Chapter 10.11 - Speed of Efflux

In the previous section, we saw the basics of venturimeter. In this section, we will see some more practical applications of the Bernoulli's equation. First we will see speed of efflux. It can be explained in 13 steps:

1. In the fig.10.39(a) below, a tank contains some liquid. Density of the liquid is ρ

Fig.10.39

2. A small hole is present at the side of the tank
• This hole
    ♦ is at a depth of h from the surface of the liquid
    ♦ is at a height of y from the bottom of the tank
• Bottom of the tank is considered as the datum
3. The liquid will be flowing out through the hole
    ♦ Such an outflow of a fluid is called efflux
    ♦ Speed of the outflow is called speed of efflux
          ✰ We have to derive a formula to obtain that speed of efflux
4. Let us mark two points
    ♦ Point A is at the surface of the liquid
    ♦ Point B is at the hole
5. Let us apply Bernoulli’s equation at the two points
• At A we get: $\mathbf\small{\rm{P_a+P_A+\frac{1}{2} \rho v_A^2+ \rho g (h+y)}}$
• At B we get: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g y}}$

6. Equating the two, we get: $\mathbf\small{\rm{P_a+P_A+\frac{1}{2} \rho v_A^2+ \rho g (h+y)=P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g y}}$

    ♦ At point A, there is no depth and it is open to atmosphere. So there is only atmospheric pressure

    ♦ At point B also, there is no depth and it is open to atmosphere. So there is only atmospheric pressure

• So the equation becomes:

$\mathbf\small{\rm{P_a+0+\frac{1}{2} \rho v_A^2+ \rho g (h+y)=P_a+0+\frac{1}{2} \rho v_B^2+ \rho g y}}$

⇒ $\mathbf\small{\rm{\frac{1}{2} \rho v_A^2+ \rho gh+\rho gy=\frac{1}{2} \rho v_B^2+ \rho g y}}$

⇒ $\mathbf\small{\rm{\frac{1}{2} \rho v_A^2+ \rho gh=\frac{1}{2} \rho v_B^2}}$

7. If the size of the hole is small when compared to the size of the tank, the level of liquid in the tank will be falling very slowly

• That means, the particles at the point A will be moving with a very small velocity
• So we can assume v_A to be equal to zero

8. So the result in (6) becomes:

$\mathbf\small{\rm{\rho gh=\frac{1}{2} \rho v_B^2}}$
• Thus we get:

Eq.10.10: $\mathbf\small{\rm{v_B=\sqrt{2gh}}}$
This equation is known as Torricelli’s law
9. Here we note an interesting point. It can be written in 2 steps:
(i) Consider a body in free fall (initial velocity zero)
    ♦ Let the height of fall be h
(ii) Just when it covers the height h, it’s velocity will be $\mathbf\small{\rm{\sqrt{2gh}}}$

• We have seen this in an earlier chapter (Details here)
10. Now consider the closed tank in fig.b
• The liquid is subjected to a pressure of P
Applying Bernoulli’s equation, we get:

• At A we get: $\mathbf\small{\rm{P+\frac{1}{2} \rho v_A^2+ \rho g (h+y)}}$
• At B we get: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g y}}$

11. Equating the two, we get: $\mathbf\small{\rm{P+\frac{1}{2} \rho v_A^2+ \rho g (h+y)=P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g y}}$

    ♦ At point B, there is no depth and it is open to atmosphere. So there is only atmospheric pressure

• So the equation becomes:

$\mathbf\small{\rm{P+\frac{1}{2} \rho v_A^2+ \rho g (h+y)=P_a+0+\frac{1}{2} \rho v_B^2+ \rho g y}}$

⇒ $\mathbf\small{\rm{P+\frac{1}{2} \rho v_A^2+ \rho gh+\rho gy=P_a+\frac{1}{2} \rho v_B^2+ \rho g y}}$

