Material and Energy Balances

A. Steady-state mass balance

Material Balances and Degrees of Freedom

The steady-state total and component mass balance, degrees-of-freedom analysis, tie components, and a disciplined recipe for solving multi-unit flowsheets.

Material balances are the grammar of chemical engineering: before you size a reactor, design a column, or rate a scrubber, you account for every kilogram and every mole that enters and leaves a control volume. On the FE the Material and Energy Balances topic carries one of the largest blocks of questions, and most of them are won or lost not on chemistry but on bookkeeping — picking the right control volume, counting unknowns against equations, and refusing to mix a mass basis with a mole basis. The balance idea itself is one line from the FE Reference Handbook (the Chemical Engineering and Thermodynamics chapters frame it as conservation applied to a defined system): what accumulates inside a boundary equals what comes in, minus what goes out, plus what reaction generates. Everything in this concept is that one statement, applied with discipline.

The general balance and its steady-state form

Draw a boundary around whatever you are analyzing — one unit, a cluster of units, or the whole plant — and write the balance for total mass or for any single species. For a non-reactive species the generation term vanishes; at steady state the accumulation term vanishes too, leaving the workhorse statement that input equals output. Total mass is always conserved (atoms cannot be created), so the total balance holds even when reactions occur; only per-species (component) balances pick up a generation term — whether written in moles or in mass — while the total mass balance never does.

\underbrace{\dfrac{dm}{dt}}_{\text{accumulation}} = \underbrace{\dot{m}_{in}}_{\text{in}} - \underbrace{\dot{m}_{out}}_{\text{out}} + \underbrace{\dot{m}_{gen}}_{\text{generation}} \;\xrightarrow{\text{steady, no rxn}}\; \dot{m}_{in} = \dot{m}_{out}

Total versus component balances

For a system with

S

species you can write one total balance plus

S

component balances, but only

S

of those

S+1

equations are independent — the total is just the sum of the components. The practical move on the exam is to choose the most convenient

S

of them. A total mass balance plus one component balance is usually faster than two component balances, because the total balance carries no composition and immediately ties the stream flows together.

\dot{m}_j^{\,in} = \dot{m}_j^{\,out}\quad(\text{each species }j),\qquad \sum_j \dot{m}_j = \dot{m}_{\text{total}}

Degrees-of-freedom analysis

Before you solve anything, count. The degrees of freedom (DOF) equal the number of unknown stream variables (flows and compositions) minus the number of independent equations you can write — independent material balances, specified compositions, given ratios, and any explicit relations.

DOF = 0

means the problem is exactly solvable;

DOF > 0

means it is underspecified (you need more data or a basis);

DOF < 0

means it is over-specified or inconsistent. Choosing a basis (e.g. '100 kg/h of feed') removes one degree of freedom and is almost always your first step when no flow is given.

DOF = N_{\text{unknowns}} - N_{\text{independent equations}}

Tie components

A tie component is a species that passes through the process untouched — it enters in only one stream and leaves in only one stream, with no reaction and no other source. Because its mass is conserved one-to-one between those two streams, it 'ties' them together and lets you find one flow directly from the other without solving the whole system. Inert solids in a dryer, ash in a furnace, and nitrogen in a combustor are classic tie components. Hunting for a tie component first is the single biggest time-saver on multi-unit problems.

\dot{m}_{\text{tie}}^{\,in} = \dot{m}_{\text{tie}}^{\,out}\;\Rightarrow\; F\,x_{\text{tie},F} = P\,x_{\text{tie},P}

Solving multi-unit flowsheets

A flowsheet with feeds, products, and recycle streams is solved the same way as one unit, only with more control-volume choices. You may draw a boundary around each unit, around groups of units, or around the entire process; the overall boundary is powerful because internal recycle and intermediate streams disappear from it. The recipe: label every stream, choose a basis, do a DOF count on each candidate control volume, and start solving where

DOF = 0

— typically the overall balance or a unit with a tie component — then march inward to the units that now have enough known streams.

Exam strategy

Work every balance problem in the same order: (1) sketch and label all streams with symbols for unknown flows and compositions; (2) choose a basis if no flow is given; (3) count degrees of freedom on each control volume and attack the one at

DOF = 0

; (4) look for a tie component before writing simultaneous equations; (5) keep one consistent basis — mass or mole — and never let mass fractions and mole fractions touch in the same equation. Compositions in each stream must sum to one, which is an equation you can spend. Finish with a quick check that total in equals total out.

Key equations

General material balance

\dfrac{dm}{dt} = \dot{m}_{in} - \dot{m}_{out} + \dot{m}_{gen}

Accumulation = in − out + generation. Per unit time for a defined control volume; $\dot m$ in kg/s (or mol/s for a species).

Steady-state total balance

\sum \dot{m}_{in} = \sum \dot{m}_{out}

Total mass is conserved even with reaction; accumulation and net generation of total mass are zero.

Component balance (no reaction)

\sum_{\text{in}} F_k\,x_{j,k} = \sum_{\text{out}} P_l\,x_{j,l}

For each species $j$ : mass (or mole) of $j$ in = out. $x_{j,k}$ is the fraction of $j$ in stream $k$ .

Composition closure

\sum_j x_j = 1

Mass or mole fractions in any single stream sum to unity — one usable equation per stream.

Tie-component relation

F\,x_{\text{tie},F} = P\,x_{\text{tie},P}

An inert that enters and leaves in one stream each directly links the two stream flows $F$ and $P$ .

