Question

100 kg/hr of aqueous mixture containing 74 %wt of sucrose is cooled from 60 ˚C to 20 ˚C. Due to the decrease in temperature, some of the sugar precipitates out from the mixture as crystals and are subsequently removed via a separator unit.  In the separation of the crystals, for every 1 kg of crystals removed, 0.05 kg of the aqueous mixture will be removed together with the crystals.  Calculate the mass flow rate of the remaining saturated aqueous solution (in kg/hr).

Assume that the solubility of sucrose as a function of temperature in ˚C is given by the equation: %wt sucrose = 63.2 + 0.146T + 0.0006T2

Answer 1

Initial flow rate: 100 kg/hr at 60°C.

Given:

Solutibility of Sucrose in the aqueos medium as weight% = 63,2 + 0.146T+ 0.0006T²

Therefore at 60°C,

Solubility= 63.2 + 0.146(60) + 0.0006(60x60)

Solubility, Weight%= 63.2 + 8.76 + 2.16

Solubility = 74.12%

Which implies 74.12% of the sucrose will dissolve at 60°C in the given medium

Given; Incoming feed 74% by weight of sucrose,

Therefore, 74% of 100 kg/hr is sucrose.= 0.74 x 100

Which implies, mass flow rate of sucrose = 74 kg/hr

Mass flow rate of water= 26 kg/hr

out of which 74% is soluble at 60°C, mass of dissovled sucrose at 60°C = 0.74 x 0.74 = 51.8 kg/hr

Therefore 51.8 kg/hr of the Sucrose is dissolved at 60°C.

Remaining amount of sucrose will be crystallized out.

Therefore, mass of crystals at 60°C= 74 -51.8 = 22.2 kg/hr

Also give, for every 1 kg of crystal formed, 0.05 kg of water is taken up by the crystal as hydration.

Which implies that 22.2 kg of crystal will take up = 22.2 x 0.05 = 1.11 kg/hr of water

Therefore, the feed solution contains (per hour), 22.2 kg of crystals with 1.11 kg of water and 51.8 kg of sucrose is dissolved in the remaining 24.89 kg of water.

Then, the temperature is reduced to 20°C, which reduces the solubility of sucrose.

As solubility is the capacity to dissolve, a decrease in solubility will increase the formations of crystals.

By using the given equation, let us find the solubility at 20°C.

solubility = 63.2 + 0.146(20) + 0.0006(20 x 20)

Solubility= 63.2 + 2.92 + 0.24= 66.36 %

Therefore, out of 74 kg of sucrose only 66.36 percent of it will dissolve, which gives us:

Mass of Dissolved sucrose= 74 x 0.6636 = 49.1064 kg/hr

Therefore, Mass of crystals formed = 74 -49.1064 = 24.893 kg/hr

Water of hydration (similar to above) = 0.05 x 24.893 = 1.2446 kg/hr

Mass of remaining water = 26 - 1.2446 = 24.7554 kg/hr

Therefore the desired answer is 24.7554 kg/hr