Automotive Engineering

1. Radiator Design
Automotive Engineering
Dr. Ahmed Rezk
1

2. Learning Outcomes
At the end of this lecture, you will be able to:
• Explain the considerations of radiator design.
• Identify the parameters affecting radiator thermal performance.
• Determine the total surface area of a radiator using the basic
concepts of thermodynamics and heat transfer.
2

3. Introduction
Double pipe heat
exchanger, the base case.
X
X
An example of radiator fitted in a
conventional passengers car.
An example of cross flow
heat exchanger.
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Radiator: it is a heat exchanger used to remove the heat generated by the conventional and/or
electrified propulsion.
An example of radiator fitted
in an electric passengers car.
The heat engine
concept.Engine cooling cycle.

4. Radiator Design
Radiator design considerations
• The thermal performance and sizing of the radiator.
• Radiator positioning according to space, frontal area and vehicle aerodynamics
limitations.
• Determining the pressure drop across the coolant and air sides for the circulation
pump and fans sizing.
4

5. Radiator Thermal Performance Analysis
From the first law of thermodynamics, the heat transfer rate of each fluid is.
𝑄 = 𝑚 𝑖ℎ,𝑖𝑛 − 𝑖ℎ,𝑜𝑢𝑡 = 𝑚𝐶 𝑝 ℎ
𝑇ℎ,𝑖𝑛 − 𝑇ℎ,𝑜𝑢𝑡 → 1. 𝑎
𝑄 = 𝑚 𝑖 𝑐,𝑜𝑢𝑡 − 𝑖 𝑐,𝑖𝑛 = 𝑚𝐶 𝑝 𝑐
𝑇𝑐,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛 → 1. 𝑏
𝑄 = 𝑈 ∙ 𝑨 ∙ ∆𝑇 𝑚 → 2
Where:
• 𝑄 = The rate of heat transfer (W).
• 𝑖 = fluid specific enthalpy (J/kg).
• 𝑚 = fluid mass flow rate (kg/s).
• 𝐶 𝑝 = specific heat at constant pressure (J/kg.K).
• 𝑇 = Fluid temperature (K).
• 𝑈 = Overall heat transfer coefficient (W/m2.K)
• 𝐴 = Radiator area (m2).
• ∆𝑇 𝑚 = Mean temperature difference (K).
• Suffix - ℎ = Hot fluid stream.
• Suffix - 𝐶 = Cold fluid stream.
• Suffix - 𝑖𝑛 = Inlet.
• Suffix - 𝑜𝑢𝑡 = outlet.
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Overall energy balance
Q
Q

6. Overall Heat Transfer Coefficient
For clean, un-finned heat exchanger, the overall heat
transfer coefficient is determined as;
𝑅𝑖 =
1
ℎ𝑖 𝐴𝑖
𝑅 𝑜 =
1
ℎ 𝑜 𝐴 𝑜
𝑅 𝑊
𝑇ℎ 𝑇𝑐
→ 𝑄 →
𝑇ℎ
𝑇𝑐
𝑈𝐴 = 𝑈 𝑜 𝐴 𝑜 = 𝑈𝑖 𝐴𝑖 =
1
𝑅𝑡𝑜𝑡𝑎𝑙
=
1
1
ℎ𝑖 𝐴𝑖
+ 𝑅 𝑊𝑎𝑙𝑙 +
1
ℎ 𝑜 𝐴 𝑜
→ 3. 𝑎
For plane Wall
𝑅 𝑊𝑎𝑙𝑙 =
𝑡
𝐾 𝑤𝑎𝑙𝑙 𝐴 𝑤𝑎𝑙𝑙
For Tubular Wall
𝑅 𝑊𝑎𝑙𝑙 =
ln 𝑟𝑜 𝑟𝑖
2𝜋𝐾 𝑤𝑎𝑙𝑙 𝐿
Where:
• 𝑅 = Thermal Resistance (K/W).
• ℎ = Convective heat transfer coefficient (W/m2.K).
• 𝐾 = Thermal conductivity (W/m.K).
• 𝑟 = Radius (m).
• 𝐿 = Length (m).
• Suffix - 𝑜 = Outside.
• Suffix - 𝑖 = Inside.
Heat Transfer Resistances
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7. Poll
Q#1: What is/are the most dominant thermal resistance in radiators?
A. Wall thermal resistance.
B. Internal (coolant side) convective resistance.
C. External (air side) convective resistance.
D. All of the above.
E. B and C.
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8. The Mean Temperature Difference ∆𝑇 𝑚
From the energy balance.
For the base case double pipe parallel and
counter flow HEX, the mean temperature
difference that drives the heat transfer is
determined as;
Where:
• 𝐿𝑀𝑇𝐷 = Logarithmic mean temperature difference (K).*
• ∆𝑇1,2 = Temperature difference at HEX terminals.
𝑄 = 𝑈 ∙ 𝑨 ∙ ∆𝑇 𝑚 → 2
∆𝑇 𝑚= 𝐿𝑀𝑇𝐷 =
∆𝑇1 − ∆𝑇2
ln ∆𝑇1 ∆𝑇2
=
∆𝑇2 − ∆𝑇1
ln ∆𝑇2 ∆𝑇1
→ 4
Double pipe counter flow heat exchanger.
The base case.
𝑇ℎ,𝑖𝑛
𝑇ℎ,𝑜𝑢𝑡
𝑇𝑐,𝑖𝑛
𝑇𝑐,𝑜𝑢𝑡
X
X
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* The derivation of LMTD is shown in the slide #15.
∆𝑇1= 𝑇ℎ,𝑖𝑛-𝑇𝑐,𝑜𝑢𝑡 ∆𝑇2= 𝑇ℎ,𝑜𝑢𝑡-𝑇𝑐,𝑖𝑛

