Mass balances

Draw a boundary round part of a process (the system) and account for what crosses it. The general balance is: input − output + generation − consumption = accumulation. Generation and consumption come from reactions; accumulation is build-up inside. At steady state with no reaction, it collapses to input = output.

You can write that balance for the total mass and for each substance separately. With two substances you get three equations, but only two are independent (the total is the sum of the parts). The number of unknowns you can solve for is the number of independent equations. That bookkeeping is called a degrees of freedom analysis: if unknowns outnumber equations, you need more information.

A mixer: 100 kg/h of a 20% sugar solution joins 50 kg/h of water. Total: out = 150 kg/h. Sugar: 0.20 × 100 = 20 kg/h in, so 20 kg/h out, and the product is 20 ÷ 150 = 13.3% sugar.

An evaporator boils water off to concentrate a solution: 1000 kg/h of brine at 10% salt is to be concentrated to 25%. Salt only leaves in the product, so the salt balance does most of the work: 100 kg/h of salt = 0.25 × P, so P = 400 kg/h, and the total balance gives 1000 − 400 = 600 kg/h of water evaporated.

A distillation column splits a feed into a top product (distillate, D) and a bottom product (B). Feed 100 mol/h at 40% benzene, distillate 95% benzene, bottoms 5%. Total: D + B = 100. Benzene: 0.95D + 0.05B = 40. Substituting B = 100 − D gives 0.90D = 35, so D = 38.9 mol/h and B = 61.1 mol/h. The benzene recovered at the top is 0.95 × 38.9 ÷ 40 = 92% of what came in.