An evaporator is fed with10000 kg/hr of a solution containing 1% solute by weigh

Question

An evaporator is fed with10000 kg/hr of a solution containing 1% solute by weight. It is to be concentrated to 1.5% solute by weight. The feed is at a temperature of 37oC. The water is evaporated by heating with steam available at a pressure of 1.34 atm absolute, corresponding to a temperature of 108.3oC. The operating pressure in the vapor space is 1 atm absolute. Boiling point elevation and other effects can be neglected. The condensate leaves at the condensing temperature. All the physical properties of the solution may be taken to be same as that of water. What is the quantity of steam required per hour? What is the quantity of steam required per hour?

Data:
Enthalpy of feed = 38.1 kcal/kg
Enthalpy of solution inside the evaporator (at 100oC) = 644 kcal/kg
Enthalpy of vapor at 100oC = 644 kcal/kg
Latent heat of vaporization of steam = 540 kcal/kg

Explanation / Answer

Calculations:

Given: 0000 kg/hr of solution containing 1% solute by weight.

Let the feed stream is equal to FS

And Concentrated solution stream Is CS

water evaporated = W

the concentration of solute = x

We know that the concentration of solute in feed is equal to concentration in vapour

so

FS X x = CS X x

10000 x 0.01 = CS x 0.015

On calcualting

CS = 6666.67 kg/hr.

and water evaporated (W) = FS - CS = 10000 - 6666.67 = 3333.33 kg/hr.

We know enthalpy is constant

Feed Stream enthalpy (HF) +Steam required X LAtent heat of stream = Amount of water X Enthalpy of water+ enthalpy of products X Concentrated stream

10000 x 38.1 + Steam required x 540 = 3333.3 x 644 + 6666.7 x 98

Steam required = 4479.6 kg/hr

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.