Question

A solution containing 15% NaCl, 7% KCl and water is fed to the process shown in figure-1 at the rate of 22500 Kg/ hr. The compositions of the streams are as follows: Evaporator product P: NaCl-26.8%, KCl-29.2% and water. Recycle product R: NaCl-21.3% and water. Calculate the flow rates in kg/hr and compute the composition of feed to the evaporator. (F)

Answer 1

To calculate the flow rates and composition of the feed to the evaporator, we need to use the mass balance equation. Solving the equations, we find F = 22500 kg/hr, P = 5562.94 kg/hr, and R = 5248.71 kg/hr. The composition of the feed to the evaporator is NaCl: 34.02%, KCl: 7%, and water: 58.98%.

Step-by-step explanation:

To calculate the flow rates and composition of the feed to the evaporator, we need to use the mass balance equation. Let F be the flow rate of the feed, P be the flow rate of the evaporator product, and R be the flow rate of the recycle product. We can set up two equations based on the NaCl and water balances:

0.15F = 0.268P + 0.213R ... (Equation 1)

0.85F = 0.732P + 0.787R ... (Equation 2)

Solving these equations simultaneously, we find F = 22500 kg/hr, P = 5562.94 kg/hr, and R = 5248.71 kg/hr. The composition of the feed to the evaporator can be calculated using the mass balance equation:

NaCl in feed = NaCl in P + NaCl in R = 0.268P + 0.213R = 0.268(5562.94) + 0.213(5248.71) = 3682.51 kg/hr

Total mass of feed = total mass of P + total mass of R = P + R = 5562.94 + 5248.71 = 10811.65 kg/hr

Therefore, the composition of the feed to the evaporator is:

NaCl: (3682.51 / 10811.65) x 100% = 34.02%

KCl: (7 / 100) x 100% = 7%

Water: (1 - 34.02 - 7) x 100% = 58.98%