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College Physics for AP® Courses

12.3 The Most General Applications of Bernoulli’s Equation

College Physics for AP® Courses12.3 The Most General Applications of Bernoulli’s Equation

Learning Objectives

By the end of this section, you will be able to:

  • Calculate using Torricelli's theorem.
  • Calculate power in fluid flow.

The information presented in this section supports the following AP® learning objectives and science practices:

  • 5.B.10.2 The student is able to use Bernoulli's equation and/or the relationship between force and pressure to make calculations related to a moving fluid. (S.P. 2.2)
  • 5.B.10.3 The student is able to use Bernoulli's equation and the continuity equation to make calculations related to a moving fluid. (S.P. 2.2)

Torricelli's Theorem

Figure 12.8 shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance hh size 12{h} {} 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

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2.P1+12ρv12+ρgh1=P2+12ρv22+ρgh2. size 12{P rSub { size 8{1} } + { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{1} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{1} } =P rSub { size 8{2} } + { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{2} } } {}
12.49

Both P1P1 size 12{P rSub { size 8{1} } } {} and P2P2 size 12{P rSub { size 8{2} } } {} equal atmospheric pressure (P1P1 size 12{P rSub { size 8{1} } } {} is atmospheric pressure because it is the pressure at the top of the reservoir. P2P2 size 12{P rSub { size 8{2} } } {} 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

12ρv12+ρgh1=12ρv22+ρgh2.12ρv12+ρgh1=12ρv22+ρgh2. size 12{ { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{1} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{1} } = { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{2} } } {}
12.50

Solving this equation for v22v22 size 12{v rSub { size 8{2} } rSup { size 8{2} } } {}, noting that the density ρ ρ cancels (because the fluid is incompressible), yields

v22=v12+2g(h1−h2).v22=v12+2g(h1−h2). size 12{v rSub { size 8{2} } rSup { size 8{2} } =v rSub { size 8{1} } rSup { size 8{2} } +2g \( h rSub { size 8{1} } - h rSub { size 8{2} } \) } {}
12.51

We let h=h1−h2h=h1−h2 size 12{h=h rSub { size 8{1} } - h rSub { size 8{2} } } {}; the equation then becomes

v22=v12+2ghv22=v12+2gh size 12{v rSub { size 8{2} } rSup { size 8{2} } =v rSub { size 8{1} } rSup { size 8{2} } +2 ital "gh"} {}
12.52

where hh size 12{h} {} is the height dropped by the water. This is simply a kinematic equation for any object falling a distance hh size 12{h} {} 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.

Part a of the figure shows a photograph of a dam with water gushing from a large tube at the base of a dam. Part b shows the schematic diagram for the flow of water in a reservoir. The reservoir is shown in the form of a triangular section with a horizontal opening along the base little near to the base. The water is shown to flow through the horizontal opening near the base. The height which it falls is shown as h two. The pressure and velocity of water at this point are P two and v two. The height to which the water can fall if it falls from a height h above the opening is given by h 2. The pressure and velocity of water at this point are P one and v one.
Figure 12.8 (a) Water gushes from the base of the Studen Kladenetz dam in Bulgaria. (credit: Kiril Kapustin; http://www.ImagesFromBulgaria.com) (b) In the absence of significant resistance, water flows from the reservoir with the same speed it would have if it fell the distance hh size 12{h} {} without friction. This is an example of Torricelli's theorem.
Figure shows a fire engine that is stationed next to a tall building. A floor of the building ten meters above the ground has caught fire. The flames are shown coming out. A fire man has reached close to the fire caught area using a ladder and is spraying water on the fire using a hose attached to the fire engine.
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 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.

All preceding applications of Bernoulli's equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli's equation in which pressure, velocity, and height all change. (See Figure 12.9.)

