Mass balance: practice problems with step-by-step solutions

Free Mass balance practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: mixing two streams, blending to a target, evaporator.

Free practice problemEasy
MixerStream 1450 kg/h · 25% saltStream 2700 kg/h · 15% saltStream 3ṁ₃ = ? kg/h

Two salt solutions are mixed at steady state. Find the outlet flow rate ṁ3.

ṁ1 = 450 kg/hx1 = 25 %ṁ2 = 700 kg/hx2 = 15 %
kg/h
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What you practise

  1. Easymixing two streams
  2. Mediumblending to a target
  3. Hardevaporator

Key formulas

Steady state
in = out
Total balance
ṁ1 + ṁ2 = ṁ3
Component balance
ṁ1x1 + ṁ2x2 = ṁ3x3
Evaporator
F·xF = P·xP, W = F − P

Learn Mass balance from scratch

1What a mass balance is

Mass cannot appear or disappear. Whatever flows into a process unit (a mixer, a tank, an evaporator) must flow out again, or stay inside.

Many units run at steady state: the flows do not change with time and nothing builds up inside. Then the rule is simple:

in = out

Think of a bath with the tap open and the plug out. If the water level stays the same, the water leaving each minute equals the water entering.

Flows are written as mass flow rates ṁ, for example in kg/h. Read ṁ as “m-dot”: the dot means “per unit of time”.

Self-check A mixer at steady state receives 300 kg/h and 200 kg/h. How much leaves it?
  1. 100 kg/h
  2. 250 kg/h
  3. 500 kg/h
  4. 600 kg/h

Answer: 500 kg/h. At steady state nothing builds up, so out = in = 300 + 200 = 500 kg/h. 250 kg/h is the average, which has no meaning here.

2Total and component balances

A stream is often a mixture, for example salt dissolved in water. The mass fraction x is the share of one component: 20% salt means 0.20 kg of salt in every kg of solution.

The mass of that component in a stream is ṁ·x. Without a chemical reaction, each component also obeys in = out. So you can write two balances:

ṁ1 + ṁ2 = ṁ3
ṁ1·x1 + ṁ2·x2 = ṁ3·x3

The first is the total balance, the second the component balance (here for salt). Two independent equations can find two unknowns.

Example: 300 kg/h at 10% salt mixes with 200 kg/h at 20% salt. Salt in: 30 + 40 = 70 kg/h. Total out: 500 kg/h. So x3 = 70/500 = 14%.

Self-check Mix 300 kg/h of 10% salt solution with 200 kg/h of 20% salt solution. What is the salt fraction of the mixture?
  1. 14%
  2. 15%
  3. 30%
  4. 7%

Answer: 14%. Salt in = 300·0.10 + 200·0.20 = 70 kg/h, and 70/500 = 14%. The plain average of the fractions, 15%, ignores that the 10% stream is larger.

3The evaporator: a tie component

An evaporator boils water off a solution to make it more concentrated. The vapour W is pure water, so all the solute leaves with the product P.

F1000 kg/h · 10%P40%W0%

The solute passes through unchanged, from F to P only. Such a component is a tie component: its balance links two flows directly.

F·xF = P·xP, W = F − P

In the figure: P = 1000 · 0.10/0.40 = 250 kg/h, and W = 1000 − 250 = 750 kg/h of water boils off.

Self-check An evaporator takes 1000 kg/h of a 10% sugar solution and makes a 40% product. How much water is evaporated?
  1. 250 kg/h
  2. 300 kg/h
  3. 4000 kg/h
  4. 750 kg/h

Answer: 750 kg/h. Sugar balance: P = 1000·0.10/0.40 = 250 kg/h. Then W = 1000 − 250 = 750 kg/h. 250 kg/h is the product, not the water removed.

4The method step by step

  1. Draw a box for the unit with every stream as an arrow. Write each known flow and fraction next to it.
  2. Mark the unknowns. Count them: you need as many independent balances.
  3. Write the total balance: sum of flows in = sum of flows out.
  4. Write the component balance for the solute: sum of ṁ·x in = sum of ṁ·x out. Use fractions as decimals (20% = 0.20) or as percent on both sides.
  5. Solve. Start with a tie component if there is one.
  6. Check: put the numbers back into both balances.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Stream 1 is 400 kg/h of a 10% salt solution. How much of stream 2 (40% salt) must be added each hour so that the blended product contains 20% salt? Find ṁ2.

1

Draw the unit

ṁ₁ = 400 kg/h10%ṁ₂ = ?40%ṁ₃20%

There are two unknowns, ṁ2 and ṁ3, so we need two balances.

2

Total balance

ṁ3 = 400 + ṁ2
3

Salt balance

400 · 0.10 + ṁ2 · 0.40 = (400 + ṁ2) · 0.20
40 + 0.40·ṁ2 = 80 + 0.20·ṁ2
4

Solve

0.20·ṁ2 = 40 ⇒ ṁ2 = 200 kg/h

Then ṁ3 = 400 + 200 = 600 kg/h.

