Section 7.5
Section 7.5 Power Required to Remain Afloat
7.4
Archimedes’ Principle Archimedes’ principle states that a body partially or wholly submerged in a fluid is buoyed upward by a force that is equal in magnitude to the weight of the displaced fluid. The derivation of this principle is found in basic physics texts. We will now use Archimedes’ principle to calculate the power required to remain afloat in water and to study the buoyancy of fish.
7.5
Power Required to Remain Afloat
Whether an animal sinks or floats in water depends on its density. If its density is greater than that of water, the animal must perform work in order not to sink. We will calculate the power P required for an animal of volume V and density ρ to float with a fraction f of its volume submerged. This problem is similar to the hovering flight we discussed in Chapter 6, but our approach to the problem will be different.
Because a fraction f of the animal is submerged, the animal is buoyed up by a force FB given by FB gf Vρw
(7.9)
where ρw is the density of water. The force FB is simply the weight of the displaced water.
The net downward force FB on the animal is the difference between its weight gVρ and the buoyant force; that is, FD gVρ − gVfρw gV (ρ − fρw)
(7.10)
To keep itself floating, the animal must produce an upward force equal to FD. This force can be produced by pushing the limbs downward against the water. This motion accelerates the water downward and results in the upward reaction force that supports the animal.
If the area of the moving limbs is A and the final velocity of the accelerated water is v, the mass of water accelerated per unit time in the treading motion is given by (see Exercise 7-1) m Avρw
(7.11)
Because the water is initially stationary, the amount of momentum imparted to the water each second is mv. (Remember that here m is the mass accelerated per second.)
Momentum given to the water per second mv Chapter 7 Fluids
This is the rate of change of momentum of the water. The force producing this change in the momentum is applied to the water by the moving limbs.
The upward reaction force FR, which supports the weight of the swimmer, is equal in magnitude to FD and is given by FR FD gV (ρ − fρw) mv
(7.12)
Substituting Eq. 7.11 for m, we obtain ρwAv2 gV (ρ − fρw) or
gV (ρ − fρw)
v
(7.13)
Aρw
The work done by the treading limbs goes into the kinetic energy of the accelerated water. The kinetic energy given to the water each second is half the product of the mass accelerated each second and the squared final velocity of the water. This kinetic energy imparted to the water each second is the power generated by the limbs; that is, KE/sec Power generated by the limbs,P 1 mv2
2
Substituting equations for m and v, we obtain (see Exercise 7-1)
3
W 1 − fρw
ρ
P 1
(7.14)
2
Aρw Here W is the weight of the animal (W gVρ).
It is shown in Exercise 7-2 that a 50-kg woman expends about 7.8 W to keep her nose above water. Note that, in our calculation, we have neglected the kinetic energy of the moving limbs. In Eq. 7.14 it is assumed that the density of the animal is greater than the density of water. The reverse case is examined in Exercise 7-3.
7.6
Buoyancy of Fish
The bodies of some fish contain porous bones or air-filled swim bladders that decrease their average density and allow them to float in water without an expenditure of energy. The body of the cuttlefish, for example, contains a
7.4
Archimedes’ Principle Archimedes’ principle states that a body partially or wholly submerged in a fluid is buoyed upward by a force that is equal in magnitude to the weight of the displaced fluid. The derivation of this principle is found in basic physics texts. We will now use Archimedes’ principle to calculate the power required to remain afloat in water and to study the buoyancy of fish.
7.5
Power Required to Remain Afloat
Whether an animal sinks or floats in water depends on its density. If its density is greater than that of water, the animal must perform work in order not to sink. We will calculate the power P required for an animal of volume V and density ρ to float with a fraction f of its volume submerged. This problem is similar to the hovering flight we discussed in Chapter 6, but our approach to the problem will be different.
Because a fraction f of the animal is submerged, the animal is buoyed up by a force FB given by FB gf Vρw
(7.9)
where ρw is the density of water. The force FB is simply the weight of the displaced water.
The net downward force FB on the animal is the difference between its weight gVρ and the buoyant force; that is, FD gVρ − gVfρw gV (ρ − fρw)
(7.10)
To keep itself floating, the animal must produce an upward force equal to FD. This force can be produced by pushing the limbs downward against the water. This motion accelerates the water downward and results in the upward reaction force that supports the animal.
If the area of the moving limbs is A and the final velocity of the accelerated water is v, the mass of water accelerated per unit time in the treading motion is given by (see Exercise 7-1) m Avρw
(7.11)
Because the water is initially stationary, the amount of momentum imparted to the water each second is mv. (Remember that here m is the mass accelerated per second.)
Momentum given to the water per second mv Chapter 7 Fluids
This is the rate of change of momentum of the water. The force producing this change in the momentum is applied to the water by the moving limbs.
The upward reaction force FR, which supports the weight of the swimmer, is equal in magnitude to FD and is given by FR FD gV (ρ − fρw) mv
(7.12)
Substituting Eq. 7.11 for m, we obtain ρwAv2 gV (ρ − fρw) or
gV (ρ − fρw)
v
(7.13)
Aρw
The work done by the treading limbs goes into the kinetic energy of the accelerated water. The kinetic energy given to the water each second is half the product of the mass accelerated each second and the squared final velocity of the water. This kinetic energy imparted to the water each second is the power generated by the limbs; that is, KE/sec Power generated by the limbs,P 1 mv2
2
Substituting equations for m and v, we obtain (see Exercise 7-1)
3
W 1 − fρw
ρ
P 1
(7.14)
2
Aρw Here W is the weight of the animal (W gVρ).
It is shown in Exercise 7-2 that a 50-kg woman expends about 7.8 W to keep her nose above water. Note that, in our calculation, we have neglected the kinetic energy of the moving limbs. In Eq. 7.14 it is assumed that the density of the animal is greater than the density of water. The reverse case is examined in Exercise 7-3.
7.6
Buoyancy of Fish
The bodies of some fish contain porous bones or air-filled swim bladders that decrease their average density and allow them to float in water without an expenditure of energy. The body of the cuttlefish, for example, contains a