Download 9.42 - IPFW ETCS

Problem 9.42
(Difficulty 2)
9.42 European Intercity Express trains operate at speeds of up to 280
𝑘𝑘
.
ℎ𝑟
Suppose that a train is 120 𝑚
long. Treat the sides and top of the train as a smooth flat plate 9 𝑚 wide. When the train moves through
still air at sea level, calculate the possible length of the laminar boundary layer and thickness of this
layer at its down-stream end. What is the thickness of the boundary layer at the rear end of the train?
What is the viscous drag force on the train and what power must be expended to overcome this
resistance at maximum speed? At 50% of maximum?
Find The drag force and other boundary layer parameters
Solution: Use the boundary layer relations to find the force and other parameters
The velocity is:
𝑈 = 280
The density and the viscosity of the air are:
𝜌 = 1.225
The Reynolds number is defined as:
𝑘𝑘
𝑎𝑎𝑎 𝜇 = 1.789 × 10−5 𝑃𝑃 ∙ 𝑠
𝑚3
The length of the laminar boundary layer is:
𝐿=
𝑘𝑘
𝑚
= 77.8
ℎ𝑟
𝑠
𝑅𝑅 =
𝜌𝜌𝜌
𝜇
𝜇𝜇𝜇 1.789 × 10−5 𝑃𝑃 ∙ 𝑠 × 500000
=
= 0.0939𝑚
𝑘𝑘
𝑚
𝜌𝜌
1.225 3 × 77.8
𝑠
𝑚
For the thickness of the laminar boundary layer we have:
𝛿
30
=�
𝑥
𝑅𝑅
At the downstream end we have:
30
30
𝛿 = 𝐿 � = 0.0939 𝑚 × �
= 7.27 × 10−4 𝑚
𝑅𝑅
500000
At the end of the train we have for the Reynolds number:
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.225 𝑚3 × 77.8 𝑠 × 120 𝑚
=
= 6.39 × 108
𝑅𝑅 =
𝜇
1.789 × 10−5 𝑃𝑃 ∙ 𝑠
The boundary layer thickness at the end is:
𝛿=𝑥
0.382
1
𝑅𝑅 5
𝛿 0.382
=
1
𝑥
𝑅𝑅 5
= 120 𝑚 ×
The drag force can be calculated by:
The drag coefficient for turbulent flow is
1
(6.39 × 108 )5
= 0.795 𝑚
1
𝐹𝐷 = 𝐶𝐷 × 𝜌𝜌𝑈 2
2
𝐶𝐷 =
And the force is
𝐹𝐷 = 0.0013 × 0.5 × 1.225
The power to overcome this drag is:
0.382
0.0742
1
𝑅𝑅 5
= 0.0013
𝑘𝑘
𝑚 2
× 120𝑚 × 9 𝑚 × �77.8 � = 5.21 𝑘𝑘
3
𝑚
𝑠
𝑃 = 𝐹𝐷 𝑈 = 5.21 𝑘𝑘 × 77.8
For 50% of the maximum velocity we have:
𝑚
= 405 𝑘𝑘
𝑠
1
𝐹𝐷 = 𝐶𝐷 × 𝜌𝜌𝑈 2
2
The Reynolds number is
𝑘𝑘
𝑚
𝜌𝜌𝜌 1.225 𝑚3 × 0.5 × 77.8 𝑠 × 120 𝑚
=
= 3.195 × 108
𝑅𝑅 =
𝜇
1.789 × 10−5 𝑃𝑃 ∙ 𝑠
And the drag force is
𝐶𝐷 =
The force is
𝐹𝐷 = 0.0015 × 0.5 × 1.225
The power to overcome this drag is:
0.0742
1
𝑅𝑅 5
= 0.0015
𝑘𝑘
𝑚 2
×
120𝑚
×
9
𝑚
×
�0.5
×
77.8
� = 1.502 𝑘𝑘
𝑚3
𝑠
𝑃 = 𝐹𝐷 𝑈 = 1.502 𝑘𝑘 × 0.5 × 77.8
𝑚
= 58.4 𝑘𝑘
𝑠