5

Check the answer

Salt in: 40 + 200 · 0.40 = 120 kg/h. Salt out: 600 · 0.20 = 120 kg/h. Both balances hold.

The target 20% lies between 10% and 40%, as it must for a blend.

Answer

Add ṁ2 = 200 kg/h of the 40% solution.

6Common mistakes

  • Averaging the fractions instead of weighting them by flow.
  • Writing only the total balance and then guessing the second unknown.
  • Mixing percent and decimals in the same equation.
  • In the evaporator, giving the product P when the water evaporated W is asked, or the other way round.
  • Turning the tie-component ratio upside down: P = F·xF/xP, so P is smaller than F.
Self-check Mix 100 kg/h of a 10% solution with 300 kg/h of a 30% solution. What is the outlet fraction?
  1. 20%
  2. 25%
  3. 33.3%
  4. 40%

Answer: 25%. Solute out = 100·0.10 + 300·0.30 = 100 kg/h, and 100/400 = 25%. The plain average of 10% and 30% is 20%, which ignores that the 30% stream is three times larger.

7How exam questions look

  • Easy: two streams are mixed. Find the outlet flow (total balance) or its fraction (component balance).
  • Medium: blending to a target fraction. Find the flow of the second stream with both balances, as in the example.
  • Hard: an evaporator. Use the solute as tie component to find P, then W = F − P.

A shortcut for blending: ṁ2 = ṁ1·(x1 − x3)/(x3 − x2). In the example: 400·(10 − 20)/(20 − 40) = 200 kg/h.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
MixerStream 1300 kg/h · 30% saltStream 2250 kg/h · 40% saltStream 3x₃ = ? %

Two salt solutions are mixed at steady state. Find the mass fraction of salt in the outlet x3.

ṁ1 = 300 kg/hx1 = 30 %ṁ2 = 250 kg/hx2 = 40 %

Find x3 (%)

Show hints
  1. At steady state, what goes in comes out: write a total balance and a balance for the dissolved component.
  2. Total: ṁ₁ + ṁ₂ = ṁ₃. Component: ṁ₁x₁ + ṁ₂x₂ = ṁ₃x₃.
  3. ṁ3 = 550 kg/h.
Show the answer and the step-by-step solution

Steady-state mass balance: in = out, for the total mass and separately for each component (no reaction).

Total balance

ṁ3 = ṁ1 + ṁ2 = 300 + 250 = 550 kg/h

Salt balance

ṁ1x1 + ṁ2x2 = ṁ3x3
x3 = (300·30 + 250·40)/550 = 34.55 %

Answer

x3 = 34.55 %
Example 2Medium
MixerStream 1850 kg/h · 50% sugarStream 2ṁ₂ = ? · 15% sugarProduct40% sugar

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 40% sugar? Find ṁ2.

ṁ1 = 850 kg/hx1 = 50 %x2 = 15 %x3 = 40 %

Find ṁ2 (kg/h)

Show hints
  1. Two unknowns (ṁ₂ and ṁ₃) need two balances: total and component.
  2. Substitute ṁ₃ = ṁ₁ + ṁ₂ into ṁ₁x₁ + ṁ₂x₂ = ṁ₃x₃.
  3. Rearranged: ṁ₂ = ṁ₁(x₁ − x₃)/(x₃ − x₂).
Show the answer and the step-by-step solution

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
sugar: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 850·(50 − 40)/(40 − 15) = 340 kg/h

Answer

ṁ2 = 340 kg/h
Example 3Hard
EvaporatorFeed F2000 kg/h · 10% NaOHProduct P65% NaOHVapour W (pure water)W = ? kg/h

An evaporator concentrates a NaOH solution by boiling off pure water. Find the water evaporated W in kg/h.

F = 2000 kg/hxF = 10 %xP = 65 %

Find W (kg/h)

Show hints
  1. The vapour contains no solute, so all the solute in the feed leaves with the product.
  2. Solute balance: F·xF = P·xP. Then total balance: F = W + P.
  3. P = 307.7 kg/h.
Show the answer and the step-by-step solution

Tie component: a component that passes through unchanged in one stream (here the solute) links the flows directly. Then the total balance gives the rest.

Solute balance

F·xF = P·xP → P = 2000·10/65 = 307.7 kg/h

Total balance

W = F − P = 2000 − 307.7 = 1692 kg/h

Answer

W = 1692 kg/h

Questions and answers

What is Mass balance?

Mass cannot appear or disappear. Whatever flows into a process unit (a mixer, a tank, an evaporator) must flow out again, or stay inside.

Many units run at steady state: the flows do not change with time and nothing builds up inside. Then the rule is simple:

How do I solve Mass balance problems step by step?

Steady-state mass balance: in = out, for the total mass and separately for each component (no reaction).

What do the Mass balance exercises cover?
  • Easy: mixing two streams
  • Medium: blending to a target
  • Hard: evaporator

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Printable worksheet

Mass balance worksheet with answers
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