Hydrodynamics deals with fluids in motion, there are two types of fluid motion:
1) Steady flow 2) turbulent flow
Chapter Three
Hydrodynamics fluid
1] Steady flow ( streamline flow ):
The fluid flows in steady layers and the layers moves in slide, smooth and soft that is means that
the transfer of the fluid from point to another point continuously- it’s called (stream lines).Or (laminar flow).
2] The turbulent flow:
When the velocity of flow exceeds a certain limit, it is characterized by small eddy circles.
The same thing happens to gases as a result of diffusion from a small space to large space or from high pressure to low pressure.
1- The velocity of the liquid at any specific point along the path of the liquid remains constant and does not change as time passes.
2- The flow is irrotational, i.e., there is no vortex motion.
3- No frictional forces between the layers of the liquid.
4- The flow rate of the liquid should be constant along its path because the amount of liquid entering the tube equals that emerging out from it in the same period of time where the liquid is incompressible and its density remains constant.
Conditions of a steady flow:
LESSON ONE
Fluid flow
Stream line:
It is the imaginary line showing the path of a liquid particle during its flow.
The properties of the stream lines:
1- They don't intersect.
2- The tangent at any point of a stream line gives the direction of the instantaneous velocity of the liquid at that point.
3- The stream lines come closer together at high velocity, and far a part at low velocity so the number of stream lines crossing perpendicularly a unity area (density of stream lines) expressing the velocity of flow of the liquid at that point.
It is the volume of the liquid flowing through a certain area per unit time.
Qv =Vol
T = A. L
T = A. Velm3/sec.
The rate of volume flow (Qv):
It is the mass of the liquid flowing through a certain area per unit time Qm = ρ. Qv = ρ. A. Vel Kg/sec
The rate of mass flow (Qm):
The quantity of the liquid that flow through a definite cross-sectional area of the tube in one second is known as flow rate.
Depending on how you define the quantity of the liquid there is:
Flow rate
The volume of liquid that flows steadily through a definite cross- sectional area of a tube in one second.
The mass of liquid that flows steadily through a definite cross-sectional area of a tube in one second.
Calculating the flow rate at any cross-sectional area:
By assuming a quantity of liquid of density (), volume (V01) and mass (m) flowing at velocity (v) for a distance (x) in time (t) through the cross- sectional are of the tube (A) as in figure.
Note:
- Density is the mass of a unit volume of the material (𝜌 = 𝑚
𝑉01) - Density is measured in kg/m3. - The density of any material is constant in a fixed temperature so it is considered as a characteristic property for the material.
From the definition of volume flow rate: From the definition of mass flow rate:
From the previous we can conclude:
The fluid flows steadily.
The amount of liquid entering the tube = That emerging out from it .
The flow rate (volume or mass) is constant at any cross-sectional area or inside the tube according to the law of conservation of mass and that leads to the continuity equation.
Deducing the continuity equation (relation between the flow speed of liquid and the cross-sectional area of the tube):
Imagine a tube carrying a fluid that flows steadily.
Then assume two cross-sectional areas (A1 , A2) that are perpendicular to the streamlines as shown in figure:
In the first cross-section (A1) In the second cross-section (A2)
Qv = A1v1 Qv = A2v2
Qm = A1v1 Qm = A2v2
The flow rate (volume or mass) is constant inside the tube in case of steady flow.
The conditions of the steady flow inside a tube:
1- The liquid should fill the tube completely.
2- The amount of liquid that enters the tube at one terminal equals the amount of liquid that emerges from the other terminal in the same time interval.
3- The velocity of the liquid remains constant at any certain point in the tube as time passes.
At A , area = A1
Velocity of liquid = V1
∴Qv at A = A1V1 At B , area = A2
Velocity of liquid = V2
∴Qv at B = A2V2 Then
The mass flow rate at A is Qm1 =ρA1V1
The mass flow rate at B is Qm2 =ρA2V2
Flow is steady: The mass flow rate is constant
∴Qm1= Qm2
ρA1V1 = ρA2V2
∴A1V1 = A2V2
∴ A1 A2 = V2
V1 Proof
A •
B
Explaining continuity on the basis of conservation of mass:
1) Consider a small amount
(∆m) of liquid entering from one side (∆m) = ρ. ∆Vol
∆Vol = A1. ∆x1
∆x1 = V1 . ∆T (∆m)1 =ρ. A1.V1∆T
2) Since the liquid is incompressible
The same amount will emerge from the other side (∆m) =ρ. A V ∆T3) (∆m)1 = (∆m)2
(∆m)1 =ρ. A1.V1∆T = ρ. A2.V2∆T
A1 V2 QV
A
V 1
= V Slope = V
A = VA Q
1 A
The tube is branched into (n) branches of the same cross-sectional area.
A1v1 = nA2v2
𝑟12𝑣1 = 𝑛𝑟22𝑣2 If
The tube is branched into a number of branches of different cross-sectional areas.
A1v1 = A2v2 + A3v3 + A4v4
𝑟12𝑣1 = 𝑟22𝑣2+ 𝑟32𝑣3+ 𝑟42𝑣4
,So
The graphical representation of the continuity equation:
∵ Qv = Av
∴ Slope = ∆𝑣
∆(𝐴1) = 𝑄𝑣
A1v1 = A2v2 + A3v3 + A4v4
𝑟12𝑣1 = 𝑟22𝑣2+ 𝑟32𝑣3+ 𝑟42𝑣4
Life applications of continuity equation:
1- The flow of blood in blood capillaries:
The speed of blood flow is faster in the main artery than in blood capillaries because the speed of blood flow varies inversely with the total cross-sectional area of the blood vessels (vein, artery or capillary). As the total cross-sectional area of the vessels increases, the velocity of flow decreases (𝑣 ∝ 1
𝐴). Blood flow is slowest in the capillaries, which allows time for exchange of gases (oxygen and carbon dioxide) and nutrients.
