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Transcript
```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.*
*We will see later how to deal with cases where these assumptions are not valid.
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.
Q
V
 A1  v1  A2  v2
t
Here A=area of the cross-section of the fluid’s container,
and the small v is the speed of the fluid.
*We will see later how to deal with cases where these assumptions are not valid.
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.
Q
V
 A1  v1  A2  v2
t
Here A=area of the cross-section of the fluid’s container,
and the small v is the speed of the fluid.
Bernoulli’s Equation - this is conservation of energy per unit volume for a flowing fluid.
p1  gy1  1 v12  p2  gy2  1 v 22
2
2
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:
*We will see later how to deal with cases where these assumptions are not valid.
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.
A1  v1  A2  v2
A
v2  1  v1
A2
slower here
faster here
1•
2•
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.
slower here
A1  v1  A2  v2
A
v2  1  v1
A2
faster here
1•
v1  1.3 m
v2  ?
s
A1 
  r12
2•
2
 9.6cm 
 

 2 
A1    (0.048m)2  0.00724m2
2
 2.5cm 
A2    r22    

 2 
A2    (0.0125m)2  0.00049m2
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.
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.
slower here
A1  v1  A2  v2
A
v2  1  v1
A2
faster here
1•
v1  1.3 m
v2  ?
s
A1 
  r12
2•
2
 9.6cm 
 

 2 
A1    (0.048m)2  0.00724m2
2
 2.5cm 
A2    r22    

 2 
A2    (0.0125m)2  0.00049m2
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:
v2 
0.00724
 1.3 m  19.2 m
s
s
0.00049
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.
slower here
A1  v1  A2  v2
A
v2  1  v1
A2
faster here
1•
v1  1.3 m
v2  ?
s
A1 
  r12
2•
2
 9.6cm 
 

 2 
A1    (0.048m)2  0.00724m2
2
 2.5cm 
A2    r22    

 2 
A2    (0.0125m)2  0.00049m2
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:
0.00724
v2 
 1.3 m  19.2 m
s
s
0.00049
Using the shortcut, we get:
2
 9.6 
v2  
 19.2 m
  1.3 m
s
s
 2.5 
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 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
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 that is 11 m lower
than the first if the pipe diameter at the second point is twice that of the first.
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.
1•
y1=11 m
y2=0
2•
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 that is 11 m lower
than the first if the pipe diameter at the second point is twice that of the first.
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.
1•
y1=11 m
Here’s Bernoulli’s equation – we need to find the speed
at point 2 using continuity, then plug in the numbers.
p1  gy1  1 v12  p2  gy2  1 v2
2
2
2
y2=0
2•
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 that is 11 m lower
than the first if the pipe diameter at the second point is twice that of the first.
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.
1•
y1=11 m
Here’s Bernoulli’s equation – we need to find the speed
at point 2 using continuity, then plug in the numbers.
p1  gy1  1 v12  p2  gy2  1 v2
2
2
2
Continuity Equation:
A1  v1  A2  v2
v2 
y2=0
2•
This is the ratio of the AREAS – it is the
square of the ratio of the diameters
A1
 v1  v2  1  v1  0.75 m
4
s
A2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 that is 11 m lower
than the first if the pipe diameter at the second point is twice that of the first.
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.
1•
y1=11 m
Here’s Bernoulli’s equation – we need to find the speed
at point 2 using continuity, then plug in the numbers.
p1  gy1  1 v12  p2  gy2  1 v2
2
2
2
Continuity Equation:
A1  v1  A2  v2
v2 
y2=0
2•
This is the ratio of the AREAS – it is the
square of the ratio of the diameters
A1
 v1  v2  1  v1  0.75 m
4
s
A2
Plugging in the numbers to the Bernoulli Equation:
p1  gy1  1 v12  p2  gy2  1 v2
2
2
2
40,000Pa  (1000
kg
3
m
kg
kg
)(9.8 m2 )(11m)  1 (1000 3 )(3 m)2  p2  0  1 (1000 3 )(0.75 m)2
s
2
m
s
2
m
s
p2  152,000Pa
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?
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?
There is a lot going on in this problem, but it is really just like the last one. In fact, it’s easier
if we assume the artery is horizontal. We’ll use Bernoulli’s Equation to find the speed just
after the blockage, then continuity will tell us the ratio of the areas.
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?
There is a lot going on in this problem, but it is really just like the last one. In fact, it’s easier
if we assume the artery is horizontal. 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
p1  gy1  1 v12  p2  gy2  1 v 22
2
2
solve for this speed
V2 = ? cm/s
these will be 0
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?
