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TRIPLE EFFECT EVAPORATOR EXAMPLE Estimate the requirements of steam and heat transfer surface, and the evaporating temperatures in each effect, for a triple effect evaporator evaporating 500 kg/hr of a 10% solution up to a 30% solution. Steam is available at 200 kPa gauge and the pressure in the evaporation space in the final effects is 60 kPa absolute. Assume that the overall heat transfer coefficients are 2270, 200 and 1420 J/m-s-degC in the first, second and third effects respectively. Neglect sensible heat. Assume no boiling point elevation, and also equal heat transfer in each effect. Solution Mass balance, kg/hr

Feed Product Evaporation

Solids Liquids Total 50 450 500 50 117 167 333

Heat balance From steam tables, the condensing temperature of steam at 200 kPa gauge is 134 degC and the latent heat is 2164 kJ/kg. Evaporating temperature in final effect under pressure of 60 kPa (abs) is 86 degC, as there is no boiling point rise and latent heat is 2294 kJ/kg. Equating the heat transfer in each effect,

and ( Now if

then

and

)

So that (

)

(

)

Consequently, (

)

And so the evaporating temperature and latent heat from steam tables:

In first effect is (134 – 12.9) = 121.1 degC; latent heat is 2200 kJ/kg In second effect is (121.1 – 14.6) = 106.5 degC; latent heat is 2240 kJ/kg In third effect is (106.5 – 20.6) = 85.9 degC; latent heat is 2294 kJ/kg

Equating the quantitites evaporated in each eefect and neglecting the sensible heat changes, if , and are the respective quantities evaporated in effects 1, 2 and 3 and is the quantity of steam condensed per hour in effect 1, then, (

)(

)

(

)(

)

(

)(

)

(

)(

)

The sum of the quantities evaporated in each effect must equal the total evaporated in all three effects so that:

(

)

(

)

(

)

Therefore, (

(

)

(

))

Steam consumption It required 115 kg of steam to evaporate based on our calculation with water, that is

Heat exchanger surface Writing the heat balance on the first effect:

(

Then,

Therefore,

)

to evaporate a total of 333 kg

DOUBLE EFFECT EVAPORATOR EXAMPLE A 5% aqueous solution of a high molecular weight solute has to be concentrated to 40% in a forwardfeed double effect evaporator at the rate of 8000 kg/hr. the feed temperature is 40 degC. Saturated steam at 3.5 kg/cm2 is available for heating. A vacuum of 600 mmHg is maintained in the second effect. Calculate the area requirements, if calendria of equal area are used. The overall heat transfer coefficients are 550 and 370 kcal/h-m2-degC in the first and the last effect respectively. The specific heat of the concentrated liquor is 0.87 kcal/kg-degC. SOLUTION

Therefore,

Solving the equations of ms and ms2 simultaneously for the revised calculation,

Thus,

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Feed Product Evaporation

Solids Liquids Total 50 450 500 50 117 167 333

Heat balance From steam tables, the condensing temperature of steam at 200 kPa gauge is 134 degC and the latent heat is 2164 kJ/kg. Evaporating temperature in final effect under pressure of 60 kPa (abs) is 86 degC, as there is no boiling point rise and latent heat is 2294 kJ/kg. Equating the heat transfer in each effect,

and ( Now if

then

and

)

So that (

)

(

)

Consequently, (

)

And so the evaporating temperature and latent heat from steam tables:

In first effect is (134 – 12.9) = 121.1 degC; latent heat is 2200 kJ/kg In second effect is (121.1 – 14.6) = 106.5 degC; latent heat is 2240 kJ/kg In third effect is (106.5 – 20.6) = 85.9 degC; latent heat is 2294 kJ/kg

Equating the quantitites evaporated in each eefect and neglecting the sensible heat changes, if , and are the respective quantities evaporated in effects 1, 2 and 3 and is the quantity of steam condensed per hour in effect 1, then, (

)(

)

(

)(

)

(

)(

)

(

)(

)

The sum of the quantities evaporated in each effect must equal the total evaporated in all three effects so that:

(

)

(

)

(

)

Therefore, (

(

)

(

))

Steam consumption It required 115 kg of steam to evaporate based on our calculation with water, that is

Heat exchanger surface Writing the heat balance on the first effect:

(

Then,

Therefore,

)

to evaporate a total of 333 kg

DOUBLE EFFECT EVAPORATOR EXAMPLE A 5% aqueous solution of a high molecular weight solute has to be concentrated to 40% in a forwardfeed double effect evaporator at the rate of 8000 kg/hr. the feed temperature is 40 degC. Saturated steam at 3.5 kg/cm2 is available for heating. A vacuum of 600 mmHg is maintained in the second effect. Calculate the area requirements, if calendria of equal area are used. The overall heat transfer coefficients are 550 and 370 kcal/h-m2-degC in the first and the last effect respectively. The specific heat of the concentrated liquor is 0.87 kcal/kg-degC. SOLUTION

Therefore,

Solving the equations of ms and ms2 simultaneously for the revised calculation,

Thus,

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