2- The design of gas burners holes:
The holes of the gas burners in the stoves are designed to be small so that the gas rushes out with high speed (𝑣 ∝ 1) to control the direction of flames.
Remarks
Solved problems ems
a) A1V1 = A2V2
𝜋 (0.01)2 × (0.1) = 𝜋 (0.005)2 × V2 V2 = (0.01)
2×0.1
(0.005)2 = 0.4 m/sec.
b) The rate of volume flow = Qv = A1V1 = A2V2
= 𝜋 × (0.01)2 × 0.1 = 3.14 × 10-5 m3/sec
∴The volume flow per min
= 3.14 × 10-5 × 60 = 188.4 × 10-5 m3/min The rate of mass flow
1] A water pipe 2cm diameter is at the entrance of an apartment building, the velocity of water in it is 0.1 m/sec, then the pipe tapers to 1cm diameter, calculate:
a) The velocity of water in the narrow pipe?
b) The quantity of water (mass and volume) flowing every minute across any section of the pipe (𝜌𝑤 = 103 kg/m3)
Solution:
Turbulent flow
• The flow of a fluid becomes turbulent which is characterized by small eddy currents as in figure when:
- The velocity of the fluid exceeds a certain limit.
- A gas transfers from small space to a wider space or from high pressure to low pressure.
a)Qv at A = AAVA = πrA2. VA
= π × (0.25)2 × 2 = 0.3926 m3/sec.
b)Qv at A = Qv at B = Qv at C + Qv at D 1) ∴ 0.3926 = π rB2. VB
0.3926 = 3.14 × (0.15)2 × VB
∴VB = 5.556 m/sec
2) 0.3926 = π rc2. VC + π. rD2 VD
0.3926 = 3.14 × (0.1)2 × 4 + 3.14 × (0.08)2 × VD
∴ VD = 13.28 m/sec . 2) In the figure shown:
The radius of tube at (A) = 25cm, at
(B) = 15cm, at (C) = 10 cm and at (D) = 8cm.
Calculate:
a) The rate of volume flow at (A) if water enters the tube at (A) by a velocity
= 2m/sec
b) The velocity of water at (B), and at (D) if the velocity of flow at (C) = 4m/sec?
Solution:
A
D C
B
A1V1 = NA2V2 π (0.7 × 10-2)2 × 0.33 = 30 × π × (0.35 × 10-2 )2 V2 ∴ V2 = 0.044 m/sec.
Comment:
1) Blood in arteries is slower than aorta because total area of arteries is larger than area of aorta
NA2 > A1 ∴V2 < V1 𝑉 ∝ 1
𝐴
2) From the same reasons blood velocity in capillaries is much smaller than arteries because their number is very high, the very low speed blood in capillaries to gives time for tissues to exchange oxygen and CO and providing it with food.
3) In normal adults the average speed of blood through aorta of radius 0.7 cm is 0.33 m/sec, from the aorta the blood goes through 30 major arteries each of radius 0.35 cm, calculate the speed of blood in arteries? Comment on your answer?
Solution:
r1 = 1cm v1 = 0.1 m/s r2 = 0.5 cm = 1000 kg/m3 (a)
∴ 𝑟12v1 = 𝑟22v2
∴ 𝑟12v1 = 𝑟22v2
(0.01)2 x 0.1 = (0.005)2 v2 = 0.4 m/s
(b) V01 = Qv t = A1v1 t = 𝑟12v1t
= 22
7x(0.01)2 x 0.1 x 60
= 1.89 x 10-3 m3
m = V01 = 1000 x 1.89 x 10-3 = 1.89 kg
4) A water pipe of diameter 2cm enters a house such that the flow speed of water inside it is 0.1 m/s, then its diameter becomes 1cm, Calculate:
(a) The speed of water in the narrow part.
(b) The amount of water (volume and mass) that flows every minute through any corss-section of the pipe.
(knowing that: density of water = 1000 kg/m3) Solution:
v1 = 0.33 m/s r1 = 0.7 cm r2 = 0.35 cm n = 30
5) If the average speed of blood in the aorta artery for a person is 0.33 m/s and radius of aorta is 0.7 cm and it distributes blood to a number of main arteries, radius of each is 0.35 cm, if the number of the main arteries is 30, Calculate the average blood speed inside each of them.
Solution:
A1 = A A2 = 2
3 A A3 = 2 A v1 = v v2 = 1
2 v 2Av3 = Av + (2
3𝐴𝑥1
2𝑣) 2v3 = v + 1
3 v = 4
3 v v3 = 2
3v
7) The opposite figure represents a steady flow of a liquid inside a tube if:
v1 = 2v2 = v, A1 = 𝟑
𝟐 A2 = 𝟏
𝟐 A3 = A
, Calculate the value of v3 in terms of v.
