SME
Heat Transfer · Problem type
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.”
Diagram
Counterflow: ΔT₁ = Tₕᵢ − Tco = 70 K and ΔT₂ = Tₕₒ − Tci = 50 K.
ΔTₗₘ
=
ΔT₁ − ΔT₂
ln(ΔT₁ ÷ ΔT₂)
=
59.44
K,
at
the
dashed
bracket.
Cₘᵢₙ is the oil’s: it changes 50 K for the water’s 30 K on the same heat.
q = 50 kW
This is an example. Type your own values to start a new problem.
q (kW)
=
ṁₕ (kg/s)
×
cₚₕ (kJ/(kg·K))
×
(
Tₕᵢ (°C)
−
Tₕₒ (°C)
)
q
=
ṁc (kg/s)
×
cpc (kJ/(kg·K))
×
(
Tco (°C)
−
Tci (°C)
)
Formulas
q
=
ṁₕ
×
cₚₕ
×
(Tₕᵢ
−
Tₕₒ)
50 = 0.5 × 2 × (120 − 70)
q
=
ṁc
×
cpc
×
(Tco
−
Tci)
50 = 0.3987 × 4.18 × (50 − 20)
Assumptions
•
The balance holds for any arrangement: counterflow, parallel or cross flow.
•
Steady flow; no heat lost to the surroundings.
•
No phase change, so each stream’s heat is ṁcpΔT.
•
Each stream’s specific heat is constant.
Step-by-step
WE KNOW
ṁₕ = 0.5 kg/s, cₚₕ = 2 kJ/(kg·K), Tₕᵢ = 120 °C, Tₕₒ = 70 °C, cpc = 4.18 kJ/(kg·K), Tci = 20 °C, Tco = 50 °C
FIND
heat rate (q), water mass flow (ṁc)
Step 1 · Find heat rate (q)
Use q = ṁₕ × cₚₕ × (Tₕᵢ − Tₕₒ)
The heat the oil gives: its flow times cₚ times its temperature change.
q
=
0.5
×
2
×
(120
−
70)
q
=
0.5
×
2
×
50
q
=
50
kW
Step 2 · Find water mass flow (ṁc)
Use q = ṁc × cpc × (Tco − Tci)
Divide the heat by cₚ and the temperature change.
ṁc
=
q
cpc × (Tco − Tci)
ṁc
=
50
4.18 × (50 − 20)
ṁc
=
50
4.18 × 30
ṁc
=
50
125.4
ṁc = 0.3987 kg/s
Check
Put the numbers back in. Both sides should match.
50 = 0.5 × 2 × (120 − 70) ✓
50 = 0.3987 × 4.18 × (50 − 20) ✓
← Previous lesson
Heat exchangers
Next lesson →
Effectiveness–NTU: the heat without the outlet temperatures
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Heat exchangers
Main lesson
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.”
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.”