Question

An aqueous solution of sodium hydroxide contains 15% NaOH by mass. It is desired to produce an 5% NaOH solution by diluting a stream of the 15% solution with a stream of pure water. Calculate the following: A) The ratios (g H,O/g feed solution) and (g product / g feed solution). B) Determine the feed rates of 15% solution and diluting water needed to produce 2500 Ib./min of the 5% solution.

Answer 1

Step-by-step 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

`(20)/(100) * F + 0 = (8)/(100) * 100`

P =
`(20)/(100) * 100 * (100)/(8)`

= 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,
`\frac{\text{grams of pure water}}{\text{grams feed solution}}` =
`(W)/(F)`

=
`(150)/(100)`

= 1.5

`\frac{\text{grams of product}}{\text{grams feed solution}}` =
`(P)/(F)`

=
`(250)/(100)`

= 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 (
`(P)/(F)`) = 2.5

So, F =
`(P)/(2.5)`

=
`(2400)/(2.5)`

= 960 lb/min

Also,
`(W)/(F)` = 1.5

W =
`1.5 * F`

=
`1.5 * 960`

= 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.

Answer 2

feed rate of 15% sol = f = 833.33 lb/min

Diluting water w = 1666.6 lb/min

Step-by-step 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

`(15)/(100) * f + 0 = (5)/(100) * P`

`(15)/(100) * 100 = (5)/(100) * P`

P = 300 g/min

putting P value in 1 eq

w get w = 200 g/min

hence,

`(g\ pure\ water)/(g\ feed\ solution) = (w)/(f) = (200)/(100) = 2`

`(g\ product)/(g\ feed\ solution) = (P)/(f) = (300)/(100) = 3`

b) it is given 5% solution p = 2500 lb/min

`(P)/(f) = 3`

`F = (P)/(3) = (2500)/(3) = 833.33 lb/min`

`(w)/(f) = 2`

`w = 2* 833.33 = 1666.66 lb/min`

feed rate of 15% sol = f = 833.33 lb/min

Diluting water w = 1666.6 lb/min