**What is the terminal velocity of an elevator in an elevator shaft?**

Let’s consider what’s going on physically. What happens if the elevator were flush against the walls of the elevator shaft so that no air from underneath could escape? When you first sever the cables, the elevator begins to fall. As it falls, it compresses the air underneath and this air starts building up pressure. The pressure eventually grows large enough to counteract the gravitational force pulling the elevator down. At this point, the elevator stops its descent and rests on a cushion of air. This will happen when

P = m g / A,

where P is the difference between the air pressure above and below the elevator, m = 2000 kg is the mass of the (maximum occupancy) elevator, g = 9.8 m/s

^{2}is the acceleration of gravity, and A = 4.0 m^{2}is the cross-sectional area of the elevator shaft.Now imagine there’s a small area A’ = 0.5 m

^{2}on the sides of the elevator shaft that the air underneath can leak through. This area is pretty small so only a small amount of air can pass through it at any given time. The elevator still falls until the pressure counteracts the gravitational force, but this time the elevator is falling with a constant velocity. The rate at which some volume of air passes through the leak must be equal to the rate at which the elevator’s volume sinks down. Written mathematically,A · v

_{elevator}= A’ · v_{air}where v

_{elevator}is the terminal velocity of the elevator and v_{air}is the speed at which air passes through the leak.Air will flow from the bottom to the top because of the pressure difference P. Using Bernoulli’s equation, we can estimate the air speed in the leak,

P = ρ · v

_{air}^{2}/ 2,or, equivalently,

v

_{air}= ( 2 · P / ρ )^{1/2},where ρ = 1.2 kg/m

^{3}is the density of air. Using the equations above, we can solve algebraically for the terminal velocity of the elevator,v

_{elevator}= [ 2 · m · g / (A · ρ) ]^{1/2}(A’ / A)= {2 · (2000 kg) · (9.8 m/s

^{2}) / [(4.0 m^{2}) · ( 1.2 kg/m^{3})]}^{1/2}(0.5 m^{2}) / (4.0 m^{2})= 11 m/s.

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