An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an aqueous solution of this salt. The fresh feed to the process contains 19.6 wt% K2SO4. The wet filter cake consists of solid K2SO4 crystals and a 40.0 wt% K2SO4 solution, in a ratio 10 kg crystals/kg solution. The filtrate, also a 40.0% solution, is recycled to join the fresh feed. Of the water fed to the evaporator, 45.0% is evaporated. The evaporator has a maximum capacity of 175 kg water evaporated/s. Calculate the maximum production rate of solid K2SO4, the rate at which fresh feed must be supplied to achieve this production rate.

feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6

Explanation:

Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.

Overall balance:

Water:

Using substitution method, we have:

= 220.77 kg/s

 = 4.16 kg/s

The maximum production rate of solid crystal is = 10*4.16 = 416 kg/s

Around evaporator:

kg/s

Around the mixing point:

Solid crystal:

Using the last two equations, we can obtain:

kg/s

The rate of supplying fresh feed to obtain the production rate is:

= 352.5/220.77 = 1.6

what is the average time it took the car to travel 0.25 meters?

⇒ 2.23 seconds

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⇒ 3.13 seconds