Fluids - Hydrodynamics

Physics 6B

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

With the following assumptions, we can find a few simple formulas to describe flowing fluids:

Incompressible – the fluid does not change density due to the pressure exerted on it.

No Viscosity - this means there is no internal friction in the fluid.

Laminar Flow – the fluid flows smoothly, with no turbulence.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

With the following assumptions, we can find a few simple formulas to describe flowing fluids:

Incompressible – the fluid does not change density due to the pressure exerted on it.

No Viscosity - this means there is no internal friction in the fluid.

Laminar Flow – the fluid flows smoothly, with no turbulence.

With these assumptions, we get the following equations:

Continuity – this is conservation of mass for a flowing fluid.

2211 vAvAtV

⋅=⋅=∆∆

Bernoulli’s Equation - this is conservation of energy per unit volume for a flowing fluid.

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

Here A=area of the cross-section of the fluid’s container, and the small v is the speed of the fluid.

Notice that there is a potential energy term and a kinetic energy term on each side.

Some examples will help clarify how to use these equations: Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 1: Water travels through a 9.6cm diameter fire hose with a speed of 1.3m/s. At the end of the hose, the water flows out through a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 1: Water travels through a 9.6cm diameter fire hose with a speed of 1.3m/s. At the end of the hose, the water flows out through a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle?

We use continuity for this one. We have most of the information, but don’t forget we need the cross-sectional areas, so we need to compute them from the given diameters.

1 • 2 •1

2

12

2211

vAA

v

vAvA

⋅=

⋅=⋅ slower here

faster here

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 1: Water travels through a 9.6cm diameter fire hose with a speed of 1.3m/s. At the end of the hose, the water flows out through a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle?

We use continuity for this one. We have most of the information, but don’t forget we need the cross-sectional areas, so we need to compute them from the given diameters.

1 • 2 •

22

sm

1

cm6.9rA

3.1v

⋅π=⋅π=

=

12

12

2211

vAA

v

vAvA

⋅=

⋅=⋅

22

2

cm5.2rA

?v

⋅π=⋅π=

=

slower here

faster here

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

221

211

m00724.0)m048.0(A

2cm6.9

rA

=⋅π=

⋅π=⋅π=

222

222

m00049.0)m0125.0(A

2cm5.2

rA

=⋅π=

⋅π=⋅π=

Note: We didn’t really need to change the units of the areas -as long as both of them are the same, the units will cancel out.

Example 1: Water travels through a 9.6cm diameter fire hose with a speed of 1.3m/s. At the end of the hose, the water flows out through a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle?

We use continuity for this one. We have most of the information, but don’t forget we need the cross-sectional areas, so we need to compute them from the given diameters.

1 • 2 •

22

sm

1

cm6.9rA

3.1v

⋅π=⋅π=

=

12

12

2211

vAA

v

vAvA

⋅=

⋅=⋅

22

2

cm5.2rA

?v

⋅π=⋅π=

=

slower here

faster here

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

221

211

m00724.0)m048.0(A

2cm6.9

rA

=⋅π=

⋅π=⋅π=

222

222

m00049.0)m0125.0(A

2cm5.2

rA

=⋅π=

⋅π=⋅π=

Note: We didn’t really need to change the units of the areas -as long as both of them are the same, the units will cancel out.

Plugging in the numbers, we get:

sm

sm

2 2.193.100049.000724.0

v =⋅=

Example 1: Water travels through a 9.6cm diameter fire hose with a speed of 1.3m/s. At the end of the hose, the water flows out through a nozzle whose diameter is 2.5 cm. What is the speed of the water coming out of the nozzle?

We use continuity for this one. We have most of the information, but don’t forget we need the cross-sectional areas, so we need to compute them from the given diameters.

1 • 2 •

22

sm

1

cm6.9rA

3.1v

⋅π=⋅π=

=

12

12

2211

vAA

v

vAvA

⋅=

⋅=⋅

22

2

cm5.2rA

?v

⋅π=⋅π=

=

slower here

faster here

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

221

211

m00724.0)m048.0(A

2cm6.9

rA

=⋅π=

⋅π=⋅π=

222

222

m00049.0)m0125.0(A

2cm5.2

rA

=⋅π=

⋅π=⋅π=

Note: We didn’t really need to change the units of the areas -as long as both of them are the same, the units will cancel out.

Plugging in the numbers, we get:

sm

sm

2 2.193.100049.000724.0

v =⋅=

Using the shortcut, we get:

sm

sm

2

2 2.193.15.26.9

v =⋅

=

Example 2: At one point in a pipeline, the water’s speed is 3 m/s and the gauge pressure is 40 kPa. Find the gauge pressure at a second point on the line that is 11 m lower than the first if the pipe diameter at the second point is twice that of the first.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

1 •

2 •

y1=11 m

We need Bernoulli’s Equation for this one (really it’s just conservation of energy for fluids).

