In an isothermal process 05 liters of air 1 atm 25 degree C

In an isothermal process, 0.5 liters of air (1 atm, 25 degree C) per second is pumped into a spherical balloon. Tension in the rubber skin or the balloon is given by sigma = kA, where k = 10 kN/m^3, and A is the surface area of the balloon in m^2. Compute the time required to inflate the balloon radius from 10 cm to 30 cm.

Solution

Given:

Pressure, P = 1 atm

Temperature, T = 25 °C

Tension, = KA, where K = 10 KN/m³

Let radius of balloon at initial position be R₁ = 10 cm = 0.1 m

Let radius of balloon at final position be R₂ = 30 cm = 0.3 m

Volumetric Flow rate = 0.5 L/sec = 0.0005 m³/sec

It is known that volumetric flow rate ‘Q’ is given as:

                                Q = density x area x velocity

                                   = density x (volume/time)                        [Assuming density of air as 1.225 Kg/m³)

                                0.0005 = 1.225 x (4/3 x x R₁³) / T₁

                                T₁ = {1.225 x (4/3 x x 0.1³)} / 0.0005

                                T₁ = 10.2 seconds

Similarly for Radius, R₂

                                Q = density x area x velocity

                                   = density x (volume/time)                        [Assuming density of air as 1.225 Kg/m³)

                                0.0005 = 1.225 x (4/3 x x R₂³) / T₂

                                T₂ = {1.225 x (4/3 x x 0.3³)} / 0.0005

                                T₂ = 277.0 seconds

So time required to inflate the balloon from radius R₁ to R₂ could be given as:

                                                T = T₂ – T₁

                                                = 277.0 – 10.2

                                                = 266.8 seconds or 4.44 minutes