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A spherical hot air balloon is initially filled with air at 120 kPa and 20 degree Celsius with a velocity of 3 m/s through a 1 m diameter opening. How many minutes will it take to inflate this balloon to a 17 m diameter when the pressure and temperature of the air in the balloon remain the same as the air entering the balloon?

The aim of this question is to understand the rate of change in volume or rate of change of mass. It also introduces the basic formulae of volume, area, and volumetric flow rate.

The mass flow rate of a fluid is defined as the unit mass passing through a point in unit time. It can be mathematically defined by the following formula:

\[ \dot{ m } \ = \ \dfrac{ \Delta m }{ \Delta t } \]

Where m is the mass while t is the time. The relationship between mass and volume of a body is mathematically described by the following formula:

\[ m \ = \ \rho V \]

Where $ \rho $ is the density of the fluid and V is the volume. the volume of a sphere is defined by the following formula:

\[ V \ = \ \dfrac{ 4 }{ 3 } \pi r^3 \ = \ \dfrac{ 1 }{ 6 } \pi D^3 \]

Where $ r $ is the radius and $ D $ is the diameter of the sphere.

Expert Answer

We know that:

\[ \dot{ m } \ = \ \dfrac{ \Delta m }{ \Delta t } \]

Since:

\[ m \ = \ \rho V \]

So:

\[ \Delta m \ = \ \rho \Delta V \]

\[ \dot{ m } \ = \ \rho \dot{ V } \]

Substituting these values in the above equation:

\[ \rho \dot{ V } \ = \ \dfrac{ \rho \Delta V }{ \Delta t } \]

\[ \dot{ V } \ = \ \dfrac{ \Delta V }{ \Delta t } \]

Rearranging:

\[ \Delta t \ = \ \dfrac{ \Delta V }{ \dot{ V } } \]

\[ \Delta t \ = \ \dfrac{ V_2 \ – \ V_1 }{ \dot{ V } } \]

Since:

\[ \dot{ V } \ = \ A v \]

The above equation becomes:

\[ \Delta t \ = \ \dfrac{ V_2 \ – \ V_1 }{ A v } \]

Substituting values for $ V $ and $ A $:

\[ \Delta t \ = \ \dfrac{ \frac{ \pi }{ 6 } D_2^3 \ – \ D_1^3 }{ \frac{ \pi }{ 4 } D^2 v } \]

\[ \Delta t \ = \ \dfrac{ 2 \bigg ( D_2^3 \ – \ D_1^3 \bigg ) }{ 3 D^2 v } … \ … \ … \ ( 1 ) \]

Substituting values:

\[ \Delta t \ = \ \dfrac{ 2 \bigg ( ( 17 )^3 \ – \ ( 5 )^3 \bigg ) }{ 3 ( 1 )^2 ( 3 ) } \]

\[ \Delta t \ = \ 1064 \ s \]

\[ \Delta t \ = \ 17.7 \ min \]

Numerical Result

\[ \Delta t \ = \ 17.7 \ min \]

Example

How much time will it take to inflate the hot air balloon if the diameter of the filling hose pipe was changed from 1 m to 2 m?

Recall equation (1):

\[ \Delta t \ = \ \dfrac{ 2 \bigg ( D_2^3 \ – \ D_1^3 \bigg ) }{ 3 D^2 v } \]

Substituting values:

\[ \Delta t \ = \ \dfrac{ 2 \bigg ( ( 17 )^3 \ – \ ( 5 )^3 \bigg ) }{ 3 ( 2 )^2 ( 3 ) } \]

\[ \Delta t \ = \ 266 \ s \]

\[ \Delta t \ = \ 4.43 \ min \]

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