Dynamic Fluids. Concept Checker Ideal fluids have three main components. What are they?

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Presentation transcript:

Dynamic Fluids

Concept Checker Ideal fluids have three main components. What are they?

Answer: Steady Flow: the velocity of the fluid particles at any point in time is constant Incompressible Nonviscous

Concept Checker 2 Explain in your own words why tornados can pull the roof off of your house.

Answer. The wind above the roof is moving faster, meaning it has a lower pressure Air in the house is moving slower, meaning it has a higher pressure The high pressure wants to be where the low pressure is, so it pushes the roof up

Determine the Mass Flow Rate Density and velocity of a fluid is given as 920kg/m 3 and 5m/s, this fluid is flowing through an area of 25cm 2. Calculate the mass flow rate?

Solution: Convert 25 cm to.25 m The formula for mass flow rate is, m = ρρVA m = 920 × 5 × 0.25 = 1150kg/s

Practice Problem 2: Water is flowing through a channel that is 23 m wide with a speed of 3 m/s. The water then flows into six identical channels that have a width of 5.0 m. The depth of the water does not change as it flows into the Six channels. What is the speed of the water in one of the smaller channels?

Solution: The volume flow Q is constant mean Av=Av since density is constant throughout Av= 6Av V = Av/6A V= 23(3) / (6 * 5) = 2.3 m/s Why is V lower ????

Practice Problem 3: Oil (ρ = 925 kg/m3) is flowing through a pipeline at a constant speed when it encounters a vertical bend in the pipe raising it 4.0 m. The cross sectional area of the pipe does not change. What is the difference in pressure (PB – PA) in the portions of the pipe before and after the rise?

Solution: P1 +.5 dv^2 + dgy = P2 +.5 dv^2 + dgy V1= v2 P 1 + dgy = P2 + dgy P2-p1 =dg (y2 – y1) Delta p = 925 * 9.81 * (0- 4) = -3.6 *10 ^4 Pa WHY IS PRESSURE NEGATIVE????

Last Problem: A large tank is filled with water to a depth of 15 meters. A spout is located 10 meters above the bottom of the tank is then opened. With what speed will the water emerge from the spout?

Answer: V is the same as if the water fell from a distance h V= (2gh)^.5 v=( 2*9.81*5)^.5 H = 5 because the tank is 15 meters and the spout is at 5, so it is fall that 5 meter distance to come out of the spout v= 9.9 m/s