There is a lot going on in this problem, but it is really just like the last one. In fact, it’s easier
if we assume the artery is horizontal. 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
p1  gy1  1 v12  p2  gy2  1 v 22
2
2
solve for this speed
V2 = 102 cm/s
these will be 0
kg
kg
12,000Pa  0  1 (1060 3 )(0.3 m)2  11,500Pa  0  1 (1060 3 )v2
2
2
s
2
m
m
v2  1.02 m  102 cm
s
s
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?
There is a lot going on in this problem, but it is really just like the last one. In fact, it’s easier
if we assume the artery is horizontal. 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
p1  gy1  1 v12  p2  gy2  1 v 22
2
2
solve for this speed
V2 = 102 cm/s
these will be 0
kg
kg
12,000Pa  0  1 (1060 3 )(0.3 m)2  11,500Pa  0  1 (1060 3 )v2
2
2
s
2
m
m
v2  1.02 m  102 cm
s
s
Now use continuity:
A1  v1  A2  v2
A2
v
30
 1 
 0.30  30%
A1 v2 102
So the artery is 70% blocked (the blood is flowing through
Prepared
a cross-section that is only 30% of the unblocked
area)by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
The Bernoulli ‘Effect’
Fast Flow=Low Pressure ↔ Slow Flow=High Pressure
Airplane Wing
Atomizer
Hurricane Damage
Curveballs, Backspin, Topspin
Motorcycle Jacket
Attack of the Shower Curtain
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
p1  gy1  1 v12  p2  gy2  1 v 22
2
2
these will be 0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
p1  gy1  1 v12  p2  gy2  1 v 22
2
2
these will be 0
p1  p2  12 (1.29 mkg3 )(50 ms )2
p1  p2  1612 .5 mN2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
p1  gy1  1 v12  p2  gy2  1 v 22
2
2
these will be 0
p1  p2  12 (1.29 mkg3 )(50 ms )2
p1  p2  1612 .5 mN2
• Part b) is just a straightforward application of the definition of pressure.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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.
p1  gy1  1 v12  p2  gy2  1 v 22
2
2
these will be 0
p1  p2  12 (1.29 mkg3 )(50 ms )2
p1  p2  1612 .5 mN2
• Part b) is just a straightforward application of the definition of pressure.
Assuming the roof is flat, we just multiply:
P
F
F
 1612.5 mN2 
 F  322,500N
A
10m  20m
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For Campus Learning
Assistance Services at UCSB
Viscosity and Turbulence
Earlier we made assumptions that our fluid has low viscosity (i.e. friction) and
laminar (not turbulent) flow. But many fluids do not behave so nicely. To measure
the importance of viscosity on a flowing fluid we can calculate its Reynolds number.
𝜌𝑙𝑣
𝑅𝑒 =
𝜂
Viscosity of the fluid
If the Reynolds number is low, then viscosity plays a significant role, and the fluid will exhibit
laminar flow, but Bernoulli’s equation will not be satisfied. In this case we can use a different
formula to relate the flow rate to the pressure difference. This is Poiseuille’s equation.
𝜋𝑅 4
𝑄=
Δ𝑝
8𝜂𝐿
In cases where the Reynolds number is high (above 2000 or so) the flow will become
turbulent and complicated.
Prepared by Vince Zaccone
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Assistance Services at UCSB
Example 5: Viscous water flow (from textbook)
Water flows at 0.500 mL/s through a horizontal tube that is 30.0cm long and has
an inside diameter of 1.50mm. Determine the pressure difference required to drive
this flow if the viscosity of water is 1.00mPa·s.
Is it reasonable to assume laminar flow in this case?
Hint: calculate the Reynolds number.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 5: Viscous water flow (from textbook)
Water flows at 0.500 mL/s through a horizontal tube that is 30.0cm long and has
an inside diameter of 1.50mm. Determine the pressure difference required to drive
this flow if the viscosity of water is 1.00mPa·s.
Is it reasonable to assume laminar flow in this case?
Hint: calculate the Reynolds number.
For now, let’s assume laminar flow so we can use Poiseuille’s equation. Be careful with units.
