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Fluid Dynamics

Fluid Dynamics. Streamline A.K.A. Laminar Flow Laminar = in layers Fluid flows smoothly We will study this kind of flow. Turbulent Flow The fluid travels in small and unpredictable circular paths. These circular paths are called eddy currents “Eddies” for short. Two Types of Fluid Flow.

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Fluid Dynamics

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  1. Fluid Dynamics

  2. Streamline A.K.A. Laminar Flow Laminar = in layers Fluid flows smoothly We will study this kind of flow Turbulent Flow The fluid travels in small and unpredictable circular paths. These circular paths are called eddy currents “Eddies” for short Two Types of Fluid Flow

  3. Flow Rates • A tub of large diameter compacts to a tube of small diameter. When a fluid is flowing is a streamline manner, how does the speed of the fluid in the small section compare to that of the large? V2 V1

  4. To Solve this we need to use two assumptions • Assumption 1: The amount of liquid that flows through the large section in one second must be the same in the smaller section. • If 4 kg of liquid pass through the large pipe in 1 second, then 4 kg of liquid must also pass through the small pipe in that same second. • Assumption 2: The liquid does not expand or compress when it’s pressure changes • The liquid’s density is constant. • If the density is constant then assumption 1 can be applied to the fluid's volume, as well as mass. • If 2 gal. of liquid pass through the large pipe in 1 seconds, then 2 gal. kg of liquid must also pass through the small pipe in 1 second.

  5. = Vol1/(1 sec) Vol2/(1 sec) For the two volumes to be equal the fluid in the pipe with the smaller cross sectional area must travel a farther distance (Must be moving faster than the fluid in the wider pipe) Comparing volumes We know that the amount of volume of fluid that flows per second is the same in both tubes Vol1/(1 sec) Vol2/(1 sec)

  6. A2 A1 L2 L1 A2 L2 A1 L1 = (1 sec) (1 sec) Remember this is true only if the fluid's density is constant. = A2 V2 A1 V1 Equation of Continuity Vol1/(1 sec) Vol2/(1 sec) = Vol1/(1 sec) Vol2/(1 sec) Vel = Length/time

  7. When a fluid is moving it exerts less pressure on the sides of its Container, Then if it were at rest. V Fluid Pressure and flow. The higher the speed of the fluid, the lower the pressure the fluid exerts on the container’s sides.

  8. The Molecular View • Even if a fluid as a whole is considered to be at rest, the molecules are in a constant state of motion. • If the fluid is at rest the purely random motion of the molecules cause all the sides of the container to be stuck by moving molecules with nearly the same frequency and force. • If the fluid is moving, however, most of the molecules motion will be parallel to the container's surface. • This means that the molecules collide with the container sides less often, and at a shallower angle (reducing the pressure)

  9. Bernoulli’s Principle • This principle relates the speed that a fluid has inside a container to the pressure that fluid creates. • Bernoulli’s Principle: That when a fluid has a high velocity it has a low pressure, and when a fluid has a low velocity it has a high pressure. • In meteorology how does the weather in a high pressure system to that of a low pressure system?

  10. V P2 P2 P1 P1 Spray Bottles P2 P1 P1

  11. Beach Balls and Fans If you have a fan that is pointing upwards and turned on. What will happen to a beach ball if you place it in the fan’s wind? Why? Ball Fan

  12. Bernoulli’s Equation • This equation relates the pressure of a fluid to both its speed, and depth. • Pressure on fluid +(1/2)rV2 + ragh = Constant • P1 +(1/2)rV12 + ragh1 = P2 +(1/2)rV22 + ragh2 • Bernoulli's equation is an extension of the conservation of energy. • Bernoulli's Equation will often work in conjunction with the equation of continuity. • A1V1 = A2V2

  13. Sample 1 Dia = 20cm Find the speed of the liquid as it flows out of the tube (the top of the tube is open to the air). 1m Dia = 5cm .4m

  14. P1 +(1/2)rV12 + ragh1 = P2 +(1/2)rV22 + ragh2 (1 atm) +(1/2)rV12 + rg(1.4m) = (1 atm) +(1/2)rV22 + rg(.4m) (1/2)rV12 + rg(1.4m) = (1/2)rV22 + rg(.4m) (1/2)V12 + g(1.4m) = (1/2)V22 + g(.4m) Substitute and solve Dia = 20cm V22 = V12 + 2 g(1.0 m) Point 1 A1V1 = A2V2 1m [p(.2m)2]V1 = p[(.05m)2 ]V2 Dia = 5cm (.05m)2 V1 = ----------------* V2 (.2m)2 .4m Point 2

  15. A2 V1 = ----* V2 (0)V2 = 0m/s A1 ~ >> Dia1 Dia2 ~ Torricelli theorem Dia1 If the Area of the top is significantly greater than the area of the bottom then the equation A1V1 = A2V2 yields the following APPROXOMATION Dh Dia2 In short the top shows almost no movement when compared to the bottom.

  16. >> Dia1 Dia2 Torricelli theorem Dia1 This APROXOMATION simplifies Bernoulli's equation to the following: V22 = V12 + 2 g(Dh) Dh V22 = 2 g(Dh) Dia2

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