Aim 36 - Dynamic Fluids
Volume Flow Rate
ANSWER Q = A1v1 = PI(r2)v = PI(.4 m2)(6 m/s)
= 2.4 m3/sec
Q = V/t = A(d)/t = A(v)
Fluid Speed and Cross-Section
narrower
SLOW FAST
Which body of water is faster?
https://commons.wikimedia.org/wiki/File:Muskingum_River_Marietta.jpg
http://commons.wikimedia.org/wiki/File:Lyre_River.JPG
Density does not typically change in liquids. This means that where a pipe is wider, the flow is slower.
Mass Flow Rate or Mass Continuity
A1v1 = A2v2
Not on Handout
25
Water flows at a constant speed though one section of a pipe. When it enters another section with half the cross sectional area, what happens to the speed of the water?
A
The speed is reduced to one half the original.
B
The speed is doubled.
C
The speed stays the same.
D
The speed quadruples
E
I need help
https://njctl.org/video/?v=IwWHMBOKoLA
Answer
B
Not on Handout
Fluid Flow Problem
Ratio of Speeds
A1v1 = A2v2
Fluid Splitting
4. An ideal fluid flows with a speed of 12 cm/s through a pipe of diameter 5 cm. The pipe splits into three smaller pipes, each with a diameter of 2.0 cm. What is the speed of the fluid in the smaller pipes?
- Initial speed: V1 =12 cm/s
- Pipe splits into three smaller pipes r2 = 2 cm
Aorta and Capillaries
A1v1 = xA2v2
[PI](r12)v1 = X[PI](r22)v2
(r12)v1 = X(r22)v2
(.01 m2).40 m/s = X(7 x 10-6 m)2(.0006 m/s)
1.4 x 109 capillaries
5. The radius of the aorta is 1 cm and that of the capillaries is 7 µm. If the average speed of blood in the aorta is 40 cm/s and in the capillaries is about 0.6 mm/s, what is the number of capillaries in the human body?
Aim 37 - How does the speed of a fluid affect the pressure?
decreases.
B) II Narrowest, Minimum Pressure
Forces on Wing
Faster / Low P
Slower / Lower P
Blood Flow
T-Shaped Tube Problem
Answer: ( A ) v goes up when tube narrows,
P goes down.
Bernoulli's Equation
Bernoulli's Equation
(B) As the velocity goes up P goes down
Fluids in Motion & Bernoulli's Equation
Since we are dealing with a fluid, let's look at each term per unit volume. Divide each term by volume V (upper-case).
The total mechanical energy of the moving fluid remains constant unless work is done by a net force, or pressure difference.
Continued on the next slide...
and since
W = Fd
Fluids in Motion & Bernoulli's Equation
Since the volume of water in the pipe equals A times d...
A
d
V
Bernoulli's
Equation
Aim 38: What is Bernoulli’s Equation?
What factors affect static pressure? ___________________________
What factor affects dynamic pressure? __________________________
Energy in a fluid is conserved
Daniel Bernoulli 1700-1782
Bernoulli’s equation is the Conservation of ________________ equation for fluids.
Depth and Atmospheric Pressure
ENERGY
Height, velocity & Atmospheric P
r1 = .025 m v1 = 1 m/s P1 = 200 x 103 Pa
v2 = ? h2 = 6 m r2 = .01 m P2 = ? ⍴ = 1000 kg/m3
v1A1 = v2A2 v1A1/A2 = v2
(1 m/s)PI(r12)/(PI)r22 = (1 m/s)(.025 m)2/(.01 m)2
= 6.25 m/s
P1 + ⍴gh1 + 1/2⍴v12 = P2 + ⍴gh2 + 1/2⍴v22
200 x 103 Pa + ½(1000 kg/m3)(1 m/s)2 =
P2 + (1000 kg/m3)(10 m/s2)6m + ½ (1000 kg/m3)(6.25 m/s)2
P2 = 1.2 x 105 Pa
1.
Lift Force Calculation
2. What is the lift force due to Bernoulli’s principle on a pair of wings of total area 100 m2 if the air passes over the top and bottom surfaces at speeds of 300 m/s and 260 m/s, respectively? (ρair = 1.29 kg/m3)
F = ΔPA
P1 + ½⍴v12 = P2 + ½ ⍴v22 (No height change)
ΔP = ½ ⍴(v22 - v12) = ½(1.29 kg/m3)[(300 m/s)2 - (260 m/s)2]
= 14448 Pa�
F = ΔPA = 14448 Pa(100 m2) = 1.4 x 105 N
Bernoulli’s Equation
3. A pump, submerged at the bottom of a well that is 35 m deep, is used to pump water uphill to a house that is 50 m above the top of the well, as shown above. The density of water is 1,000 kg/m3. Neglect the effects of friction, turbulence, and viscosity.
