Page 1 Chapter 11 Flow Over Bodies: Drag and Lift Flow over Flat Plates 11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), the lower the thickness of the boundary layer. 11-44C 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. 11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate. 11-46C 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. 11-47E 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 ? = 55.3 lbm/ft 3 and ? = 7.751×10 –3 ft 2 /s. 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 × = × = = - ? VL L Oil 6 ft/s 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 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1 5 . 0 4 5 . 0 = × × = = - - L f C L = 15 ft Noting that the pressure drag is zero and thus C for a flat plate, the drag force acting on the top surface of the plate per unit width becomes f D C = 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 . 0 2 V 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. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-22 Page 2 Chapter 11 Flow Over Bodies: Drag and Lift Flow over Flat Plates 11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), the lower the thickness of the boundary layer. 11-44C 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. 11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate. 11-46C 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. 11-47E 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 ? = 55.3 lbm/ft 3 and ? = 7.751×10 –3 ft 2 /s. 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 × = × = = - ? VL L Oil 6 ft/s 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 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1 5 . 0 4 5 . 0 = × × = = - - L f C L = 15 ft Noting that the pressure drag is zero and thus C for a flat plate, the drag force acting on the top surface of the plate per unit width becomes f D C = 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 . 0 2 V 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. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-22 Chapter 11 Flow Over Bodies: Drag and Lift 11-48 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. v 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. Properties The dynamic viscosity is independent of pressure, and for air at 25 °C it is µ = 1.849 ×10 -5 kg/m ·s. The air density at 25 °C = 298 K and 83.4 kPa is 3 3 kg/m 9751 . 0 K) K)(298 /kg m kPa (0.287 kPa 83.4 = · · = = RT P ? Analysis (a) If the air flows parallel to the 8 m side, the Reynolds number becomes 6 5 3 10 531 . 2 s kg/m 10 849 . 1 m) m/s)(8 6 ( ) kg/m 9751 . 0 ( Re × = · × = = - µ ?VL L 11-23 which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be 003189 . 0 10 531 . 2 1742 - ) 10 531 . 2 ( 074 . 0 Re 1742 - Re 074 . 0 6 5 / 1 6 5 / 1 = × × = = L L f C 2.5 m 8 m 6 m/s Air Noting that the pressure drag is zero and thus for a flat plate, the drag force acting on the top surface of the plate becomes f D C C = N 1.12 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 )( kg/m 9751 . 0 ( ) m 5 . 2 8 ( 003189 . 0 2 V A C F f D ? (b) If the air flows parallel to the 2.5 m side, the Reynolds number is 5 5 3 10 910 . 7 s kg/m 10 849 . 1 m) m/s)(2.5 6 ( ) kg/m 9751 . 0 ( Re × = · × = = - µ ?VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be 002691 . 0 10 910 . 7 1742 - ) 10 910 . 7 ( 074 . 0 Re 1742 - Re 074 . 0 5 5 / 1 5 5 / 1 = × × = = L L f C Then the drag force acting on the top surface of the plate becomes N 0.94 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 )( kg/m 9751 . 0 ( ) m 5 . 2 8 ( 002691 . 0 2 V A C F f D ? Discussion Note that the drag force is proportional to density, which is proportional to the pressure. Therefore, the altitude has a major influence on the drag force acting on a surface. Commercial airplanes take advantage of this phenomenon and cruise at high altitudes where the air density is much lower to save fuel. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Page 3 Chapter 11 Flow Over Bodies: Drag and Lift Flow over Flat Plates 11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), the lower the thickness of the boundary layer. 