· Energy balance for effect II:

(3. 3)

Enthalpy values: reference temperature = 60 °C (temp of solution leaving effect II)

H₁ = (112.6 - 60)(0.94 kcal/ kg °C)= 49.4 kcal/ kg

(Specific heat of the solution leaving the 1 st effect is taken as the mean value of the sp. heat of feed and the concentrated liquor, i.e., 0.94)

H₂ = 0 kcal/ kg (w.r.t. the reference temperature of 60 °C)

λ_s1 = 531 kcal/ kg; λ_s2= 563 kcal/ kg

∴ (8000 - m_s1)(49.4) + m_s1 (531) = (8000- m_s1 - m_s2)(0) + m_s2 (563)

⇒ m_s2 = 0.855 m_s1 + 702.6 (3.C)

Solving Eqs. 3.A, 3.B and 3.C for m_s, m_s1 and m_s2:

m_s = 4713; m_s1 = 3395 and m_s2 = 3605 kg/ h

Areas: ;

The areas in the two effects are not equal. Revised calculation is required.

The revised temperature difference in 1^st effect, taken for the calculation (you may also continue the calculation with this revised value (3.13)

Temperature of the saturated vapor from the first effect = 148 -60= 108 °C

Corresponding evaporator drum pressure = 1.2116 bar

Latent heat of the vapor leaving the 1^st effect at 108 °C= 530 kcal/ kg (λ_s1)

Corresponding pressure in the vapor drum = 1.3317 bar

Revised calculation: Energy balance for effect I (reference temperature 108 °C)

∴ (8000)(-72.6) + m_s(506) = (8000- m_s1)(0) + m_s1(530) (3.2)

⇒ m_s = 1.047 m_s1 + 1075 (3.D)

Revised calculation: Energy balance for effect II (reference temperature 60 °C)

∴ (8000 - m_s1)(0.94)(108-60) + m_s1 (530) = (8000- m_s1 - m_s2)(0) + m_s2 (563) (3.3)

⇒ m_s2 = 0.8612 m_s1 +641 (3.E)

Solving 3.A, 3.D and 3.E, m_s= 4652; m_s1= 3417 and m_s2 = 3583 kg/ h

Areas: ;

These areas are fairly close, use A₁= A₂= 104 m² plus overdesign. About 10% excess area will provide a reasonable overdesign.