Chapter 17.1, Problem 17.24P

(a)

To determine

Find the change in the angular velocity of the turbine disk.

Answer to Problem 17.24P

The change in angular velocity of the turbine disk is $\underset{\_}{\Delta \omega =-0.250\text{rpm}}$.

Explanation of Solution

Given information:

The mass of the turbine disk is $m_{\text{turbine}}=30\text{kg}$.

The centroidal radius of gyration of the turbine disk is $k_{\text{turbine}}=175\text{mm}$.

The angular velocity of the small blade is ${\omega }_1=60\text{rpm}$.

The weight of the small blade is $W_{\text{blade}}=0.5\text{N}$.

The centroidal radius of the blade is $k_{\text{blade}}=300\text{mm}$

The angle the turbine disk rotates is $\theta =90°$.

Position 1:

Find the mass of inertia about point O $\left({\overline{I}}_O\right)$ using the equation.

$\begin{array}{c}{\overline{I}}_O=m_{\text{turbine}}{k}_{\text{turbine}}^{\text{2}}-m_{\text{blade}}{k}_{\text{blade}}^{\text{2}}\\ =m_{\text{turbine}}{k}_{\text{turbine}}^{\text{2}}-\frac{W_{\text{blade}}}{g}{k}_{\text{blade}}^{\text{2}}\end{array}$

Here, the acceleration due to gravity is g.

Consider the acceleration due to gravity is $g=9.81{\text{m/s}}^{\text{2}}$.

Substitute 30 kg for $m_{\text{turbine}}$, 175 mm for $k_{\text{turbine}}$, 0.5 N for $W_{\text{blade}}$, $9.81{\text{m/s}}^{\text{2}}$ for g, and 300 mm for $k_{\text{blade}}$.

$\begin{array}{c}{\overline{I}}_O=30\times {\left(175\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2-\frac{0.5}{9.81}\times {\left(300\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2\\ =0.91875-4.58716\times {10}^{-3}\\ =0.91416\text{kg}\cdot {\text{m}}^2\end{array}$

Find the location of mass center $\left(\overline{x}\right)$ using the relation.

$\begin{array}{c}\overline{x}=-\frac{m_{\text{blade}}{r}_{\text{blade}}}{m_{\text{turbine}}-m_{\text{blade}}}\\ =-\frac{\left(\frac{W_{\text{blade}}}{g}\right)r_{\text{blade}}}{m_{\text{turbine}}-\left(\frac{W_{\text{blade}}}{g}\right)}\end{array}$

Position 1 $\left(\theta =0°\right)$:

The angular velocity at the position 1 is ${\omega }_1=60\text{rpm}$.

Find the total kinetic energy $\left(T_1\right)$ using the equation.

$T_1=\frac{1}{2}{\overline{I}}_O{\omega }_1^2$

Substitute $0.91416\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_O$ and 60 rpm for ${\omega }_1$.

$\begin{array}{c}{T}_1=\frac{1}{2}\times 0.91416\times {\left(60\text{rpm}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\text{s}}\right)}^2\\ =18.04480\text{J}\end{array}$

Find the total potential energy $\left(V_1\right)$ using the equation.

$V_1=m_{\text{turbine}}g{h}_1-m_{\text{blade}}g{h}_1$

Here, the conditional center of gravity is $h_1$.

In this case, the center of gravity lies at the point O. so, $h_1=0$.

Substitute 0 for $h_1$.

$\begin{array}{c}{V}_1=m_{\text{turbine}}g\left(0\right)-m_{\text{blade}}g\left(0\right)\\ =0\end{array}$

Position 2 $\left(\theta =90°\right)$:

Find the total kinetic energy $\left(T_2\right)$ using the equation.

$T_2=\frac{1}{2}{\overline{I}}_O{\omega }_2^2$

Substitute $0.91416\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_O$.

$\begin{array}{c}{T}_2=\frac{1}{2}\times 0.91416\times {\omega }_2^2\\ =0.45708{\omega }_2^2\end{array}$

Find the total potential energy $\left(V_2\right)$ using the equation.

$\begin{array}{c}{V}_2=m_{\text{turbine}}g{h}_2-m_{\text{blade}}g{h}_2\\ =m_{\text{turbine}}g{h}_2-W_{\text{blade}}{h}_2\end{array}$

Here, the conditional center of gravity is $h_2$.

In this case, the center of gravity lies at the point O. so,

$\begin{array}{c}{h}_2=-\overline{x}\\ =\frac{\left(\frac{W_{\text{blade}}}{g}\right)r_{\text{blade}}}{m_{\text{turbine}}-\left(\frac{W_{\text{blade}}}{g}\right)}\end{array}$.

