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
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| 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 vA to be equal to zero
• So we can assume vA 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:
• 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 $\mathbf\small{\rm{\sqrt{2gh}}}$
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:
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: $\mathbf\small{\rm{P_a+P_B+\frac{1}{2} \rho v_B^2+ \rho g y}}$
• 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 vA to be equal to zero
• So we can assume vA 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: $\mathbf\small{\rm{v_B=\sqrt{2gh+\frac{2(P-P_a)}{\rho}}}}$• Thus we get:
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 PA(abs) and PB(abs) are the absolute pressures at A and B
2. We can split up the pressures
♦ Where PA(abs) and PB(abs) are the absolute pressures at A and B
2. We can split up the pressures
PA(abs) at A can be split up as: (Pa + PA)
PB(abs) at B can be split up as: (Pa + PB)
PB(abs) at B can be split up as: (Pa + PB)
♦ Where Pa is the atmospheric pressure, PA and PB are the gauge pressures at A and B
3. In this situation, the Pa 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
• 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 ?
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:
♦ 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 PA(abs) and PB(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
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:
4. So we can write:
♦ PA(abs) = Pa + 2000
♦ PB(abs) = Pa
5. Since blood flows slowly, we can put vA =vB = 0
6. Thus the equation written in (2) becomes:
5. Since blood flows slowly, we can put vA =vB = 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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