Example 12.5

Calculating Pressure: A Fire Hose Nozzle

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 1.62×106N/m21.62×106N/m2 size 12{1 "." "62" times "10" rSup { size 8{6} } `"N/m" rSup { size 8{2} } } {}. 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

P 1 + 1 2 ρv 1 2 +ρ gh1 =P2+ 12 ρv22 +ρ gh2 , P 1 + 1 2 ρv 1 2 +ρ gh1 =P2+ 12 ρv22 +ρ gh2 , size 12{P rSub { size 8{1} } + { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{1} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{1} } =P rSub { size 8{2} } + { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{2} } } {}
12.53

where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds v1v1 size 12{v rSub { size 8{1} } } {} and v2v2 size 12{v rSub { size 8{2} } } {}. Since Q = A 1 v 1 Q = A 1 v 1 size 12{Q=A rSub { size 8{1} } v"" lSub { size 8{1} } } {} , we get

v1=QA1=40.0×10−3m3/sπ(3.20×10−2m)2=12.4344m/s.v1=QA1=40.0×10−3m3/sπ(3.20×10−2m)2=12.4344m/s. size 12{v rSub { size 8{1} } = { {Q} over {A rSub { size 8{1} } } } = { {"40" "." 0 times "10" rSup { size 8{ - 3} } " m" rSup { size 8{3} } "/s"} over {π \( 3 "." "20" times "10" rSup { size 8{ - 2} } " m" \) rSup { size 8{2} } } } ="12" "." 4" m/s"} {}
12.54

Similarly, we find

v2=56.588 m/s.v2=56.588 m/s. size 12{v rSub { size 8{2} } ="56" "." 6" m/s"} {}
12.55

(This rather large speed is helpful in reaching the fire.) Now, taking h1h1 size 12{h rSub { size 8{1} } } {} to be zero, we solve Bernoulli's equation for P2P2 size 12{P rSub { size 8{2} } } {}:

P2=P1+12ρ v 1 2 − v 2 2 −ρgh2.P2=P1+12ρ v 1 2 − v 2 2 −ρgh2. size 12{P rSub { size 8{2} } =P rSub { size 8{1} } + { {1} over {2} } ρ \( v rSub { size 8{1} rSup { size 8{2} } } - v rSub { size 8{2} rSup { size 8{2} } } \) - ρ ital "gh" rSub { size 8{2} } } {}
12.56

Substituting known values yields

P2=1.62×106N/m2+12(1000kg/m3)(12.434m/s)2−(56.588m/s)2− (1000kg/m3)(9.80m/s2)(10.0m)≈1800N/m2.P2=1.62×106N/m2+12(1000kg/m3)(12.434m/s)2−(56.588m/s)2− (1000kg/m3)(9.80m/s2)(10.0m)≈1800N/m2. size 12{P rSub { size 8{2} } =1 "." "62" times "10" rSup { size 8{6} } " N/m" rSup { size 8{2} } + { {1} over {2} } \( "1000"" kg/m" rSup { size 8{3} } \) left [ \( "12" "." 4" m/s" \) rSup { size 8{2} } - \( "56" "." 6" m/s" \) rSup { size 8{2} } right ] - \( "1000"" kg/m" rSup { size 8{3} } \) \( 9 "." 8" m/s" rSup { size 8{2} } \) \( "10" "." 0" m" \) =0} {}
12.57

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.

Making Connections: Squirt Toy

There is a diagram of a blue syringe with black text and arrows showing measurements. From left to right are the following labels: A red arrow pointing toward the plunger shows 2.2 N above the arrow. In the center of the syringe is 1.0 cm with an arrow pointing up and down to the edges of the barrel. On the far right is 1.5mm pointing to two black lines running from the top and bottom of the syringe opening.
Figure 12.10

A horizontally oriented squirt toy contains a 1.0-cm-diameter barrel for the water. A 2.2-N force on the plunger forces water down the barrel and into a 1.5-mm-diameter opening at the end of the squirt gun. In addition to the force pushing on the plunger, pressure from the atmosphere is also present at both ends of the gun, pushing the plunger in and also pushing the water back in to the narrow opening at the other end. Assuming that the water is moving very slowly in the barrel, with what speed does it emerge from the toy?