11-44C 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. 11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate. 11-46C 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. 11-47E 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 ? = 55.3 lbm/ft 3 and ? = 7.751×10 –3 ft 2 /s. 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 × = × = = - ? VL L Oil 6 ft/s 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 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1 5 . 0 4 5 . 0 = × × = = - - L f C L = 15 ft Noting that the pressure drag is zero and thus C for a flat plate, the drag force acting on the top surface of the plate per unit width becomes f D C = 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 . 0 2 V 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. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-22 Chapter 11 Flow Over Bodies: Drag and Lift 11-48 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. v 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. Properties The dynamic viscosity is independent of pressure, and for air at 25 °C it is µ = 1.849 ×10 -5 kg/m ·s. The air density at 25 °C = 298 K and 83.4 kPa is 3 3 kg/m 9751 . 0 K) K)(298 /kg m kPa (0.287 kPa 83.4 = · · = = RT P ? Analysis (a) If the air flows parallel to the 8 m side, the Reynolds number becomes 6 5 3 10 531 . 2 s kg/m 10 849 . 1 m) m/s)(8 6 ( ) kg/m 9751 . 0 ( Re × = · × = = - µ ?VL L 11-23 which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be 003189 . 0 10 531 . 2 1742 - ) 10 531 . 2 ( 074 . 0 Re 1742 - Re 074 . 0 6 5 / 1 6 5 / 1 = × × = = L L f C 2.5 m 8 m 6 m/s Air Noting that the pressure drag is zero and thus for a flat plate, the drag force acting on the top surface of the plate becomes f D C C = N 1.12 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 )( kg/m 9751 . 0 ( ) m 5 . 2 8 ( 003189 . 0 2 V A C F f D ? (b) If the air flows parallel to the 2.5 m side, the Reynolds number is 5 5 3 10 910 . 7 s kg/m 10 849 . 1 m) m/s)(2.5 6 ( ) kg/m 9751 . 0 ( Re × = · × = = - µ ?VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be 002691 . 0 10 910 . 7 1742 - ) 10 910 . 7 ( 074 . 0 Re 1742 - Re 074 . 0 5 5 / 1 5 5 / 1 = × × = = L L f C Then the drag force acting on the top surface of the plate becomes N 0.94 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 )( kg/m 9751 . 0 ( ) m 5 . 2 8 ( 002691 . 0 2 V A C F f D ? Discussion Note that the drag force is proportional to density, which is proportional to the pressure. Therefore, the altitude has a major influence on the drag force acting on a surface. Commercial airplanes take advantage of this phenomenon and cruise at high altitudes where the air density is much lower to save fuel. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 11 Flow Over Bodies: Drag and Lift 11-49 Wind is blowing parallel to the side wall of a house. The drag force acting on the wall is to be determined for two different wind velocities. vEES 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 wall surface is smooth (the actual wall surface is usually very rough). 5 The wind blows parallel to the wall. Properties The density and kinematic viscosity of air at 1 atm and 5 °C are ? = 1.269 kg/m 3 and ? = 1.382×10 –5 m 2 /s . Air 55 km/h 4 m 10 m Analysis The Reynolds number is 7 2 5 10 105 . 1 /s m 10 382 . 1 m) m/s)(10 6 . 3 / 55 ( Re × = × = = - ? VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is 002730 . 0 10 105 . 1 1742 - ) 10 105 . 1 ( 074 . 0 Re 1742 - Re 074 . 0 7 5 / 1 7 5 / 1 = × × = = L L f C Noting that the pressure drag is zero and thus C f D C = for a flat plate, the drag force acting on the wall surface is N 16.2 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 . 3 / 55 )( kg/m 269 . 1 ( ) m 4 10 ( 00273 . 0 2 V A C F f D ? (b) When the wind velocity is doubled to 110 km/h, the Reynolds number becomes 7 2 5 10 211 . 2 /s m 10 382 . 1 m) m/s)(10 6 . 3 / 110 ( Re × = × = = - ? VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient and the drag force become 002435 . 0 10 211 . 2 1742 - ) 10 211 . 2 ( 074 . 0 Re 1742 - Re 074 . 0 7 5 / 1 7 5 / 1 = × × = = L L f C N 57.7 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 . 