Exercise set

Chemical Material and Energy Balance Exercises

Worked chemical material and energy balance exercises for stream closure, components, recycle, purge, accumulation, heat duty, utilities and release checks.

Branch: Chemical Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises focus on material and energy balances used in chemical process design, troubleshooting and release review. The goal is not only to close arithmetic totals. The goal is to define the boundary, basis, stream measurement, composition convention, accumulation term, heat duty and validation evidence well enough that a plant decision can be trusted.

Assume simplified nominal values unless an exercise states otherwise. Real process balances require verified stream tags, density and composition data, phase state, sampling method, instrument uncertainty, recycle returns, purge destinations, vents, drains, inventory movement and utility conditions.

Release Evidence Notes

A balance is release evidence only when the same boundary is used for every stream. Record the time period, whether flows are mass, molar or volumetric, whether composition is on a mass or mole basis, which phase is measured, how density was corrected, which recycle and purge streams cross the boundary and whether inventory is changing. If one unmeasured vent, drain, leak, foam carryover, vapor loss or side draw is omitted, the balance is a hypothesis rather than evidence.

Engineering Boundary Notes

Material balances and energy balances answer different questions but must be consistent. A stream can close by mass while failing a component balance. A heat duty can look plausible while the cooling utility is unavailable. A recycle loop can close at steady state while transient accumulation is still unsafe.

Common Release Mistakes

mixing kg/h, kmol/h, volume flow and concentration without conversion;
using nominal density when temperature or composition changed;
assuming steady state during startup, drain-down, solvent swap or tank fill;
omitting purge, vent, sample, seal flush, wash or entrainment streams;
reporting heat duty without utility inlet temperature, fouling allowance or flow margin;
treating a small positive closure as acceptable without uncertainty and action limits.

Scenario Map

Scenario	Main calculation	Engineering decision
Total balance	$\sum \dot{m}_{in}=\sum \dot{m}_{out}$	Find missing stream or accumulation.
Component balance	species in, out and reaction terms	Check product, impurity or reagent accounting.
Recycle and purge	inert or contaminant steady state	Prevent buildup and emissions surprises.
Transient inventory	accumulation over time	Decide startup, hold or drain action.
Heat duty	$\dot{Q}=\dot{m}c_p\Delta T$	Verify heater, cooler or utility capacity.
Guarded closure	error plus uncertainty	Decide pass, retest or investigation.

Validation Package Checklist

calculation basis and boundary diagram in words;
stream tags, measurement type and time averaging;
density, molecular-weight and composition conversion basis;
inventory state for tanks, vessels and lines;
recycle, purge, vent, drain and sample paths;
heat-duty calculation with utility temperature and flow evidence;
uncertainty allowance and action rule for closure error.

Exercise 1: Unknown Waste Stream

A unit receives liquid feed at $2400\ \text{kg/h}$ and solvent at $1600\ \text{kg/h}$ . Measured outlet streams are product at $3650\ \text{kg/h}$ and vent at $75\ \text{kg/h}$ . The unit is at steady state. Estimate the unknown waste stream.

Solution

\dot{m}_{in}=2400+1600=4000\ \text{kg/h}

\dot{m}_{known}=3650+75=3725\ \text{kg/h}

\dot{m}_{waste}=4000-3725=275\ \text{kg/h}

Engineering Comment

The result assumes no accumulation and no unmeasured outlet. A missing $275\ \text{kg/h}$ must be tied to a real stream, not hidden in a spreadsheet residual.

Plausibility Check

The waste stream is positive because the known outlet is less than the inlet.

Exercise 2: Component Balance on a Solute

A liquid feed is $1200\ \text{kg/h}$ with $8.0\%$ solute by mass. Product leaves at $950\ \text{kg/h}$ with $9.5\%$ solute. A waste stream carries the remaining solute. Compute solute in the waste.

Solution

Solute in:

\dot{m}_{s,in}=0.080(1200)=96.0\ \text{kg/h}

Solute in product:

\dot{m}_{s,p}=0.095(950)=90.25\ \text{kg/h}

Waste solute:

\dot{m}_{s,w}=96.0-90.25=5.75\ \text{kg/h}

Engineering Comment

Component closure can fail even when total mass appears reasonable. The waste stream must be sampled on the same mass-fraction basis.

Plausibility Check

Waste solute is small because the product carries most of the solute.

Exercise 3: Volumetric Flow to Mass Flow

A solvent stream has volumetric flow:

Q=3.2\ \text{m}^3/\text{h}

Density at operating temperature is:

\rho=860\ \text{kg/m}^3

Compute mass flow.

Solution

\dot{m}=\rho Q=860(3.2)=2752\ \text{kg/h}

Engineering Comment

The density must match the operating temperature and composition. Using a room-temperature density can bias a production balance.

Plausibility Check

The value is close to $3.2$ tonnes per hour because the density is slightly below water.