Substitute $\frac{\left(\frac{W_{\text{blade}}}{g}\right)r_{\text{blade}}}{m_{\text{turbine}}-\left(\frac{W_{\text{blade}}}{g}\right)}$ for $h_2$.

$V_2=\left(m_{\text{turbine}}g-W_{\text{blade}}\right)\frac{\left(\frac{W_{\text{blade}}}{g}\right)r_{\text{blade}}}{m_{\text{turbine}}-\left(\frac{W_{\text{blade}}}{g}\right)}$

Substitute 30 kg for $m_{\text{turbine}}$, $9.81{\text{m/s}}^2$ for g, 0.5 N for $W_{\text{blade}}$, and 300 mm for $r_{\text{blade}}$.

$\begin{array}{c}{V}_2=\left(30\times 9.81-0.5\right)\frac{\left(\frac{0.5}{9.81}\right)\times 300\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}{30-\left(\frac{0.5}{9.81}\right)}\\ =293.8\times \frac{0.015291}{29.94903}\\ =0.15\text{N}\cdot \text{m}\end{array}$

Write the equation of conservation of energy as follows;

$T_1+V_1=T_2+V_2$

Substitute 18.04480 J for $T_1$, 0 for $V_1$, $0.45708{\omega }_2^2$ for $T_2$, and $0.15\text{N}\cdot \text{m}$ for $V_2$.

$\begin{array}{c}18.04480+0=0.45708{\omega }_2^2+0.15\\ {\omega }_2=6.25702\text{rad/s}\times \frac{1\text{rev}}{2\pi \text{rad}}\times \frac{60\text{s}}{1\min }\\ =59.75\text{rpm}\end{array}$

Find the change in angular velocity $\left(\Delta \omega \right)$ using the relation.

$\Delta \omega ={\omega }_2-{\omega }_1$

Substitute 59.75 rpm for ${\omega }_2$ and 60 rpm for ${\omega }_1$.

$\begin{array}{c}\Delta \omega =59.75-60\\ =-0.250\text{rpm}\end{array}$

Therefore, the change in angular velocity of the turbine disk is $\underset{\_}{\Delta \omega =-0.250\text{rpm}}$.

(b)

To determine

Find the change in the angular velocity of the turbine disk.

Answer to Problem 17.24P

The change in angular velocity of the turbine disk is $\underset{\_}{\Delta \omega =0.249\text{rpm}}$.

Explanation of Solution

Given information:

The mass of the turbine disk is $m_{\text{turbine}}=30\text{kg}$.

The centroidal radius of gyration of the turbine disk is $k_{\text{turbine}}=175\text{mm}$.

The angular velocity of the small blade is ${\omega }_1=60\text{rpm}$.

The weight of the small blade is $W_{\text{blade}}=0.5\text{N}$.

The centroidal radius of the blade is $k_{\text{blade}}=300\text{mm}$

The angle the turbine disk rotates is $\theta =270°$.

Position 1:

Find the mass of inertia about point O $\left({\overline{I}}_O\right)$ using the equation.

$\begin{array}{c}{\overline{I}}_O=m_{\text{turbine}}{k}_{\text{turbine}}^{\text{2}}-m_{\text{blade}}{k}_{\text{blade}}^{\text{2}}\\ =m_{\text{turbine}}{k}_{\text{turbine}}^{\text{2}}-\frac{W_{\text{blade}}}{g}{k}_{\text{blade}}^{\text{2}}\end{array}$

Here, the acceleration due to gravity is g.

Consider the acceleration due to gravity is $g=9.81{\text{m/s}}^{\text{2}}$.

Substitute 30 kg for $m_{\text{turbine}}$, 175 mm for $k_{\text{turbine}}$, 0.5 N for $W_{\text{blade}}$, $9.81{\text{m/s}}^{\text{2}}$ for g, and 300 mm for $k_{\text{blade}}$.

$\begin{array}{c}{\overline{I}}_O=30\times {\left(175\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2-\frac{0.5}{9.81}\times {\left(300\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2\\ =0.91875-4.58716\times {10}^{-3}\\ =0.91416\text{kg}\cdot {\text{m}}^2\end{array}$

Find the location of mass center $\left(\overline{x}\right)$ using the relation.

$\begin{array}{c}\overline{x}=-\frac{m_{\text{blade}}{r}_{\text{blade}}}{m_{\text{turbine}}-m_{\text{blade}}}\\ =-\frac{\left(\frac{W_{\text{blade}}}{g}\right)r_{\text{blade}}}{m_{\text{turbine}}-\left(\frac{W_{\text{blade}}}{g}\right)}\end{array}$

Position 1 $\left(\theta =0°\right)$:

The angular velocity at the position 1 is ${\omega }_1=60\text{rpm}$.