3 / 110 )( kg/m 269 . 1 ( ) m 4 10 ( 002435 . 0 2 V A C F f D ? Treating flow over the side wall of a house as flow over a flat plate is not quite realistic. When flow hits a bluff body like a house, it separates at the sharp corner and a separation bubble exists over most of the side panels of the house. Therefore, flat plat boundary layer equations are not appropriate for this problem, and the entire house should considered in the solution instead. Discussion Note that the actual drag will probably be much higher since the wall surfaces are typically very rough. Also, we can solve this problem using the turbulent flow relation (instead of the combined laminar- turbulent flow relation) without much loss in accuracy. Finally, the drag force nearly quadruples when the velocity is doubled. This is expected since the drag force is proportional to the square of the velocity, and the effect of velocity on the friction coefficient is small. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-24 Page 4 Chapter 11 Flow Over Bodies: Drag and Lift Flow over Flat Plates 11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), the lower the thickness of the boundary layer. 11-44C 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. 11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate. 11-46C 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. 11-47E 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 ? = 55.3 lbm/ft 3 and ? = 7.751×10 –3 ft 2 /s. 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 × = × = = - ? VL L Oil 6 ft/s 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 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1 5 . 0 4 5 . 0 = × × = = - - L f C L = 15 ft Noting that the pressure drag is zero and thus C for a flat plate, the drag force acting on the top surface of the plate per unit width becomes f D C = 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 . 0 2 V 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. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-22 Chapter 11 Flow Over Bodies: Drag and Lift 11-48 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. v 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. Properties The dynamic viscosity is independent of pressure, and for air at 25 °C it is µ = 1.849 ×10 -5 kg/m ·s. The air density at 25 °C = 298 K and 83.4 kPa is 3 3 kg/m 9751 . 0 K) K)(298 /kg m kPa (0.287 kPa 83.4 = · · = = RT P ? Analysis (a) If the air flows parallel to the 8 m side, the Reynolds number becomes 6 5 3 10 531 . 2 s kg/m 10 849 . 1 m) m/s)(8 6 ( ) kg/m 9751 . 0 ( Re × = · × = = - µ ?VL L 11-23 which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be 003189 . 0 10 531 . 2 1742 - ) 10 531 . 2 ( 074 . 0 Re 1742 - Re 074 . 0 6 5 / 1 6 5 / 1 = × × = = L L f C 2.5 m 8 m 6 m/s Air Noting that the pressure drag is zero and thus for a flat plate, the drag force acting on the top surface of the plate becomes f D C C = N 1.12 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 )( kg/m 9751 . 0 ( ) m 5 . 2 8 ( 003189 . 0 2 V A C F f D ? (b) If the air flows parallel to the 2.5 m side, the Reynolds number is 5 5 3 10 910 . 7 s kg/m 10 849 . 1 m) m/s)(2.5 6 ( ) kg/m 9751 . 0 ( Re × = · × = = - µ ?VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be 002691 . 0 10 910 . 7 1742 - ) 10 910 . 7 ( 074 . 0 Re 1742 - Re 074 . 0 5 5 / 1 5 5 / 1 = × × = = L L f C Then the drag force acting on the top surface of the plate becomes N 0.94 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 )( kg/m 9751 . 0 ( ) m 5 . 2 8 ( 002691 . 0 2 V A C F f D ? Discussion Note that the drag force is proportional to density, which is proportional to the pressure. Therefore, the altitude has a major influence on the drag force acting on a surface. Commercial airplanes take advantage of this phenomenon and cruise at high altitudes where the air density is much lower to save fuel. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 11 Flow Over Bodies: Drag and Lift 11-49 Wind is blowing parallel to the side wall of a house. The drag force acting on the wall is to be determined for two different wind velocities. vEES 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 wall surface is smooth (the actual wall surface is usually very rough). 