Department of Mechanical Engineering

... 1] A jet of alcohol strikes the vertical plates shown in the figure. A force F=425 N is required to hold the plate stationary. Assuming there are no losses in the nozzle, estimate a) The mass flow rate of alcohol. b) The absolute pressure at section 1. 2] For the container shown, derive a formula fo ...

... 1] A jet of alcohol strikes the vertical plates shown in the figure. A force F=425 N is required to hold the plate stationary. Assuming there are no losses in the nozzle, estimate a) The mass flow rate of alcohol. b) The absolute pressure at section 1. 2] For the container shown, derive a formula fo ...

States of Matter Part 3

... inbetween them. b. Strong wind between buildings. c. Big trucks passing small cars. d. Landscapers use trees to channel and create wind. ...

... inbetween them. b. Strong wind between buildings. c. Big trucks passing small cars. d. Landscapers use trees to channel and create wind. ...

Aula Teórica 17

... Adverse pressure gradient • In case of the adverse pressure gradient pressure force decreases the velocity and can invert the sense of the flow. • For layers close to the wall the forward shear stress (above) is larger that the backward shear (below) and thus friction contribute to keep the forward ...

... Adverse pressure gradient • In case of the adverse pressure gradient pressure force decreases the velocity and can invert the sense of the flow. • For layers close to the wall the forward shear stress (above) is larger that the backward shear (below) and thus friction contribute to keep the forward ...

UNDERVISNING I TPM VED HiB

... located in a fluid flow, also will experience a force normal to the direction of the incoming flow. This is called lift. ...

... located in a fluid flow, also will experience a force normal to the direction of the incoming flow. This is called lift. ...

Water is flowing into and discharging from a pipe Usection as shown

... Water is flowing into and discharging from a pipe Usection as shown in Fig. P6–58. At flange (1), the total absolute pressure is 200 kPa, and 30 kg/s flows into the pipe. At flange (2), the total pressure is 150 kPa. At location (3), 8 kg/s of water discharges to the atmosphere, which is at 100 kPa. ...

... Water is flowing into and discharging from a pipe Usection as shown in Fig. P6–58. At flange (1), the total absolute pressure is 200 kPa, and 30 kg/s flows into the pipe. At flange (2), the total pressure is 150 kPa. At location (3), 8 kg/s of water discharges to the atmosphere, which is at 100 kPa. ...

Levitating Beachballs - Physics Department, Princeton University

... For levitation to occur, the high-speed drag force must be large enough to counteract gravity and the pressure-gradient force. If so, we anticipate that vertical motion in the vicinity of the equilibrium point is stable. The situation for horizontal motion is less clear,since the (large) drag force ...

... For levitation to occur, the high-speed drag force must be large enough to counteract gravity and the pressure-gradient force. If so, we anticipate that vertical motion in the vicinity of the equilibrium point is stable. The situation for horizontal motion is less clear,since the (large) drag force ...

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 ...

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, ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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: ...

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 ...

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 ...

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. ...

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.