Solution:
r1 = 5 cm r2 = 0.5 cm r3 = 1 cm r4 = 2.5cm v2 = 2 m/s v3 = 0.8 m/s v4 = 0.3 m/s
∴ 𝑟12v1 = 𝑟22v2 + 𝑟32v3 + 𝑟42v4
(5x10-2)2 x v1 = [(0.5 x 10-2)2 x 2] + [(1 x 10-2)2 x 0.8] + [(2.5 x 10-2)2 x 0.3]
v1 = 0.127 m/s
= 22
7 x (5x10-2)2 x 0.127 x 30 = 0.03 m3 = 22
7 x (0.5x10-2)2 x 2 x 30 = 4.71 x 10-3 m3 = 22
7 x (1x10-2)2 x 0.8 x 30 = 7.54 x 10-3 m3 = 22
7 x (2.5x10-2)2 x 0.3 x 30 = 0.02 m3
6) A tube of diameter 10cm ends with a stopper with 3 holes of diameter 1cm, 2cm and 5 cm. If the speed of water which is emerging out from the three holes are 2m/s, 0.8 m/s and 0.3 m/s respectively, Calculate:
(a) The speed of water in the main tube.
(b) The volume of the flowing liquid in the main tube and that emerging out from the three holes within half a minute.
Solution:
tx = 15 minutes ty = 30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠
𝑉01
𝑡 = 𝑉01
𝑡𝑥 = 𝑉01 𝑡𝑦
∴ 1
𝑡
=
1𝑡𝑥
+
1𝑡𝑦
=
115
+
130 = 1
10
∴ t = 10 minutes
8) The opposite figure shows a water container of volume V01 which is being filled by two water taps x and y such that if only tap x is used, the container takes 15 minute to be filled and if only tap y is used, the container takes 30 minutes to be filled.
What are the time required to fill the container by using the two taps together.
Solution:
1. The steady flow?
It means that the flow in which the adjacent layers slide smoothly with respect to each other and the lines of the liquid follow non intersecting smooth paths called stream lines, its rate of flow inside the tube is the same at any cross-section.
2. The continuity equation?
It means that the velocity of a steady flow is inversely proportional to the cross sections area of the tube through which it flows.
3. The flow rate of a liquid inside a tube =0.5 m3/sec?
It means that a volume of 0.5m3 of liquid flows through any cross-section of the tube during 1sec.
4. The mass flow rate of a liquid in a tube = 20 kg/s?
It means that the mass of liquid passing through any cross- section area of the tube during
I second is 20kg.
From the continuity equation the velocity of flow is inversely proportional to the cross- sectional area ( V 1/A ) so narrow tips give high velocity of water flow so it can reach far distances.
1. Fire hoses are equipped with narrow tips.
As waterfalls down by gravity its velocity increases and from continuity equation (V 1/A ) , so the cross- sectional area decreases as waterfalls down.
2. Water falling from a tap as a conical shape.
Because velocity of water decreases as it rises up and since (V 1/A ) (from continuity equation) then area increases.
3. If a rubber tube from which the water flows is directed upwards the cross- section area of water increases as It goes higher
Because density of stream line increases when cross-section area of tube decreases and from continuity equation ( V 1/A ) so velocity of flow increases.
4. When the density of stream lines increases at certain point in steady flow the velocity of flow increases
The large number of blood capillaries makes its area larger than the area of the major artery and since from continuity equation ( V 1/A ) the velocity of blood in capillaries is smaller than in the major artery.
A1V1 = NA2V2 NA2 > A1 V2 < V1
5.The velocity of blood flow in capillaries is less than in major artery.
laws
1) (Q𝑣) rate of volume flow 𝑉𝑜𝑙
𝑇 = A. Vel m3/s.
∴ The volume that flows a pipe during time (t): Vol = Qv.t= A.V.t m3
2) (Q𝑚) rate of mass flow = 𝜌 Qv = . A. Vel
∴ The mass that flows a pipe during time (t): M = Qm .t = 𝜌. A. V. t Kg
3) The continuity equation: Qv1 = Qv2 A1V1 = A2V2
4) For tube is branched to N tubes of equal areas: A1V1 = N A2V2
5) For tube branching:
Qv at A= Qv at B= Qv at C+ Qv at D+ Qv at E
AA.VA = ABVB =ACVC + ADVD + AEVE
A B
C
D E
LESSON ONE
Q
1:What’s meant by?
(a) The steady flow
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(b) The stream line?
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(c) The turbulent flow?
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(d) The rate of flow of a liquid = 8 liter/sec.?
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(e) The flow rate of a liquid = 3x106 kg/sec?
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(f) The continuity equation?
Home Work
Fluid flow
Q
2:Give reasons for each of the following:
1- The blood flows slower in the capillaries rather than in the major artery?
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2- In the steady flow the liquid will flow very slowly when the cross-sectional area is large and quickly if it small
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3- The hose of firemen has narrow end.
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4- Thanks to Allah as the cross sectional area of the blood capillaries is larger than the cross sectional area of major artery.
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Q
3: A rubber hose is connected to a tap, and the water flows through in a steady flow, Explain why the cross – sectional area of the following water decrease when the end of the rubber hose is directed down, and increases when the end of the rubber hose is directed up?………..
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Q
4: Mention the conditions of steady flow, and the properties of stream lines, then deduce the relation to prove that the flow velocity of a liquid at any point in a tube is inversely proportional to cross-sectional area of the tube at such point.………..
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Q
5: Which graph from the following represent the relation between:a) The rate of flow and cross sectional area of tube (in steady flow) b) The velocity of flow and cross sectional area
c) The velocity of flow and the reciprocal cross sectional area
d) The coefficient of viscosity of a certain liquid and tangential force acts on its layers.