Degrees of freedom

DOF = N_{\text{unknowns}} - N_{\text{indep. equations}}

$DOF=0$ solvable; $>0$ underspecified (set a basis); $<0$ over-specified. Independent equations = balances + specs + relations.

Independent balances

N_{\text{indep. balances}} = S

For $S$ species, only $S$ of the (total + $S$ component) balances are independent.

Mass–mole conversion

n_j = \dfrac{m_j}{M_j}, \qquad y_j = \dfrac{n_j}{\sum_k n_k}

Convert a stream between mass basis ( $m_j$ , mass fraction) and mole basis ( $n_j$ , mole fraction $y_j$ ) using molar masses $M_j$ .

Worked examples

Evaporator concentration with a solids balance

Problem. A feed of $1000\,\text{kg/h}$ of an aqueous solution containing $20\,\%$ dissolved solids by mass is concentrated in an evaporator to a product containing $50\,\%$ solids. Find the product mass flow and the water evaporated.

Solution. Streams: feed

F=1000\,\text{kg/h}

, vapor

W

(pure water), product

P

. Solids are a tie component (they appear only in feed and product). Solids balance:

F\,x_F = P\,x_P

, so

P = \dfrac{1000(0.20)}{0.50} = 400\,\text{kg/h}

. Total balance:

W = F - P = 1000 - 400 = 600\,\text{kg/h}

. Final: product

= 400\,\text{kg/h}

, water evaporated

= 600\,\text{kg/h}

. Sanity check: solids out

= 400 \times 0.50 = 200\,\text{kg/h} = 1000 \times 0.20

in. Total in

=

total out

= 1000\,\text{kg/h}

. ✓

P = \frac{F\,x_F}{x_P} = \frac{1000(0.20)}{0.50} = 400\ \text{kg/h}

Distillation overall balance

Problem. A distillation column is fed $100\,\text{kmol/h}$ of a benzene–toluene mixture that is $45\,\text{mol}\%$ benzene. The distillate is $95\,\text{mol}\%$ benzene and the bottoms is $5\,\text{mol}\%$ benzene. Find the distillate and bottoms molar flow rates.

Solution. Draw the overall control volume: feed

F=100

(benzene fraction

z_F=0.45

), distillate

D

(benzene fraction

x_D=0.95

), bottoms

B

(benzene fraction

x_B=0.05

). Two unknowns (

D

,

B

). Total balance:

D + B = 100

. Benzene balance:

x_D\,D + x_B\,B = 0.95\,D + 0.05\,B = 0.45(100) = 45

. Solve:

D = \dfrac{F(z_F - x_B)}{x_D - x_B} = \dfrac{100(0.45-0.05)}{0.95-0.05} = 44.4\,\text{kmol/h}

, and

B = 100 - 44.4 = 55.6\,\text{kmol/h}

. Sanity check: benzene out

= 0.95(44.4) + 0.05(55.6) = 42.2 + 2.78 = 45.0\,\text{kmol/h}

= benzene in. ✓

D = \frac{F(z_F - x_B)}{x_D - x_B} = \frac{100(0.45-0.05)}{0.95-0.05} = 44.4\ \text{kmol/h}

Dryer with a bone-dry-solid tie component

Problem. Wet solids enter a dryer at $100\,\text{kg/h}$ containing $30\,\%$ water by mass. The dried product contains $5\,\%$ water. Find the product rate and the water removed.

Solution. Bone-dry solid (BDS) is the tie component. BDS in feed

= 100(1-0.30) = 70\,\text{kg/h}

, conserved. In the product BDS is

1-0.05 = 95\,\%

:

P = \dfrac{70}{0.95} = 73.7\,\text{kg/h}

. Water removed

= 100 - 73.7 = 26.3\,\text{kg/h}

. Final: product

= 73.7\,\text{kg/h}

, water removed

= 26.3\,\text{kg/h}

. Sanity check: water in product

= 0.05(73.7) = 3.68\,\text{kg/h}

; water in feed

= 30\,\text{kg/h}

; removed

= 30 - 3.68 = 26.3\,\text{kg/h}

. ✓

Common pitfalls

•Mixing mass fractions and mole fractions in the same balance. Convert the whole stream to one basis first; molar masses are the bridge.
•Counting the total balance as an extra independent equation. With $S$ species only $S$ balances are independent — the total is the sum of the components.
•Forgetting to set a basis when no flow rate is given. An unspecified flow leaves $DOF>0$ ; pick 100 kg/h or 100 mol/s and scale at the end.
•Overlooking a tie component and grinding through simultaneous equations instead. An inert that enters and leaves in one stream each gives a flow directly.
•Choosing a control volume that still contains the recycle stream when the overall boundary would cancel it. Draw the boundary that hides the unknown you don't need.
•Not using the closure $\sum x_j = 1$ as an equation — it is a free relation that often reduces DOF to zero.
•Assuming total mass is not conserved when a reaction occurs. Total mass is always conserved; only per-species (component) balances carry a generation term — whether written in moles or in mass — never the total mass balance.

References

NCEES FE Reference Handbook — Chemical Engineering
NCEES FE Reference Handbook — Thermodynamics
Felder & Rousseau, Elementary Principles of Chemical Processes — material-balance procedure and DOF analysis

B. Unsteady-state mass balance

Unsteady-State Mass Balances

The accumulation = in − out − consumption ODE for filling, draining, and well-mixed tanks, and the first-order concentration response derived straight from the balance.