .....3

An Enthalpy balance on a process stream yields

$m_{f}.H_{f}+Q=m_{v}.Hv+m.H$

From Equation (3), adding and Substracting $m_{f}.H$,we get

$m_{f}.H_{f}+Q=(m_{f}-m)Hv+m_{f}.H-m_{f}.H$$T_{l}$

$m_{f}*(H_{f}-H)+Q-(m_{f}-m)*(H_{v}-H)=0$ .....4

Where,

$H_{f}$ = enthalpy of feed(kJ/kg) $H$= enthalpy of bottom product(kJ/kg)

$H_{v}$ = enthalpy of vapour at the boiling point temperature of the feed( kJ/kg).

$Q$= rate of heat transfer across the tubes ( from the steam to the thick liquor) (kJ/sec)

The enthalpy balance on the given condensor diagram is given by

$Q=m_{s,i}(H_{s,i}-H_{so,i})$

The rate of heat transfer Q is given by the following relationship

$Q=UA(T_{l}-T)$

Where-:

$U$ = overall heat transfer coefficient

$A$ =surface area of the tubes available for heat transfer.

$T_{l}$= saturation temperature of the steam entering the chest.

$T$= boiling point temperature of thick liquor at the pressure of the vapour space

HENCE BY USING Equations.(1 -6), WE CAN MODEL THE SINGLE EFFECT EVAPORATOR AND FIND THE PARTICULAR PARAMETERS AT GIVEN CONDITIONS LIKE (FLOW RATE,ENTHALPY,MASS FRACTION,MOLE FRACTION,TEMPERATURES)

See also

• Evaporators
• Mathematical Modeling
• Evaporation

References

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