Exercise 4: Mole Flow From Mass Flow

A stream contains $450\ \text{kg/h}$ of component $A$ with molecular weight:

MW_A=72\ \text{kg/kmol}

Compute molar flow.

Solution

\displaystyle F_A=\frac{\dot{m}_A}{MW_A}=\frac{450}{72}=6.25\ \text{kmol/h}

Engineering Comment

Reaction stoichiometry uses moles, not mass. Convert before writing component reaction balances.

Plausibility Check

A heavier molecule gives fewer kmol per kg, so $6.25\ \text{kmol/h}$ is plausible.

Exercise 5: Mixing Concentration

Stream 1 is $1000\ \text{kg/h}$ at $12\%$ solute. Stream 2 is $500\ \text{kg/h}$ at $4\%$ solute. Compute mixed solute mass fraction.

Solution

\dot{m}_s=0.12(1000)+0.04(500)=140\ \text{kg/h}

\dot{m}_{total}=1500\ \text{kg/h}

\displaystyle w_s=\frac{140}{1500}=0.0933=9.33\%

Engineering Comment

The mixed concentration is flow-weighted. Averaging $12\%$ and $4\%$ directly would be wrong because the streams have different flows.

Plausibility Check

The result lies between $4\%$ and $12\%$ and is closer to $12\%$ because the larger stream is richer.

Exercise 6: Purge for Inert Control

A recycle loop receives inert material at $2.0\ \text{mol/min}$ . Inert leaves only through a purge with allowable inert mole fraction $8.0\%$ . Compute purge flow.

Solution

\displaystyle F_{purge}=\frac{F_{I,in}}{y_I}=\frac{2.0}{0.080}=25.0\ \text{mol/min}

Engineering Comment

The purge controls inert buildup but may also lose reactant or solvent. The purge destination and treatment system are part of the release basis.

Plausibility Check

$8\%$ of $25.0\ \text{mol/min}$ is $2.0\ \text{mol/min}$ , matching the inert feed.

Exercise 7: Tank Accumulation

A surge tank receives $18.0\ \text{m}^3/\text{h}$ and discharges $15.5\ \text{m}^3/\text{h}$ . The liquid density is constant. Compute volume accumulation after $2.0\ \text{h}$ .

Solution

\displaystyle \frac{dV}{dt}=18.0-15.5=2.5\ \text{m}^3/\text{h}

\Delta V=2.5(2.0)=5.0\ \text{m}^3

Engineering Comment

This is not a steady-state case. A level alarm, overflow route or operator action must be checked before using steady-state balances downstream.

Plausibility Check

The tank fills because inlet flow is larger than outlet flow.

Exercise 8: Batch Charge Reconciliation

A batch recipe requires $1200\ \text{kg}$ of solvent. The weigh tank starts with $85\ \text{kg}$ heel and ends with $30\ \text{kg}$ heel after charging. Fresh solvent added to the weigh tank is $1140\ \text{kg}$ . How much solvent entered the batch?

Solution

Available solvent:

m_{avail}=85+1140=1225\ \text{kg}

Charged solvent:

m_{charge}=1225-30=1195\ \text{kg}

Engineering Comment

The batch is $5\ \text{kg}$ below target. Whether this matters depends on recipe tolerance and concentration sensitivity.

Plausibility Check

Ending heel reduces the amount transferred to the reactor.

Exercise 9: Evaporation Loss

A dryer feed is $500\ \text{kg/h}$ at $35\%$ solids. Product is $80\%$ solids. Assume no solids loss. Compute product rate and evaporated water.

Solution

Solids flow:

\dot{m}_s=0.35(500)=175\ \text{kg/h}

Product rate:

\displaystyle \dot{m}_p=\frac{175}{0.80}=218.75\ \text{kg/h}

Evaporated water:

\dot{m}_{evap}=500-218.75=281.25\ \text{kg/h}

Engineering Comment

The heat duty and vent treatment should be based on the evaporated water, not only product rate.

Plausibility Check

The product flow is lower than the feed because most water is removed.

Exercise 10: Sensible Heating Duty

A liquid stream of $2200\ \text{kg/h}$ has:

c_p=3.6\ \text{kJ/(kg K)}

It is heated from $25^\circ\text{C}$ to $70^\circ\text{C}$ . Compute heat duty.

Solution

\dot{Q}=\dot{m}c_p\Delta T

\dot{Q}=2200(3.6)(70-25)=356400\ \text{kJ/h}

Convert to kW:

\displaystyle \dot{Q}=\frac{356400}{3600}=99.0\ \text{kW}

Engineering Comment

This is only sensible heat. Phase change, heat loss and startup metal heating may increase utility demand.

Plausibility Check

A few tonnes per hour heated by tens of kelvin gives a duty near $100\ \text{kW}$ .

Exercise 11: Cooling-Water Flow

A cooler must remove $180\ \text{kW}$ . Cooling water may rise from $24^\circ\text{C}$ to $34^\circ\text{C}$ . Use:

c_p=4.18\ \text{kJ/(kg K)}

Compute water mass flow.