Notice we set up the y-axis so point 2 is at y=0.

y2=0

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

Here’s Bernoulli’s equation – we need to find the speed at point 2 using continuity, then plug in the numbers.

Example 2: At one point in a pipeline, the water’s speed is 3 m/s and the gauge pressure is 40 kPa. Find the gauge pressure at a second point on the line that is 11 m lower than the first if the pipe diameter at the second point is twice that of the first.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

2 •2

1 •

2 •

y1=11 m

We need Bernoulli’s Equation for this one (really it’s just conservation of energy for fluids).

Notice we set up the y-axis so point 2 is at y=0.

y2=0

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

Here’s Bernoulli’s equation – we need to find the speed at point 2 using continuity, then plug in the numbers.

Example 2: At one point in a pipeline, the water’s speed is 3 m/s and the gauge pressure is 40 kPa. Find the gauge pressure at a second point on the line that is 11 m lower than the first if the pipe diameter at the second point is twice that of the first.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

2 •2

Continuity Equation:

sm

141

212

12

2211

75.0vvvAA

v

vAvA

=⋅=⇒⋅=

⋅=⋅This is the ratio of the AREAS – it is the square of the ratio of the diameters

1 •

2 •

y1=11 m

We need Bernoulli’s Equation for this one (really it’s just conservation of energy for fluids).

Notice we set up the y-axis so point 2 is at y=0.

y2=0

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

Here’s Bernoulli’s equation – we need to find the speed at point 2 using continuity, then plug in the numbers.

Example 2: At one point in a pipeline, the water’s speed is 3 m/s and the gauge pressure is 40 kPa. Find the gauge pressure at a second point on the line that is 11 m lower than the first if the pipe diameter at the second point is twice that of the first.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

2 •2

Continuity Equation:

sm

141

212

12

2211

75.0vvvAA

v

vAvA

=⋅=⇒⋅=

⋅=⋅This is the ratio of the AREAS – it is the square of the ratio of the diameters

Plugging in the numbers to the Bernoulli Equation:

Pa000,152p

)75.0)(1000(0p)3)(1000()m11)(8.9)(1000(Pa000,40

vgypvgyp

2

2sm

m

kg21

22

sm

m

kg21

sm

m

kg

222

122

212

111

3323

=

++=++

ρ+ρ+=ρ+ρ+

Example 3: A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 12 kPa, while in the region of blockage it is 11.5 kPa. Furthermore, she knows that the blood flowing through the normal artery just before the blockage is traveling at 30 cm/s, and the density of the patient’s blood is 1060 kg/m3.

What percentage of the cross-sectional area of the patient’s artery is blocked by plaque?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 3: A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 12 kPa, while in the region of blockage it is 11.5 kPa. Furthermore, she knows that the blood flowing through the normal artery just before the blockage is traveling at 30 cm/s, and the density of the patient’s blood is 1060 kg/m3.

What percentage of the cross-sectional area of the patient’s artery is blocked by plaque?

It seems like there is a lot going on in this problem, but it is really just like the last one. In fact, it’s even easier ifwe assume the artery is horizontal (they don’t mention any vertical distances so this is probably ok). We’ll use Bernoulli’s Equation to find the speed just after the blockage, then continuity will tell us the ratio of the areas.

V1 = 30 cm/s

V2 = ?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Example 3: A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 12 kPa, while in the region of blockage it is 11.5 kPa. Furthermore, she knows that the blood flowing through the normal artery just before the blockage is traveling at 30 cm/s, and the density of the patient’s blood is 1060 kg/m3.

What percentage of the cross-sectional area of the patient’s artery is blocked by plaque?

It seems like there is a lot going on in this problem, but it is really just like the last one. In fact, it’s even easier ifwe assume the artery is horizontal (they don’t mention any vertical distances so this is probably ok). We’ll use Bernoulli’s Equation to find the speed just after the blockage, then continuity will tell us the ratio of the areas.

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

these will be 0solve for this speed

V1 = 30 cm/s

V2 = ?

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

these will be 0

Example 3: A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 12 kPa, while in the region of blockage it is 11.5 kPa. Furthermore, she knows that the blood flowing through the normal artery just before the blockage is traveling at 30 cm/s, and the density of the patient’s blood is 1060 kg/m3.

What percentage of the cross-sectional area of the patient’s artery is blocked by plaque?

It seems like there is a lot going on in this problem, but it is really just like the last one. In fact, it’s even easier ifwe assume the artery is horizontal (they don’t mention any vertical distances so this is probably ok). We’ll use Bernoulli’s Equation to find the speed just after the blockage, then continuity will tell us the ratio of the areas.

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

these will be 0solve for this speed

V1 = 30 cm/s

V2 = 112 cm/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

these will be 0

scm

sm

2

22m

kg212

sm

m

kg21

10202.1v

v)1060(0Pa500,11)3.0)(1060(0Pa000,1233

==

++=++

Example 3: A medical technician is trying to determine what percentage of a patient’s artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 12 kPa, while in the region of blockage it is 11.5 kPa. Furthermore, she knows that the blood flowing through the normal artery just before the blockage is traveling at 30 cm/s, and the density of the patient’s blood is 1060 kg/m3.