9. ∆𝑇 𝑚 for Different Flow Arrangement
For other HEX arrangements (i.e. cross flow HEX), a
correction is required to adjust ∆𝑇 𝑚 which is
determined based on double pipe counter flow
arrangement as;
Factor R is determined as;
Factor P is determined as;
∆𝑇 𝑚= 𝑭 ∙ 𝐿𝑀𝑇𝐷 𝐶𝑜𝑢𝑛𝑡𝑒𝑟 𝑓𝑙𝑜𝑤 → 5
𝐹 = 𝑓𝑛 𝑅, 𝑃, 𝑓𝑙𝑜𝑤 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡
𝐹𝑎𝑐𝑡𝑜𝑟 𝑅 =
𝑇ℎ,𝑖𝑛 − 𝑇ℎ,𝑜𝑢𝑡
𝑇𝑐,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛
𝐹𝑎𝑐𝑡𝑜𝑟 𝑃 =
𝑇𝑐,𝑜𝑢𝑡 − 𝑇𝑐,𝑖𝑛
𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑖𝑛
=
∆𝑇𝑐
∆𝑇 𝑚𝑎𝑥
𝐹𝑎𝑐𝑡𝑜𝑟 𝑃
𝐹𝑎𝑐𝑡𝑜𝑟 𝑅 =
Chart for determining LMTD correction factor. 9

