Evaporation of cane sugar solutions. an evaporator is used to concentrate cane sugar solutions. a feed of 10 000 kg/d of a solution containing 38 wt % sugar is evaporated, producing a 74 wt % solution. calculate the weight of solution produced and amount of water removed.

Weight of solution produced = 5135 kg

Amount of water removed = 4865 kg

Explanation:

For the balance of mass, the incoming mass of sugar must be equal to the outgoing mass. So, the incoming mass (mi) is 38% of 10000 kg

mi = 0.38x10000 = 3800 kg

The outgoing mass (mo) must be 3800 kg, and it is 74% of the total mass (mt)

mo = 0.74xmt

0.74xmt = 3800

mt = 3800/0.74

mt = 5135 kg

This is the mass of solution produced.

The amount of water removed (wr) is the amount of water incoming (wi) less the amount of water outgoing (wo). Both will be the total mass less the mass of sugar :

wi = 10000 - 3800 = 6200 kg

wo = 5135 - 3800 = 1335 kg

wr = wi - wo

wr = 6200 - 1335

wr = 4865 kg

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.