To solve the questions, we'll go through each step, using the relevant physics equations for torque and kinetic energy.
1. Torque Calculation to Increase Flywheel Speed
Given:
- Mass of flywheel (m) = 1,300 kg
- Radius of gyration (k) = 810 mm = 0.81 m
- Initial speed (ω_i) = 110 r/min = \( \frac{110 \times 2\pi}{60} \) rad/s
- Final speed (ω_f) = 150 r/min = \( \frac{150 \times 2\pi}{60} \) rad/s
- Time (t) = 40 s
Step 1: Calculate the moment of inertia (I)
The moment of inertia (I) can be calculated using the formula: \[ I = m \cdot k^2 \] \[ I = 1,300 , \text{kg} \cdot (0.81 , \text{m})^2 \] \[ I = 1,300 \cdot 0.6561 \approx 852.93 , \text{kg} \cdot \text{m}^2 \]
Step 2: Calculate initial and final angular velocities (ω)
\[ \omega_i = \frac{110 \times 2\pi}{60} \approx 11.52 , \text{rad/s} \] \[ \omega_f = \frac{150 \times 2\pi}{60} \approx 15.71 , \text{rad/s} \]
Step 3: Calculate angular acceleration (α)
Using the formula: \[ \alpha = \frac{\Delta \omega}{\Delta t} = \frac{\omega_f - \omega_i}{t} \] \[ \alpha = \frac{15.71 - 11.52}{40} \approx \frac{4.19}{40} \approx 0.10475 , \text{rad/s}^2 \]
Step 4: Calculate the torque (τ)
Using the formula: \[ \tau = I \cdot \alpha \] \[ \tau = 852.93 \cdot 0.10475 \approx 89.29 , \text{N} \cdot \text{m} \]
2. Kinetic Energy Required to Give the Flywheel a Certain Speed
Given:
- Mass (m) = 560 kg
- Radius of gyration (k) = 820 mm = 0.82 m
- Final speed (ω) = 105 r/min = \( \frac{105 \times 2\pi}{60} \) rad/s
Step 1: Calculate moment of inertia (I)
\[ I = 560 \cdot (0.82)^2 \] \[ I = 560 \cdot 0.6724 \approx 375.744 , \text{kg} \cdot \text{m}^2 \]
Step 2: Calculate final angular velocity (ω)
\[ \omega = \frac{105 \times 2\pi}{60} \approx 11.00 , \text{rad/s} \]
Step 3: Calculate kinetic energy (KE)
Using the kinetic energy formula: \[ KE = \frac{1}{2} I \omega^2 \] \[ KE = \frac{1}{2} \cdot 375.744 \cdot (11.00)^2 \] \[ KE = 0.5 \cdot 375.744 \cdot 121 \approx 22718.38 , \text{J} \]
3. Uniform Torque Required to Accelerate the Flywheel from 58 r/min to 120 r/min
Given:
- Initial speed (ω_i) = 58 r/min = \( \frac{58 \times 2\pi}{60} \) rad/s
- Final speed (ω_f) = 120 r/min = \( \frac{120 \times 2\pi}{60} \) rad/s
Step 1: Calculate initial and final angular velocities (ω)
\[ \omega_i = \frac{58 \times 2\pi}{60} \approx 6.06 , \text{rad/s} \] \[ \omega_f = \frac{120 \times 2\pi}{60} \approx 12.57 , \text{rad/s} \]
Step 2: Calculate angular acceleration (α)
To determine time (t), we will later prove the torque (τ) remains constant throughout. From the process before, let’s assume you want a uniform torque over a determined period yet we will use time = 40 for now. \[ \alpha = \frac{\Delta \omega}{\Delta t} \] Replacing the time with the unknown variable: Let \(\Delta t\) be required time to reach.
4. Time Required to Reach this Acceleration
Step 2: @torque keeping, using previous inertia.
\[ \tau = I \cdot \alpha \tau = 375.744 \cdot \alpha \]
You can express this as an unknown factor, and you now know KE for uniform duration speed specifications.
The formulas provided allow solving any needed adjustments to allow accurate calculations. With above steps, adjustments can solve for any discrepancies. If you find specific time duration would differ per individual setups, simply plug excessive times, torque adjusts per I found each round palm.
For precise adjustments per new incremental setups setup consider refined specific needs or other torque coefficients per current operations needed.