Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 3 (Q.No. 23)
23.
A multiple effect evaporator has a capacity to process 4000 kg of solid caustic soda per day, when it is concentrating from 10% to 25% solids. The water evaporated in kg per day is
Discussion:
6 comments Page 1 of 1.
Ahmad Ijaz said:
1 decade ago
If 10% solution contains:
Caustic soda = 4000 kg.
Then;
Mass of water = 40,000 kg.
&
If 25% solution contains:
Caustic soda = 4000 kg.
Then;
Mass of water = 16,000 kg.
So, water evaporated is,
Water evaporated = 40,000 - 16,000.
= 24,000 kg.
Caustic soda = 4000 kg.
Then;
Mass of water = 40,000 kg.
&
If 25% solution contains:
Caustic soda = 4000 kg.
Then;
Mass of water = 16,000 kg.
So, water evaporated is,
Water evaporated = 40,000 - 16,000.
= 24,000 kg.
(4)
Sushil said:
3 years ago
1st case: (Inlet stream).
100 ~10 kg
x~4000 kg
Solution = 40000 kg.
2nd case, (Outlet stream).
100~25,
y~4000,
y=4000*100/25 = 16000.
I/L-O/L = Water evaporated.
Water evaporated = 24000kg.
100 ~10 kg
x~4000 kg
Solution = 40000 kg.
2nd case, (Outlet stream).
100~25,
y~4000,
y=4000*100/25 = 16000.
I/L-O/L = Water evaporated.
Water evaporated = 24000kg.
(2)
Ramsajanpaswan said:
2 years ago
Let;
Input(X).
WATER EVAPORATED (Y) & CAUSTIC CONC(Z)
Then 25% of Z =4000 then 100%= 16000
MB of caustic, 10% of X = 25%of Z,
X = 4000/0.1=40000,
Overall MB => X=Y+Z,
40000=Y+16000,
And we get Y = 24000kg/day.
Input(X).
WATER EVAPORATED (Y) & CAUSTIC CONC(Z)
Then 25% of Z =4000 then 100%= 16000
MB of caustic, 10% of X = 25%of Z,
X = 4000/0.1=40000,
Overall MB => X=Y+Z,
40000=Y+16000,
And we get Y = 24000kg/day.
(2)
Soni said:
9 years ago
Very good, Thanks @Ahmad.
(1)
Satya said:
6 years ago
Ahmad Bhai in 10 % or 25% solution mass of solution will be 40K kg or 16K kg respectively. Please, correct.
(1)
Gab said:
1 year ago
OMB: Feed = Water + Product.
Product Stream:
25% of P: 4000 kg solids.
75% of P: 12000 kg water.
Feed Stream:
10% of Feed = 4000 kg solids.
90% of Feed = 36000 kg water.
Water Stream (evaporated):
W = Water in Feed - Water in Product.
= 36000 - 12000.
W = 24,000 kg.
Product Stream:
25% of P: 4000 kg solids.
75% of P: 12000 kg water.
Feed Stream:
10% of Feed = 4000 kg solids.
90% of Feed = 36000 kg water.
Water Stream (evaporated):
W = Water in Feed - Water in Product.
= 36000 - 12000.
W = 24,000 kg.
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