Residents of the house use 0.35 m3 of water per day. The day’s pumping is completed in 2 hours during the day.
Calculate the minimum work required to pump the water used per day
Calculate the pump's minimum power rating.
(b) In the well, the water flows at 0.50 m/s and the pipe has a diameter of 3.0 cm. At the house the diameter of the pipe is 1.25 cm.
Calculate the flow velocity at the house when a faucet in the house is open.
Calculate the pressure at the well when the faucet in the house is open
W = FΔd = [⍴Vg]Δh = (1000 kg/m3)(.35m3)10m/s2(85m)
= 3.0 x 105 Joules
P = W/t = 2.97 x 105 J /7200 sec = 41 Watts
V1 = .50 m/s r1 = .015 m r2 = .00615 m
P1 + ½⍴v12 + ⍴gh1 = P2 + ½⍴v22 + ⍴gh2 = P1 + ½(1000 kg/m3) continued on next slide
Δh = 35 m + 50 m = 85 m density 1000 kg/m3
.35 m3 2 hrs to pump(3600 sec/1 hr = 7200 sec
Answer COM A1v1 = A2v2 = ( PI)(r12)v1 = ( PI)(r22)v2 = (r12)v1 = (r22)v2
= (.015 m)2 [.50 m/s] = (.00625 m)2v2 = 2.88 m/s
Patm = 1 x 105 Pa
Bernoulli’s Equation
(b) In the well, the water flows at 0.50 m/s and the pipe has a diameter of 3.0 cm. At the house the diameter of the pie is 1.25 cm.
Calculate the flow velocity at the house when a faucet in the house is open.
Calculate the pressure at the well when the faucet in the house is open
V1 = .50 m/s r1 = .015 m r2 = .00615 m
P1 + ½⍴v12 + ⍴gh1 = P2 + ½⍴v22 + ⍴gh2 =�
P1 + ½(1000 kg/m3)(.50 m/s)2 + 0 =
1 x 105 Pa + ½(1000 kg/m3)(2.88 m/s)2 + (1000 kg/m3)(10 m/s2)85 m
Answer C.O.M. A1v1 = A2v2 = ( PI)(r12)v1 = ( PI)(r22)v2 = (r12)v1 = (r22)v2
= (.015 m)2 [.50 m/s] = (.00625 m)2v2 = 2.88 m/s
Patm = 1 x 105 Pa
Aim 39 Problem Solving with Bernoulli’s Equation
A drinking fountain projects water at an initial angle of 50° above the horizontal, and the water reaches a maximum height of 0.150 m above the exit point. Assume air resistance is negligible. Calculate the speed at which the water leaves the fountain.
theta = 50 deg, dy = .150 m Vfy = 0 ay = 10 m/s2 Viy = Vsin(theta)
Viy2 = VIy2 + 2aydy 02 = Viy2 + 2(10 m/s2).150 m
Viy = 1.7 m/s Viy = Vsin(theta) 1.7 m/s = Vsin(50 deg) v = 2.2 m/s
See Next Slide
Aim 39 Problem Solving with Bernoulli’s Equation
A drinking fountain projects water at an initial angle of 50° above the horizontal, and the water reaches a maximum height of 0.150 m above the exit point. Assume air resistance is negligible. Calculate the speed at which the water leaves the fountain.
P1 + ⍴gh1 + ½ ⍴v12 = P2 + ⍴gh2 + ½ ⍴v22
P1 = (1.0 x 103 kg/m2)(10 m/s2)3.00 m + ½(1.0 x 103 kg/m2)[(2.242 m/s - .732 m/s]
= 1.32 x 105 kg/m3 Guage Pressure = P1 - Patm Patm = 1 x 105 Pa
= 3.2 x 104 Pascals
Q = Av = (PI)r2(v) = (PI)(4.00 x 10-3 m)2(2.2 m/s)
Q = 1.1 x 10-4 m3/sec
Q = 1.1 x 10-4 m3/sec = PI(7 x 10-3 m)2V1 V1 = .73 m/s
r1 = 7.00 x 10-3 m Δh = 3.00 m
Efflux speed: is the speed a fluid leaves a container.
