Calculation of Wind Tunnel Speed and Drag Force on a Full-Size Car

Answer
We need to utilize the principles of similarity to determine the speed of the
wind tunnel and the drag force on the full-size car.
Assumptions:
1. Geometrical similarity exists between the model and the prototype. 2. Air
is in-compressible. 3. The walls of the tunnel are far away from the model car
and do not interfere with the aerodynamic drag on the car.
Properties of Air:
For atmospheric pressure and T = 25◦ : ρ = 1.184 kg/m3, μ = 1.849×10−5 kg
m · s .
At T = 5◦ : ρ = 1.269 kg/m3, μ = 1.754 × 10−5 kg
m · s .
a)
For the model and prototype to be dynamically similar, their Reynolds number
must be equal, that is:
Rem = Rep
Now,
Reynolds number: Re = ρ V L
μ
where, V is velocity of the object and L is the characteristic dimension.
Since the Reynolds number for model and prototype are equal:
ρm Vm Lm
μ = ρp Vp Lp
μp
∴ Vm
Vp
= ρp
ρm
· Lp
Lm
· μm
μp
1

The model car is tested at 5◦ while the full-size car will experience a temper-
ature of 25◦. Therefore properties of air (density and viscosity) for model and
prototype have to be considered at these temperatures respectively.
The model is built at one-fifth scale, which implies:
Lp
Lm
= 5
1
Therefore,
Vm
Vp
= 1.184
1.269 · 5
1 · 1.754 × 10−5
1.849 × 10−5 = 4.425
=⇒ Vm = 4.425 × Vp = 4.425 × 75 = 331.91 km/h
Thus, the wind tunnel should run at a speed of 331.9 km/h, in order to create
the required dynamic similarity.
b)
Owing to the similarity principles the non-dimensional drag on the model and
prototype must be same:
( FD
ρ V 2 L2
)
m
=
( FD
ρ V 2 L2
)
p
where, FD is the drag force experienced.
Therefore,
FDp
FDm
= L2
p
L2
m
· V 2
p
V 2
m
· ρp
ρm
=
( 5
1
)2
· 752
331.92 · 1.184
1.269 = 1.191
∴ FDp = FDm × 1.191 = 47.643 N
Thus, the predicted drag on the full-size car will be 47.643 N .
2

Assumptions:
1. Steady and in-compressible flow. 2. Turbulent and fully developed flow. 3.
Reservoir at atmospheric pressure. 4. Velocity heads are negligible.
The properties of water at 20◦ are:
ρ = 998 kg/m3, μ = 1.002 × 10−3 kg/ms, ν = 1.004 × 10−6 m2/s
The roughness of commercial steel pipe is 45 μ − m.
a) Flow rate of water through shower head:
From entry to the shower, the piping is 11m in length, a tee (KL = 0.9),
two standard elbows (KL = 0.9 each), a fully open globe valve (KL = 10),
and a shower head (KL = 12) installed along this line. Therefore, ∑ KL =
0.9 + 2 × 0.9 + 10 + 12 = 24.7. The energy equation for a control volume
between points 1 and 2 is:
P1
ρ g + V 2
1
2 g + z1 = P2
ρ g + V 2
2
2 g + z1 + hL
Implies, P1 − P2
ρ g = P1, gauge
ρ g = z2 − z1 + hL
Therefore,
hL = 200000
998 × 9.81 − 2 = 18.428 m
Also,
hL =
(
f L
D + ∑ KL
) V 2
2g
Implies,
18.428 = (733.333f + 24.7) V 2
19.82 (1)
3