Q
6:Choose the correct answer:
1. In the steady flow of liquids, the ratio between the number of streamlines passing in the wide part of a tube to that in the narrow part of the same tube is ………
a) greater than one b) less than one c) equal to one d) indeterminable 2. In the steady flow, when the cross-sectional area of the tube decreases, the density of
the streamlines ………
a) increase b) decrease c) vanishes d) remains unchanged 3. If the cross-sectional area of the tube increases to the double in the steady flow, the
volume flow rate ………..
a) increases to the double b) decreases to the half
a
c b
(i) The flow speed at the upper floor is …………
(a) 1 m/s (b) 2 m/s (c) 3 m/s (d) 4 m/s
(ii) The volume flow rate of water at the ground floor equals …………
(a) 4x10-4 m3/s (b) 6x10-4 m3/s (c) 8x10-4 m3/s (d) 12x10-4 m3/s (iii) The mass flow rate of water at the upper floor equals …………
(a) 1.2 kg/s (b) 0.8 kg/s (c) 0.6 kg/s (d) 0.4 kg/s
5. The ratio between the mass flow rate and the volume flow rate for a liquid which flows steadily equals ………..
a) the cross-section of the tube b) the time of liquid flow c) the speed of liquid flow d) the liquid density
6. Water is rushing through a pump opening of cross-sectional area 5cm2 at a speed of 12m/s, so the mass of the water coming out from the pump within 30 minutes is …...
(a) 18.2 x 103 kg (b) 15.1 x 103 kg (c) 10.8 x 103 (d) 8.6 x 103 kg 7. The continuity equation is represented by graph ………….
8. In the opposite figure the tubes (A) and (B) are different in cross-sectional area and the steady flow rate of the liquid inside them is 0.3 m3/s. The two tubes meet to open in the tube (C) as in figure, then the volume flow rate in the tube (C) is ………...
(a) 0.1 m3/s (b) 0.3 m3/s (c) 0.6 m3/s (d) 0.9 m3/s 9. In the opposite figure, the speed of the steady flow
of water at A and C is 8 m/s respectively, then its speed at B is …….…...
(a) 16 m/sec (b) 12 m/sec (c) 8 m/sec (d) 6 m/sec
10. Water flows steadily in a tube of cross-section 10-3 m2, if the volume of the water which is coming out from the tube within 30 minutes was 18 m3, then ……….
Volume flow rate (m3/s) Speed of water flow (m/s)
(a) 0.01 10
(b) 0.01 600
(c) 0.6 10
(d) 0.6 600
11. The speed of a steady flow of water in the wide part of a tube is 1.2 m/s and in the narrow part of it is 6 m/s. Then the ratio between the diameter of the wide part and the diameter of the narrow part is …...
(a) 𝟏𝟐
𝟓 (b) 𝟕
√𝟓 (c) 5 (d) √𝟓
12. If the cross-section of the tube increases to the double in the steady flow, then the flow speed …….…...
(a) increases to the double (b) decreases to the half (c) increases four times (d) remains constant
13. If the ratio between the radii of the cross-sectional areas of a tube in the steady flow is
…….…...
(a) 𝟏
𝟒 (b) 𝟏
𝟐 (c) 2 (d) 4
14. A tube of radius r branches into a number of tubes each of radius 0.04 r. If the flow speed of the liquid in any of these tubes is five times its speed in the main tube, then the number of the small tubes is …….…...
(a) 5 (b) 125 (c) 140 (d) 150
15. The average speed of the steady flow of water in a tube of radius 2 cm is 3 m/s. The end of the tube is closed by a stopper that has 10 holes each of radius 0.3 cm, then the speed of water coming out of each hole is …….…...
(a) 13.33 m/s (b) 1.6 m/s (c) 2 m/s (d) 2.8 m/s
16. The opposite figure shows a tube that has different branches of equal cross-sectional areas where water flows through them steadily. If the volume flow rate at the branches A, B, C, E and F was 6 cm3/s, 3 cm3/s, 5 cm3/s, 4cm3/s and 3cm3/s respectively, then …….…...
Direction of water flow in branch D
Volume flow rate Of water at D
(a) to the inside 7 cm3/s
(b) to the inside 15 cm3/s
(c) to the inside 7 cm3/s
(d) to the inside 15 cm3/s
17. A water tank takes 10 minutes to be filled by using three water taps together and takes 20 minutes when using the first water tap only and an hour when using the second water tap only. Then the time required to fill the tank using the third water tap only is
…….…...
(a) 10 minutes (b) 20 minutes (c) 30 minutes (d) 60 minutes 18. Water flows steadily through a horizontal pipe which has a narrow section of radius
half the other section of the pipe, so the mass flow rate through the narrow section……
(a) decreases to quarter (b) decreases to half (c) increases to quadruple (d) remains constant 19. The opposite graph shows the relation between the
flow speed (v) and the reciprocal of the cross- sectional area
(
1𝐴
)
of the tube for two liquids. If the two liquids flow steadily, ………….(a) volume flow rates of the two liquids are equals (b) mass flow rates of the two liquids are equals.
(c) the flow speed of liquid x is three times that of y through the same cross- sectional area.
(d) the flow rate of y is double that of x.
20. The opposite diagram shows a liquid flowing steadily in a tube. If the volume flow rates in branches B and C respectively are 0.1 m3/s and 0.3 m3/s, the volume flow rate in tube A equals …….…...
(a) 0.1 m3/s (b) 0.2 m3/s (c) 0.3 m3/s (d) 0.4 m3/s 21. A hose has a cross-sectional area 25cm2 at the water source and 5 cm2 at its end. If
water flows steadily in the hose with a speed of 4 m/s, at the water source , the mass of the flowing water during 15 min equals …….…... ( = 1000 kg/mw 3 )
(a) 5 ton (b) 9 ton (c) 20 ton (d) 25 ton
1. liquid flow in a pipe of radius 2cm, with velocity 8 m/sec, calculate the radius of the tube that makes water pass with velocity 32 m/sec. [1cm].
