SME
Process Design
Equipment sizing
Use this for “A 500 kW cooler cools a stream from 150 to 90 °C with water heated from 30 to 80 °C, U = 500 W/(m²·K). Find the area.”
Diagram
Counterflow: ΔT₁ = Tₕᵢ − Tco = 70 K and ΔT₂ = Tₕₒ − Tci = 60 K.
ΔTₗₘ
=
ΔT₁ − ΔT₂
ln(ΔT₁ ÷ ΔT₂)
=
64.87
K,
at
the
dashed
bracket.
Cₘᵢₙ is the hot stream’s: it changes 60 K for the cooling water’s 50 K on the same heat.
Q = 500 kW
U
=
500
W/(
m
2
·K)
A = 15.42 m²
This is an example. Type your own values to start a new problem.
ΔT₁ (K)
=
Th,in (°C)
−
Tc,out (°C)
,
ΔT₂ (K)
=
Th,out (°C)
−
Tc,in (°C)
LMTD (K)
=
(
ΔT₁ (K)
−
ΔT₂ (K)
)
÷
ln(
ΔT₁ (K)
÷
ΔT₂ (K)
)
Q (kW)
=
U (W/(m²·K))
×
A (m²)
×
LMTD (K)
Formulas
ΔT₁ = Th,in − Tc,out
70 = 150 − 80
ΔT₂ = Th,out − Tc,in
60 = 90 − 30
LMTD
=
ΔT₁ − ΔT₂
ln(ΔT₁ ÷ ΔT₂)
64.8716
=
70 − 60
ln(70 ÷ 60)
Q
=
U
×
A
×
LMTD
500,000 = 500 × 15.4151 × 64.8716
Formulas are worked in W. The step-by-step shows the conversions.
Assumptions
•
Countercurrent flow: the hot stream enters at the end where the cold stream leaves.
•
No phase change; U stays the same along the whole exchanger.
•
Steady flow with no heat lost to the surroundings.
Step-by-step
WE KNOW
Q = 500 kW, U = 500 W/(m²·K), Th,in = 150 °C, Th,out = 90 °C, Tc,in = 30 °C, Tc,out = 80 °C
FIND
hot-inlet end difference (ΔT₁), hot-outlet end difference (ΔT₂), log-mean difference (LMTD), heat transfer area (A)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to W).
Q = 500 kW = 500,000 W (1 kW = 1000 W)
Step 2 · Find hot-inlet end difference (ΔT₁)
Use ΔT₁ = Th,in − Tc,out
At the hot inlet’s end the hot stream meets the cold outlet.
ΔT₁ = 150 − 80
ΔT₁ = 70 K
Step 3 · Find hot-outlet end difference (ΔT₂)
Use ΔT₂ = Th,out − Tc,in
At the other end the hot outlet meets the cold inlet.
ΔT₂ = 90 − 30
ΔT₂ = 60 K
Step 4 · Find log-mean difference (LMTD)
Use LMTD = (ΔT₁ − ΔT₂) ÷ ln(ΔT₁ ÷ ΔT₂)
The difference shrinks exponentially along the length, so its mean is the log mean of the ends.
70 ≠ 60: unequal ends
LMTD
=
70 − 60
ln(70 ÷ 60)
LMTD
=
10
ln(1.16667)
LMTD
=
10
0.15415
LMTD = 64.8716 K
Step 5 · Find heat transfer area (A)
Use Q = U × A × LMTD
Area
is
the
duty
over
what
each
square
meter
passes,
U
×
LMTD.
A
=
Q
U
× LMTD
A
=
500,000
500 × 64.8716
A
=
500,000
32,435.8
A
=
15.4151
m²
Check
Put the numbers back in. Both sides should match.
70 = 150 − 80 ✓
60 = 90 − 30 ✓
64.8716
=
70 − 60
ln(70 ÷ 60)
✓
500,000 = 500 × 15.41507 × 64.87159 ✓
Topic check
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