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

Free Energy balance practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: heating a stream, heat exchanger, exchanger with heat loss.

Free practice problemEasy
Heaterglycol mix · 10 °C85 °CQ = ?

A stream of glycol mix flowing at 0.5 kg/s is heated from 10 °C to 85 °C. Find the heat duty Q.

ṁ = 0.5 kg/scp = 3.6 kJ/(kg·K)Tin = 10 °CTout = 85 °C
kW
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What you practise

  1. Easyheating a stream
  2. Mediumheat exchanger
  3. Hardexchanger with heat loss

Key formulas

Sensible heat
Q = ṁ·cp·ΔT
Latent heat
Q = ṁ·hfg
Mixing
ΣṁcpTin = ΣṁcpTout
Water
cp ≈ 4.18 kJ/(kg·K)

Learn Energy balance from scratch

Before this: Mass balance

1What an energy balance is

Energy, like mass, is conserved. When a stream is heated or cooled in a process, the heat that goes in must show up as a change in the stream’s temperature, or leave somewhere else.

At steady state nothing builds up inside the equipment, so:

energy in = energy out

In this topic the streams only warm up or cool down, without boiling or reacting. The heat that changes the temperature of a stream is called sensible heat.

The heat duty Q of a heater or cooler is the heat it transfers per second, in kJ/s. One kJ/s is one kilowatt (kW).

Self-check A heater transfers 50 kJ of heat every second. What is its heat duty?
  1. 3000 kW
  2. 50 W
  3. 50 kJ
  4. 50 kW

Answer: 50 kW. A duty is a rate: 50 kJ per second is 50 kJ/s = 50 kW. 50 kJ alone is an amount of energy, not a rate.

2Sensible heat

The heat needed to warm a flowing stream depends on three things: how much flows, what it is, and how many degrees it warms up.

Q = ṁ · cp · ΔT
  • ṁ is the mass flow rate in kg/s.
  • cp is the specific heat capacity in kJ/(kg·K): the energy to warm 1 kg by 1 K. Water: 4.18. Light oil: about 2.1.
  • ΔT = Tout − Tin. A difference of 1 °C equals a difference of 1 K, so you do not add 273 here.
ṁ · 20 °C60 °CQ

Example: 0.5 kg/s of water heated from 20 °C to 60 °C needs Q = 0.5 · 4.18 · 40 = 83.6 kW.

Self-check 0.5 kg/s of water (cp = 4.18 kJ/(kg·K)) is heated from 20 °C to 60 °C. What is the heat duty?
  1. 125.4 kW
  2. 167.2 kW
  3. 41.8 kW
  4. 83.6 kW

Answer: 83.6 kW. ΔT = 60 − 20 = 40 K, so Q = 0.5 · 4.18 · 40 = 83.6 kW. Using 60 °C as ΔT (125.4 kW) forgets to subtract the inlet temperature.

3Heat exchangers

In a heat exchanger a hot stream and a cold stream flow past each other, separated by a metal wall. Heat flows from the hot side to the cold side; the fluids do not mix.

T = 140 °C90 °Ct = 20 °Ct = ?hc

With no heat lost to the surroundings, the heat released by the hot stream is the heat gained by the cold stream:

ṁh·cp,h·(Th,in − Th,out) = ṁc·cp,c·(tout − tin)

If some heat escapes, the cold stream gets less: Qc = Qh − Qloss. With 10% loss, Qc = 0.90·Qh.

4The method step by step

  1. Draw the unit with every stream. Write ṁ, cp and the temperatures you know.
  2. Find the stream where you know both temperatures. Calculate its heat: Q = ṁ·cp·ΔT.
  3. If there is a heat loss, take it off the heat released by the hot stream to find what the cold stream receives.
  4. Use that heat on the other stream to find the unknown: a temperature (tout = tin + Q/(ṁ·cp)) or a flow rate (ṁ = Q/(cp·ΔT)).
  5. Check: the cold outlet must stay below the hot inlet.

5A worked example

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

Worked example

Hot light oil (2 kg/s, cp = 2.1 kJ/(kg·K)) cools from 140 °C to 90 °C in a counter-flow heat exchanger. It heats 2 kg/s of water (cp = 4.18 kJ/(kg·K)) that enters at 20 °C. There is no heat loss. Find the water outlet temperature tout.

1

Draw the exchanger

2 kg/s · 140 °C90 °C2 kg/s · 20 °Ct = ?hc

Both temperatures of the oil are known, so start on the hot side.

2

Heat released by the oil

Q = 2 · 2.1 · (140 − 90) = 2 · 2.1 · 50 = 210 kW
3

Heat taken up by the water

No heat is lost, so the water receives all 210 kW.

Δt = Q/(ṁc·cp,c) = 210/(2 · 4.18) = 25.12 K
4

Outlet temperature

tout = 20 + 25.12 = 45.12 °C
5

Check the answer

Water side: 2 · 4.18 · (45.12 − 20) = 210 kW, the same as the oil side.