5 The wind blows parallel to the wall. Properties The density and kinematic viscosity of air at 1 atm and 5 °C are ? = 1.269 kg/m 3 and ? = 1.382×10 –5 m 2 /s . Air 55 km/h 4 m 10 m Analysis The Reynolds number is 7 2 5 10 105 . 1 /s m 10 382 . 1 m) m/s)(10 6 . 3 / 55 ( Re × = × = = - ? VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is 002730 . 0 10 105 . 1 1742 - ) 10 105 . 1 ( 074 . 0 Re 1742 - Re 074 . 0 7 5 / 1 7 5 / 1 = × × = = L L f C Noting that the pressure drag is zero and thus C f D C = for a flat plate, the drag force acting on the wall surface is N 16.2 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 . 3 / 55 )( kg/m 269 . 1 ( ) m 4 10 ( 00273 . 0 2 V A C F f D ? (b) When the wind velocity is doubled to 110 km/h, the Reynolds number becomes 7 2 5 10 211 . 2 /s m 10 382 . 1 m) m/s)(10 6 . 3 / 110 ( Re × = × = = - ? VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient and the drag force become 002435 . 0 10 211 . 2 1742 - ) 10 211 . 2 ( 074 . 0 Re 1742 - Re 074 . 0 7 5 / 1 7 5 / 1 = × × = = L L f C N 57.7 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 . 3 / 110 )( kg/m 269 . 1 ( ) m 4 10 ( 002435 . 0 2 V A C F f D ? Treating flow over the side wall of a house as flow over a flat plate is not quite realistic. When flow hits a bluff body like a house, it separates at the sharp corner and a separation bubble exists over most of the side panels of the house. Therefore, flat plat boundary layer equations are not appropriate for this problem, and the entire house should considered in the solution instead. Discussion Note that the actual drag will probably be much higher since the wall surfaces are typically very rough. Also, we can solve this problem using the turbulent flow relation (instead of the combined laminar- turbulent flow relation) without much loss in accuracy. Finally, the drag force nearly quadruples when the velocity is doubled. This is expected since the drag force is proportional to the square of the velocity, and the effect of velocity on the friction coefficient is small. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-24 Chapter 11 Flow Over Bodies: Drag and Lift 11-50E Air flows over a flat plate. The local friction coefficients at intervals of 1 ft is to be determined and plotted against the distance from the leading edge. 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. Properties The density and kinematic viscosity of air at 1 atm and 70°F are ? = 0.07489 lbm/ft 3 and ? = 0.5913 ft 2 /h = 1.643×10 –4 ft 2 /s . Analysis For the first 1 ft interval, the Reynolds number is 5 2 4 10 522 . 1 /s ft 10 643 . 1 ft) ft/s)(1 25 ( Re × = × = = - ? VL L 10 ft Air 25 ft/s which is less than the critical value of 510 5 × . Therefore, the flow is laminar. The local friction coefficient is 001702 . 0 ) 10 522 . 1 ( 664 . 0 Re 664 . 0 5 . 0 5 5 . 0 , = × = = x f C We repeat calculations for all 1-ft intervals. The results are x, ft Re C f 1 2 3 4 5 6 7 8 9 10 1.522E+05 3.044E+05 4.566E+05 6.088E+05 7.610E+05 9.132E+05 1.065E+06 1.218E+06 1.370E+06 1.522E+06 0.001702 0.001203 0.000983 0.004111 0.003932 0.003791 0.003676 0.003579 0.003496 0.003423 rho=0.07489 "lbm/ft3" nu=0.5913/3600 "ft2/s" V=25 “Local Re and C_f” Re=x*V/nu "f=0.664/Re^0.5" f=0.059/Re^0.2 12 3 4 5 6 7 8 9 10 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 x, ft f Discussion Note that the Reynolds number exceeds the critical value for x > 3 ft, and thus the flow is turbulent over most of the plate. For x > 3 ft, we used for friction coefficient. Note that C L L f C 1742/Re - Re / 074 . 0 5 / 1 = f decreases with Re in both laminar and turbulent flows. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-25 Page 5 Chapter 11 Flow Over Bodies: Drag and Lift Flow over Flat Plates 11-43C The fluid viscosity is responsible for the development of the velocity boundary layer. Velocity forces the boundary layer closer to the wall. Therefore, the higher the velocity (and thus Reynolds number), the lower the thickness of the boundary layer. 11-44C 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. 