For fluids in active flow, we need something better than the balance of forces. Fortunately, we can get a long way with one simple assumption, the conservation of matter. When fluid moves from one position to the next, it must do so in such a way that no fluid matter is destroyed.

For instance, if we inject fluid into the mouth of a tube at a rate of J in = 1 L s − 1 J_\text{in}=\SI{1}{\liter\per\second} Jin​=1 Ls−1 , we should find that 1 L \SI{1}{\liter} 1 L of the fluid comes out the other side every second, i.e. that J in = J out J_\text{in}=J_\text{out} Jin​=Jout​.

For a toy model, consider a the device below that consists of one level section of tube Γ A \Gamma_\text{A} ΓA​ with cross section of area A A A_A AA​, connected by a linker section to another level tube Γ B \Gamma_\text{B} ΓB​ with cross section A B A_B AB​.

We want to find a relationship that connects the velocity and pressure of the fluid in either section of tube.

To start, let us take our system to be the fluid that's between the discs ∂ A \partial_\text{A} ∂A​ and ∂ B \partial_\text{B} ∂B​ at time zero, which we call Σ \Sigma Σ. In order for Σ \Sigma Σ to flow to the right, there must be a net force to push it along. If the fluid pressure in Γ A \Gamma_\text{A} ΓA​, P A P_A PA​ is greater than the fluid pressure in Γ B \Gamma_{B} ΓB​, P B P_B PB​, then the fluid to the left of Σ \Sigma Σ, Σ L \Sigma_L ΣL​ will push with greater force than the fluid to the right, Σ R \Sigma_R ΣR​, and hence Σ \Sigma Σ will flow. These two forces will be P A A A P_A A_A PA​AA​ and P B A B P_B A_B PB​AB​ respectively.

Applying the work energy principle, we have W = Δ T + Δ U W = \Delta T + \Delta U W=ΔT+ΔU. For our parcel of fluid, work is performed by the two pressures in moving Σ \Sigma Σ along the tube. Consider the flow undertaken in some span of time Δ t \Delta t Δt. In Γ A \Gamma_A ΓA​, Σ \Sigma Σ will move the distance v A Δ t v_A\Delta t vA​Δt, and in Γ B \Gamma_B ΓB​, Σ \Sigma Σ will move the distance v B Δ t v_B\Delta t vB​Δt. Hence, the net work W W W on the fluid is given by the work done on our fluid by the fluid to the left Σ L \Sigma_L ΣL​: F d x = P A A A v A Δ t F dx = P_A A_A v_A \Delta t Fdx=PA​AA​vA​Δt, minus the work done by our fluid to the fluid on the right Σ R \Sigma_R ΣR​: P B A B v B Δ t P_B A_B v_B \Delta t PB​AB​vB​Δt.

W = P A A A v A Δ t − P B A B v B Δ t W = P_A A_A v_A \Delta t - P_B A_B v_B \Delta t W=PA​AA​vA​Δt−PB​AB​vB​Δt

However, we have the conservation condition J in = J out J_\text{in} = J_\text{out} Jin​=Jout​ which applies for the discs ∂ A \partial_A ∂A​ and ∂ B \partial_B ∂B​. Hence, it must be true that A A v A Δ t = A B v A Δ t A_A v_A \Delta t = A_B v_A \Delta t AA​vA​Δt=AB​vA​Δt.

In other words, the volume of fluid Δ m \Delta m Δm that flows through ∂ A \partial_A ∂A​ is equal to the mass of fluid that flows out from ∂ B \partial_B ∂B​. Hence,

W = Δ V ( P A − P B ) W = \Delta V \left(P_A-P_B\right) W=ΔV(PA​−PB​)

