$\text {If an irrotational vortex approximates the air flow in a tornado, the highest possible velocity,}$ $\text {assuming an incompressible flow with $p _ { \text {atm } } = 100 \mathrm { kPa }$, is nearest:}$
A) $400 \mathrm {~m} / \mathrm { s }$
B) $350 \mathrm {~m} / \mathrm { s }$
C) $300 \mathrm {~m} / \mathrm { s }$
D) $250 \mathrm {~m} / \mathrm { s }$

Question

Quizplus · Accepted Answer

$400 \mathrm {~m} / \mathrm { s }$
$\text {Assume the velocity at the extreme outer edges of the tornado is zero and the}$ $\text {pressure is $100 \mathrm { kPa }$. The lowest possible pressure is absolute zero, i.e., $p _ { 2 } = 0 \mathrm { kPa }$.}$$\text { Use Bernoulli's equation and find}$
$$\frac { V _ { 1 } ^ { 2 } } { 2 } + \frac { p _ { 1 } } { \rho } + g L _ { 1 } = \frac { V _ { 2 } ^ { 2 } } { 2 } + \frac { \partial _ { 2 } } { \rho } + g Z _ { 2 } . \quad \frac { 100000 } { 1.2 } = \frac { V _ { 2 } ^ { 2 } } { 2 } . \quad \therefore V _ { 2 } = \underline { 408 \mathrm {~m} / \mathrm { s } }$$
$\text {where standard conditions have been assumed so that $\rho = 1.2 \mathrm {~kg} / \mathrm { m } ^ { 3 }$. This is unrea-}$$\text {sonable since it exceeds the speed of sound, i.e., it would be a compressible flow, so}$$\text { the assumption of an incompressible flow would not be applicable. But, it does }$$\text {illustrate why the velocities are so high near the eye of a tornado.}$

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$\text {If an irrotational vortex approximates the air flow in a tornado, the highest possible velocity,}$