Bernoulli's theorem and its applications

In 1738, the Swiss scientist Daniel Bernoulli worked out a relationship for the flow of a fluid through a pipe of varying cross section, based on the streamlined flow of a liquid and the law of conservation of energy. BERNOULLI'S THEOREM states that, for an incompressible, non-viscous fluid in streamlined flow, the SUM of pressure energy, kinetic energy, and potential energy PER UNIT MASS remains constant everywhere along the flow: $\dfrac{P}{\rho}+\tfrac12 v^2+gh=\text{constant}$. PROOF. Consider a liquid flowing through a pipe from A to B, where the pipe has cross-sectional area $a_A$, fluid velocity $v_A$, pressure $P_A$, and height $h_A$ at A, and correspondingly $a_B$, $v_B$, $P_B$, $h_B$ at B. Let a volume V of liquid enter at A in a time t (equal to the volume leaving at B in the same time, by the equation of continuity). The pressure energy of this liquid at A is $E_{PA}=P_AV=\dfrac{P_A}{\rho}m$ (writing V in terms of the mass m and density $\rho$); its potential energy at A is $\text{PE}_A=mgh_A$; and its kinetic energy at A is $\text{KE}_A=\tfrac12 mv_A^2$. So the TOTAL energy of this liquid at A is $E_A=\dfrac{P_A}{\rho}m+\tfrac12 mv_A^2+mgh_A$. By the identical reasoning, the total energy of the same mass of liquid once it has flowed to B is $E_B=\dfrac{P_B}{\rho}m+\tfrac12 mv_B^2+mgh_B$. Since energy is conserved (and no energy is lost, for an ideal non-viscous fluid), $E_A=E_B$; dividing through by m gives $\dfrac{P_A}{\rho}+\tfrac12 v_A^2+gh_A=\dfrac{P_B}{\rho}+\tfrac12 v_B^2+gh_B$, i.e., $\dfrac{P}{\rho}+\tfrac12 v^2+gh=\text{constant}$ -- Bernoulli's equation, exactly as stated. This equation is a direct consequence of energy conservation, so it is strictly valid only when there is NO energy loss to friction; in practice, real (viscous) fluid flow does lose some energy as heat, because layers flowing at different velocities exert frictional (viscous) forces on one another, so Bernoulli's relation strictly applies only to non-viscous (zero-viscosity) fluids. For a HORIZONTAL pipe, where $h=0$ everywhere, the equation simplifies to $\dfrac{P}{\rho}+\tfrac12 v^2=\text{constant}$. APPLICATIONS. (a) BLOWING OFF OF ROOFS in a storm: old-style sloped hut and house roofs are explained by Bernoulli's principle -- high wind blowing fast over the outer roof surface creates a LOW pressure $P_1$ there, while the comparatively still air trapped underneath the roof stays at the higher pressure $P_2$; the resulting pressure difference $(P_2-P_1)$ creates an upward thrust strong enough to blow the roof off during a cyclone or storm, without damaging the rest of the house. (b) AEROFOIL LIFT: an aircraft wing (aerofoil) is shaped with a more curved upper surface than lower surface, so air moving past the wing travels FASTER over the top than along the bottom; by Bernoulli's principle this makes the pressure below the wing greater than the pressure above it, and this pressure difference produces the upward dynamic LIFT that keeps the aircraft airborne. (c) BUNSEN BURNER: gas leaves the burner's nozzle at high velocity, lowering the pressure in the stem around the jet (by Bernoulli's principle); this pressure drop draws outside air in through a side air vent, and the resulting air-gas mixture rising up the stem burns with the characteristic blue flame. (d) VENTURIMETER: this device measures the flow speed of an incompressible fluid moving through a pipe, working on Bernoulli's principle. It has two wider tubes A and A' of cross-sectional area A connected by a narrower tube B of cross-sectional area a, with a U-tube manometer (containing a liquid of density $\rho_m$) connected between the wide and narrow sections. As the fluid (density $\rho$) speeds up from $v_1$ in the wide tube to $v_2=\dfrac{A}{a}v_1$ in the narrow tube (by the equation of continuity), Bernoulli's equation, $P_1+\tfrac12\rho v_1^2=P_2+\tfrac12\rho v_2^2$, shows the pressure must DROP at the narrow section; this pressure drop is read off as a height difference h between the two arms of the manometer. Working through the algebra gives the pressure difference $\Delta P=P_1-P_2=\dfrac{\rho v_1^2}{2}\left(\dfrac{A^2}{a^2} …