The document outlines a material balance problem involving an evaporator and a crystallizer to process a feed stream containing KNO3 and water. It details the calculations for flow rates, compositions of streams, and other parameters, ultimately leading to a component balance analysis. The results indicate the final mass fractions and flow rates for both the evaporator and crystallizer systems.

1. Material balance Problem Without Chemical
Reaction
Evaporator - Crystallizer Unit
PREPARED BY :
19013123-019 M. FAHEEM ULLAH
University of
Gujrat

2. Numerical:
 Fresh feed stream flowing at 100 kg/h contains 20% by weight KNO3,
(K) in H2O (W). The fresh feed stream is combined with a recycle stream
and is fed to an evaporator. The concentrated liquid solution exited the
evaporator contains 50% KNO3, is fed to a crystallizer. The crystals obtained
from the crystallizer are 96% KNO3 and 4% water. The liquid from the
crystallizer constitutes the recycle stream and contains 0.6 kg KNO3 per 1.0 kg
of H2O.
 Calculate:
• All streams flow rate values
• Compositions of streams
• Mass fraction of recycle stream

3. Flow Sheet Diagram:

4. Step #2 Degree of Freedom Analysis
Evaporator:
D-O-F = Number of Unknowns – Number of independent Equations
 Number of Unknowns = Nu = 4
 Number of independent Equations = 2 (Species) + 0 + 0
 D-O-F = 4 – 2
 D-O-F= 2
Crystallizer:
 Number of unknowns = 4
 Number of independent Equations = 2(Species) + 1 + 0
 D-O-F = 1
Mixer:
 Number of unknowns = 3
 Number of independent Equations= 2(Species) + 0 + 0
 D-O-F = 1
Overall Process:
 Number of unknowns = 2
 Number of independent Equations= 2(Species) + 0 + 0
 D-O-F = 0

5. Feed
Product

6. Step #3
Mass Fraction
 BASIS:
 100 kg/h feed
System (Overall process):
Mass fraction of K and W in the Recycle stream = ?
 XK,5 = 𝟎
.
𝟔 𝒌𝒈
/
𝒉
𝟏 𝒌𝒈 𝑾
𝒉
+
𝟎
.
𝟔 𝒌𝒈 𝑲
𝒉
 XK,5 = 0.375
Mass fraction of W:
 XW,5 = 1 - XK,5
 XW,5 = 1- 0.375
 XW,5 = 0.625

7. Step #4
Component Balance
Overall Material Balance:
 100 kg/h = 𝒎 𝟔 + 𝒎 𝟒 …………. (1)
Component Balance (k)
 (0.2) (100) = 𝒎 𝟒 (0.96) …………..(2)
 𝒎 𝟒 = 20.83 kg/h
 Putting value in (1)
 100 kg/h = 𝒎 𝟔 + 𝒎 𝟒
 100 kg/h = 𝒎 𝟔 + 20.83 kg/h
 𝒎 𝟔 = 79.17 kg/h
 Lets check the Answer by balancing water Component:
Water Balance:
 (0.8)(100) = 𝒎 𝟔 + (0.04)(20.83)………………….(3)
 80 kg/h = 79.17 kg/h + (0.004)(20.83kg/h)
 80 kg/h = 80 kg/h

8. Step #5
Crystallizer
Total Material Balance
 M3 = 20.83 kg/hr+m5 …(4)
Component Balance of K:
 (0.5)(m3) = (m4)(0.96) +(0.375)(m5)
 (0.5)(m3) = (20.83)(0.96)+ (0.375)(m5) …(5)

9. Step #5
By Substitution:
Putting values of (m3) from eq. (4) in (5) :
• (0.5)(20.83 + m5)=(19.99)+(0 375)( 𝒎 𝟓)
• (10.415)+(0.5)( 𝒎 𝟓) =19.99+0.375( 𝒎 𝟓)
• 𝒎 𝟓 =76.6 kg/h
Putting values in eq. (4)
• 𝒎 𝟑 =20.83+76.6
• 𝒎 𝟑 = 97.43 kg/h

10. STEP #6
System : Mixer
Overall Material Balance:
𝒎 𝟏 + 𝒎 𝟓 = 𝒎 𝟐
100 kg/h + 76.65 kg/h = 𝒎 𝟐
𝒎 𝟐 = 176.65 kg/h
Component balance (k)
(0.2)( 𝒎 𝟏) + (𝑿 𝒌,𝟓)( 𝒎 𝟓) = (𝑿 𝒌,𝟐)( 𝒎 𝟐)
(0.2)(100) + (0.375)(76.65) = 𝑿 𝒌,𝟐(176.65)
𝑿 𝒌,𝟐 = 0.276

11. THANK YOU