A brine solution is 26% salt with 70.0 kg of water evaporated per hour. To produce 195 kg of pure salt (0% moisture) per day, how long should the process operate each day and how much brine must be fed to the evaporator per hour?
The process should operate each day for _ hours.
The amount of brine that must be fed to the evaporator is _ kg/h.

Question

contestada

Respuesta :

nuhulawal20 nuhulawal20 · Answer 1 · 2020-12-06T07:54:05+01:00

Answer:

- the process should operate each day for 7.9286 hours

- the amount of brine that must be fed to the evaporator is 94.594 kg/hr

Explanation:

Given that;

concentration of brine = 26%

so water concentration will be 100% - 26% = 74%

for evaporation of 70kg water per hour, residual is pure salt( 0% moisture)

so mass flow rate of brine = 70/(74%) = 70/0.74 = 94.5945 kg/hr

Amount of pure dry salt produced(0%) = 94.5945 - 70 = 24.5945 kg/hr

Now for production of 195kg of pure dry salt, number of hours required will be

T = 195 / 24.5945 = 7.9286 hrs

Therefore the process should operate each day for 7.9286 hours.

Total brine solution required ( 26% salt conc.) = 195/0.26 = 750 kg

Feed rate of brine solution ( 26% salt conc.) = 750 / 7.9286 = 94.594 kg/hr

Therefore the amount of brine that must be fed to the evaporator is 94.594 kg/hr