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2. Major artery of a radius 0.5 cm divided into 100 capillaries each of radius 0.2 cm as blood velocity in aorta is 0.04 m/sec, calculate the blood velocity in each capillary?
[2.5 x 10-3 m/sec.]
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3. In normal adults the average speed of blood through aorta of radius 0.7cm. is 0.33m/sec from the aorta the blood goes through 30 arteries each of radius 0.35 cm, calculate the speed of blood in each of arteries? [0.044 m/sec]
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Problems
4. If kerosene flow across pipe by rate 60 liter /min, with velocity 40m/sec, calculate the cross-sectional area of this pipe? [2.5 × 10-5 m2]
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5. A liquid flow with velocity v m/sec, and pass through a water pipe of radius r, calculate the velocity of liquid when the pipe becomes narrow such that its diameter becomes
¼r. [64v]
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6. A water pipe entering a home its radius is 1.5 cm with a speed of 0.2 m/sec, if the pipe becomes of radius 0.5cm at its end calculate:
a) The speed of water at the narrow end.
b) The volume of water that flow per minute across any cross sectional area (
=3.14) [8.478× 10-3 m3/min]………..
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7. A pipe of cross sectional area 4cm2 provides a field with water with speed of 10 m/sec, at its end there are 100 pores the area of each opening is 1mm2, what is the speed of water passing through each pore? [40m /sec]
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8. A tube of diameter 10cm where the speed of water at that section is 1 m/sec, at the end of the tube, the diameter becomes 2.5 cm, calculate:
a) The speed of water at the narrow section.
b) The mass of water that flow per minute across any of its cross sectional area . ( = 10w 3 kg/m3 , π = 3.14 ) [16 m/sec, 471 kg/min]
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9. A major artery of radius 0.5cm, the speed of the blood through it is 0.4 m /sec is divided into number of capillaries each of radius 0.2 cm, and the speed of the blood through each capillary is 0.25 m/sec, find the number of capillaries? [10 capillaries]
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10. In the shown figure, given that the radius of the tube at (A) is 30cm and the velocity of water entering at same point is 2m/sec, the velocity of water at (C) is 4m/sec , and at (E) is 3m/sec where the radius of the tubs at (B) is 20cm , at (C) is 15 cm at (D) is 10 cm and at (E) is 5cm, calculate:
a) The rate of volume of water entering at (A).
b) The velocity of water flow at (B) and (D).
[0.562m3/sec, 4.5m/sec, 8.25m/s]
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A B
C
D
E
11. In the human body, blood runs from the heart to the aorta which branches out to main arteries then to small arteries until it reaches the blood capillaries. If the radius of the aorta is about 1.2cm, the blood flow speed inside it is 40 cm/s, the average radius of a blood capillary is about 4 x 10-4cm and the speed of blood inside it is about 5x 10-4 m/s.
What is the number of blood capillaries which are connected to this artery?
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12. The water flows steadily in a tube of diameter 2 cm at a speed of 5 m/s, Calculate:
(a) The volume of water which is flowing through the cross-section of the tube in one minute.
(b) The time required to fill a tank of volume 20 m3 by using the flowing water from the tube. (where: = 3.14) (0.0942 m3, 212.31 min.)
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13. The opposite figure shows the graphical relation between the volumes of two different liquids that flow steadily in two different tubes A and B and the time. If the ratio between the densities of the liquids is 2:1 respectively, Calculate the ratio between the mass flow rates for each of them.
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14. Water flows through a hose of diameter 1.2cm at a speed of 3 m/s. Calculate the diameter of the nozzle of the hose, if it shoots out water with a speed of 27 m/s.
(0.4 cm)
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15. A water pipe of cross-section 6 x 10-4 m2 and the speed of water inside it is 3 m/s.
Calculate the speed of water when the cross-section decreases to 3 x 10-4 m2.
(6 m/s)
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16. The radius of water pipe is 1.5 cm and the speed of water inside it is 0.2 m/s, if its radius became 0.5 cm, Calculate:
(a) The speed of water at the narrow end.
(b) The volume of the flowing water every minute at any cross-section of it. (=3.14) (1.8 m/s, 8.478 x 10-3 m3)
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17. Water flows at a speed of 1 m/s through a tube of diameter 10 cm which tapers (narrow) to a diameter of 2.5 cm at its end. Calculate:
(a) The speed of water at the narrow end.
(b) The mass of the flowing water every minute through any cross-section of the tube.
(where: water density = 1000 kg/m3 , = 3.14) (16 m/s, 471 kg)
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18. A liquid flows steadily in a tube of cross-sectional area 2x10-4 m2 at a speed of 4 m/s, Calculate:
(a) The volume flow rate of the liquid.
(b) The flow speed if the radius of the rube increases to the double.
(8x10-4 m3/s, 1 m/s)
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19. Oil flows in a tube at the rate of 6 liter/minute and gets out from another tube which is connected to the first tube at a speed of 4 m/s. Calculate the cross-sectional area of the second tube. (2.5x 10-5 m2)
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20. A liquid flows in a tube where the radius of one of its ends is r and the radius of the other end is 𝑟
4. Calculate the ratio between the speed of flow of the liquid in the first cross-section of the tube and its flow speed in the second cross-section of the tube.
(1
16)
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21. Water flows steadily in a tube of diameter 2 cm at an average speed of 3 m/s, if the end of the tube is closed with a stopper of 10 openings each of radius 1mm.
Calculate the average flow speed at every opening.