10. Determining the Radiator’s Total Area
The following steps are used to determine the radiators total heat exchanger area.
• Calculate 𝑄 and the unknown outlet temperature. You always looking to cool down the
coolant by predefined ∆𝑇.
• Calculate ∆𝑇 𝑚 and find the correction factor 𝑭 based on the selected radiator design.
• Calculate the overall heat transfer coefficient 𝑈. It is acceptable assumption if you
neglect the conduction thermal resistance through the wall at the early stages of design.
• Determine the total heat transfer area 𝐴.
• You may recalculate the radiator area using less assumptions.
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11. Area Calculation Example
A counter flow double pipe oil cooler of cooling water flow rate trough the inner tube (𝑑𝑖 = 25𝑚𝑚)
is 0.2kg/s. The flow rate of oil through the outer annulus (𝑑 𝑜 = 45𝑚𝑚) is 0.1kg/s. The oil and water
enter at temperatures of 100 and 30°C, respectively. Calculate the total surface area of the heat
exchanger if the outlet temperature of the oil is to be 60°C? The inner and outer convective heat
transfer coefficient are 2250 and 38.8 W/m2K respectively. The oil and water 𝐶 𝑝 are 2131 and 4178
J/kgK respectively.
Solution & Assumptions:
1- Determine the heat transfer rate and the water outlet temperature.
2- Determine the mean temperature.
𝑄 = 𝑚𝐶 𝑝 ℎ
𝑇ℎ,𝑖𝑛 − 𝑇ℎ,𝑜𝑢𝑡 = 0.2 × 2131 100 − 60 = 8524𝑊 → 1. 𝑎
𝑇𝑐,𝑜 = 𝑇𝑐,𝑖 +
𝑄
𝑚𝐶 𝑝 𝑐
= 30 +
8524
0.1 × 4178
= 40.2°𝐶 → 1. 𝑏
∆𝑇 𝑚= 𝐿𝑀𝑇𝐷 =
∆𝑇1 − ∆𝑇2
ln ∆𝑇1 ∆𝑇2
=
59.8 − 30
ln 59.8 30
= 43.2°𝐶 → 4
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12. Area Calculation Example
2- Determine the overall heat transfer coefficient.
2.a- Normally the heat exchanger design based on the outer surface area.
2.b- The wall thermal resistance can be neglected, this mean that the wall thickness in negligible.
The U can be determined as;
3- Determine the surface area.
𝑈 =
1
1
ℎ𝑖
+
1
ℎ 𝑜
=
1
1
2250
+
1
38.8
= 38.1 𝑊 𝑚2
𝐾 → 3. 𝑎
𝑈𝐴 = 𝑈 𝑜 𝐴 𝑜 = 𝑈𝑖 𝐴𝑖 =
1
𝑅𝑡𝑜𝑡𝑎𝑙
=
1
1
ℎ𝑖 𝐴𝑖
+ 𝑅 𝑊𝑎𝑙𝑙 +
1
ℎ 𝑜 𝐴 𝑜
→ 3. 𝑎
𝐴 =
𝑄
𝑈
=
8524
38.1
= 223.7𝑚2 → 2
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13. References
• Bergman, T. L., & Incropera, F. P. (2011). Fundamentals of heat and
mass transfer (7th ed.). Hoboken, NJ: Wiley.
• Çengel, Y. A. (1998). Heat transfer : A practical approach. United
States:
• Kakaç, S., & Liu, H. (1998). Heat exchangers: Selection, rating, and
thermal design. Boca Raton, Fla ; London; CRC Press.
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14. Questions
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15. LMTD Derivation “Reading”
𝑑𝑄 = − 𝑚ℎ ∙ 𝐶 𝑝,ℎ∙ 𝑑𝑇ℎ= −𝐶ℎ ∙ 𝑑𝑇ℎ
𝑑𝑄 = 𝑚 𝑐 ∙ 𝐶 𝑝,𝑐∙ 𝑑𝑇𝑐 = 𝐶𝑐 ∙ 𝑑𝑇𝑐
𝑑𝑄 = 𝑈 ∙ ∆𝑇 ∙ 𝑑𝐴
∆𝑇 = 𝑇ℎ − 𝑇𝑐 & 𝑑 ∆𝑇 = 𝑑𝑇ℎ − 𝑑𝑇𝑐
𝑑 ∆𝑇 = −𝑑𝑄
1
𝐶ℎ
+
1
𝐶𝑐
1
2
𝑑 ∆𝑇
∆𝑇
= −𝑈
1
𝐶ℎ
+
1
𝐶𝑐 1
2
𝑑𝐴
ln
∆𝑇2
∆𝑇1
= −𝑈𝐴
1
𝐶ℎ
+
1
𝐶𝑐
ln
∆𝑇2
∆𝑇1
= −𝑈𝐴
𝑇ℎ,𝑖𝑛 − 𝑇ℎ,𝑜𝑢𝑡
𝑄
+
𝑇ℎ,𝑖𝑛 − 𝑇ℎ,𝑜𝑢𝑡
𝑄
ln
∆𝑇2
∆𝑇1
= −
𝑈𝐴
𝑄
𝑇ℎ,𝑖𝑛 − 𝑇𝑐,𝑖𝑛 − 𝑇ℎ,𝑜𝑢𝑡 − 𝑇𝑐,𝑜𝑢𝑡
𝑄 = 𝑈𝐴
∆𝑇2 − ∆𝑇1
ln ∆𝑇2 ∆𝑇1
= 𝑈𝐴
∆𝑇1 − ∆𝑇2
ln ∆𝑇1 ∆𝑇2
Incremental heat balance.
Integrating the
incremental heat to
find the total heat.
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