**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 formul**a:

\[ m \ = \ \rho V \]

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:**

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

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

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