3.1 Steady Material Balances Without Reaction

3.1 Steady Material Balances Without Reaction

Quick Answer: For a steady-state process with no chemical reaction, total mass and every nonreactive component obey In = Out. Write balances around a clear system boundary, pick a basis, and use degree-of-freedom counting so you know when the problem is closed.

Material balances are the backbone of Domain A on the UPDA/MMUP Chemical exam (material and energy balances, stoichiometry — about 20% of this guide's planning allocation). Even when a question looks like a unit-ops or reactor item, the first move is often: draw a boundary, label streams, write In = Out. This section covers nonreactive steady balances—the cases you must solve before you add extent of reaction, recycle, or energy.

The General Balance Equation

For any conserved accounting quantity (total mass, component mass, moles of an element):

[ \text{In} - \text{Out} + \text{Generation} - \text{Consumption} = \text{Accumulation} ]

Steady state means accumulation is zero (holdup is not changing with time). No reaction means generation and consumption of species are zero. Therefore:

[ \text{In} = \text{Out} ]

for total mass and for each nonreactive component. Total mass is always conserved (ignoring nuclear processes); individual species without reaction simply rearrange among streams.

Why Mass Is Preferred for Total Balances

Total mass balances never need a reaction term. Total mole balances do when reactions change the number of moles. On exam day, if the problem is nonreactive, either mass or mole balances work if compositions are consistent—but mass fractions of a stream must sum to 1, and mole fractions must sum to 1. Mixing mass-fraction data with a mole-only equation is a classic trap.

Basis Selection

A basis is a fixed quantity you choose so all other flows scale relative to it. Common choices:

Once the basis is set, every stream flow is expressed on that same time or amount basis. If the real plant runs at a different rate, scale all flows by the same factor after you finish the balance.

Mixers, Splitters, and Separators

Mixer

Two or more inlet streams combine into one outlet. Without reaction, at steady state:

• Total: (\dot{m}_1 + \dot{m}2 + \cdots = \dot{m}{\text{out}})
• Component (i): (\dot{m}1 w{i,1} + \dot{m}2 w{i,2} + \cdots = \dot{m}{\text{out}} w{i,\text{out}})

The outlet composition is the flow-weighted average of the inlets. Mixers do not change the fact that each component is conserved—they only redistribute.

Splitter

One inlet is divided into two or more outlets that have the same composition as the inlet (and each other). Only the flow rates differ. If you invent different compositions for splitter outlets without a separation mechanism, the balance is physically wrong.

[ \dot{m}{\text{in}} = \dot{m}{A} + \dot{m}{B}, \quad w{i,\text{in}} = w_{i,A} = w_{i,B} ]

Separator

A separator produces outlets with different compositions (flash drum, filter, absorber product streams treated as pure separation for balance purposes). You need enough independent composition and flow specifications—or recovery fractions—to close the DOF. Component balances still read In = Out for each species; the unknowns are how much of each species goes to each outlet.

Degree-of-Freedom (DOF) Intuition

Before solving, count:

[ \text{DOF} = (\text{unknowns}) - (\text{independent balances}) - (\text{independent process specs}) ]

• DOF = 0: uniquely solvable (exam problems are usually written this way).
• DOF > 0: under-specified; missing a recovery, split ratio, or composition.
• DOF < 0: over-specified; data may be inconsistent (or a redundant spec).

Independent balances for a nonreactive system with (C) components typically include one total mass balance and (C-1) independent component balances (the last component is dependent if compositions sum to 1). Process specs include stated flow rates, compositions, recoveries, and stream sum constraints.

Worked Example: Blending Two NaOH Streams

Problem. A Ras Laffan utilities skid blends two aqueous NaOH streams to a target intermediate strength used for scrubbing.

• Stream A: (120,\mathrm{kg/h}), 8.0 wt% NaOH
• Stream B: 22.0 wt% NaOH (flow unknown)
• Product P: 15.0 wt% NaOH, total product flow 200 kg/h

Find the required flow of Stream B and verify the water balance.

Step 1 — Basis and unknowns. Basis is already given as continuous rates. Unknown: (\dot{m}_B). Also check consistency with product rate.

Step 2 — Total mass balance (steady, no reaction).

[ \dot{m}_A + \dot{m}_B = \dot{m}_P ]

[ 120 + \dot{m}_B = 200 \implies \dot{m}_B = 80,\mathrm{kg/h} ]

Step 3 — NaOH component balance.

[ 120(0.08) + 80(0.22) \stackrel{?}{=} 200(0.15) ]

[ 9.6 + 17.6 = 27.2, \quad 200 \times 0.15 = 30.0 ]

27.2 ≠ 30.0 — the stated product rate and compositions are inconsistent if both feeds are free. In a real design problem you would free one specification. Suppose instead product flow is unknown and we require 15 wt% NaOH with A fixed at 120 kg/h and B free:

Total: (120 + \dot{m}_B = \dot{m}_P)

NaOH: (120(0.08) + \dot{m}_B(0.22) = 0.15,\dot{m}_P)

Substitute (\dot{m}_P = 120 + \dot{m}_B):

[ 9.6 + 0.22,\dot{m}_B = 0.15(120 + \dot{m}_B) = 18 + 0.15,\dot{m}_B ]

[ 0.22,\dot{m}_B - 0.15,\dot{m}_B = 18 - 9.6 = 8.4 ]

[ 0.07,\dot{m}_B = 8.4 \implies \dot{m}_B = 120,\mathrm{kg/h}, \quad \dot{m}_P = 240,\mathrm{kg/h} ]

Check water:

• Water in: (120(0.92) + 120(0.78) = 110.4 + 93.6 = 204,\mathrm{kg/h})
• Water out: (240(0.85) = 204,\mathrm{kg/h}) ✓

Exam lesson: When numbers do not close, a specification is wrong or DOF was miscounted. UPDA items usually give a consistent set; if options conflict, re-check which variable is free.

Systematic Exam Workflow

1. Sketch the unit and label every stream with flow and composition symbols.
2. State assumptions: steady, no reaction, no leak, compositions on mass or mole basis.
3. Count DOF mentally: can you solve with the given numbers?
4. Write total balance, then component balances for all but one independent species.
5. Solve algebraically; avoid mixing wt% and mol% mid-equation.
6. Close by checking an unused component or that fractions sum to 1.

Common Nonreactive Traps

• Treating a separator like a splitter (forcing equal compositions).
• Using mole fractions in a mass balance without converting.
• Forgetting that air-free or dry basis compositions need a separate wet/dry conversion.
• Solving for one stream then not scaling when the basis was 100 kg of feed but the question asks plant-scale product.

Master nonreactive steady balances first. Section 3.2 adds reaction (generation/consumption); Section 3.3 adds recycle, bypass, and purge—which still rest on the same In = Out logic at every boundary you draw.