SME
Heat Transfer · Problem type
The same oil cooler in parallel flow
Use this for “Oil cools from 120 °C to 70 °C and water warms from 20 °C to 50 °C in a parallel-flow exchanger. The duty is 50 kW and U = 300 W/(m²·K). Find the area.”
Diagram
Parallel flow: ΔT₁ = Tₕᵢ − Tci = 100 K and ΔT₂ = Tₕₒ − Tco = 20 K.
ΔTₗₘ
=
ΔT₁ − ΔT₂
ln(ΔT₁ ÷ ΔT₂)
=
49.71
K,
at
the
dashed
bracket.
Cₘᵢₙ is the oil’s: it changes 50 K for the water’s 30 K on the same heat.
In parallel flow the cold outlet stays below the hot outlet.
q = 50 kW
U
=
300
W/(
m
2
·K)
A = 3.353 m²
This is an example. Type your own values to start a new problem.
Tₕᵢ
Oil inlet temperature
example · °C
Tₕₒ
Oil outlet temperature
example · °C
Tci
Water inlet temperature
example · °C
Tco
Water outlet temperature
example · °C
ΔT₁
Temperature difference at the oil inlet’s end
calculated · K
ΔT₂
Temperature difference at the oil outlet’s end
calculated · K
ΔTₗₘ
Log-mean temperature difference
calculated · K
q
Heat rate
example
U
Overall heat transfer coefficient
example · W/(m²·K)
A
Heat transfer area
calculated · m²
Formulas
ΔT₁ = Tₕᵢ − Tci
100 = 120 − 20
ΔT₂ = Tₕₒ − Tco
20 = 70 − 50
ΔTₗₘ
=
ΔT₁ − ΔT₂
ln(ΔT₁ ÷ ΔT₂)
49.7068
=
100 − 20
ln(100 ÷ 20)
q
=
U
×
A
×
ΔTₗₘ
50,000 = 300 × 3.353 × 49.7068
Formulas are worked in W. The step-by-step shows the conversions.
Assumptions
•
Parallel flow: both streams enter at the same end and run the same way.
•
Steady flow; no heat lost to the surroundings.
•
U is the same all along the exchanger; no phase change.
•
Each stream’s specific heat is constant.
Step-by-step
WE KNOW
Tₕᵢ = 120 °C, Tₕₒ = 70 °C, Tci = 20 °C, Tco = 50 °C, q = 50 kW, U = 300 W/(m²·K)
FIND
temperature difference at the oil inlet’s end (ΔT₁), temperature difference at the oil outlet’s end (ΔT₂), log-mean temperature difference (ΔTₗₘ), 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 = 50 kW = 50,000 W (1 kW = 1000 W)
Step 2 · Find temperature difference at the oil inlet’s end (ΔT₁)
Use ΔT₁ = Tₕᵢ − Tci
Both streams enter at this end: oil inlet minus water inlet.
ΔT₁ = 120 − 20
ΔT₁ = 100 K
Step 3 · Find temperature difference at the oil outlet’s end (ΔT₂)
Use ΔT₂ = Tₕₒ − Tco
Both streams leave at this end: oil outlet minus water outlet.
ΔT₂ = 70 − 50
ΔT₂ = 20 K
Step 4 · Find log-mean temperature difference (ΔTₗₘ)
Use ΔTₗₘ = (ΔT₁ − ΔT₂) ÷ ln(ΔT₁ ÷ ΔT₂)
The difference between the streams changes exponentially along the length; its mean is the log mean of the two ends.
ΔTₗₘ
=
100 − 20
ln(100 ÷ 20)
ΔTₗₘ
=
80
ln(5)
ΔTₗₘ
=
80
1.6094
ΔTₗₘ = 49.7068 K
Step 5 · Find heat transfer area (A)
Use q = U × A × ΔTₗₘ
The
heat
over
U
and
ΔTₗₘ:
the
area
the
duty
needs.
A
=
q
U
× ΔTₗₘ
A
=
50,000
300 × 49.7068
A
=
50,000
14,912.04
A
=
3.353
m²
Check
Put the numbers back in. Both sides should match.
100 = 120 − 20 ✓
20 = 70 − 50 ✓
49.7068
=
100 − 20
ln(100 ÷ 20)
✓
50,000 = 300 × 3.353 × 49.7068 ✓
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Effectiveness–NTU: the heat without the outlet temperatures
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Transient conduction
Topic check
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Related lessons
Heat exchangers
Main lesson
The energy balance: the water flow an oil cooler needs
Use this for “Oil at 0.5 kg/s (cₚ = 2 kJ/(kg·K)) cools from 120 °C to 70 °C. Water (cₚ = 4.18) warms from 20 °C to 50 °C. Find the water flow.”
Effectiveness–NTU: the heat without the outlet temperatures
Use this for “A counterflow exchanger has UA = 841 W/K. The oil’s Cₘᵢₙ is 1000 W/K and Cᵣ = 0.6. Oil enters at 120 °C, water at 20 °C. Find the heat rate.”