HAN Automotive
SEV Lesson 5:
Aerodynamics
Saving Energy in a Vehicle
Note: most figures in this PowerPoint are from the reader SEV, others are directly sourced
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Static and dynamic pressure
The relation between pressure and force
Bernoulli’s Law
Laminar and Turbulent Flows
The influence of the border layer around objects
Methods to lower the air resistance
The drag force coefficient
Calculate the power needed to overcome air resistance
Stationary Mass Flow
Static and Dynamic pressure
Static Pressure
𝑝𝑔𝑟𝑎𝑣 = 𝑝ℎ𝑦𝑑𝑟 =  ∙ 𝑔 ∙ ℎ
Dynamic Pressure
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𝑝𝑑𝑦𝑛 =  𝑣 2 [𝑃𝑎]
https://aviation.stackexchange.com/questions/36656/how-can-dynamic-and-static-pressure-be-explained?rq=1
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Bernoulli’s Law
air=1.3 [kg/m3]
Wind = 10 [m/s]
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Wind
Glass
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Glass-plate = 2 [m2]
Calculate the Force
on the glass plate
because of the wind
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𝑝1 +  𝑔 ℎ1 +  𝑣12 = 𝑝2 +  𝑔 ℎ2 +  𝑣22
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2
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105 + 1.3 ∙ 9.81 ∙ 0 + ∙ 1.3 ∙ 102 = 𝑝2 + 1.3 ∙ 9.81 ∙ 0 + ∙ 1.3 ∙ 02
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𝑝2 = 100 065 𝑃𝑎
This means for the Force on the glass plate:
𝐹
𝑝 = 𝐴 => 𝐹 = 𝑝 ∙ 𝐴 = 65 ∙ 2 = 130 [𝑁]
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How high can we get the water ?
Try to calculate the maximum height,
while pumping with maximum vacuum.
water = 1000 [kg/m3]
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Remember:
𝑝1 +  𝑔 ℎ1 +
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
2
𝑣12
= 𝑝2 +  𝑔 ℎ2 +
105 + 0 + 0 = 0 + 1000 ∙ 9.81 ∙ ℎ2 + 0
ℎ2 = 10.19 [𝑚]
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1

2
𝑣22
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Venturi => the carburetor
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𝑝1 +  𝑔 ℎ1 +  𝑣12 = 𝑝2 +  𝑔 ℎ2 +  𝑣22
Who can prove with Bernoulli’s principle that the
pressure in the venturi is lower?
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𝑝1 + 𝑣12 = 𝑝2 + 𝑣22
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2
𝑝1 + 𝑠𝑚𝑎𝑙𝑙 = 𝑝2 + 𝑏𝑖𝑔
𝑝2 < 𝑝1
Aerodynamics in practice
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Border layer (grenslaag [NL])
https://www.rideapart.com/articles/255508/when-motorcycle-helmets-look-like-golf-balls/
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the speed of the medium at the border layer of the object is always zero
Polishing the surfaces of objects to lessen the drag forces is not always true
Within the border layer the friction causes a drag force
The dimples on the surface of a golf ball makes it easier for the air to leave the
surface, the border layer becomes thinner and the friction is lower
Pressure around the object
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A few tips
A few tips :
• Avoid sharp edges at the front
• when you need to use obstacles
(like wipers…) make a dimple or
edge in your construction to release
the border layer.
• Let the frontal area increase slowly
at the front, but also at the rear!
• Look at good practices on the
internet.
The drag force coefficient has the
minimum value when you can reach
the speed of sound. If not, then review
the tips above
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Drag force example
A Mc Laren F1 has a power of 465 [kW] (around 630 [hp]),
and has a top speed of 380 [km/h].
How much power is needed to overcome only the air resistance?
We need some numbers of course:
First we calculate the speed in [m/s] :
𝑘𝑔
𝑎𝑖𝑟 = 1.3 𝑚3
𝑐𝑑 = 0.32 [−]
𝐴 = 1.8 [𝑚2 ]
380
𝑣=
= 105.5 [𝑚/𝑠]
3.6
Now we fill in the formula for the Air resistance:
𝐹𝑎𝑖𝑟
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= 𝑐𝑑 ∗ 𝐴 ∗  𝑣 = 0.32 ∗ 1.8 ∗ ∗ 1.3 ∗ 105.52 = 4.17 [𝑘𝑁]
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2
The power is Force times velocity:
𝑃 = 𝐹 ∗ 𝑣 = 4.17 ∗ 105.5 = 440 [𝑘𝑊]
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stationary mass flow
Definitions :
𝑚ሶ
𝑘𝑔
mass flow or mass flux with unit
𝑠
𝑉ሶ volume flow or volume flux with unit
𝜌 density with unit
𝑘𝑔
𝑚3
𝐴 cross-sectional area with unit 𝑚2
For a cilindrical pipe
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𝐴=
1
2
𝜋𝑑
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𝑚3
𝑠
stationary mass flow
There is stationary flow if the mass which passes per unit of time
remains constant.
For a cilindrical pipe
𝑚ሶ = 𝜌 𝐴 𝑣 =
1
2
𝜋∙𝑑
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𝜌 𝑣 = constant
In the case of a liquid the volume flow is also constant because a liquid
is incompressible (density remains the same)
𝑚ሶ = 𝜌 𝐴 𝑣 = 𝜌 𝜑𝑉
𝑉ሶ = 𝐴 𝑣 =
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1
2 𝑣 = constant
𝜋∙𝑑
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A liquid flows through a pipe that narrows
𝑚ሶ 1 = 𝑚ሶ 2 ⇒
𝜌1 ∙ 𝐴1 ∙ 𝑣1 = 𝜌2 ∙ 𝐴2 ∙ 𝑣2
Because liquids are incompressible there follows 𝜌1 = 𝜌2
So
𝐴1 ∙ 𝑣1 = 𝐴2 ∙ 𝑣2
𝑣2 =
14
𝑑1 2
𝑣1
𝑑2
⟹
⟹
1
𝜋𝑑12
4
𝑣1 =
𝑑1 > 𝑑2
1
𝜋𝑑22
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⟹
𝑣2
⟹
𝑣2 > 𝑣1
𝑑12 𝑣1 = 𝑑22 𝑣2
A liquid flows through a pipe that narrows
𝑣2 =
𝑑1 2
𝑣1
𝑑2
twice as narrow ⟹ four times faster
three times narrower ⟹ nine times faster
four times narrower ⟹ sixteen times faster
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A liquid through a pipe that splits into two pipes
𝑚ሶ 1 = 𝑚ሶ 2 + 𝑚ሶ 3
𝑚ሶ 1 =
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⇒
𝜌 ∙ 14𝜋𝑑2 ∙ 𝑣
2
𝑚ሶ 1 = 𝜌 ∙ 𝐴 ∙ 𝑣
2
𝜌 ∙ 14𝜋𝑑2 ∙ 𝑣
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+
+ 𝜌∙𝐴∙𝑣
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Homework Aerodynamics
Please study chapter 5 of the reader:
Saving Energy in a Vehicle-Reader 2022.pdf:
And do the exercises of week 5
see Education / Onderwijs Online
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