The Most General Applications of Bernoulli’s Equation Quiz

PDF page 548: resistance is negligible, the speed is just what it would be if the water fell a distance from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is

PDF page 548: Both and equal atmospheric pressure ( is atmospheric pressure because it is the pressure at the top of the

PDF page 548: reservoir. must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving

PDF page 548: Solving this equation for , noting that the density cancels (because the fluid is incompressible), yields

PDF page 548: 12.31 We let ; the equation then becomes 12.32 where is the height dropped by the water. This is simply a kinematic equation for any object falling a distance with negligible resistance. In fluids, this last equation is called Torricelli’s theorem. Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.

PDF page 548: 12.31 We let ; the equation then becomes 12.32 where is the height dropped by the water. This is simply a kinematic equation for any object falling a distance with negligible resistance. In fluids, this last equation is called Torricelli’s theorem. Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.

PDF page 549: FIGURE 12.9 Pressure in the nozzle of this fire hose is less than at ground level for two reasons: the water has to go uphill to get to the

PDF page 549: FIGURE 12.9 Pressure in the nozzle of this fire hose is less than at ground level for two reasons: the water has to go uphill to get to the

PDF page 549: nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes, by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.

PDF page 549: nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes, by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.

PDF page 550: Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of . The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle? Strategy Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant. Solution Bernoulli’s equation states

PDF page 550: Discussion This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure is very close to atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.

PDF page 550: All three terms have units of energy per unit volume, as discussed in the previous section. Now, considering units, if we multiply energy per unit volume by flow rate (volume per unit time), we get units of power. That is, . This means that if we multiply Bernoulli’s equation by flow rate , we get power. In equation form, this is