contestada

The largest and the smallest balls used in the experiment are with diameter 9.52 mm, and 2.38 mm respectively. For a glycerin with viscosity 1.0 Pa.s, what is the time necessary for each ball to reach a velocity 95% of the terminal velocity? Density of the ball material is 1.42 g/cm^3. Round the result to three decimal places.

Respuesta :

ofentsemaphalla

Answer:

The answer is = 0.0080

Explanation:

From the stokes law,

           v = g * D^2 *(d p - d m) / (18 V)

Here, v is the terminal Velocity, D is the diameter Of a Particle, V is the Viscosity of Medium, dp is the density Of Particle, dm is the density Of Medium.

The density of ball is 1.42 gm/cc

           v = 9.81 x 9.52 x 10^-3 ^2 (1420 - 1300) / 18 x 1.0

              =5.92x10^-4

          v(t) =0 .99 x V trm

                =0 .99 x  0.0059

                  = 0.00059 m/s

                v(t)/Vterm =1-e^(-t/r)

                         0.99 = 1 - e ^ (-t / 9.52 x 10^-3)

e ^ (-t / 9.52 x 10^-3) = 0.01

Take natural log

         -t / 9.52 x 10^-3 = -4.6

                                t = 0.0438 s

For the smaller ball

v = 9.81 x 2.38 x 10^-3 ^2 (1420 - 1300) / 18 x 1.0

             = 0.000037 m/s

                 v(t)/Vterm=1-e^(-t/r)

                         0.99 = 1 - e ^ (-t / 2.38 x 10^-3)

e ^ (-t / 2.38 x 10^-3) = 0.01

Take natural log

-t / 2.38 x 10^-3 = -4.6

                       t = 0.0109 s

Reynolds number is,

R= rho vd / mu

                = 1420 x  0.00059 x 9.52 x 10^-3 x / 1.0

               = 0.0080

Otras preguntas