11-45C The local friction coefficient decreases with downstream distance in laminar flow over a flat plate. 11-46C 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. 11-47E 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 ? = 55.3 lbm/ft 3 and ? = 7.751×10 –3 ft 2 /s. 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 × = × = = - ? VL L Oil 6 ft/s 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 . 0 ) 10 161 . 1 ( 328 . 1 Re 328 . 1 5 . 0 4 5 . 0 = × × = = - - L f C L = 15 ft Noting that the pressure drag is zero and thus C for a flat plate, the drag force acting on the top surface of the plate per unit width becomes f D C = 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 . 0 2 V 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. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-22 Chapter 11 Flow Over Bodies: Drag and Lift 11-48 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. v 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. Properties The dynamic viscosity is independent of pressure, and for air at 25 °C it is µ = 1.849 ×10 -5 kg/m ·s. The air density at 25 °C = 298 K and 83.4 kPa is 3 3 kg/m 9751 . 0 K) K)(298 /kg m kPa (0.287 kPa 83.4 = · · = = RT P ? Analysis (a) If the air flows parallel to the 8 m side, the Reynolds number becomes 6 5 3 10 531 . 2 s kg/m 10 849 . 1 m) m/s)(8 6 ( ) kg/m 9751 . 0 ( Re × = · × = = - µ ?VL L 11-23 which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be 003189 . 0 10 531 . 2 1742 - ) 10 531 . 2 ( 074 . 0 Re 1742 - Re 074 . 0 6 5 / 1 6 5 / 1 = × × = = L L f C 2.5 m 8 m 6 m/s Air Noting that the pressure drag is zero and thus for a flat plate, the drag force acting on the top surface of the plate becomes f D C C = N 1.12 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 )( kg/m 9751 . 0 ( ) m 5 . 2 8 ( 003189 . 0 2 V A C F f D ? (b) If the air flows parallel to the 2.5 m side, the Reynolds number is 5 5 3 10 910 . 7 s kg/m 10 849 . 1 m) m/s)(2.5 6 ( ) kg/m 9751 . 0 ( Re × = · × = = - µ ?VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is determined to be 002691 . 0 10 910 . 7 1742 - ) 10 910 . 7 ( 074 . 0 Re 1742 - Re 074 . 0 5 5 / 1 5 5 / 1 = × × = = L L f C Then the drag force acting on the top surface of the plate becomes N 0.94 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 )( kg/m 9751 . 0 ( ) m 5 . 2 8 ( 002691 . 0 2 V A C F f D ? Discussion Note that the drag force is proportional to density, which is proportional to the pressure. Therefore, the altitude has a major influence on the drag force acting on a surface. Commercial airplanes take advantage of this phenomenon and cruise at high altitudes where the air density is much lower to save fuel. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. Chapter 11 Flow Over Bodies: Drag and Lift 11-49 Wind is blowing parallel to the side wall of a house. The drag force acting on the wall is to be determined for two different wind velocities. vEES 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 wall surface is smooth (the actual wall surface is usually very rough). 5 The wind blows parallel to the wall. Properties The density and kinematic viscosity of air at 1 atm and 5 °C are ? = 1.269 kg/m 3 and ? = 1.382×10 –5 m 2 /s . Air 55 km/h 4 m 10 m Analysis The Reynolds number is 7 2 5 10 105 . 1 /s m 10 382 . 1 m) m/s)(10 6 . 3 / 55 ( Re × = × = = - ? VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient is 002730 . 0 10 105 . 1 1742 - ) 10 105 . 1 ( 074 . 0 Re 1742 - Re 074 . 0 7 5 / 1 7 5 / 1 = × × = = L L f C Noting that the pressure drag is zero and thus C f D C = for a flat plate, the drag force acting on the wall surface is N 16.2 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 . 3 / 55 )( kg/m 269 . 1 ( ) m 4 10 ( 00273 . 0 2 V A C F f D ? (b) When the wind velocity is doubled to 110 km/h, the Reynolds number becomes 7 2 5 10 211 . 2 /s m 10 382 . 1 m) m/s)(10 6 . 3 / 110 ( Re × = × = = - ? VL L which is greater than the critical Reynolds number. Thus we have combined laminar and turbulent flow, and the friction coefficient and the drag force become 002435 . 0 10 211 . 2 1742 - ) 10 211 . 2 ( 074 . 0 Re 1742 - Re 074 . 