Question 3.1: A-D: Physical Insight Gained from Solving Bernoulli’s Equati......

(A) Well, they need a lift force larger than the plane+ passenger and baggage weights.

Sketch:

Rational:

• Bernoulli states \rm{\frac{p}{\gamma } +\frac{v^2}{2g}+z=\cancel c }; here, with \rm{z_\infty =z_{lower}\approx z_{upper}}:

p_{e}+\frac{\rho}2\mathrm{v}_{e}^{~2}=p_{u}+\frac{\rho}2\mathrm{v}_{u}^{~2}=p_{\infty }+\frac{\rho}2\mathrm{v}_{\infty }^{2}                     (E.3.1.1)

• Observation

Above the wing (or airfoil) streamlines cluster and the air velocity is very high so that with z ≈ constant, \rm{p_{upper}} is low according to Bernoulli. The opposite occurs below the wing, i.e., \rm{v_{lower}\downarrow p_{lower}\uparrow }. Hence, for an effective airfoil surface:

\mathrm{F}_{\mathrm{lift}}=\left(\mathrm{p}_{1}-\mathrm{p}_{\mathrm{u}}\right)\mathrm{A}_{\mathrm{surface}}\gt \mathrm{W}                   (E.3.1.2)

(B) Perhaps because the pressure field above the pitched roof is much lower than the atmospheric pressure in the attic.

Sketch:

Rational:

• Similar to (A), the pressure field on the roof due to the locally very high wind velocity is very low so that the pressure force inside the house, \rm{p_{atm}*A_{ceiling}}, blows the roof off the house.

(C) In general,

\left.{\frac{\mathrm{v}_{1}^{2}}{2\mathrm{g}}}+{\frac{\mathrm{p}_{1}}{\mathbf{\rho }\mathrm{g}}}+\mathrm{z}_{1}={\frac{\mathrm{v}_{2}^{2}}{2\mathrm{g}}}+{\frac{\mathrm{p}_{2}}{\mathrm{\rho }\mathrm{g}}}+\mathrm{z}_{2}\right.                  (E.3.1.3a)

With the given information,

\mathrm{{p}}_{2}={\frac{\mathrm{\rho }}{2}}\,\mathrm{v}_{1}^{2}                   (E.3.1.3b)

and hence

\mathrm{F_{R}=\int p d A\approx p_{2}~A_{window}}                    (E.3.1.4)

For example, for a 65 mph wind, \rm{F_R\approx 1\,kN} for a typical window. Similarly, the form drag, part of the total drag on submerged bodies

\rm{F}_{\mathrm{Drag}}^{\mathrm{total}}={F}_{\mathrm{net\; pressure}}^{\mathrm{form}}+{F}_{\mathrm{wall \;shear}}^{\mathrm{friction}}                       (E.3.1.5)

can be estimated using Bernoulli’s equation, i.e.,

\rm F_{\mathrm{Pr\,essure}}=\Delta p A_{\mathrm{projected}}                        (E.3.1.6a)

where                        \Delta{\rm p}={\rm p}_{stagnation}-{\rm p}_{\mathrm{wake}}                     (E.3.1.6b)

(D) • (Bernoulli)    \rm\frac{p_0}{\rho} +\frac{v_0^2}{2} +gz_0=\frac{p_1}{\rho} +\frac{v_1^2}{2} +gz_1 can be reduced to:

\rm\frac{v_0^2}{2} =\frac{v_1^2}{2} +g\,h(t)\Rightarrow v_1=\sqrt{v_0^2-2gh}                      (E.3.1.7a, b)

• (Continuity)        \rm v_0A_0=v_1A_1\Rightarrow A_1=A_0\frac{v_0}{v_1}

• (Momentum RTT)

can be reduced to:

\hspace{140 pt} \approx 0\\\rm-Mg-Ma=v_{rel}[-\rho(v_{rel}A_1)]+{\Large\nearrow }v_2\dot m_2    (E.3.1.8b)

with \rm v_{rel}=v_1-\frac{dh}{dt} and \rm a=\frac{d^2h}{dt^2} , we have:

\rm M\left\lgroup g+\frac{d^2h}{dt^2} \right\rgroup =\rho\left\lgroup v_1-\frac{dh}{dt} \right\rgroup ^2A_1                              (E.3.1.9a)

Substituting \rm v_1 and \rm A_1 yields:

\rm\ddot h-\frac{\rho\,v_0A_0}{M\sqrt{v_0^2-2gh} } \left(\sqrt{v_0^2-2gh}-\dot h \right) ^2+g=0                                (E.3.1.9b)

Now, at equilibrium height, \rm h=h_0,\,\dot h=\ddot h=0. Thus,

\rm-\rho\sqrt{v_0^2-2gh_0} A_0v_0+Mg=0                                  (E.3.1.10a)

or

h_0=\frac{\nu _0^2}{2{\rm g}} \left[1-\left\lgroup\frac{{\rm g}M}{\rho\,\nu _0^2A_0} \right\rgroup^2 \right] :=\frac{\nu _0^2}{2{\rm g}} \left[1-\left\lgroup\frac{M{\rm g}}{\dot m\nu _0} \right\rgroup^2 \right]                         (E.3.1.10b, c)

Graph:

Comments:
(i) Although frictionless flow was assumed, the ODE for h(t) is nonlinear which implies oscillations as well as a decrease in amplitude.
(ii) This problem solution couples Bernoulli’s equation with mass and momentum conservation and shows the interactive nature of fluid mechanics (see also Chap. 8).

(iii)The Bernoulli equation is even more powerful when extended to include frictional and form losses; for example, in pipe flow with entrance effects, changes in cross sectional area, valves, etc (see Eq. (2.31)). This is further discussed in Chap. 4.

{\frac{\mathrm{v}_{1}^{2}}{2\mathrm{g}}}+{\frac{\mathrm{p}_{1}}{\gamma}}+\mathrm{z}_{1}={\frac{\mathrm{v}_{2}^{2}}{2\mathrm{g}}}+{\frac{\mathrm{p}_{2}}{\gamma}}+\mathrm{z}_{2}+\mathrm{h}_{\mathrm{f}}                                 (2.31)