⇒ $\mathbf\small{\rm{P+\frac{1}{2} \rho v_A^2+ \rho gh=P_a+\frac{1}{2} \rho v_B^2}}$

12. If the size of the hole is small when compared to the size of the tank, the level of liquid in the tank will be falling very slowly

• That means, the particles at the point A will be moving with a very small velocity
• So we can assume v_A to be equal to zero

13. So the result in (11) becomes:

$\mathbf\small{\rm{P+\rho gh=P_a+\frac{1}{2} \rho v_B^2}}$

⇒ $\mathbf\small{\rm{\frac{1}{2} \rho v_B^2=P-P_a+\rho gh}}$
• Thus we get:

Eq.10.11: $\mathbf\small{\rm{v_B=\sqrt{2gh+\frac{2(P-P_a)}{\rho}}}}$

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Now we will see some solved examples

 

Solved example 10.23

Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.

Solution:

Bernoulli’s equation is applicable only when the flow is steady (streamline flow). The flow of water through a rapid is a turbulent flow. So We cannot use Bernoulli’s equation

 

Solved example 10.24

Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain

Solution:

1. When we apply Bernoulli’s equation, we can write:

$\mathbf\small{\rm{P_{A(abs)}+\frac{1}{2} \rho v_A^2+ \rho g h_A=P_{B(abs)}+\frac{1}{2} \rho v_B^2+ \rho g h_B}}$
    ♦ Where P_A(abs) and P_B(abs) are the absolute pressures at A and B
2. We can split up the pressures

P_A(abs) at A can be split up as: (P_a + P_A)
P_B(abs) at B can be split up as: (P_a + P_B)

    ♦ Where P_a is the atmospheric pressure,  P_A and P_B are the gauge pressures at A and B

3. In this situation, the P_a on both sides will cancel each other. So we can use gauge pressure instead of atmospheric pressure

4. But this type of cancellation occurs only if both points A and B are connected to the atmosphere
• If the container is closed at any one side, and a pressure is applied, we have to use the absolute pressure at that side

 

Solved example 10.25

During blood transfusion the needle is inserted in a vein where the gauge pressure
is 2000 Pa. At what height must the blood container be placed so that blood may
just enter the vein ?

Solution:

1. Imagine a horizontal line through the vein. Let this line be the datum
    ♦ Let us denote the point in the vein as A and the point of the container as B
    ♦ Let the container be placed at a height h above the datum
2. Applying Bernoulli’s equation, we get:

$\mathbf\small{\rm{P_{A(abs)}+\frac{1}{2} \rho v_A^2+ \rho g \times 0=P_{B(abs)}+\frac{1}{2} \rho v_B^2+ \rho g h}}$

⇒ $\mathbf\small{\rm{P_{A(abs)}+\frac{1}{2} \rho v_A^2=P_{B(abs)}+\frac{1}{2} \rho v_B^2+ \rho g h}}$ 

    ♦ Where P_A(abs) and P_B(abs) are the absolute pressures at A and B
3. The pressure at the vein is given as 2000 Pa
• Usually, the blood pressure is reported as gauge pressure
    ♦ This is because, the device used to measure blood pressure shows zero reading at atmospheric pressure

    ♦ So any pressure measured by the device will be above atmospheric pressure
4. So we can write:

    ♦ P_A(abs) = P_a + 2000

    ♦ P_B(abs) = P_a
5. Since blood flows slowly, we can put v_A =v_B = 0
6. Thus the equation written in (2) becomes:

$\mathbf\small{\rm{P_a+2000+0=P_a+0+ \rho g h}}$

⇒ $\mathbf\small{\rm{2000=\rho g h}}$

⇒ $\mathbf\small{\rm{h=\frac{2000}{1060 \times 9.8}}}$ = 0.193 m

Blood flow and heart attack

The relation between blood flow and heart attack can be explained in steps:

1. Sometimes, plaque gets deposited on the inner walls of the artery

• So the blood has to flow through a narrow area

2. This creates a heavier load on the heart because, now heart has to pump with greater effort.

3. We know that, when the area of flow decreases, velocity increases

• Also, when velocity increases, pressure decreases

4. So, when the blood flows with a greater velocity through the narrow path, pressure inside the artery in that narrow portion decreases

• So the external pressure around the artery will cause it to shrink

5. This creates even greater load on the heart

• With greater effort from the heart, the blood flows through the shrunk artery

• When flowing through the newly shrunk portion, the velocity again increases

6. This causes further decrease in pressure and further shrinkage of artery

• Such a repeating process will lead to heart attack

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In the next section, we will see more practical applications of Bernoulli's equation



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Chapter 10.10 - The Venturimeter

In the previous section, we saw the basics of Bernoulli's equation. In this section, we will see some practical applications of the equation. First we will see venturimeter. It can be explained in 12 steps:

1. Fig.10.37(a) below shows a simple water supply system

• Water is taken out from the yellow tank

    ♦ First the water flows through the red pipe

    ♦ Then it flows through the magenta pipe

    ♦ A horizontal magenta pipe with a tap is taken out from the vertical magenta pipe

Fig.10.37

2. A special device is inserted in the horizontal magenta pipe

• This special device is called venturimeter. It is shown in yellow color

• It's main features can be written in 4 steps:

(i) Water enters the venturimeter from the left end

    ♦ At the point of entrance, the venturimeter has the same diameter as the magenta pipe

    ♦ This uniform diameter continues upto a short distance to the right

(ii) But as we proceed further to the right, the diameter gradually reduces to a minimum

(iii) As we proceed further, the diameter again increases gradually and become equal to the diameter of the magenta pipe

(iv) This can be clearly seen in fig.b

    ♦ Fig.b is the enlarged view of the venturimeter in fig.a

3. Position of the venturimeter:

• The venturimeter

    ♦ is at a vertical distance of z₁ from the top surface of water

    ♦ is at a height of h₁ from the datum

4. Two points A and B are marked inside the venturimeter

    ♦ A is at the wider section

    ♦ B is at the narrow section

5. Let us apply Bernoulli’s equation at the two points

• At A we get: $\mathbf\small{\rm{P_a+P_A+\frac{1}{2} \rho_w v_A^2+ \rho_w g h_1}}$

• At B we get: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho_w v_B^2+ \rho_w g h_1}}$

    ♦ Where ρ_w is the density of water

6. Equating the two, we get: $\mathbf\small{\rm{P_A+\frac{1}{2} \rho_w v_A^2+ \rho_w g h_1=P_B+\frac{1}{2} \rho_w v_B^2+ \rho_w g h_1}}$

⇒ $\mathbf\small{\rm{P_A+\frac{1}{2} \rho_w v_A^2=P_B+\frac{1}{2} \rho_w v_B^2}}$

(Note that, since the venturimeter is horizontal, h₁ is same for both points)

■ It is clear that:

    ♦ If velocity increase at any one of the points, the pressure at that point will decrease
    ♦ If velocity decrease at any one of the points, the pressure at that point will increase
7. Rearranging the equation in (6), we get:

$\mathbf\small{\rm{P_A-P_B=\frac{1}{2} \rho_w \left (v_B^2-v_A^2 \right)}}$
8. So we want the pressure difference (P_A - P_B) between the two points A and B
• For that, a U-tube manometer as shown in fig.b is used. It can be explained in 4 steps:
(i) In fig.b, we see that:
    ♦ One end of the manometer is connected to the wide section of the venturimeter
    ♦ The other end is connected to the narrow section of the venturimeter
(ii) Now we apply the equation of continuity: A_Av_A = A_Bv_B
    ♦ It is clear that, when area decreases, velocity increases
    ♦ Velocity at B will be greater than the velocity at A
(iii) Consequently, the pressure at B will be lesser than the pressure at A
(iv) The greater pressure at A will push the mercury upwards
• As a result, there will be a level difference between the mercury surfaces in the two limbs of the manometer
• This level difference is denoted as ‘h’ in fig.b
• The pressure difference between A and B can be calculated using ‘h’