The water warms less than the oil cools because water has twice the heat capacity. And 45 °C is well below the oil inlet of 140 °C.

Answer

tout ≈ 45.1 °C.

6Common mistakes

  • Using one temperature instead of the difference ΔT.
  • Using the cp of water for the oil, or the other way round.
  • Adding the heat loss to the cold side instead of taking it off.
  • Mixing kg/h and kg/s. With kg/s and kJ/(kg·K) the duty comes out in kW.
  • Giving a cold outlet temperature above the hot inlet: this is impossible and shows a slip.
Self-check A hot stream releases 200 kW and 10% is lost. How much water (cp = 4.18) can it heat from 20 °C to 50 °C?
  1. 0.86 kg/s
  2. 1.44 kg/s
  3. 1.59 kg/s
  4. 1.75 kg/s

Answer: 1.44 kg/s. The water receives 0.90 · 200 = 180 kW, so ṁ = 180/(4.18 · 30) = 1.44 kg/s. Ignoring the loss gives 1.59 kg/s, and using 50 °C instead of ΔT = 30 K gives 0.86 kg/s.

7How exam questions look

  • Easy: heating one stream. Find the duty Q = ṁ·cp·ΔT.
  • Medium: a heat exchanger with no loss. Find Q from the hot side, then the water outlet temperature, as in the example.
  • Hard: an exchanger that loses 5 to 15% of the heat released. Find the water flow rate needed.

Always find Q on the stream where both temperatures are known. Then the other stream gives the one remaining unknown.

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

Worked examples

Example 1Easy
Heaterglycol mix · 25 °C70 °CQ = ?

A stream of glycol mix flowing at 1 kg/s is heated from 25 °C to 70 °C. Find the heat duty Q.

ṁ = 1 kg/scp = 3.6 kJ/(kg·K)Tin = 25 °CTout = 70 °C

Find Q (kW)

Show hints
  1. Sensible heat with no phase change: Q = ṁ·cp·ΔT.
  2. A temperature difference in °C equals the same difference in K.
  3. ΔT = 45 K.
Show the answer and the step-by-step solution

Sensible heat: Q̇ = ṁ·cp·ΔT. With ṁ in kg/s and cp in kJ/(kg·K), Q̇ comes out in kJ/s = kW.

Energy balance

ΔT = 70 − 25 = 45 K
Q = ṁ·cp·ΔT = 1·3.6·45 = 162 kW

Answer

Q = 162 kW
Example 2Medium
hot sidecold sideglycol mix · 150 °C120 °Cwater · 15 °Ct = ? °C

Hot glycol mix cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.

ṁh = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 120 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C

Find tout (°C)

Show hints
  1. Heat lost by the hot stream equals heat gained by the cold stream.
  2. Find Q from the hot side first, since you know both of its temperatures.
  3. Q = 108 kW.
Show the answer and the step-by-step solution

Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).

Heat released by the hot stream

Q = 1·3.6·(150 − 120) = 108 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 108/(1·4.18) = 40.84 °C

Answer

tout = 40.84 °C
Example 3Hard
hot sidecold sidelight oil · 140 °C90 °Cwater · 30 °C75 °C · ṁ = ?10% lost

Hot light oil heats water in an exchanger that loses 10% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 1.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 140 → 90 °Ct: 30 → 75 °Ccp,c = 4.18 kJ/(kg·K)loss = 10 %

Find ṁc (kg/s)

Show hints
  1. Start with the heat released by the hot stream.
  2. Only 90% of it reaches the water.
  3. Heat to the water = 141.8 kW.
Show the answer and the step-by-step solution

Energy balance with losses: heat released = heat absorbed + heat lost.

Heat released

Qh = 1.5·2.1·(140 − 90) = 157.5 kW

Heat that reaches the water

Qc = (1 − 0.1)·157.5 = 141.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 141.8/(4.18·45) = 0.7536 kg/s

Answer

ṁc = 0.7536 kg/s

Questions and answers

What is Energy balance?

Energy, like mass, is conserved. When a stream is heated or cooled in a process, the heat that goes in must show up as a change in the stream’s temperature, or leave somewhere else.

At steady state nothing builds up inside the equipment, so:

How do I solve Energy balance problems step by step?

Sensible heat: Q̇ = ṁ·cp·ΔT. With ṁ in kg/s and cp in kJ/(kg·K), Q̇ comes out in kJ/s = kW.

What do the Energy balance exercises cover?
  • Easy: heating a stream
  • Medium: heat exchanger
  • Hard: exchanger with heat loss

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.

Topics before and after this one

Learn first

Mass balance

Continue with

Heat exchangers (LMTD)

Courses with this topic

Thermodynamics 1Material and Energy BalancesFE Chemical exam practice

More topics in Chemical

Mass balanceIdeal gas lawReaction kineticsReactor sizing (CSTR & PFR)Heat exchangers (LMTD)

Printable worksheet

Energy balance worksheet with answers
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