Respuesta :
Answer:
feed rate of 15% sol = f = 833.33 lb/min
Diluting water w = 1666.6 lb/min
Explanation:
Assume feed solution of 100 gm
a) considering mass balanced
Total mass in = total mass out
feed + pure water = product
100 + w = P
P- w = 100 ...........1
mass balanced for NaOH
Total NaOH in = total NaOH out
f +w = NaOH in P
[tex]\frac{15}{100} \times f + 0 = \frac{5}{100} \times P[/tex]
[tex]\frac{15}{100} \times 100 = \frac{5}{100} \times P[/tex]
P = 300 g/min
putting P value in 1 eq
w get w = 200 g/min
hence,
[tex]\frac{g\ pure\ water}{g\ feed\ solution} = \frac{w}{f} = \frac{200}{100} = 2[/tex]
[tex]\frac{g\ product}{g\ feed\ solution} = \frac{P}{f} = \frac{300}{100} = 3[/tex]
b) it is given 5% solution p = 2500 lb/min
[tex]\frac{P}{f} = 3[/tex]
[tex]F = \frac{P}{3} = \frac{2500}{3} = 833.33 lb/min[/tex]
[tex]\frac{w}{f} = 2[/tex]
[tex]w = 2\times 833.33 = 1666.66 lb/min[/tex]
feed rate of 15% sol = f = 833.33 lb/min
Diluting water w = 1666.6 lb/min
Explanation:
We will assume that the flow rate of feed (F) = 100 g/min
(a) The overall mass balance will be as follows.
Total mass IN = Total mass OUT
as, Feed + Pure water = Product
100 + W = P
or, P - W = 100 .......... (1)
Whereas NaOH mass balance will be as follows.
Total NaOH mass IN = Total NaOH OUT
NaOH in feed + NaOH in water = NaOH in product
[tex]\frac{20}{100} \times F + 0 = \frac{8}{100} \times 100[/tex]
P = [tex]\frac{20}{100} \times 100 \times \frac{100}{8}[/tex]
= 250 g/min
Now, we will put the value of P in equation (1) as follows.
P - W = 100
250 - W = 100
W = 250 - 100
= 150 g/min
Therefore, [tex]\frac{\text{grams of pure water}}{\text{grams feed solution}}[/tex] = [tex]\frac{W}{F}[/tex]
= [tex]\frac{150}{100}[/tex]
= 1.5
[tex]\frac{\text{grams of product}}{\text{grams feed solution}}[/tex] = [tex]\frac{P}{F}[/tex]
= [tex]\frac{250}{100}[/tex]
= 2.5
(b) As it is given that 8% solution (P) = 2400 lb/min.
As it is calculated that the ratio of P and F ([tex]\frac{P}{F}[/tex]) = 2.5
So, F = [tex]\frac{P}{2.5}[/tex]
= [tex]\frac{2400}{2.5}[/tex]
= 960 lb/min
Also, [tex]\frac{W}{F}[/tex] = 1.5
W = [tex]1.5 \times F[/tex]
= [tex]1.5 \times 960[/tex]
= 1440 lb/min
Therefore, we can conclude that the feed rate of 20% solution = F = 960 lb/min.
When diluted with water then W = 1440 lb/min.
