Question

1191 One day Daniel Bernoulli was sitting in a Starbucks sipping on a Venti Mocha Frappuccino. He pondered some of the many applications of his newly developed principle. Grab a Starbucks of your own, and help him to answer the following questions. Assume air weighs 0.081 lb/ft3 NOTES IMAGES DISCUSS UNITS STATS HELP Part Description Answer Chk History In a hurricane, roofs of structures are really lifted off, and not blown off, due to the decreased pressure as the wind blows over the roof Determine the required wind speed to lift a roof that weighs 39 psf Assume the roof is not tied down (include units with answer) 50.82 m 1737 pts, 108% 296 try penalty Hints: 1,0 A. # tries: 1 Show Details Format Check Determine the minimum required take-off speed of a 28000 pound UScairways jet with a wing area of 1380 ft2. Use a lift coefficient of K-0.12 (include units with answer) 107.68 m/s Format Check 1737 pts, 108% 2% try penalty Hints: 2.0 B. # tries: 1 Show Details Daniel's good friend Giovanni Venturi joined him for another round of fraps. They were discussing ways to determine the velocity of water inn a pipe. If the pressure drops 690 5.62 m kPa as water flowing in a 40 cm diameter pipe is forced to flow through a 10 cm diameter portion, what is the velocity of the water in the 40 cm section of pipe? include units with ans 17.97 pts.108% 2% try penalty Hints: 1,0 C. Format # tries: 1 Show Details

The answers that I got are all wrong.

Answer 1

a)

To get the wind speed we need to find the drop in pressure above and below the roof. If it gets so large that it goes above the weight of the roof the roof will go off.

Now,

pressure exerted by the roof = 39psf = 1.87kPa =1.87kN / m²

density of air =0.081 lb/ft³ =1.3kg/m³

Now, Pressure difference due to moving fluid

$p_{o}=\frac{1}{2}\rho v^{2}+ p$

p_o is pressure at rest

p is pressure of moving air

v is velocity

$\rho$ is density

Now,

$p_{o}- p=\frac{1}{2}\rho v^{2}$

Difference of Pressure of fluid at rest (which is under the roof) and Pressure of moving fluid ( which is above the roof) should be equal to the pressure exerted by the roof (i.e., due to its own weight)

So,

$p_{due\: to\: roof}=\frac{1}{2}\rho v^{2}$

$\Rightarrow 1870=\frac{1}{2}*1.3v^{2}$

$\Rightarrow v=\sqrt{\frac{1870*2}{1.3}}$

$\Rightarrow v=53.63ms^{-1}$, which is the required solution

b)

Lift force is given by

$L=\frac{1}{2} k \rho v^{2}A$

k is lift coefficient = 0.12

A is area of project wing = 1380 *2 ft² = 128.2 *2 m² (since there are two wings)

Now, mass of the aircraft = 28000 pounds = 12700.6 kg

Weight of the aircraft = 12700.6 * 9.8 N =124466 N

For the aircraft to fly weight of the aircraft should be countered by lift force.

$124466=\frac{1}{2} *0.12* 1.3 v^{2} *128.2*2$

$v=\sqrt{\frac{124466*2}{128.2*0.12* 1.3*2} }$

$v=78.89ms^{-1}$

c)

Pressure drops by 690kPa

We can use bernoulli's equation

$\Rightarrow p_{o}- p=\frac{1}{2}\rho \left ( v_{2}^{2} -v_{1}^{2}\right )$

by equation of continuity

A₁v₁=A₂v₂

_{$\Rightarrow p_{o}- p=\frac{1}{2}\rho \left ( \left (\frac{A_{1}v_{1}}{A_{2}} \right )^{2} -v_{1}^{2}\right )$}

_{$\Rightarrow p_{o}- p=\frac{1}{2}\rho \left ( \left (\frac{A_{1}}{A_{2}} \right )^{2} -1 \right )v_{1}^{2}$}

_{$\Rightarrow p_{o}- p=\frac{1}{2}\rho \left ( \left (\frac{d_{1}}{d_{2}} \right )^{2} -1 \right )v_{1}^{2}$}

density of water = 1000kg/m³

$\Rightarrow 690*10^{3}=\frac{1}{2} *1000 \left ( \left (\frac{40}{10} \right )^{2} -1 \right )v_{1}^{2}$

$\Rightarrow 690*2= \left ( 16 -1 \right )v_{1}^{2}$

$\Rightarrow v_{1}=\sqrt{\frac{690*2}{15}}$

$\Rightarrow v_{1}=9.6ms^{-1}$

which is the required soltuion

If any doubt feel free to comment