It is related to the __________ of the water.
Torricelli’s Law - The speed of water coming out of a hole in a ��bucket is
A cylindrical tank containing water of density 1000 kg/m3 is filled to a height of 0.70 m and placed on a stand as shown in the cross-section above. A hole of radius 0.0010 m in the bottom of the tank is opened. Water then flows through the hole and through an opening in the stand and is collected in a tray 0.30 m below the hole. At the same time, water is added to the tank at an appropriate rate so that the water level in the tank remains constant.
DEPTH
(b) Calculate the volume rate at which water flows out from the hole.
(c) Calculate the volume of water collected in the tray in t = 2.0 minutes.
V = Qt = [1.16 X 10-5 m3/sec][120 sec]
= .0014 m3
mgh = ½mv2 V = (2gh).5 = (2(10 m/s2)h).5
mgh = ½mv2 V = (2gh).5 =
(2(10 m/s2).70 m).5
= 3.7 m/s
Q = Av = (PI)(.0010)2 (3.7 m/s)
1.2 X 10-5 m3/sec
2007B4.
The large container shown in the cross-section is filled with a liquid of density 1.1 x 103 kg/m3. A small hole of area 2.5 x 10-6 m2 is opened in the side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into a beaker placed to the right of the container. At the same time, liquid is also added to the container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0 minutes is 7.2 x 10–4 m3.
Calculate the volume rate of flow of liquid from the hole in m3/s.
Calculate the speed of the liquid as it exits from the hole.
Calculate the height h of liquid needed above the hole to cause the speed you determined in part (b).
(d) Suppose that there is now less liquid in the container so that the height h is reduced to h/2. In relation to the collection beaker, where will the liquid hit the tabletop?
____ Left of the beaker ____ In the beaker ____ Right of the beaker. Justify your answer.
Q = V/t = 7.2 x 10–4 m3/120 sec = �= 6 x 10-6 m3/sec
Q = Av = 6 x 10-6 m3/sec =[2.5 x 10-6 m2]v
v = 2.4 m/s
v = (2gh).5 = 2.4 m/s
h = .29 m
⍴ = 1.1 x 10 kg/m3 A = 2.5 x 10-6 m2 t = 120 sec
X
10. A pump, submerged at the bottom of a well that is 30 meters deep, is used to pump water up to faucets in a house. The highest level faucet is 10 m above the top of the well. If the average gauge pressure at the pump is 403,700 Pa, the radius of the pipe connected to the pump is 0.03m, and the radius of the faucet opening is 0.01m. Find the flow speed when the highest faucet in the house is open.
Velocity at the top of the well
P1 = 1/2(Density)(Vf2 - 0) + (Density)g(h2 - h1)
403,700 Pa = 1/2(1000 kg/m3)Vf2 + (1000 kg/m3)(10 m/s2)10 m
Vf = 24.645 m/s
v1A1 = v2A2 24.645 m/s(PI)(.03m 2) + v2(PI)(.01 m2) V2 = 15 m/s
4. A Venturi meter measures the speed of water flow in a pipe of a cross-sectional area of 0.01 m2. The pipe is constricted (of cross-sectional area 0.002 m2). Two vertical tubes, open to the atmosphere, rise from two points, one of which is in the constriction.
Speed lower at 1 so P1 higher, so h1 higher
A1v1=A2v2 0.01v1=0.002v2 v2=5v1
P1 + ½⍴v12 = P2 + ½ ⍴v22 ⍴gh1 + ½⍴v12 = ⍴gh2 + ½⍴v22
gh1 + ½v12 = gh2 + ½v22 10m/s2(1.5 m) + ½v12 = 10 m/s2(.6m) + ½v22
15 + ½ V12 = 6 + ½(25)V12 V2 = 4.2 m/s V1 = .832 m/s
A1 = .01 m2 A2 = .02 m2 P = F/A
(B) V greater at bottom, so P less
C
A1v1 = A2v2
(2.0 x 10-4 m2 )1.0 m/s = (1.2 x 10-4 m2)V2
V2 = 1.66 m/s
Vi = 0 ay = 10 m/s2 d = ? Vf = 1.66 m/s
Vf2 = Vi2 + 2ad
(1.66 m/s)2 = 0 + 2(10 m/s2)d
d = .14 m