• Once we know the pressure difference, we can calculate the rate of flow

• The following steps from (9) to (12) show the derivation of the equation for rate of flow:

9. The pressure difference can be calculated in 3 steps:

(i) In fig.b, let the dashed line be the datum. Then we get:

• Pressure at A = P_a + P_A + Pressure corresponding to 'x meters of water'

• Pressure at B = P_a + P_B + Pressure corresponding to '(x-h) meters of water' + Pressure corresponding to 'h meters of mercury'

(ii) Equating the two pressures, we get:

[P_a + P_A + Pressure corresponding to 'x meters of water']

= [P_a + P_B + Pressure corresponding to '(x-h) meters of water' + Pressure corresponding to 'h meters of mercury']

= [P_a + P_B + Pressure corresponding to 'x meters of water' - Pressure corresponding to h meters of water' + Pressure corresponding to 'h meters of mercury']

(iii) Thus we get:

[P_A] = [P_B - Pressure corresponding to 'h meters of water' + Pressure corresponding to 'h meters of mercury']

⇒ [P_A - P_B] = [Pressure corresponding to 'h meters of mercury' - Pressure corresponding to 'h meters of water']

⇒ [P_A - P_B] = [ρ_mgh - ρ_wgh] = [ρ_m - ρ_w]gh N m^-2

    ♦ Where ρ_m and ρ_w are the densities of mercury and water respectively   

10. Next we write v_A in terms of v_B

• From (8.ii), we have: $\mathbf\small{\rm{v_A=\frac{A_B V_B}{A_A}}}$

11. substituting (9) and (10) in (7), we get:

$\mathbf\small{\rm{[\rho_m - \rho_w] gh=\frac{1}{2} \rho_w \left (v_B^2-(\frac{A_B v_B}{A_A})^2 \right)}}$

⇒ $\mathbf\small{\rm{v_B^2=2\left (\frac{\rho_m - \rho_w}{\rho_w} \right )gh \left(\frac{A_A^2}{A_A^2-A_B^2} \right )}}$

⇒ $\mathbf\small{\rm{v_B=\left [\sqrt{\left (\frac{\rho_m}{\rho_w}-1 \right )\left (\frac{2gh}{A_A^2-A_B^2} \right ) }\;\right ]\times A_A}}$

12. Multiplying v_B by A_B, we get the rate of flow. So we get:

Eq.10.8: Rate of flow through the venturimeter =

$\mathbf\small{\rm{\left [\sqrt{\left (\frac{\rho_m}{\rho_w} -1 \right )\left (\frac{2gh}{A_A^2-A_B^2} \right ) }\;\right ]\times A_A A_B}}$

 

Solved example 10.20

A venturimeter has 30 cm diameter at the wide section and 15 cm diameter at the narrow section. If the level difference in the manometer is 20 cm mercury, what is the rate of flow of water through the venturimeter?
Solution:

• Area at the wide section A_A = 0.25 π d_A² = 0.07065 m^₂
• Area at the narrow section A_B = 0.25 π d_B² = 0.01766 m^₂
• Density of water ρ_w = 1000 kg m^_-3
• Density of mercury ρ_m = 13600 kg m^_-3

• h = 20 cm = 0.20 m

• g = 9.81 m s_^-2
• Substituting the known values in Eq.10.8, we get:

Rate of flow through the venturimeter = 0.12570 m_³ s_^-1

Solved example 10.21

Figures 10.38(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is in correct? Why?

