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
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| 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 ρw 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:
♦ 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 (PA - PB) 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: AAvA = ABvB
♦ 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’
8. So we want the pressure difference (PA - PB) 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: AAvA = ABvB
♦ 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 = Pa + PA + Pressure corresponding to 'x meters of water'
• Pressure at B = Pa + PB + Pressure corresponding to '(x-h) meters of water' + Pressure corresponding to 'h meters of mercury'
(ii) Equating the two pressures, we get:
[Pa + PA + Pressure corresponding to 'x meters of water']
= [Pa + PB + Pressure corresponding to '(x-h) meters of water' + Pressure corresponding to 'h meters of mercury']
= [Pa + PB + 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:
[PA] = [PB - Pressure corresponding to 'h meters of water' + Pressure corresponding to 'h meters of mercury']
⇒ [PA - PB] = [Pressure corresponding to 'h meters of mercury' - Pressure corresponding to 'h meters of water']
⇒ [PA - PB] = [ρmgh - ρwgh] = [ρm - ρw]gh N m-2
♦ Where ρm and ρw are the densities of mercury and water respectively
10. Next we write vA in terms of vB
• 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 vB by AB, 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:
Solution:
• Area at the wide section AA = 0.25 π dA2 = 0.07065 m2
• Area at the narrow section AB = 0.25 π dB2 = 0.01766 m2
• Density of water ρw = 1000 kg m-3
• Density of mercury ρm = 13600 kg m-3
• Area at the narrow section AB = 0.25 π dB2 = 0.01766 m2
• Density of water ρw = 1000 kg m-3
• Density of mercury ρm = 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:
• Substituting the known values in Eq.10.8, we get:
Rate of flow through the venturimeter = 0.12570 m3 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
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
♦ When velocity increases, pressure decreases
♦ When velocity decreases, pressure increases
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| Fig.10.38 |
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 mm2. The narrower part has an area a = 4 mm2. 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
• 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:
♦ 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 (PA - PB) between the two points A and B
8. So we want the pressure difference (PA - PB) 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 vA in terms of vB
• 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 vB by AB, 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 m3 s-1
13. So we have:
• Rate of flow = AAvA= 9.83 ×10-7 m3 s-1
• So vA = $\mathbf\small{\rm{\frac{9.83 \times 10^{-7}}{8 \times 10^{-6}}}}$ = 0.122 m s-1
In the next section, we will see speed of efflux
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