Chapter 7, Problem 7.75P

To determine

(a)

Find the angle $\theta$ if stream velocity is equal to $5m/s$ ?

Answer to Problem 7.75P

$\theta =72.21°$

Explanation of Solution

Given information:

The balloon is filled with helium gas

The diameter is equal to $50cm$

The weight is $0.2N$

The helium pressure is $120kPa$

Atmospheric pressure $1atm$ at $20°C$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho V^{2}A}{2}$

In above equation,

$\rho$ - Density of the fluid

$A$ - Characteristic area

$V$ - Velocity

$C_D$ - Drag co-efficient

The buoyancy force is defined as,

$F_{Bouyancy}=\left(Vol\right)\rho g$

The specific heat capacity $R$ is defined as,

$R=\frac{P}{\rho T}$

$P$ - Pressure

$\rho$ - Density

$T$ - Temperature

Assume, the air at $20°C$ will have,

$\begin{array}{l}{\rho }_{air}=1.2kg/m^3\\ \mu =1.8\times {10}^{-5}kg/ms\end{array}$

Assume,

The specific heat capacity of helium is equal to,

$R=2077J/kg°K$

Calculation:

Calculate the density of helium,

$\begin{array}{l}R=\frac{P}{\rho T}\\ 2077J/kg°K=\frac{120000Pa}{{\rho }_{He}\left(293°K\right)}\\ {\rho }_{He}=0.197kg/m^3\end{array}$

For vertical force balance,

$F_{Bouyancy}-W=0$

Where,

$W$ - Weight

Calculate the buoyancy force,

$\left(Vol\right)\rho g=\left(\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right)\left({\rho }_{air}-{\rho }_{He}\right)g$

Substitute for known values,

$\begin{array}{l}{F}_{Bouyancy}=\left(\frac{4}{3}\pi {\left(\frac{0.5m}{2}\right)}^3\right)\left(1.2kg/m^3-0.197kg/m^3\right)\left(9.81m/s^2\right)\\ F_{Bouyancy}=0.644N\end{array}$

Calculate the vertical force,

$\uparrow {\displaystyle \sum F}=F_{Bouyancy}-W=0.644N-0.2N=0.444N$

Calculate the Reynolds’s number,

$\mathrm{Re}=\frac{\rho Vd}{\mu }=\frac{\left(1.2kg/m^3\right)\left(5m/s\right)\left(0.5m\right)}{1.8\times {10}^{-5}kg/ms}=166666.67$

Therefore, laminar flow

According to table 7.3 the drag co-efficient is equal to $C_D=0.47$

Calculate the drag force,

$F_{drag}=C_D\frac{\rho V^{2}A}{2}=\left(0.47\right)\left(\frac{\left(1.2kg/m^3\right){\left(5m/s\right)}^2\left(\pi {\left(0.25m\right)}^2\right)}{2}\right)=1.384N$

Calculate the tilted angle,

According o the given information,

$\tan \theta =\frac{F_{drag}}{F_{Vertical}}=\frac{1.384N}{0.444N}$

Therefore,

$\theta ={\tan }^{-1}\left(\frac{1.384N}{0.444N}\right)=72.21°$

Conclusion:

The tilted angle is equal to $72.21°$ if the airstream velocity is equal to $5m/s$.

To determine

(b)

Find the angle $\theta$ if stream velocity is equal to $20m/s$ ?

Answer to Problem 7.75P

$\theta =87.3°$

Explanation of Solution

Given information:

The balloon is filled with helium gas

The diameter is equal to $50cm$

The weight is $0.2N$

The helium pressure is $120kPa$

Atmospheric pressure $1atm$ at $20°C$

According to sub-part a,

We have found that the net vertical force is equal to,

$\uparrow {\displaystyle \sum F}=F_{Bouyancy}-W=0.444N$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho V^{2}A}{2}$

In above equation,

$\rho$ - Density of the fluid

$A$ - Characteristic area

$V$ - Velocity

$C_D$ - Drag co-efficient

Assume, the air at $20°C$ will have,

$\begin{array}{l}{\rho }_{air}=1.2kg/m^3\\ \mu =1.8\times {10}^{-5}kg/ms\end{array}$

Calculation:

Calculate the Reynolds’s number,

$\mathrm{Re}=\frac{\rho Vd}{\mu }=\frac{\left(1.2kg/m^3\right)\left(20m/s\right)\left(0.5m\right)}{1.8\times {10}^{-5}kg/ms}=666666.67$

Therefore, turbulent flow

According to table 7.3 the drag co-efficient is equal to $C_D=0.2$

Calculate the drag force,

$F_{drag}=C_D\frac{\rho V^{2}A}{2}=\left(0.2\right)\left(\frac{\left(1.2kg/m^3\right){\left(20m/s\right)}^2\left(\pi {\left(0.25m\right)}^2\right)}{2}\right)=9.424N$

Calculate the tilted angle,

According o the given information,

$\tan \theta =\frac{F_{drag}}{F_{Vertical}}=\frac{9.424N}{0.444N}$

Therefore,

$\theta ={\tan }^{-1}\left(\frac{9.424N}{0.444N}\right)=87.3°$

Conclusion:

The tilted angle is equal to $87.3°$ if the airstream velocity is equal to $20m/s$.