(30 m/s)
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22. Blood flows through a main artery of radius 0.5cm with a speed of 0.4 m/s, then it branches into many capillaries each of radius 0.2 cm where the flow speed of blood in each one becomes 0.25 m/s. Find the number of the blood capillaries. [10 capillaries]
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23. The blood flows in a main artery with a speed of 0.24 m/s and then it branches into 120 capillaries each of a diameter that equals 1
4 the diameter of the artery. Calculate the flow speed of the blood in each capillary.
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24. In the opposite figure:
Calculate the speed of the liquid at (C), if the inner diameter of the tube is 20 cm at (A), 8 cm at (B) and 10 cm at (C), if the speed of liquid is 20 cm/s at (A) and 30 cm/s at (B).
(60.8 m/s)
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25. The opposite graph illustrates the relation between the liquid flow speed (v) at a point in a tube and the reciprocal of the cross-sectional area of the tube (1
𝐴) at that point.
From the graph find:
(a) The volume flow rate.
(b) The mass of the flowing liquid within 30 minutes if the liquid density is 1000 kg/m3. (0.004
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26. A garden hose, ends with a nozzle of diameter 2.74 cm, is used to fill a container of volume 25 liter within 1.5 minutes. Calculate the speed of the water coming out of the nozzle. If the nozzle is closed by a stopper that has an opening of a diameter that equals one third of that of the hose, Calculate the speed of the water coming out of that
opening. (0.47 m/s, 4.23 m/s)
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27. An oil pump pumps 55 m3 of oil within 35 s in a cylindrical tank of diameter 11 m and height 10 m. If the oil’s density is 820 kg/m3, Calculate: (188.57 kg/s, 10.08 min)
(a) The mass flow rate of the oil.
(b) The time required to fill the oil tank.
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0 5 10 15 20 25 30 35 40
0 0.2 0.4 0.6 0.8 1
Y-Values
V(m/s)
1/A (cm-2)
28. The opposite figure shows the steady flow of water in a tube. If v1=2 v2 = v Calculate v3 in terms of v. (𝑣
2)
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29. The following table shows the data which describes the flow of water in the opposite figure:
Section Radius (cm) Flow speed (m/s)
A 30 2
B 20 vB
C 15 3
D 10 vD
E 5 15
Calculate:
(a) The volume flow rate at (A).
(b) The flow speed of water at both of cross-section (where : = 3.14)
30. Three taps, the first fills a tank within one hour, the second fills the same tank within
1
2 an hour and the third fills it within 1
4 an hour. Calculate the time required to fill the tank by using the three taps together. (1
7 hour)
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"Is a property of fluids (liquids or gases) due to friction forces between its layers which resist sliding of them one over the other and resist the motion of bodies inside fluids".
Viscosity:
1-We hang two funnels each on a stand and put
a beaker under each, we pour alcohol in one funnel and glycerin in the other, we observe that the velocity of alcohol is higher than that of glycerin
2-Take two equal volume of water and honey in a glass container, and stir them by a glass rod by same velocity then remove the rod, we notice that:
a) Water is easier to stir, which means that water resistance to the glass rod is less than the resistance of honey.
b) The motion in honey stops almost immediately after we remove the rod, while it continues for a little while longer in water.
3-Take two similar measuring cylinder and fill them to end ,one of them with water and the other with glycerin, drop two similar metal balls in the two liquids and use a stop watch to determine the time through which each ball reaches the bottom
You notice that: the ball takes longer time in glycerin Experim Experiments to show viscosity:-
ents to show viscosity:-
alcohol glycerin
water glycerin
The viscosity LESSON TWO
1- Some liquids such as water and alcohol flow easily and have little resistance to the motion of solids through them they have small viscosity .
2- Some liquids such as honey and glycerin, flow slowly and have high resistance to the motion of solids through them, they have high viscosity
Conclusion:
High viscosity Low viscosity
Low ability to flow.
High resistance to body motion through it
Honey & glycerin.
High ability to flow.
Water & alcohol.
Low resistance to body motion through it.
1- Let us imagine a section of fluid between two parallel plates, one at rest while the other is under tangential force F and moves with velocity V.
2- The fluid layer adjacent to the rest plate stays at rest due to the adhesive force between this layer and the rest plate. V
= zero
3- The fluid layer adjacent to moving plates with velocity V due to the adhesive force with the moving plate.
Explain the concept of viscosity:
To keep the upper plate moving at uniform velocity V , a force F is required to overcome the friction force between the layers of fluid this force depends on:
1 − Area of a moving plate FαA
2- The difference of velocities between the two plates FαV 3- The distance between the two plate Fα 1 𝑑ൗ
∴ Fα𝐴.𝑉
𝑑
F = η𝑣𝑖𝑠 = 𝐴.𝑉
𝑑 Newton
➔ η𝑣𝑖𝑠is the coefficient of viscosity
∴ η𝑣𝑖𝑠 = 𝐹. 𝑑
𝐴. 𝑉N. sec/m2 or pascal. sec = kg/ m. sec. = kg. m . sec.
Coefficient of viscosity:
It is the tangential force acting on unit area to produce a unit change in velocity between two liquid layers separated by a unit distance"
𝛈𝐯𝐢𝐬Coefficient of viscosity:
𝐅𝐚𝐜𝐭𝐨𝐫𝐬 𝐚𝐟𝐟𝐞𝐜𝐭𝐢𝐧𝐠 𝐨𝐧 𝛈𝐯𝐢𝐬
a) kind of liquid b) temperature
➔ So it is a physical property of fluids
Metallic parts in machines have to be lubricated from time to time this process leads to:
a) Reduction of heat generated by friction
b) Protecting the machine part from corrosion (wear)
If we use water (low viscosity), it will rapidly flow away from the machine part as a result of weak adhesive force during its motion, so it is necessary to use liquids with high viscosity, so they remain in contact with moving machine parts.
