1 http://www.physics.usyd.edu.au/teach_res/jp/fluids09 web notes: lect6.ppt.

  • Published on
    31-Mar-2015

  • View
    214

  • Download
    2

Embed Size (px)

Transcript

  • Slide 1

1 http://www.physics.usyd.edu.au/teach_res/jp/fluids09 web notes: lect6.ppt Slide 2 2 Ideal fluid Real fluid Slide 3 3 (1) Surface of liquid (2) Just outside hole v 2 = ? m.s -1 y1y1 y2y2 Draw flow tubes v 1 ~ 0 m.s -1 p 1 = p atm p 2 = p atm h = (y 1 - y 2 ) What is the speed with which liquid flows from a hole at the bottom of a tank? Slide 4 4 Assume liquid behaves as an ideal fluid and that Bernoulli's equation can be applied p 1 + v 1 2 + g y 1 = p 2 + v 2 2 + g y 2 A small hole is at level (2) and the water level at (1) drops slowly v 1 = 0 p 1 = p atm p 2 = p atm g y 1 = v 2 2 + g y 2 v 2 2 = 2 g (y 1 y 2 ) = 2 g h h = (y 1 - y 2 ) v 2 = (2 g h) Torricelli formula (1608 1647) This is the same velocity as a particle falling freely through a height h Slide 5 5 (1) (2) FF mm h v 1 = ? What is the speed of flow in section 1 of the system? Slide 6 6 Assume liquid behaves as an ideal fluid and that Bernoulli's equation can be applied for the flow along a streamline p 1 + v 1 2 + g y 1 = p 2 + v 2 2 + g y 2 y 1 = y 2 p 1 p 2 = F (v 2 2 - v 1 2 ) p 1 - p 2 = m g h A 1 v 1 = A 2 v 2 v 2 = v 1 (A 1 / A 2 ) m g h = F { v 1 2 (A 1 / A 2 ) 2 - v 1 2 } = F v 1 2 {(A 1 / A 2 ) 2 - 1} Slide 7 7 How does a siphon work? Q: What do we know? Continuous flow Pressure in top section > 0 otherwise there will be a vacuum p C 0 Focus on falling water not rising water p atm - p C 0 p atm g y C yCyC p C Slide 8 8 C B A D yAyA yByB yCyC Assume that the liquid behaves as an ideal fluid, the equation of continuity and Bernoulli's equation can be used. y D = 0 p A = p atm = p D Slide 9 9 p C + v C 2 + g y C = p D + v D 2 + g y D From equation of continuity v C = v D p C = p D + g (y D - y C ) = p atm + g (y D - y C ) The pressure at point C can not be negative p C 0 and y D = 0 p C = p atm - g y C 0 y C p atm / ( g) For a water siphon p atm ~ 10 5 Pa g ~ 10 m.s -1 ~ 10 3 kg.m -3 y C 10 5 / {(10)(10 3 )} m y C 10 m Consider points C and D and apply Bernoulli's principle. Slide 10 10 p A + v A 2 + g y A = p D + v D 2 + g y D v D 2 = 2 (p A p D ) / + v A 2 + 2 g (y A - y D ) p A p D = 0 y D = 0 assume v A 2

Recommended

View more >