𝑄=
𝜋𝑅 4
8𝜂𝐿
−3
Δ𝑝 → Δ𝑝 =
8𝜂𝐿𝑄 8(10
=
𝜋𝑅4
𝑃𝑎∙𝑠)(0.3𝑚)(0.5
−3
𝜋(1.5∙102 𝑚)4
∙
−6 𝑚3
10 𝑠 )
= 1207 𝑃𝑎
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 5: Viscous water flow (from textbook)
Water flows at 0.500 mL/s through a horizontal tube that is 30.0cm long and has
an inside diameter of 1.50mm. Determine the pressure difference required to drive
this flow if the viscosity of water is 1.00mPa·s.
Is it reasonable to assume laminar flow in this case?
Hint: calculate the Reynolds number.
For now, let’s assume laminar flow so we can use Poiseuille’s equation. Be careful with units.
𝑄=
𝜋𝑅 4
8𝜂𝐿
−3
Δ𝑝 → Δ𝑝 =
8𝜂𝐿𝑄 8(10
=
𝜋𝑅4
𝑃𝑎∙𝑠)(0.3𝑚)(0.5
−3
𝜋(1.5∙102 𝑚)4
∙
−6 𝑚3
10 𝑠 )
= 1207 𝑃𝑎
To find the Reynolds number we need the flow speed. We can get that from our original
formulation of flow rate: Q=Av.
3
0.5𝑐𝑚
𝑄
𝑠
𝑐𝑚
𝑚
𝑣=
=
=
28.3
=
0.283
0.15𝑐𝑚
2
𝑠
𝑠
𝜋𝑅
𝜋 ∙ ( 2 )2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 5: Viscous water flow (from textbook)
Water flows at 0.500 mL/s through a horizontal tube that is 30.0cm long and has
an inside diameter of 1.50mm. Determine the pressure difference required to drive
this flow if the viscosity of water is 1.00mPa·s.
Is it reasonable to assume laminar flow in this case?
Hint: calculate the Reynolds number.
For now, let’s assume laminar flow so we can use Poiseuille’s equation. Be careful with units.
𝑄=
𝜋𝑅 4
8𝜂𝐿
−3
Δ𝑝 → Δ𝑝 =
8𝜂𝐿𝑄 8(10
=
𝜋𝑅4
𝑃𝑎∙𝑠)(0.3𝑚)(0.5
∙
−3
𝜋(1.5∙102 𝑚)4
−6 𝑚3
10 𝑠 )
= 1207 𝑃𝑎
To find the Reynolds number we need the flow speed. We can get that from our original
formulation of flow rate: Q=Av.
3
0.5𝑐𝑚
𝑄
𝑠
𝑐𝑚
𝑚
𝑣=
=
=
28.3
=
0.283
0.15𝑐𝑚
2
𝑠
𝑠
𝜋𝑅
𝜋 ∙ ( 2 )2
Now we can use the formula for Reynolds number.
𝑘𝑔
−3 𝑚)(0.283𝑚)
1000𝑚
𝜌𝑙𝑣
3 (1.5 ∙ 10
𝑠
𝑅𝑒 =
=
= 424.5
𝜂
10−3 𝑃𝑎 ∙ 𝑠
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example 5: Viscous water flow (from textbook)
Water flows at 0.500 mL/s through a horizontal tube that is 30.0cm long and has
an inside diameter of 1.50mm. Determine the pressure difference required to drive
this flow if the viscosity of water is 1.00mPa·s.
Is it reasonable to assume laminar flow in this case?
Hint: calculate the Reynolds number.
For now, let’s assume laminar flow so we can use Poiseuille’s equation. Be careful with units.
𝑄=
𝜋𝑅 4
8𝜂𝐿
−3
Δ𝑝 → Δ𝑝 =
8𝜂𝐿𝑄 8(10
=
𝜋𝑅4
𝑃𝑎∙𝑠)(0.3𝑚)(0.5
∙
−3
𝜋(1.5∙102 𝑚)4
−6 𝑚3
10 𝑠 )
= 1207 𝑃𝑎
To find the Reynolds number we need the flow speed. We can get that from our original
formulation of flow rate: Q=Av.
3
0.5𝑐𝑚
𝑄
𝑠
𝑐𝑚
𝑚
𝑣=
=
=
28.3
=
0.283
0.15𝑐𝑚
2
𝑠
𝑠
𝜋𝑅
𝜋 ∙ ( 2 )2
Now we can use the formula for Reynolds number.
𝑅𝑒 =
𝜌𝑙𝑣
=
𝜂
𝑘𝑔
1000𝑚
3
(1.5 ∙ 10−3 𝑚)(0.283𝑚
𝑠)
10−3 𝑃𝑎 ∙ 𝑠
= 424.5
This is well
below 2000, so
laminar flow is
reasonable. Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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