What percentage of the cross-sectional area of the patient’s artery is blocked by plaque?

It seems like there is a lot going on in this problem, but it is really just like the last one. In fact, it’s even easier ifwe assume the artery is horizontal (they don’t mention any vertical distances so this is probably ok). We’ll use Bernoulli’s Equation to find the speed just after the blockage, then continuity will tell us the ratio of the areas.

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

these will be 0solve for this speed

V1 = 30 cm/s

V2 = 112 cm/s

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

these will be 0

scm

sm

2

22m

kg212

sm

m

kg21

10202.1v

v)1060(0Pa500,11)3.0)(1060(0Pa000,1233

==

++=++

%3030.010230

vv

AA

vAvA

2

1

1

2

2211

====

⋅=⋅

Now use continuity:

So the artery is 70% blocked (the blood is flowing through a cross-section that is only 30% of the unblocked area)

The Bernoulli ‘Effect’

Fast Flow=Low Pressure ↔ Slow Flow=High Pressure

Airplane Wing

Atomizer

Hurricane Damage

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Hurricane Damage

Curveballs, Backspin, Topspin

Motorcycle Jacket

Attack of the Shower Curtain

Example 4: Hurricane

a) If the speed of the wind is 50 m/s (that’s about 100 miles per hour), and the density of the air is 1.29 kg/m3, Find the reduction in air pressure due to the wind.

b) If the area of the roof measures 10m x 20m, what is the net upward force on the roof?

Example 4: Hurricane

• For part a) we need to use Bernoulli’s equation.

a) If the speed of the wind is 50 m/s (that’s about 100 miles per hour), and the density of the air is 1.29 kg/m3, Find the reduction in air pressure due to the wind.

b) If the area of the roof measures 10m x 20m, what is the net upward force on the roof?

Example 4: Hurricane

• For part a) we need to use Bernoulli’s equation.

We can assume (as in the last example) that y1=y2=0.

We can also assume that the wind is not blowing inside.

Take point 1 to be inside the house, and point 2 to be outside.

a) If the speed of the wind is 50 m/s (that’s about 100 miles per hour), and the density of the air is 1.29 kg/m3, Find the reduction in air pressure due to the wind.

b) If the area of the roof measures 10m x 20m, what is the net upward force on the roof?

Take point 1 to be inside the house, and point 2 to be outside.

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

these will be 0

Example 4: Hurricane

• For part a) we need to use Bernoulli’s equation.

We can assume (as in the last example) that y1=y2=0.

We can also assume that the wind is not blowing inside.

Take point 1 to be inside the house, and point 2 to be outside.

a) If the speed of the wind is 50 m/s (that’s about 100 miles per hour), and the density of the air is 1.29 kg/m3, Find the reduction in air pressure due to the wind.

b) If the area of the roof measures 10m x 20m, what is the net upward force on the roof?

Take point 1 to be inside the house, and point 2 to be outside.

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

these will be 0

2

3

mN

21

2sm

m

kg21

21

5.1612pp

)50)(29.1(pp

=−

+=

Example 4: Hurricane

• For part a) we need to use Bernoulli’s equation.

We can assume (as in the last example) that y1=y2=0.

We can also assume that the wind is not blowing inside.

Take point 1 to be inside the house, and point 2 to be outside.

a) If the speed of the wind is 50 m/s (that’s about 100 miles per hour), and the density of the air is 1.29 kg/m3, Find the reduction in air pressure due to the wind.

b) If the area of the roof measures 10m x 20m, what is the net upward force on the roof?

Take point 1 to be inside the house, and point 2 to be outside.

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

these will be 0

• Part b) is just a straightforward application of the definition of pressure.

2

3

mN

21

2sm

m

kg21

21

5.1612pp

)50)(29.1(pp

=−

+=

Example 4: Hurricane

a) If the speed of the wind is 50 m/s (that’s about 100 miles per hour), and the density of the air is 1.29 kg/m3, Find the reduction in air pressure due to the wind.

b) If the area of the roof measures 10m x 20m, what is the net upward force on the roof?

• For part a) we need to use Bernoulli’s equation.

We can assume (as in the last example) that y1=y2=0.

We can also assume that the wind is not blowing inside.

Take point 1 to be inside the house, and point 2 to be outside.Take point 1 to be inside the house, and point 2 to be outside.

222

122

212

111 vgypvgyp ρ+ρ+=ρ+ρ+

these will be 0

2

3

mN

21

2sm

m

kg21

21

5.1612pp

)50)(29.1(pp

=−

+=

• Part b) is just a straightforward application of the definition of pressure.Assuming the roof is flat, we just multiply:

N500,322Fm20m10

F5.1612

AF

P 2mN =⇒

⋅=⇒=