Question 1.4: Power Capacity of Punch Press Flywheel A high-strength steel......
Power Capacity of Punch Press Flywheel
A high-strength steel flywheel of outer and inner rim diameters d_o and d_i , and length in axial direction of l, rotates at a speed of n (Figure 1.7). It is to be used to punch metal during two-thirds of a revolution of the flywheel. What is the average power available?
Given
d_o=0.5 m , n=1000 rpm , \rho=7860 kg / m ^3 (Table B.1)
Assumptions
1. Friction losses are negligible.
2. Flywheel proportions are d_i=0.75 d_o \text { and } l=0.18 d_o .
3. The inertia contributed by the hub and spokes is omitted: the flywheel is considered as a rotating ring free to expand.
Through the use of Equations (1.9) and (1.10), we obtain
W=T \theta (1.9)
E_k=\frac{1}{2} I \omega^2 (1.10)
T \theta=\frac{1}{2} I \omega^2 (1.18)
where
\theta=\frac{2}{3}(2 \pi)=4 \pi / 3 rad\omega=1000(2 \pi / 60)=104.7 rad / s
\begin{aligned} I & =\frac{\pi}{32}\left(d_o^4-d_i^4\right) l \rho \quad(\text { Case 5, Table A.5 }) \\ & =\frac{\pi}{32}\left[(0.5)^4-(0.375)^4\right](0.09)(7860)=2.967 kg \cdot m ^2 \end{aligned}
Introducing the given data into Equation (1.18) and solving T=3882 N · m. Equation (1.15) is therefore
kW =\frac{F V}{1000}=\frac{T n}{9549} (1.15)
\begin{aligned} kW =\frac{T n}{9549} & =\frac{3882(1000)}{9549} \\ & =406.5 \end{aligned}
Comment: The braking torque required to stop a similar disk in a two thirds revolution would have an average value of 3.88 kN · m (see Section 16.5).
| TABLE A.5 Mass and Mass Moments of Inertia of Solids |
|
 |
\begin{aligned} & m=\frac{\pi d^2 L \rho}{4} \\ & I_y=I_z=\frac{m L^2}{12} \end{aligned} |
 |
\begin{aligned} & m=\frac{\pi d^2 t \rho}{4} \\ & I_x=\frac{m d^2}{8} \\ & I_y=I_z=\frac{m d^2}{16} \end{aligned} |
 |
\begin{aligned} & m=a b c \rho \\ & I_x=\frac{m}{12}\left(a^2+b^2\right) \\ & I_y=\frac{m}{12}\left(a^2+c^2\right) \\ & I_z=\frac{m}{12}\left(b^2+a^2\right) \end{aligned} |
 |
\begin{aligned} & m=\frac{\pi d^2 L \rho}{4} \\ & I_x=\frac{m d^2}{8} \\ & I_y=I_z=\frac{m}{48}\left(3 a^2+4 L^2\right) \end{aligned} |
 |
\begin{aligned} & m=\frac{\pi d^2 L \rho}{4}\left(3 d_o^2-d_i^2\right) \\ & I_x=\frac{m}{4}\left(d_o^2+d_i^2\right) \\ & I_y=I_z=\frac{m}{48}\left(3 d_o^2+3 d_i^2+4 L^2\right) \end{aligned} |
| Notes: \rho , mass density; m , mass; I , mass moment of inertia. | |