Final Exam Time: 120 min Course: 58:160, Fall 2006 Name
... 3) A thin plastic panel (3 mm thick) is lowered from a ship to a construction site on the ocean floor. The plastic panel weighs 500 N in air and is lowered at a constant rate of 6m/s. Assuming that the panel remains vertically oriented, calculate the tension in the cable. ( Note: Buoyancy is not neg ...
... 3) A thin plastic panel (3 mm thick) is lowered from a ship to a construction site on the ocean floor. The plastic panel weighs 500 N in air and is lowered at a constant rate of 6m/s. Assuming that the panel remains vertically oriented, calculate the tension in the cable. ( Note: Buoyancy is not neg ...
transitions in a soft-walled channel.
... is studied at Reynolds numbers less than 1000 for the hard-walled laminar-turbulent transition. Here, the Reynolds number is defined as Re=(ρV h/η ), where ρ and η are the fluid density and viscosity, h is the channel height and V is the average flow velocity. Due to the fluid wall coupling, the wal ...
... is studied at Reynolds numbers less than 1000 for the hard-walled laminar-turbulent transition. Here, the Reynolds number is defined as Re=(ρV h/η ), where ρ and η are the fluid density and viscosity, h is the channel height and V is the average flow velocity. Due to the fluid wall coupling, the wal ...
Biomechanical principles of motion through air and water.
... • There are three shapes a flight path can form, depending on the angle od release: vertical shape (straight up and down), parabolic (when projectile is between 0-90 degrees), parabola (object projected at 0 degrees or perfectly horizontal). ...
... • There are three shapes a flight path can form, depending on the angle od release: vertical shape (straight up and down), parabolic (when projectile is between 0-90 degrees), parabola (object projected at 0 degrees or perfectly horizontal). ...
Lecture Outline (WORD)
... around closed lows and closed highs on upper air maps. (Again the figure shows upper air isobars rather than height contours). o Notice that for either a close upper high or a closed upper low, the net force must be directed inward in order for the air to curve. o The same thing must happen when air ...
... around closed lows and closed highs on upper air maps. (Again the figure shows upper air isobars rather than height contours). o Notice that for either a close upper high or a closed upper low, the net force must be directed inward in order for the air to curve. o The same thing must happen when air ...
Fluid Dynamics
... The onset of turbulence occurs when the Reynolds number, Re >2000. Reynolds number is defined as ...
... The onset of turbulence occurs when the Reynolds number, Re >2000. Reynolds number is defined as ...
Lecture 3 - fluid motion - BYU Physics and Astronomy
... when they look so different? And what it’s the value of the constant in the second equation, anyway? What should you tell him? ...
... when they look so different? And what it’s the value of the constant in the second equation, anyway? What should you tell him? ...
Chapter-9 The Behavior of Fluids
... This observation is now known as Archimedes' Principle and gave him the means to solve the problem. He was so excited that he ran naked through the streets of Syracuse shouting "Eureka! eureka!" (I have found it!). The fraudulent goldsmith was brought to justice. ...
... This observation is now known as Archimedes' Principle and gave him the means to solve the problem. He was so excited that he ran naked through the streets of Syracuse shouting "Eureka! eureka!" (I have found it!). The fraudulent goldsmith was brought to justice. ...
Homework #2
... at 300℃ and 15.5MPa. MIT people suggest the annular fuel(shown in the righthanded side in the below figure) in which the central channel exists to cool the fuel. In the actual design process, the pressure drop in the external channel should be the same as that in the internal channel. Estimate the p ...
... at 300℃ and 15.5MPa. MIT people suggest the annular fuel(shown in the righthanded side in the below figure) in which the central channel exists to cool the fuel. In the actual design process, the pressure drop in the external channel should be the same as that in the internal channel. Estimate the p ...
8.Conclusions (1/2)
... similar effect on the front wing and shows the marked change in the Cl as the AOA approaches 12. After this, there is a drop in that value of about 10%, indicating the suspected “stall” condition. In a similar vein, there is a marked monotonous increase in the Cd as the AOA increases, which is antic ...
... similar effect on the front wing and shows the marked change in the Cl as the AOA approaches 12. After this, there is a drop in that value of about 10%, indicating the suspected “stall” condition. In a similar vein, there is a marked monotonous increase in the Cd as the AOA increases, which is antic ...
Problem 1. Water flows steadily from a large closed tank as shown in
... Solution. Solution of the Navier-Stokes equation for flow between the parallel plates looks like that: ...
... Solution. Solution of the Navier-Stokes equation for flow between the parallel plates looks like that: ...
Problem Sheet 3
... At t = 0 the cylinder is suddenly brought to rest. Assuming that the flow remains twodimensional, find ω for t > 0. Hence determine ∇2 ψ in t > 0 and verify that the flow can ...
... At t = 0 the cylinder is suddenly brought to rest. Assuming that the flow remains twodimensional, find ω for t > 0. Hence determine ∇2 ψ in t > 0 and verify that the flow can ...