Solution:

1. Bernoulli's equation tells us that:

$\mathbf\small{\rm{P+\frac{1}{2} \rho v^2+ \rho_w g h}}$ = A constant

2. In our present case, the venturimeter is horizontal. So we need not consider the third term. We can write:

$\mathbf\small{\rm{P+\frac{1}{2} \rho v^2}}$ = A constant 

3. So it is clear that,
    ♦ When velocity increases, pressure decreases
    ♦ When velocity decreases, pressure increases

Fig.10.38

4. From the equation of continuity, we know that:
At B, the velocity will be higher than the velocity at A
5. So the pressure at B will be lower than at A
So the liquid level at B will be lower than that at A
6. Thus we get:
The fig.a is incorrect

 

Solved example 10.22

The flow of blood in a large artery of an anesthetised dog is diverted through a venturimeter. The wider part of the meter has a cross-sectional area equal to that of the artery. A = 8 mm_². The narrower part has an area a = 4 mm_². The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery?

Solution:

• In this problem, we are given the pressure difference P directly
• So we will derive a new equation like the one we derived in Eq.10.8

1. Two points A and B are marked inside the venturimeter (see fig.10.38.b)

    ♦ A is at the wider section

    ♦ B is at the narrow section

2. Let us apply Bernoulli’s equation at the two points

• At A we get: $\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2+ \rho g h_1}}$

• At B we get: $\mathbf\small{\rm{P_B+\frac{1}{2} \rho v_B^2+ \rho g h_1}}$

3. Equating the two, we get: $\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2+ \rho g h_1=P_B+\frac{1}{2} \rho v_B^2+ \rho g h_1}}$

⇒ $\mathbf\small{\rm{P_A+\frac{1}{2} \rho v_A^2=P_B+\frac{1}{2} \rho v_B^2}}$

■ It is clear that:

    ♦ If velocity increase at any one of the points, the pressure at that point will decrease
    ♦ If velocity decrease at any one of the points, the pressure at that point will increase
7. Rearranging the equation in (6), we get:

$\mathbf\small{\rm{P_A-P_B=\frac{1}{2} \rho \left (v_B^2-v_A^2 \right)}}$
8. So we want the pressure difference (P_A - P_B) between the two points A and B

    ♦ But it is directly given in the question as 24 Pa

    ♦ We will denote it as ΔP

9. Next we write v_A in terms of v_B

• From equation of continuity, we have: $\mathbf\small{\rm{v_A=\frac{A_B V_B}{A_A}}}$

10. substituting (8) and (9) in (7), we get:

$\mathbf\small{\rm{\Delta P=\frac{1}{2} \rho \left (v_B^2-(\frac{A_B v_B}{A_A})^2 \right)}}$

⇒ $\mathbf\small{\rm{v_B^2=\frac{2\, \Delta P\,A_A^2}{\rho(A_A^2-A_B^2)}}}$

⇒ $\mathbf\small{\rm{v_B=\left [\sqrt{\frac{2\, \Delta P}{\rho(A_A^2-A_B^2)}}\;\right ]\times A_A}}$

11. Multiplying v_B by A_B, we get the rate of flow. So we get:

Eq.10.9: Rate of flow through the venturimeter =

$\mathbf\small{\rm{\left [\sqrt{\frac{2\, \Delta P}{\rho(A_A^2-A_B^2)}}\;\right ]\times A_A \times A_B}}$

12. Substituting the known values in Eq.10.9, we get:

Rate of flow of blood through the venturimeter =

$\mathbf\small{\rm{\left [\sqrt{\frac{2\, (24)}{(1060)(8^2-4^2)\times 10^{-12}}}\;\right ]\times 8 \times 4\times 10^{-12}}}$ = 9.83 ×10_^-7 m_³ s_^-1

13. So we have: 

• Rate of flow = A_Av_A= 9.83 ×10^_-7 m^₃ s^_-1

• So v_A = $\mathbf\small{\rm{\frac{9.83 \times 10^{-7}}{8 \times 10^{-6}}}}$ = 0.122 m s^_-1

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In the next section, we will see speed of efflux

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