Solution

First, find the cross-sectional areas for each part of the toy. The wider part is

A 1 = π r 1 2 = π ( d 1 2 ) 2 = π (0.005) 2 =7.85× 10 −5   m 2 A 1 = π r 1 2 = π ( d 1 2 ) 2 = π (0.005) 2 =7.85× 10 −5   m 2
12.58

Next, we will find the area of the narrower part of the toy:

A 2 = π r 2 2 = π ( d 2 2 ) 2 = π (0.0015) 2 =7.07× 10 −6   m 2 A 2 = π r 2 2 = π ( d 2 2 ) 2 = π (0.0015) 2 =7.07× 10 −6   m 2
12.59

The pressure pushing on the barrel is equal to the sum of the pressure from the atmosphere (1.0 atm=101,300 N/m21.0 atm=101,300 N/m2) and the pressure created by the 2.2-N force.

P 1 =101,300+ ( Force A 1 ) P 1 =101,300+ ( Force A 1 )
12.60
P 1 =101,300+( 2.2 N 7.85x 10 −5   m 2 )=129,300 N/ m 2 P 1 =101,300+( 2.2 N 7.85x 10 −5   m 2 )=129,300 N/ m 2
12.61

The pressure pushing on the smaller end of the toy is simply the pressure from the atmosphere:

P 2 =101,300 N/ m 2 P 2 =101,300 N/ m 2
12.62

Since the gun is oriented horizontally (h1=h2h1=h2), we can ignore the potential energy term in Bernoulli's equation, so the equation becomes:

P 1 +  1 2 ρ v 1 2 =  P 2 +  1 2 ρ v 2 2 P 1 +  1 2 ρ v 1 2 =  P 2 +  1 2 ρ v 2 2
12.63

The problem states that the water is moving very slowly in the barrel. That means we can make the approximation that v1 ≈0v1 ≈0, which we will justify mathematically.

129,300+( 0.500 )( 1000 ) (0) 2 =101,300+(0.500)(1000) v 2 2 129,300+( 0.500 )( 1000 ) (0) 2 =101,300+(0.500)(1000) v 2 2
12.64
v 2 2 = (129,300−101,300) 500   v 2 2 = (129,300−101,300) 500  
12.65
v 2 2 = (129,300−101,300) 500   v 2 2 = (129,300−101,300) 500  
12.66
v 2 = 7.5 m/s v 2 = 7.5 m/s
12.67

How accurate is our assumption that the water velocity in the barrel is approximately zero? Check using the continuity equation:

v 1 =( A 2 A 1 ) v 2 =( 7.07× 10 −6 7.85× 10 −5 )( 7.5 )=0.17 m/s v 1 =( A 2 A 1 ) v 2 =( 7.07× 10 −6 7.85× 10 −5 )( 7.5 )=0.17 m/s
12.68

How does the kinetic energy per unit volume term for water in the barrel fit into Bernoulli's equation?

129,300+( 0.500 )( 1000 ) ( 0.17 ) 2 = 101,300+(0.500)(1000) (7.5) 2 129,300+( 0.500 )( 1000 ) ( 0.17 ) 2 = 101,300+(0.500)(1000) (7.5) 2
12.69
129,300+14=101,300+28,000 129,300+14=101,300+28,000
12.70

As you can see, the kinetic energy per unit volume term for water in the barrel is very small (14) compared to the other terms (which are all at least 1000 times larger). Another way to look at this is to consider the ratio of the two terms that represent kinetic energy per unit volume:

K 2 K 1 =  1 2 ρ v 2 2 1 2 ρ v 1 2 =  v 2 2 v 1 2 K 2 K 1 =  1 2 ρ v 2 2 1 2 ρ v 1 2 =  v 2 2 v 1 2
12.71

Remember that from the continuity equation

v 2 v 1 =  A 2 A 1 =  π ( d 2 2 ) 2 π ( d 1 2 ) 2 =  d 2 2 d 1 2 v 2 v 1 =  A 2 A 1 =  π ( d 2 2 ) 2 π ( d 1 2 ) 2 =  d 2 2 d 1 2
12.72

