CE 319 F Daene Mc Kinney Elementary Mechanics

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CE 319 F Daene Mc. Kinney Elementary Mechanics of Fluids Bernoulli Equation

CE 319 F Daene Mc. Kinney Elementary Mechanics of Fluids Bernoulli Equation

Euler Equation • • Fluid element accelerating in l direction & acted on by

Euler Equation • • Fluid element accelerating in l direction & acted on by pressure and weight forces only (no friction) Newton’s 2 nd Law

Ex (5. 1) • • Given: Steady flow. Liquid is decelerating at a rate

Ex (5. 1) • • Given: Steady flow. Liquid is decelerating at a rate of 0. 3 g. Find: Pressure gradient in flow direction in terms of specific weight. Flow l 30 o

EX (5. 3) • • vertical Given: g = 10 k. N/m 3, p.

EX (5. 3) • • vertical Given: g = 10 k. N/m 3, p. B-p. A=12 k. Pa. Find: Direction of fluid acceleration. A 1 m B

HW (5. 7) • Ex (5. 6) What pressure is needed to accelerate water

HW (5. 7) • Ex (5. 6) What pressure is needed to accelerate water in a horizontal pipe at a rate of 6 m/s 2?

Ex (5. 10) • • Given: Steady flow. Velocity varies linearly with distance through

Ex (5. 10) • • Given: Steady flow. Velocity varies linearly with distance through the nozzle. Find: Pressure gradient ½-way through the nozzle V 1/2=(80+30)/2 ft/s = 55 ft/s d. V/dx = (80 -30) ft/s /1 ft = 50 ft/s/ft

HW (5. 11)

HW (5. 11)

Bernoulli Equation • Consider steady flow along streamline • s is along streamline, and

Bernoulli Equation • Consider steady flow along streamline • s is along streamline, and t is tangent to streamline

Ex (5. 47) Point 1 • • • Given: Velocity in outlet pipe from

Ex (5. 47) Point 1 • • • Given: Velocity in outlet pipe from reservoir is 6 m/s and h = 15 m. Find: Pressure at A. Solution: Bernoulli equation Point A

Example • • Given: D=30 in, d=1 in, h=4 ft Find: VA • Solution:

Example • • Given: D=30 in, d=1 in, h=4 ft Find: VA • Solution: Bernoulli equation Point 1 Point A

Example – Venturi Tube • • • Given: Water 20 o. C, V 1=2

Example – Venturi Tube • • • Given: Water 20 o. C, V 1=2 m/s, p 1=50 k. Pa, D=6 cm, d=3 cm Find: p 2 and p 3 Solution: Continuity Eq. D D d 2 1 3 Nozzle: velocity increases, pressure decreases • Bernoulli Eq. Diffuser: velocity decreases, pressure increases Similarly for 2 3, or 1 3 Pressure drop is fully recovered, since we assumed no frictional losses Knowing the pressure drop 1 2 and d/D, we can calculate the velocity and flow rate

Ex (5. 48) • • • Given: Velocity in circular duct = 100 ft/s,

Ex (5. 48) • • • Given: Velocity in circular duct = 100 ft/s, air density = 0. 075 lbm/ft 3. Find: Pressure change between circular and square section. Solution: Continuity equation Air conditioning (~ 60 o. F) • Bernoulli equation

Ex (5. 49) • • Given: r = 0. 0644 lbm/ft 3 V 1=

Ex (5. 49) • • Given: r = 0. 0644 lbm/ft 3 V 1= 100 ft/s, and A 2/A 1=0. 5, gm=120 lbf/ft 3 Find: Dh Solution: Continuity equation • Bernoulli equation • Heating (~ 170 o. F) • Manometer equation

HW (5. 51)

HW (5. 51)

Stagnation Tube

Stagnation Tube

Stagnation Tube in a Pipe 2 Flow 1

Stagnation Tube in a Pipe 2 Flow 1

Pitot Tube

Pitot Tube

Pitot Tube Application (p. 170) V 1 z 1 -z 2 2 l y

Pitot Tube Application (p. 170) V 1 z 1 -z 2 2 l y

HW (5. 69)

HW (5. 69)

HW (5. 75)

HW (5. 75)

HW (5. 84)

HW (5. 84)

HW (5. 93)

HW (5. 93)