Now, the work done on Σ \Sigma Σ is equal to the change in energy of the fluid. The kinetic energy of our fluid Σ \Sigma Σ should be the same as before with the exception that the quantity Δ m \Delta m Δm of liquid which initially moved with velocity v A v_A vA​ in Γ A \Gamma_A ΓA​ is now traveling with velocity v B v_B vB​ in Γ B \Gamma_B ΓB​, and therefore Δ T = 1 2 ρ Δ V v B 2 − 1 2 ρ Δ V v A 2 \Delta T = \frac12 \rho\Delta V v_B^2 - \frac12 \rho\Delta V v_A^2 ΔT=21​ρΔVvB2​−21​ρΔVvA2​. So, we have Δ V ( P A − P B ) = 1 2 ρ Δ V ( v B 2 − v A 2 ) \Delta V \left(P_A - P_B\right) = \frac12 \rho \Delta V \left( v_B^2 - v_A^2 \right) ΔV(PA​−PB​)=21​ρΔV(vB2​−vA2​), or P A + 1 2 ρ v A 2 = P B + 1 2 ρ v B 2 P_A + \frac12\rho v_A^2 = P_B + \frac12\rho v_B^2 PA​+21​ρvA2​=PB​+21​ρvB2​, which is Bernoulli's relation for fluid flow in an arbitrary tube of level height.

In this derivation, the tubes were kept at equal level for simplicity's sake. It is trivial to recalculate our relation for the case when the two tube sections are of differing heights in a gravitational field, as occurs for the plumbing system in an apartment building. In this case, the work-energy principle is given by W = Δ T + Δ U W = \Delta T + \Delta U W=ΔT+ΔU and we have thus have the full Bernoulli relation

P A + 1 2 ρ v A 2 + ρ g h A = P B + 1 2 ρ v B 2 + ρ g h B P_A + \frac12\rho v_A^2 + \rho gh_A = P_B + \frac12\rho v_B^2 + \rho gh_B PA​+21​ρvA2​+ρghA​=PB​+21​ρvB2​+ρghB​

Note that our calculation did not depend in any way on the particular setup that we used (the two tubes and linker section). The same calculation applies to a tube of arbitrary shape which carries out an arbitrary trajectory through a gravitational field. Thus, the relation can be used to connect any two cross sections of a fluid's flow.

Bernoulli's relation has a number of applications, particularly in the use of hydraulics.

Fork-lift operator Suppose a sleepy forklift operator parked clumsily on a hill without applying the emergency brake. Suddenly, the forklift slips and starts rolling backward down the hill. In order to stop before slamming into a schoolbus full of teddy bears, it will be necessary to apply a normal force of 150 N to the disk brake. Suppose the force is delivered by a piston of surface area 2 m 2 ^2 2 which is connected by a hydraulic system to a foot lever of surface area 0.02 m 2 ^2 2. With what force must the operator step on the lever in order to apply the required force? Assume there is no significant height difference from the brake to the lever. In this problem, the fluid is transmitting the force from the operators foot to the brake pad. Brake pads hover very close to the disc they press, so we can assume that the fluid velocity is negligible. Indeed, once the foot is clamped down on the lever, the arrangement will be static. In this case, Bernoulli's relation simplifies to P foot = P brake P_\text{foot} = P_\text{brake} Pfoot​=Pbrake​ As pressure is given by the force per unit area, we have F foot A foot = F brake A brake \frac{F_\text{foot}}{A_\text{foot}} = \frac{F_\text{brake}}{A_\text{brake}} Afoot​Ffoot​​=Abrake​Fbrake​​ yielding F foot = F brake × A foot A brake 1.5 N F_\text{foot} = F_\text{brake} \times \frac{A_\text{foot}}{A_\text{brake}} 1.5 \text{N} Ffoot​=Fbrake​×Abrake​Afoot​​1.5N The asymmetry in the piston areas gives the operator a significant mechanical advantage over the braking system, making it very easy for the operator to apply the huge force required.

Submit your answer I drink from a glass of water with a vertical straw. What's the longest straw I can use and still drink water if the ambient pressure is 1 atm 1~\mbox{atm} 1 atm? Give your answer in meters.

Details and Assumptions: 1 atm = 101 , 325 Pa 1~\mbox{atm}=101,325~\mbox{Pa} 1 atm = 1 0 1 , 3 2 5 Pa .

. The acceleration of gravity is − 9.8 m/s 2 -9.8~\mbox{m/s}^2 − 9 . 8 m/s 2 .

. The density of water is 1 g/cm 3 1~\mbox{g/cm}^3 1 g/cm 3 .