Unsteady State Balances Questions and Answers

Answer: A

The first two are correct. The third is wrong, and the point matters because start-up is where a plant is at its most hazardous and steady state calculations say nothing about it. Most incidents happen away from normal operation.

Answer: B

Setting the accumulation term to zero recovers the algebraic balance, so the steady state is the long time limit of the unsteady solution. Checking that limit is a useful test of a solution. The long time limit should match the algebraic answer.

Answer: C

Correct answer: the state of the system as the clock starts

An unsteady state balance is a differential equation, and its solution contains a constant that only the initial condition can fix. It states what is in the system at \( t=0 \) - the 50 kilograms of salt in the earlier item, the starting temperature of a batch. The flow rates and stoichiometry appear in the equation itself, and the final steady state is an output of the solution rather than an input to it.

Answer: D

Removing ninety-five per cent takes about three time constants, and the time constant here is twenty minutes. Three time constants is a useful rule of thumb for any exponential approach. Five of them take the residue below one per cent.

Answer: A

The first two are correct, and the wall term matters in a glass lined reactor. The third is wrong, since the starting temperature must be stated exactly as a starting composition must. Wall mass appears alongside the contents in the term.