Question

A mixing beater consists of three thin rods, each 10.0 cm long. The rods diverge from a central hub, separated from each other by 120°, and all turn in the same plane. A ball is attached to the end of each rod. Each ball has cross-sectional area 4.00and is so shaped that it has a drag coefficient of 0.600. Calculate the power input required to spin the beater at 1000 rev/min (a) in air and (b) in water.

Short answer

1. The power in air is $P=0.496 \mathrm{W}$
2. The power in water is$P=413 \mathrm{W}$

Step-by-step solution

Resistance force

The sum of the vectors of one force or many forces, the direction of which is opposite to the motion of the body, and may indicate: friction during sliding and/or rolling. Drag (physics), during movement through the fluid (see fluid dynamics), is called resistance force.

Given Information

Resistance force:$R=\frac{1}{2}D\rho A{v}^2$

Where,

role="math" localid="1663740303399" $\begin{array}{rcl}\nu & =& r\omega \\ \tau & =& \mathrm{radius}·\mathrm{Force}\\ P& =& \tau \omega \end{array}$

Given in question as:

role="math" localid="1663740386142" $\begin{array}{rcl}D& =& 0.600\\ \rho & =& 1.20\frac{\mathrm{kg}}{{\mathrm{m}}^3}\\ & =& 1000\frac{\mathrm{kg}}{{\mathrm{m}}^3}\\ A& =& 4.00{\mathrm{cm}}^2\end{array}$

Convert to ${\text{m}}^2$,

$A=0.000400 {\mathrm{m}}^2$

$\begin{array}{rcl}r& =& 10.0\mathrm{cm}\\ & =& 0.100 \mathrm{m}\end{array}$

And,

$\begin{array}{rcl}\omega & =& 1000\frac{\mathrm{rev}}{\min }\\ & =& 104.7 \frac{\mathrm{rev}}{\mathrm{s}}\end{array}$

Finding torque

Resistance force for a total of three balls is,

$F{r}_{\mathrm{total}}=\frac{3}{2}D\rho A{v}^2$

Torque is$\tau =\mathrm{Force}·\mathrm{radius}$

$\begin{array}{c}\tau =F{r}_{\mathrm{iotal}}\times r\\ =\frac{3}{2}D\rho A{v}^{2}r\end{array}$

Substitute $\nu =r\omega$.

$\tau =\frac{3}{2}D\rho A{r}^3{\omega }^2$

Finding the power in air

(a)

$\begin{array}{c}P=\tau \omega \\ =\frac{3}{2}D\rho A{r}^3{\omega }^3\end{array}$

Substitute the found date in the above equation,

$\begin{array}{rcl}P& =& \frac{3}{2}\left(0.6\right)\left(1.2 \frac{\mathrm{kg}}{{\mathrm{m}}^3}\right)\left(0.0001 {\mathrm{m}}^2\right)(0.1 \mathrm{m}{)}^3{\left(101.7 \frac{\mathrm{rad}}{\mathrm{s}}\right)}^3\\ P& =& 0.496\mathrm{W}\end{array}$

The power in air isrole="math" localid="1663740774253" $P=0.496\mathrm{W}$.

Finding the power in the water

(b)

$\begin{array}{c}P=\frac{3}{2}D\rho A{r}^3{\omega }^3\\ P=\frac{3}{2}(0.6)\left(1000 \frac{\mathrm{kg}}{{\mathrm{m}}^3}\right)\left(0.0004 {\mathrm{m}}^2\right)(0.1 \mathrm{m}{)}^3{\left(104.7 \frac{\mathrm{rad}}{\mathrm{s}}\right)}^3\\ P=413 \mathrm{W}\end{array}$

The power in water is$P=413 \mathrm{W}$.