Solution

\displaystyle \dot{m}=\frac{\dot{Q}}{c_p\Delta T}

\displaystyle \dot{m}=\frac{180\ \text{kJ/s}}{4.18(10)}=4.31\ \text{kg/s}

Engineering Comment

The utility check should include fouling, control-valve authority and supply temperature during summer operation.

Plausibility Check

Water removing $180\ \text{kW}$ with only $10\ \text{K}$ rise requires several kg/s.

Exercise 12: Steam Condensation Requirement

A heater needs $250\ \text{kW}$ . Condensing steam provides latent heat:

h_{fg}=2100\ \text{kJ/kg}

Estimate steam flow.

Solution

\displaystyle \dot{m}_{steam}=\frac{250\ \text{kJ/s}}{2100\ \text{kJ/kg}}=0.119\ \text{kg/s}

In hourly units:

0.119(3600)=429\ \text{kg/h}

Engineering Comment

This ignores condensate subcooling and distribution losses. The steam trap and condensate return must handle the actual load.

Plausibility Check

High latent heat means a few hundred kg/h of steam can supply hundreds of kW.

Exercise 13: Heat-Exchanger Area Screen

A heat exchanger must transfer $320\ \text{kW}$ . Use:

U=650\ \text{W/(m}^2\text{K)},\quad \Delta T_{lm}=28\ \text{K}

Estimate required area.

Solution

\displaystyle A=\frac{\dot{Q}}{U\Delta T_{lm}}

\displaystyle A=\frac{320000}{650(28)}=17.6\ \text{m}^2

Engineering Comment

This is a clean-service screen. Fouling allowance, correction factor and pressure-drop limits may increase required area.

Plausibility Check

Hundreds of kW at moderate $U$ and temperature difference often require tens of square metres.

Exercise 14: Balance Closure Error

Measured inlets total $12500\ \text{kg/h}$ . Measured outlets total $12400\ \text{kg/h}$ . Compute relative closure error.

Solution

\displaystyle e_m=\frac{12500-12400}{12500}=0.008=0.8\%

Engineering Comment

The sign indicates apparent accumulation or missing outlet. Whether $0.8\%$ is acceptable depends on meter uncertainty and process consequence.

Plausibility Check

$100\ \text{kg/h}$ over $12500\ \text{kg/h}$ is less than $1\%$ .

Exercise 15: Guarded Balance Closure

Use the closure error from Exercise 14. The uncertainty allowance is $0.6\%$ and the allowable guarded error is $1.0\%$ . Use:

e_{guarded}=|e_m|+U_e

Solution

e_{guarded}=0.8\%+0.6\%=1.4\%

Since $1.4\%>1.0\%$ , the balance fails.

Engineering Comment

The correct action is not to force closure. Investigate inventory movement, density correction, analyzer bias, vents, drains, recycle measurement and leaks.

Plausibility Check

Adding uncertainty makes the acceptance test stricter than the raw closure error.

Exercise 16: Slurry Solids Balance

A slurry feed is $10.0\ \text{t/h}$ at $18\%$ solids. Thickened product is $42\%$ solids. Assume no solids loss. Compute product flow.

Solution

Solids flow:

\dot{m}_s=0.18(10.0)=1.80\ \text{t/h}

Product flow:

\displaystyle \dot{m}_p=\frac{1.80}{0.42}=4.29\ \text{t/h}

Engineering Comment

The liquid removed is large and may control pump sizing, tailings water return or wastewater load.

Plausibility Check

The product is more concentrated, so product mass flow is less than feed mass flow.

Exercise 17: Neutralization Reagent Demand

An acidic wastewater contains $0.85\ \text{kmol/h}$ acid equivalents. The neutralizing base is $90\%$ pure and reacts one-to-one. Compute required base feed in kmol/h of commercial material.

Solution

Pure base required:

F_b=0.85\ \text{kmol/h}

Commercial material:

\displaystyle F_{comm}=\frac{0.85}{0.90}=0.944\ \text{kmol/h}

Engineering Comment

The purity correction matters for inventory and pH control. Real neutralization also needs mixing, heat release and endpoint measurement.

Plausibility Check

Because the reagent is not pure, commercial feed must exceed stoichiometric pure base.

Exercise 18: Balance Release Gate

A production balance has raw closure error $0.45\%$ , uncertainty allowance $0.35\%$ and action limit $1.0\%$ . The heat-duty calculation has $12\%$ spare utility capacity. Decide release status.

Solution

e_{guarded}=0.45\%+0.35\%=0.80\%

Since $0.80\%<1.0\%$ , material closure passes. Utility spare capacity is also positive:

12\%>0\%

Engineering Comment

The calculation supports release if stream tags, density corrections, inventory state and utility temperatures are documented. Without that evidence, the numeric pass is incomplete.

Plausibility Check

Both guarded material closure and utility margin are positive, so the decision is a documented pass rather than retest.

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