Find the total kinetic energy $\left(T_1\right)$ using the equation.

$T_1=\frac{1}{2}{\overline{I}}_O{\omega }_1^2$

Substitute $0.91416\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_O$ and 60 rpm for ${\omega }_1$.

$\begin{array}{c}{T}_1=\frac{1}{2}\times 0.91416\times {\left(60\text{rpm}\times \frac{2\pi \text{rad}}{1\text{rev}}\times \frac{1\min }{60\text{s}}\right)}^2\\ =18.04480\text{J}\end{array}$

Find the total potential energy $\left(V_1\right)$ using the equation.

$V_1=m_{\text{turbine}}g{h}_1-m_{\text{blade}}g{h}_1$

Here, the conditional center of gravity is $h_1$.

In this case, the center of gravity lies at the point O. so, $h_1=0$.

Substitute 0 for $h_1$.

$\begin{array}{c}{V}_1=m_{\text{turbine}}g\left(0\right)-m_{\text{blade}}g\left(0\right)\\ =0\end{array}$

Position 3 $\left(\theta =270°\right)$:

Find the total kinetic energy $\left(T_3\right)$ using the equation.

$T_3=\frac{1}{2}{\overline{I}}_O{\omega }_3^2$

Substitute $0.91416\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_O$.

$\begin{array}{c}{T}_3=\frac{1}{2}\times 0.91416\times {\omega }_2^2\\ =0.45708{\omega }_3^2\end{array}$

Find the total potential energy $\left(V_3\right)$ using the equation.

$\begin{array}{c}{V}_3=m_{\text{turbine}}g{h}_3-m_{\text{blade}}g{h}_3\\ =m_{\text{turbine}}g{h}_3-W_{\text{blade}}{h}_3\end{array}$

Here, the conditional center of gravity is $h_3$.

In this case, the center of gravity lies at the point O. so,

$\begin{array}{c}{h}_3=\overline{x}\\ =-\frac{\left(\frac{W_{\text{blade}}}{g}\right)r_{\text{blade}}}{m_{\text{turbine}}-\left(\frac{W_{\text{blade}}}{g}\right)}\end{array}$.

Substitute $\frac{\left(\frac{W_{\text{blade}}}{g}\right)r_{\text{blade}}}{m_{\text{turbine}}-\left(\frac{W_{\text{blade}}}{g}\right)}$ for $h_3$.

$V_3=\left(m_{\text{turbine}}g-W_{\text{blade}}\right)\left(-\frac{\left(\frac{W_{\text{blade}}}{g}\right)r_{\text{blade}}}{m_{\text{turbine}}-\left(\frac{W_{\text{blade}}}{g}\right)}\right)$

Substitute 30 kg for $m_{\text{turbine}}$, $9.81{\text{m/s}}^2$ for g, 0.5 N for $W_{\text{blade}}$, and 300 mm for $r_{\text{blade}}$.

$\begin{array}{c}{V}_2=\left(30\times 9.81-0.5\right)\left(-\frac{\left(\frac{0.5}{9.81}\right)\times 300\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}{30-\left(\frac{0.5}{9.81}\right)}\right)\\ =-293.8\times \frac{0.015291}{29.94903}\\ =-0.15\text{N}\cdot \text{m}\end{array}$

Write the equation of conservation of energy as follows;

$T_1+V_1=T_3+V_3$

Substitute 18.04480 J for $T_1$, 0 for $V_1$, $0.45708{\omega }_2^2$ for $T_2$, and $-0.15\text{N}\cdot \text{m}$ for $V_2$.

$\begin{array}{c}18.04480+0=0.45708{\omega }_3^2-0.15\\ {\omega }_3=6.309247\text{rad/s}\times \frac{1\text{rev}}{2\pi \text{rad}}\times \frac{60\text{s}}{1\min }\\ =60.249\text{rpm}\end{array}$

Find the change in angular velocity $\left(\Delta \omega \right)$ using the relation.

$\Delta \omega ={\omega }_3-{\omega }_1$

Substitute 60.249 rpm for ${\omega }_3$ and 60 rpm for ${\omega }_1$.

$\begin{array}{c}\Delta \omega =60.249-60\\ =0.249\text{rpm}\end{array}$

Therefore, the change in angular velocity of the turbine disk is $\underset{\_}{\Delta \omega =0.249\text{rpm}}$.