Application of viscosity:
1- Lubrication :
-1
∴ η𝑣𝑖𝑠
= 𝐹. 𝑑
𝐴. 𝑉 N. sec
/m2 or pascal. sec
= kg / m. sec.
= kg. m
− 1. sec.
-1
∴ η𝑣𝑖𝑠
=𝐹. 𝑑
𝐴. 𝑉 N. sec
/m2 or pascal. sec
= kg / m. sec.
= kg. m
− 1. sec.
➔ (S.R.T) sedimentation rate test
▪ Is the final speed of the falling red blood corpuscles (R.B.Cs)through plasma.
When a ball undergoes a free fall in a liquid, it is under three forces, its weight, buoyancy of the liquid and friction between the ball and the liquid due to
viscosity, it is found that such a ball attains a final velocity which is proportional to its radius square. Vfinal α r2
This is applied in medicine by taking a blood sample and measuring its precipitation rate.
So, the doctor may then decide if the size of red blood cells is normal or not.
a) In rheumatic fever, gout , and rheumatic heart:
➔ Red blood cells adhere together so their volume and radius increases and accordingly sedimentation (precipitation) rate increases.
b) In Anemia
➔ Red blood cells are broken down so their volume and radius decreases so the sedimentation rate decreases below its normal level.
|
3- In medicine :
For low and moderate car speeds, the air resistance (R) due to its viscosity is directly proportional to car speed R α V.
For high car speed, the air resistance is directly proportional to square of car's speed R α v2. And the fuel consumption increases.
So, we must decrease the car speed to economize fuel consumption Application of viscosity:
2- moving vehicles :
Solved problems ems
ηvis =F. d A. V V = F. d
A. ηvis =2.5 × 2 × 10−3
0.01 × 4 = 0.125m/sec
1)A plane plate of area 0.01m2 is separated from another parallel plate by a layer of liquid 2 mm thick, its coefficient of viscosity is
4N.sec/m2, if a force of 2.5 N acts tangentially on the 1st plate, find its velocity?
Solution:
ηvis =F. d A. V d = ηvis. A. V
F =0.78 × 2 × 40 × 10−2 × 4
200 = 0.0125m
2)Metallic plate of dimensions 2 m, 40 cm moves with velocity 4m/sec, on a smooth surface covered with a layer of glycerin, if a force of viscosity is 200N and the coefficient of viscosity is 0.78 kg/m.sec, calculate the thickness glycerin layer?
Solution:
F2
F1
2.5 cm
2.5 cm
F2
F1 1 cm
4 cm
1. The coefficient of viscosity of a lubricant =4 kgm-1 s-1 ?
It means that 4 N is the tangential force needed to act on a unit area of the lubricant to produce a unit change in velocity between two layers separated by a distance 1 m.
I second is 20kg.
a) The force required to move the middle plate against viscosity both sides:
F = F1 + F2 ∴ F1 = F2 = ηvis×A. V 𝑑 𝐹 = 2 × ηvis×A.V
𝑑
𝐹 = 2 × 0.785 × 0.8
2.5 × 10−2 = 50.24𝑁 b) in the 2nd case:
F = F1 + F2 F = ηvis ×A. V
𝑑1 + ηvis×A. V 𝑑2 = 0.785 0.8
2.5 × 10−2 + 0.785 0.8
2.5 × 10−2 = 62.8 + 15.7 = 78.5 N
➔ So it is easier to move a body between two fixed plates if it is in the middle.
3)Two parallel plate of distance 5 cm a part, the fluid between them is glycerin with coefficient of viscosity 0.785 N.Sec/m2 calculate the force required to move a thin plate of 0.8m2 area between them with velocity 1m/sec in the following cases:
a) If the plate is in the middle between the two fixed plates.
b) If the plate is 1cm a part from one of the two fixed plates.
➔ Then explain what did you deduce?
Solution:
f 1 f 2
f 1 f 2
2.5cm 2.5cm
4cm 1cm
Oil has high viscosity so it adheres well to the mechanical parts and it helps to decrease friction, wear and heat.
1. Oil is used to lubricate mechanical parts.
Because it has low viscosity and can't adhere well to the mechanical parts.
2. Water is not used to lubricate machines.
At low speed the air resistance due to its viscosity α v, at high speed the air resistance increases and becomes α 𝑣2 and the fuel consumption increases 3. It is not recommended to drive cars with high speeds .
Because the viscosity of oil decreases with temperature.
4. Lubricating oil in summer has to be more viscous than in winter.
Because the coefficient of viscosity is directly proportional to the friction force with molecules of the fluid : ηvis ∝ F
5. The resistance of a fluid to the motion of a solid body increases as the viscosity of the fluid increases.
In rheumatic fever RBC's stick together making large volume and as the terminal velocity α 𝑅2, the precipitation rate increases.
6. Precipitation rate increases in case of rheumatic fever.
Anemia breaks down RBC's decreasing its volume and as the terminal velocity α 𝑅2,, precipitation rate decreases.
7. Precipitation rate decreases in case of anemia.
For rheumatic fever, rheumatic heart, gout precipitation rate is higher than normal, but for anemia, jaundice, the precipitation rate is low.