2014
... c) Suppose now that the fluid has viscosity. From the NavierStokes equations and continuity, develop a differential equation for the pressure variation in the small gap between wall and disk. Assume u z = 0 in the gap. Describe briefly how viscosity modifies the inviscid solution. For exampl ...
... c) Suppose now that the fluid has viscosity. From the NavierStokes equations and continuity, develop a differential equation for the pressure variation in the small gap between wall and disk. Assume u z = 0 in the gap. Describe briefly how viscosity modifies the inviscid solution. For exampl ...
AMEE 202 Midterm S14_1 Group 2
... Answer all questions from Section A and two questions from Section B. All necessary work must be shown. Wherever needed: H 2O 1000 kg/m3 , H2O 1.15 10-3 Pa s, air 1.225 kg/m 3 , air 1.79 10 5 kg /(m s ) g=9.81 m/s2 , patm 1.01 105 N/m 2 , 1.0 in= 0.0254 m, 1 ft=12 in, R ...
... Answer all questions from Section A and two questions from Section B. All necessary work must be shown. Wherever needed: H 2O 1000 kg/m3 , H2O 1.15 10-3 Pa s, air 1.225 kg/m 3 , air 1.79 10 5 kg /(m s ) g=9.81 m/s2 , patm 1.01 105 N/m 2 , 1.0 in= 0.0254 m, 1 ft=12 in, R ...
Page 1 of 8 King Saud University Mech. Eng. Department College of
... Hence the force exerted on the jet by the gate is in the negative x-direction and the force exerted by the jet on the gate is equal to – . For the gate to stay vertical the moment due to the force on the gate by the jet and the moment due to the force by the body of water on the right has to vanish ...
... Hence the force exerted on the jet by the gate is in the negative x-direction and the force exerted by the jet on the gate is equal to – . For the gate to stay vertical the moment due to the force on the gate by the jet and the moment due to the force by the body of water on the right has to vanish ...
ME19b. SOLUTIONS. Jan. 5, 2010. Due Jan. 14
... two departing jets are βb and (1 − β)b as indicated in the figure. It is assumed that the flow is planar and that the pressure in surrounding air is everywhere atmospheric. 1. Find the lift and drag on the wedge per unit length normal to the sketch as functions of ρ, U , b, β, α and θ. Note that dra ...
... two departing jets are βb and (1 − β)b as indicated in the figure. It is assumed that the flow is planar and that the pressure in surrounding air is everywhere atmospheric. 1. Find the lift and drag on the wedge per unit length normal to the sketch as functions of ρ, U , b, β, α and θ. Note that dra ...
ME 750A: Spring 2005 HW Due on Wednesday, March 9
... ME 750A: Spring 2005 HW Due on Wednesday, March 9 1. It has been suggested that the velocity field near the core of a tornado may be approximated by V = – er (q/r) + e (K/r). Does this represent incompressible flow? Is this an irrotational flow? [FM 5.16, 5.78] 2. An incompressible fluid of negligi ...
... ME 750A: Spring 2005 HW Due on Wednesday, March 9 1. It has been suggested that the velocity field near the core of a tornado may be approximated by V = – er (q/r) + e (K/r). Does this represent incompressible flow? Is this an irrotational flow? [FM 5.16, 5.78] 2. An incompressible fluid of negligi ...
Control surface Control Volume
... EXP1)The balloon is being filled through section 1, where the area is A1, velocity is V1,and fluid density is ρ1. The average density within the balloon is ρb(t). Find an expression for the rate of change of system mass within the balloon at this ...
... EXP1)The balloon is being filled through section 1, where the area is A1, velocity is V1,and fluid density is ρ1. The average density within the balloon is ρb(t). Find an expression for the rate of change of system mass within the balloon at this ...
OH 5: Fluid Dynamics
... More air flows over the upper curved surface than the lower flat surface, such that the difference in velocity across the surfaces results in a pressure difference between the two sides The external force resulting from the pressure difference is perpendicular to the direction of flow velocity, ...
... More air flows over the upper curved surface than the lower flat surface, such that the difference in velocity across the surfaces results in a pressure difference between the two sides The external force resulting from the pressure difference is perpendicular to the direction of flow velocity, ...
Fluid System - Model paper 1
... Where a: cross-sectional area. w: velocity of jet relative to the motion of plate. v: absolute velocity of jet. Force exerted on the fluid by the vane Fxis equal to ...
... Where a: cross-sectional area. w: velocity of jet relative to the motion of plate. v: absolute velocity of jet. Force exerted on the fluid by the vane Fxis equal to ...
Coandă effect
The Coandă effect /ˈkwaːndə/ is the tendency of a fluid jet to be attracted to a nearby surface. The principle was named after Romanian aerodynamics pioneer Henri Coandă, who was the first to recognize the practical application of the phenomenon in aircraft development.