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Problem 9.42 (Difficulty 2) 9.42 European Intercity Express trains operate at speeds of up to 280 ππ . βπ Suppose that a train is 120 π long. Treat the sides and top of the train as a smooth flat plate 9 π 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: π = 280 The density and the viscosity of the air are: π = 1.225 The Reynolds number is defined as: ππ πππ π = 1.789 × 10β5 ππ β π π3 The length of the laminar boundary layer is: πΏ= ππ π = 77.8 βπ π π π = πππ π πππ 1.789 × 10β5 ππ β π × 500000 = = 0.0939π ππ π ππ 1.225 3 × 77.8 π π For the thickness of the laminar boundary layer we have: πΏ 30 =οΏ½ π₯ π π At the downstream end we have: 30 30 πΏ = πΏ οΏ½ = 0.0939 π × οΏ½ = 7.27 × 10β4 π π π 500000 At the end of the train we have for the Reynolds number: ππ π πππ 1.225 π3 × 77.8 π × 120 π = = 6.39 × 108 π π = π 1.789 × 10β5 ππ β π The boundary layer thickness at the end is: πΏ=π₯ 0.382 1 π π 5 πΏ 0.382 = 1 π₯ π π 5 = 120 π × The drag force can be calculated by: The drag coefficient for turbulent flow is 1 (6.39 × 108 )5 = 0.795 π 1 πΉπ· = πΆπ· × πππ 2 2 πΆπ· = And the force is πΉπ· = 0.0013 × 0.5 × 1.225 The power to overcome this drag is: 0.382 0.0742 1 π π 5 = 0.0013 ππ π 2 × 120π × 9 π × οΏ½77.8 οΏ½ = 5.21 ππ 3 π π π = πΉπ· π = 5.21 ππ × 77.8 For 50% of the maximum velocity we have: π = 405 ππ π 1 πΉπ· = πΆπ· × πππ 2 2 The Reynolds number is ππ π πππ 1.225 π3 × 0.5 × 77.8 π × 120 π = = 3.195 × 108 π π = π 1.789 × 10β5 ππ β π And the drag force is πΆπ· = The force is πΉπ· = 0.0015 × 0.5 × 1.225 The power to overcome this drag is: 0.0742 1 π π 5 = 0.0015 ππ π 2 × 120π × 9 π × οΏ½0.5 × 77.8 οΏ½ = 1.502 ππ π3 π π = πΉπ· π = 1.502 ππ × 0.5 × 77.8 π = 58.4 ππ π