8. It is possible for a doctor to know some diseases by precipitation rate.
The existence of friction forces due to viscosity between each liquid layer and the
adjacent one resists sliding of liquid layers with respect to each other. This
produces a relative change in velocity between two adjacent layers. And since the middle of river has the maximum distance from the banks as from the bottom, the velocity of layers range from zero at the bank or bottom to maximum value at the middle (ηvis =F.d
A.V = constant, :. v α d)
9. Speed of water at the middle of rivers is higher than at the sides.
Because due to viscosity of air ,the friction forces resisting the fall of droplets is directly proportional to v2,so as velocity increases ,the resisting friction increases until it becomes equal to the weight of droplet, the speed of fall becomes constant 10. Rain droplets are falling with uniform velocity instead of free fall
acceleration.
* Viscosity:
F = ηvis×A. V
𝑑 Newton ηvis = F. d
A. V N. sec/m2 or kg/m. sec.
laws
LESSON TWO
Q
1:What’s meant by?
(a) The viscosity?
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(b) The coefficient of viscosity of a liquid = 0.3 kg/m.sec?
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Q
2:Give reasons for each:
a- It is necessary to lubricate machines from time to time.
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b- Water is not use to lubricate metallic machines, while oils with high viscosity are used.
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c- Sedimentation rate decreases in case of anemia, and increases in case of rheumatic fever.
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Home Work
The Viscosity
d- When the velocity of car exceeds a certain limit, leading to a noticeable increase in the fuel consumption.
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e- If a solid body moves in a fluid it loses a part of its momentum.
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Q
3:What happens when? Mention the reasons for each of the
following:
a- The velocity of a car increases above a certain limit.
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b- The temperature of lubricating oils increase.
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c- The red blood corpuscles adhere and their radius increase.
d- The opposite figure shows the position of a metallic ball after 4s from dropping it into four different fluids (a, b, c and d).
Rank the fluids by their viscosities with explanation?
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Q
4:Choose the correct answer:
1.The unit of the coefficient of viscosity is …….
a) kg.m-1.sec-2 b) kg.m-2sec c) kg.m-1.sec-1 2.Measuring sedimentation rate is an application for ………
a) Archimedes' principle b) viscosity c) Pascal's principle
3.The sedimentation rate in the case of rheumatic fever is
a) below normal b) above normal c) normal
4.When the tangential force between two liquid layers increases the coefficient of viscosity of this liquid
a) decreases b) increases c) remains constant
5.Speed of falling of a ball in a viscous liquid is ………. that of a ball of smaller volume a) grater than b) less than c) equal
6.Speed of water at the middle of river is ……… that at sides a) higher than b) lower than c) equal 7. In the high speeds of a car, the air resistance due to its viscosity is ………..
a) directly proportional to the speed of the car.
b) inversely proportional to the speed of the car.
c) directly proportional to square the speed of the car.
d) inversely proportional to square the speed of the car.
8. A plate of surface area 0.25 m2 slides at a speed of 0.6 m/s above a layer of viscous liquid of thickness 5 mm. If the viscosity coefficient of the liquid is 0.95 N.s/m2, the tangential force that acts on the plate equals ………..
a) 14.25 N b) 21.375 N c) 28.5 N d) 42.75 N
9. A layer of viscous liquid, its thickness is 4.5 mm and its viscosity coefficient is 1.8 kg/m.s, is placed between two parallel plane plates. If a force of 32N acted on the upper plate, so it moved at a speed of 1 m/s. The surface area of the upper plate is…..
a) 0.02 m2 b) 0.04 m2 c) 0.06 m2 d) 0.08 m2 10. The resistance of liquids to the motion of objects inside them is a result of…..
a) The liquid density.
b) The liquid viscosity.
c) The pressure inside the liquid.
d) the motion of the liquids from point to another.
11. A circular plate of radius 70 cm slides in a speed of 0.1 m/s on a ceramic floor covered with a layer of viscous liquid of thickness 2.5 mm and viscosity coefficient 2.5 N.s/m2, then the tangential force acting on the plate is ………..
a) 154 N b) 132N c) 124 N d) 112 N
12. A rectangular plate of dimensions 50 cm, 25cm is affected by a tangential force of 15N which moves it at a speed of 0.8 m/s on a layer of viscous liquid of thickness 9.375 mm, so the viscosity coefficient of the liquid is ………..
a) 0.42 kg/m.s b) 0.85 kg/m.s c) 1.41 kg/m.s d) 2.31 kg/m.s 13. If the speed difference between two liquid layers decreased when a tangential force
acted on the upper layer then the viscosity coefficient ……….
a) vanishes b) decreases c) increases d) remains constant 14. When the temperature of a liquid decreases, its viscosity coefficient ….…..
a) increases.
15. In the relatively low or medium speeds of a car, the air resistance due to its viscosity is ….…..
a) directly proportional to square the speed of the car.
b) directly proportional to the speed of the car.
c) inversely proportional to square the speed of the car.
d) inversely proportional to the speed of the car.
1. A metallic flat plate of surface area 0.01 m2 is separated from another large plate by a liquid layer 2mm height, if a force of 2.5 N acts on the first plate which moves by velocity of 12.5 cm/sec. Calculate the coefficient of viscosity? [4N.sec/m2]
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2. In the opposite figure:
If a tangential force of 10 N acted upon the upper plate to move at a speed of 3 m/s, Calculate the viscosity coefficient of the liquid.
[4N.sec/m2]
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3. A square plate of side length 12 cm slides above another static plate with a liquid layer of thickness 2mm in between. If the viscosity coefficient of the liquid is 0.2 N.s/m2 and the speed of the plate is 0.01 m/s, Calculate the tangential force acting on the plate.
[0.0144 N]
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4. A layer of a viscous liquid of thickness 8cm is put between two parallel horizontal plane plates. If the viscosity coefficient of the liquid is 0.8 kg/m.s Find the force required to move a thin plate of area 0.5 m2 parallel to the two plates with a speed of