0 7 5 / 1 7 5 / 1 = × × = = L L f C N 57.7 = ? ? ? ? ? ? ? ? · × × = = 2 2 3 2 2 m/s kg 1 N 1 2 m/s) 6 . 3 / 110 )( kg/m 269 . 1 ( ) m 4 10 ( 002435 . 0 2 V A C F f D ? Treating flow over the side wall of a house as flow over a flat plate is not quite realistic. When flow hits a bluff body like a house, it separates at the sharp corner and a separation bubble exists over most of the side panels of the house. Therefore, flat plat boundary layer equations are not appropriate for this problem, and the entire house should considered in the solution instead. Discussion Note that the actual drag will probably be much higher since the wall surfaces are typically very rough. Also, we can solve this problem using the turbulent flow relation (instead of the combined laminar- turbulent flow relation) without much loss in accuracy. Finally, the drag force nearly quadruples when the velocity is doubled. This is expected since the drag force is proportional to the square of the velocity, and the effect of velocity on the friction coefficient is small. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-24 Chapter 11 Flow Over Bodies: Drag and Lift 11-50E Air flows over a flat plate. The local friction coefficients at intervals of 1 ft is to be determined and plotted against the distance from the leading edge. 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. Properties The density and kinematic viscosity of air at 1 atm and 70°F are ? = 0.07489 lbm/ft 3 and ? = 0.5913 ft 2 /h = 1.643×10 –4 ft 2 /s . Analysis For the first 1 ft interval, the Reynolds number is 5 2 4 10 522 . 1 /s ft 10 643 . 1 ft) ft/s)(1 25 ( Re × = × = = - ? VL L 10 ft Air 25 ft/s which is less than the critical value of 510 5 × . Therefore, the flow is laminar. The local friction coefficient is 001702 . 0 ) 10 522 . 1 ( 664 . 0 Re 664 . 0 5 . 0 5 5 . 0 , = × = = x f C We repeat calculations for all 1-ft intervals. The results are x, ft Re C f 1 2 3 4 5 6 7 8 9 10 1.522E+05 3.044E+05 4.566E+05 6.088E+05 7.610E+05 9.132E+05 1.065E+06 1.218E+06 1.370E+06 1.522E+06 0.001702 0.001203 0.000983 0.004111 0.003932 0.003791 0.003676 0.003579 0.003496 0.003423 rho=0.07489 "lbm/ft3" nu=0.5913/3600 "ft2/s" V=25 “Local Re and C_f” Re=x*V/nu "f=0.664/Re^0.5" f=0.059/Re^0.2 12 3 4 5 6 7 8 9 10 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 x, ft f Discussion Note that the Reynolds number exceeds the critical value for x > 3 ft, and thus the flow is turbulent over most of the plate. For x > 3 ft, we used for friction coefficient. Note that C L L f C 1742/Re - Re / 074 . 0 5 / 1 = f decreases with Re in both laminar and turbulent flows. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-25 Chapter 11 Flow Over Bodies: Drag and Lift 11-51 Air flows on both sides of a continuous sheet of plastic. The drag force air exerts on the plastic sheet in the direction of flow is to be determined. 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 Both surfaces of the plastic sheet are smooth. 5 The plastic sheet does not vibrate and thus it does not induce turbulence in air flow. Properties The density and kinematic viscosity of air at 1 atm and 60 °C are ? = 1.059 kg/m 3 and ? = 1.896×10 –5 m 2 /s . Analysis The length of the cooling section is Air V 8 = 3 m/s m 0.5 = s) 2 ( m/s] ) 60 / 15 [( sheet = ? = t V L 15 m/min Plastic sheet The Reynolds number is 5 2 5 10 899 . 1 /s m 10 896 . 1 m) m/s)(1.2 (3 Re × = × = = - ? VL L which is less than the critical Reynolds number. Thus the flow is laminar. The area on both sides of the sheet exposed to air flow is 2 m 1.2 = m) m)(0.5 2 . 1 ( 2 2 = = wL A Then the friction coefficient and the drag force become 003048 . 0 ) 10 899 . 1 ( 328 . 1 Re 328 . 1 5 . 0 5 5 . 0 = × = = L f C N 0.0174 = = = 2 m/s) )(3 kg/m (1.059 ) m 2 . 1 )( 003048 . 0 ( 2 2 3 2 2 V A C F f D ? Discussion Note that the Reynolds number remains under the critical value, and thus the flow remains laminar over the entire plate. In reality, the flow may be turbulent because of the motion of the plastic sheet. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11-26Read More

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