Chapter 7, Problem 7.55P

To determine

(a)

The towing power if the cylinder is parallel to tow direction.

Answer to Problem 7.55P

$Power=121.25kW$

Explanation of Solution

Given information:

The diameter of the cylinder is equal to $1.5m$

The length of the cylinder is $22m$

The velocity is equal to $5m/s$

The fluid is water at $20°C$

The power $P$ is defined as,

$P=FV$

Where,

$F$ - Drag force

$V$ - Velocity

The drag force is defined as,

$F=C_D\left(\frac{\rho V^{2}A}{2}\right)$

In above equation,

$\rho$ - Density of the fluid

$A$ - Characteristic area

The Reynolds’s based on length number is defined as,

${\mathrm{Re}}_L=\frac{\rho VL}{\mu }$

Where,

$L$ - Length

$\mu$ - Dynamic viscosity

Assume, the water at $20°C$ will have,

$\rho =998kg/m^3$

$\mu =0.001kg/ms$

Calculation:

Calculate the Reynolds’s number,

${\mathrm{Re}}_L=\frac{\rho VL}{\mu }=\frac{\left(998kg/m^3\right)\left(5m/s\right)\left(22m\right)}{0.001kg/ms}=1.0978\times {10}^8$

$L/D$ is equal to,

$\frac{L}{D}=\frac{22}{1.5}\approx 15$

According to the definitions, the drag co-efficient is equal to

$C_D=1.1$

Calculate the drag force,

$F=C_D\left(\frac{\rho V^{2}A}{2}\right)=\left(1.1\right)\left(\frac{\left(998kg/m^3\right){\left(5m/s\right)}^2\left(\pi {\left(0.75m\right)}^2\right)}{2}\right)=24249.66N$

Calculate the power,

$\begin{array}{l}P=FV=\left(24249.66N\right)\left(5m/s\right)=121248.3W\\ =121.25kW\end{array}$

Conclusion:

The towing power required when the cylinder is parallel to the tow direction is equal to $121.25kW$.

To determine

(b)

The towing power if the cylinder is normal to tow direction.

Answer to Problem 7.55P

$Power=823.35kW$

Explanation of Solution

Given information:

The diameter of the cylinder is equal to $1.5m$

The length of the cylinder is $22m$

The velocity is equal to $5m/s$

The fluid is water at $20°C$

The power $P$ is defined as,

$P=FV$

Where,

$F$ - Drag force

$V$ - Velocity

The drag force is defined as,

$F=C_D\left(\frac{\rho V^{2}A}{2}\right)$

In above equation,

$\rho$ - Density of the fluid

$A$ - Characteristic area

The Reynolds’s based on length number is defined as,

${\mathrm{Re}}_d=\frac{\rho Vd}{\mu }$

Where,

$d$ - Diameter

$\mu$ - Dynamic viscosity

Assume, the water at $20°C$ will have,

$\rho =998kg/m^3$

$\mu =0.001kg/ms$

Calculation:

Calculate the Reynolds’s number,

${\mathrm{Re}}_d=\frac{\rho Vd}{\mu }=\frac{\left(998kg/m^3\right)\left(5m/s\right)\left(1.5m\right)}{0.001kg/ms}=7.485\times {10}^6$

According to the definitions, the drag co-efficient is equal to

$C_D=0.4$

Calculate the drag force,

$F=C_D\left(\frac{\rho V^{2}A}{2}\right)=\left(0.4\right)\left(\frac{\left(998kg/m^3\right){\left(5m/s\right)}^2\left(1.5m\right)\left(22m\right)}{2}\right)=164670N$

Calculate the power,

$\begin{array}{l}P=FV=\left(164670N\right)\left(5m/s\right)=823350W\\ Power=823.35kW\end{array}$

Conclusion:

The towing power required when the cylinder is normal to the tow direction is equal to $823.35kW$.