FTFS Chap15 P043 - Chapter 15 Flow Over Bodies: Drag and...

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Unformatted text preview: Chapter 15 Flow Over Bodies: Drag and Lift Flow over Flat Plates 15-43C The fluid viscosity is responsible for the development of the velocity boundary layer. For the idealized inviscid fluids (fluids with zero viscosity), there will be no velocity boundary layer. 15-44C The friction coefficient represents the resistance to fluid flow over a flat plate. It is proportional to the drag force acting on the plate. The drag coefficient for a flat surface is equivalent to the mean friction coefficient. 15-45C The friction coefficient will change with position in laminar flow over a flat plate (it will decrease along the plate in the flow direction). 15-46C The average friction coefficient in flow over a flat plate is determined by integrating the local friction coefficient over the entire plate, and then dividing it by the length of the plate. Or, it can be determined experimentally by measuring the drag force, and dividing it by the dynamic pressure. 15-47E Light oil flows over a flat plate. The total drag force per unit width of the plate is to be determined. Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 The surface of the plate is smooth. Properties The density and kinematic viscosity of light oil at 75°F are ρ = 55.3 lbm/ft 3 and ν = 7.751×10 –3 ft 2 /s (Table A-19E). Analysis Noting that L = 15 ft, the Reynolds number at the end of the plate is 4 2 3 10 161 . 1 /s ft 10 751 . 7 ft) ft/s)(15 6 ( Re × = × = =- ν L L V which is less than the critical Reynolds number. Thus we have laminar flow over the entire plate, and the average friction coefficient is determined from 01232 . ) 10 161 . 1 ( 328 . 1 Re 328 . 1 5 . 4 5 . = × × = =-- L f C Noting that the pressure drag is zero and thus f D C C = for a flat plate, the drag force acting on the top surface of the plate per unit width becomes lbf 5.87 = ⋅ × × = = 2 2 3 2 2 ft/s lbm 32.2 lbf 1 2 ft/s) 6 )( lbm/ft 8 . 56 ( ) ft 1 15 ( 01232 . 2 V ρ A C F f D The total drag force acting on the entire plate can be determined by multiplying the value obtained above by the width of the plate. Discussion The force per unit width corresponds to the weight of a mass of 5.87 lbm. Therefore, a person who applies an equal and opposite force to the plate to keep it from moving will feel like he or she is using as much force as is necessary to hold a 5.87 lbm mass from dropping. 15-22 6 ft/s Oil L = 15 ft Chapter 15 Flow Over Bodies: Drag and Lift 15-48 Air flows over a plane surface at high elevation. The drag force acting on the top surface of the plate is to be determined for flow along the two sides of the plate. √ Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds number is Re cr = 5 × 10 5 . 3 Air is an ideal gas. 4 The surface of the plate is smooth....
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.

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FTFS Chap15 P043 - Chapter 15 Flow Over Bodies: Drag and...

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