In the manufacturing of soluble coffee, after roasting and grinding the coffee, a syrup is prepared from the mixture of the ground beans with hot water. This process is called extraction. The extract that comes out of the extractor contains 30% of solids. After extraction, the solution obtained (extract) is sent to an evaporator, where it is concentrated. The concentrated extract goes to a "spray" type dryer, where soluble coffee beans are obtained with 5% water. As during the evaporation process much more aroma is lost than during the drying process, an interesting alternative is to subdivide the extract into two streams, in which one goes through the evaporator and the other is sent directly to the dryer. Consider the process feed of 1000 kg/h. In this way, if 1/5 of the extract is deviated, determine:

a) the amount of coffee and the concentration of solids in the outlet of the evaporator, so that the dryer feed has 55% of solids;
b) the amount of coffee fed into the dryer;
c) the amount of soluble coffee obtained after drying (product).

The process feed is 1000 kg/h of syrup with 30% solids. 1/5 of this (200 kg/h) is diverted, and the remaining 800 kg/h is fed to the evaporator. The exit stream from the evaporator is mixed with the 200 kg/h and the resulting stream is 55% solids. This is fed to the dyer, and the final result is 5% solids.

a) The amount of solids going into the evaporator = the amount of solids coming out of the evaporator.

0.30 (800 kg/h) = xy

The solids going into the junction = the solids coming out of the junction.

0.30 (200 kg/h) + xy = 0.55 (200 kg/h + x)

Substitute.

0.30 (200 kg/h) + 0.30 (800 kg/h) = 0.55 (200 kg/h + x)

0.30 (1000 kg/h) = 0.55 (200 kg/h + x)

300 kg/h = 110 kg/h + 0.55x

x = 345.45 kg/h

The concentration is:

0.30 (800 kg/h) = y (345.45 kg/h)

y = 0.695

b) 200 + x = 545.45 kg/h

c) Solids going into the dryer = solids coming out of the dryer

0.55 (545.45 kg/h) = 0.95 z

z = 315.79 kg/h