Thus, the ratio of the kinetic energy per unit volume terms depends on the fourth power of the ratio of the diameters:

K 2 K 1 =  ( v 2 v 1 ) 2 =  ( d 2 2 d 1 2 ) 2 =  ( d 2 d 1 ) 4   K 2 K 1 =  ( v 2 v 1 ) 2 =  ( d 2 2 d 1 2 ) 2 =  ( d 2 d 1 ) 4  
12.73

In this case, the diameter of the barrel (d2) is 6.7 times larger than the diameter of the opening at the end of the toy (d1), which makes the kinetic energy per unit volume term for water in the barrel (6.7)4≈2000(6.7)4≈2000 times smaller. We can usually neglect such small terms in addition or subtraction without a significant loss of accuracy.

Power in Fluid Flow

Power is the rate at which work is done or energy in any form is used or supplied. To see the relationship of power to fluid flow, consider Bernoulli's equation:

P+12ρv2+ρgh=constant.P+12ρv2+ρgh=constant. size 12{P+ { {1} over {2} } ρv rSup { size 8{2} } +ρ ital "gh"="constant"} {}
12.74

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, (E/V)(V/t)=E/t(E/V)(V/t)=E/t size 12{ \( E/V \) \( V/t \) =E/t} {}. This means that if we multiply Bernoulli's equation by flow rate QQ size 12{Q} {}, we get power. In equation form, this is

P+12ρv2+ρghQ=power.P+12ρv2+ρghQ=power. size 12{ left (P+ { {1} over {2} } ρv rSup { size 8{2} } +ρ ital "gh" right )Q="power"} {}
12.75

Each term has a clear physical meaning. For example, PQPQ size 12{ ital "PQ"} {} is the power supplied to a fluid, perhaps by a pump, to give it its pressure PP size 12{P} {}. Similarly, 12ρv2Q12ρv2Q size 12{ { { size 8{1} } over { size 8{2} } } ρv rSup { size 8{2} } Q} {} is the power supplied to a fluid to give it its kinetic energy. And ρghQρghQ size 12{ρ ital "ghQ"} {} is the power going to gravitational potential energy.

Making Connections: Power

Power is defined as the rate of energy transferred, or E/tE/t size 12{E/t} {}. Fluid flow involves several types of power. Each type of power is identified with a specific type of energy being expended or changed in form.

Example 12.6

Calculating Power in a Moving Fluid

Suppose the fire hose in the previous example is fed by a pump that receives water through a hose with a 6.40-cm diameter coming from a hydrant with a pressure of 0.700×106N/m20.700×106N/m2 size 12{0 "." "700" times "10" rSup { size 8{6} } `"N/m" rSup { size 8{2} } } {}. What power does the pump supply to the water?

Strategy

Here we must consider energy forms as well as how they relate to fluid flow. Since the input and output hoses have the same diameters and are at the same height, the pump does not change the speed of the water nor its height, and so the water's kinetic energy and gravitational potential energy are unchanged. That means the pump only supplies power to increase water pressure by 0.92×106N/m20.92×106N/m2 size 12{0 "." "92" times "10" rSup { size 8{6} } `"N/m" rSup { size 8{2} } } {} (from 0.700×106N/m20.700×106N/m2 size 12{0 "." "700" times "10" rSup { size 8{6} } `"N/m" rSup { size 8{2} } } {} to 1.62×106N/m21.62×106N/m2 size 12{1 "." "62" times "10" rSup { size 8{6} } `"N/m" rSup { size 8{2} } } {}).

Solution

As discussed above, the power associated with pressure is

power = PQ = 0.920×106N/m2 40.0×10−3m3/s. = 3.68×104W=36.8kW power = PQ = 0.920×106N/m2 40.0×10−3m3/s. = 3.68×104W=36.8kW .
12.76

Discussion

Such a substantial amount of power requires a large pump, such as is found on some fire trucks. (This kilowatt value converts to about 50 hp.) The pump in this example increases only the water's pressure. If a pump—such as the heart—directly increases velocity and height as well as pressure, we would have to calculate all three terms to find the power it supplies.

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