Scribd
Upload
267 views4 pages

Triple Effect Evaporator Design

The document describes a problem to design a triple effect evaporator system to concentrate a 25,000 kg/h solution from 10% to 50% solids. Steam is available at 1.877 kg/cm2 and the last effect operates at 66 cm Hg vacuum. Energy balances are performed to determine that 7,190 kg/h will evaporate in the 3rd effect, 6,630 kg/h in the 2nd effect, and 6,180 kg/h in the 1st effect. A steam consumption of 9,488 kg/h is required. The number of tubes needed can be calculated based on the maximum heat transfer area of 164.2 m2.

Uploaded by

Kusmakar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
267 views4 pages

Triple Effect Evaporator Design

The document describes a problem to design a triple effect evaporator system to concentrate a 25,000 kg/h solution from 10% to 50% solids. Steam is available at 1.877 kg/cm2 and the last effect operates at 66 cm Hg vacuum. Energy balances are performed to determine that 7,190 kg/h will evaporate in the 3rd effect, 6,630 kg/h in the 2nd effect, and 6,180 kg/h in the 1st effect. A steam consumption of 9,488 kg/h is required. The number of tubes needed can be calculated based on the maximum heat transfer area of 164.2 m2.

Uploaded by

Kusmakar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 4

Problem on Multiple Effect Evaporator

You are required to design a calendria type triple effect evaporator system to concentrate 25000
kg/h of solution at 38oC containing 10% solids to a product which contains 50% solids. Steam is
available at 1.877 kg/cm2 and the last effect will be assumed to operate at a vacuum of 66 cm Hg
referred to a 76 cm barometer. Water is available at 29 oC for use in a barometric condenser.
Assume negligible BPE, average specific heat of 1 in all effects. The condensate from each
effect leaving at its saturation temperature, negligible radiation losses, average pressure drop
equal in each effect and height and diameter of each tube 150cm and 5 cm respectively. Find out
the the number of tubes required for the purpose, steam consumption and evaporation per kg of
steam if U1, U2 and U3 = 2930, 1220 & 610 kcal/h m2 oC , respectively.

Solution:

Feeed, F= 25000 kg/h, xf = 0.1 and xp= 0.5

Here,

P is final product which is also the product from III effect

Product from I effect is feed to II effect and Vapor from I effect is steam for II effect

Product from II effect is feed to III effect and Vapor from II effect is steam for III effect
Overall Material Balance:

F = E + P = E + P3, where E = E1+ E2 + E3

F. xf = P3. xp ; 25000 x 0.1 = 0.5 x P3 ; thus, P3 = 5000kg/h & E = 20,000 kg/h

Pressure Distribution in Effects:

Pressure in III Effect, P3=76-66 = 10cm Hg (given) = 0.139 kg/cm2

Steam pressure Ps == 1.877 kg/cm2(given)

Total pressure drop = Ps - P3 = 1.877 – 0.139 = 1.738 kg/cm2

Assuming equal pressure drop in each effect 1.738/3 = 0.58 kg/cm2

P1 = Ps - ∆𝑷𝟏 = 1.877 – 0.58 = 1.297 kg/cm2

P2 = 1.297 – 0.58 = 0.717 kg/cm2

P3 = 0.717 – 0.58 = 0.139 kg/cm2

Material Balance in effects

𝐹 = 𝐸1 + 𝑃1 𝑜𝑟 𝑃1 = 𝐹 − 𝐸1

𝑃1 = 𝐸2 + 𝑃2 𝑜𝑟 𝑃2 = 𝑃1 − 𝐸2

Also, E = E1+ E2 + E3 𝐸1 = 20000 − 𝐸2 − 𝐸3

Therefore, 𝑃1 = 25000 − (20000 − 𝐸2 − 𝐸3 ) = 5000 + 𝐸2 + 𝐸3

𝑃2 = 5000 + 𝐸2 + 𝐸3 − 𝐸2 = 5000 + 𝐸3

From Steam Table

First Trial, (Base Temp for energy balance = 0oC)

I Effect II Effect III Effect


Steam 117.8oC 106.5 oC 90 oC
Liquid 106.5 oC 90 oC 51.6 oC
Vapour 106.5 oC 90 oC 51.6 oC
Temp oC Enthalpy(kcal/kg)
λ HE
117.8 527.5 645.3
106.5 534.8 641.3
90 545.3 635.3
51.6 568.9 619.6

ℎ𝐹 = Cp x∆𝑇 = 1 x 38 = 38 kcal/kg ; ℎ𝑃1 = 106.5 kcal/kg; ℎ𝑃2 =90 kcal/kg; ℎ𝑃3 = 51.6 kcal/kg

Energy Balance:

I effect

ℎ𝐹 𝐹 + 𝑆 λ𝑠 = 𝐸1 𝐻𝐸1 + 𝑃1 ℎ𝑃1

20000x 38 + S 527.5 = (20000-𝐸2 − 𝐸3 ) 641.5 + (5000 + 𝐸2 + 𝐸3 ) 106.5

527.5 S + 536 (𝐸2 + 𝐸3 ) = 12412500

II Effect

𝑃1 ℎ𝑃1 + 𝐸1 λ𝐸1 = 𝐸2 𝐻𝐸2 + 𝑃2 ℎ𝑃2

III Effect

𝑃2 ℎ𝑃2 + 𝐸2 λ𝐸2 = 𝐸3 𝐻𝐸3 + 𝑃3 ℎ𝑃3

Substitute all the values and get the equations to solve,

𝐸3 = 7190 kg/h

𝐸2 = 6630 kg/h

𝐸1 = 6180 kg/h

S = 9488 kg/ h

∆𝑇1 = 117.8 − 106.5 = 11.3

∆𝑇2 = 106.5 − 90 = 16.5

∆𝑇3 = 90 − 51.6 = 38.5


𝑆 λ𝑠
𝐴1 = = 151.2
𝑈1 ∆𝑇1

𝐸1 λ1
𝐴2 = = 164.4
𝑈2 ∆𝑇2

𝐸2 λ2
𝐴3 = = 154.4
𝑈3 ∆𝑇3

Evap per kg steam = 20,000/ 9488 = 2.11

For design Maximum area considered

𝜋𝐷𝑜 𝐿𝑁 = 164.2

N can be calculated

You might also like