A low-altitude meteorological research balloon, temperature sensor, and radio transmitter together weigh 2.5 lb. When inflated with helium, the balloon is spherical with a diameter of 4 ft. The volume of the transmitter can be neglected when compared to the balloon's size. The balloon is released from ground level and quickly reaches its terminal ascent velocity. Neglecting variations in the atmosphere's density, how long does it take the balloon to reach an altitude of 1000 ft?

12 mins

Explanation:

The summation of the forces in vertical direction

= Fb - Fd - w = 0 ∴ Fd = Fb - w  ----- ( 1 )

Fb ( buoyant force ) = Pair * g * Vballoon ------- ( 2 )

Pair = air density , Vballoon = volume of balloon

Vballoon = ,  where D = 4  ∴  Vballoon = 33.51 ft^3

g = 32.2 ft/s^2

From property tables

Pair = 2.33 * 10^-3 slug/ft^3

μ ( dynamic viscosity ) = 3.8 * 10^-7 slug/ft.s

Insert values into equation 2

Fb = ( 2.33 * 10^-3 ) * ( 32.2 ) *( 33.51 )  = 2.514 Ib

∴ Fd = 2.514 - 2.5 = 0.014 Ib ( equation 1 )

Assuming that flow is Laminar  and  RE < 1

Re = (Pair * vd) / μair   -------- ( 3 )

where:  Pair = 2.33 * 10^-3 slug/ft^3 ,  vd = ( 987 * 4 ) ft^2/s ,   μair = 3.8 * 10^-7 slug/ft.s

Insert values into equation 3

Re = 2.4 * 10^7 (  this means that the assumption above is wrong )

Hence we will use drag force law

Assume Cd = 0.5

Express  Fd using the relation below

Fd = 1/2* Cd * Pair * AV^2

therefore V = 1.39 ft/s

Recalculate  Reynold's number using  v = 1.39 ft/s

Re = 34091

from the figure Cd ≈ 0.5 at Re = 34091  

Finally calculate the rise time ( time taken to reach an altitude of 1000 ft )

t = h/v

  = 1000 / 1.39 = 719 seconds ≈ 12 mins

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