Suppose MEG is concentrated from 19.73 % to 60 % in this long tube vertical falling film evaporator.
Feed:
H2O =99277KG
MEG =25508KG
DEG =4201KG
TEG =300KG
Product:
H2O =?
MEG =25508KG
DEG =4201KG
TEG =300KG
Total material out from evaporator= 25508 / 0.6
= 42513 kg
H2o at Outlet = 42513-25508-4201-300
= 12504 kg
H2O Evaporator = 99277-12504
= 86773 kg
D = D1+D2+D3 =86773 Kg ________(1)
1st Effect:
ms λs = FCP(T1 – T2) + D1 λ2
Steam temp. Ts = 486 K
λs =1886 kJ/kg
Cp = 1.98 KJ/kg.k
ms (1886)=129286(1.98)(455-468)+D1(202.2)
ms = -1764.49 +D1(1.071)______(2)
2nd Effect:
D1 λ1 = (F-D1)Cp(T2-T1)+D2 λ2
λ1 = 2020.2 kj/kg
λ2 = 2179 kj/kg
Cp =4.23 kJ/kg
(2020.2)D1 = (129286-D1)*4.23*(405- 455)+D2(2179)
D1 = -15118.03+D2*(1.205) ______(3)
3rd effect:
D2 λ2 = (F-D1-D2)Cp(T3-T2)+D3 λ3
λ3 =2326 KJ/KG
λ2 =2179 KJ/KG
Cp =4.03 KJ/KG.K
(2179)D2 = (129286-D1-D2)*4.03*(350- 405)+D3(2326)
From =1
D3 = 86773-D1-D2
2179 *D2 = (129386-D1-D2)(-221.65)+(86773-D1-D2)*2326
4283.35*D2 = 173177756.1- 2104.35*D1___________(4)
From = 3
4283.35*D2 = 173177754.1 - 2104.35 * (-15118.03+1.21*D2)
D2 = 22194 kg/hr
From = 4
D1 = 33927kg/hr
From = 1
D3 = 33927 kg/hr
From = 2
ms = 27931 kg/hr
Component | Mass (kg) IN | Mass % Inx | Mass(kg) OUT | Mass % OUT |
Water | 99277 | 76.8 | 12504 | 29.4 |
MEG | 25508 | 19.73 | 25508 | 60.0 |
DEG | 4201 | 3.25 | 4201 | 9.9 |
TEG | 300 | 0.25 | 300 | 0.7 |
Total | 129286 | 100.0 | 42513 | 29.4 |
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