Question 11.AE.4: Design a flywheel storage system which can provide for about......
Design a flywheel storage system which can provide for about 6 minutes a power of 100 MW, that is, energy of 10 MWh. The flywheel power plant consists (see Figures E11.4.1a, b) of a flywheel, mechanical gear, synchronous machine, inverterrectifier set and a step-up transformer. The individual components of this plant must be designed as follows:
a) For the flywheel (made from steel) as shown in Figure E11.4.1a (h=0.9 m, R_{1o}=1.5 m, R_{1i}=1.3 m, R_{2o}=0.50 m, R_{2i}=0.10 m, b=0.2 m), compute the ratio of inertia (J) to the weight (W) of the flywheel, that is (J/W).
The axial moment of inertia of a flywheel with 4 spokes is
J_{\text {wheel }}=\frac{1}{2} \cdot \pi \cdot \gamma \cdot h \cdot\left(R_{1 o}^4-R_{1 \mathrm{i}}^4\right)+\frac{4}{3} \cdot h \cdot b \cdot \gamma \cdot\left(R_{1 \mathrm{i}}^3-R_{2 \mathrm{o}}^3\right)+\frac{1}{2} \cdot \pi \cdot \gamma \cdot h \cdot\left(R_{2 \mathrm{o}}^4-R_{2 \mathrm{i}}^4\right)\left[\mathrm{kgm}^2\right]b) For the given values h=0.9 m, R_{1o}=1.5 m, R_{1i}=1.3 m, R_{2o}=0.50 m, R_{2i}=0.10 m, and b=0.2 m calculate for the wheel-type configuration the stored energy E_{stored\ rated} provided the flywheel rotates at n_{flywheel\ rated}=21,210 rpm.
a) Axial moment of inertia of flywheel:
J_{\text {wheel }}=\frac{1}{2} \cdot \pi \cdot \gamma \cdot h \cdot\left(R_{1 \mathrm{o}}^4-R_{1 \mathrm{i}}^4\right)+\frac{4}{3} \cdot h \cdot b \cdot \gamma \cdot\left(R_{1 \mathrm{i}}^3-R_{2 \mathrm{o}}^3\right)+\frac{1}{2} \cdot \pi \cdot \gamma \cdot h \cdot\left(R_{2 \mathrm{o}}^4-R_{2 \mathrm{i}}^4\right),
\begin{aligned} J_{\text {wheel }}= & \frac{1}{2} \pi \cdot 7.86 \cdot 10^3 \cdot 0.9\left[(1.5)^4-(1.3)^4\right]+\frac{4}{3} 0.9 \cdot 0.2 \cdot 7.86 \cdot 10^3\left[(1.3)^3-(0.5)^3\right] \\ & +\frac{1}{2} \pi \cdot 7.86 \cdot 10^3 \cdot 0.9\left[(0.5)^4-(0.1)^4\right]=29.117 \cdot 10^3 \mathrm{kgm}^2 \end{aligned}weight of flywheel: W_{\text {wheel }}=(\gamma \cdot V) g=\gamma \cdot\left[\pi \cdot h\left(R_{1 \mathrm{o}}^2-R_{1 \mathrm{i}}^2\right)+4 h \cdot b\left(R_{1 \mathrm{i}}-R_{2 \mathrm{o}}\right)\right. \left.+\pi h\left(R_{2 \mathrm{o}}^2-R_{2 \mathrm{i}}^2\right)\right] \cdot g,
\begin{aligned} W_{\text {wheel }} & =7.86 \cdot 10^3 \cdot\left[\pi \cdot 0.9\left(1.5^2-1.3^2\right)+4 \cdot 0.9 \cdot 0.2(1.3-0.5)+\pi 0.9\left(0.5^2-0.1^2\right)\right] \cdot 9.81 \\ & =218.753 \cdot 10^3 \mathrm{~N} \end{aligned},
ratio inertia/weight: \left(\frac{J_{\text {wheel }}}{W_{\text {wheel }}}\right)=\frac{29.117}{218.75}=0.133 \frac{\mathrm{kgm}^2}{\mathrm{~N}}.
b) Stored energy at rated speed: The rated angular velocity is \omega_{Alywheel\_rated }=2 \pi\left(\frac{n_{flywheel\_rated }}{60}\right)=2221.04 \mathrm{rad} / \mathrm{s},
\begin{aligned} E_{stored\_rated } & =\frac{1}{2} J_{\text {wheel }}\left(\omega_{\text {flywheel }}\right)^2=\frac{1}{2} 29.117 \cdot 10^3(2221.04)^2=71.82 \cdot 10^9 \mathrm{Ws} \\ & =19.95 \mathrm{MWh} \end{aligned}Note the flywheel must not be completely discharged. The angular velocity at 9.95 MWh is
\begin{aligned} 9.95 \mathrm{MWh} & =\frac{1}{2} 29.117 \cdot 10^3\left(\omega_{flywheel\_minimum }\right)^2 \text { or } \omega_{flywheel\_minimum }=\sqrt{\frac{2 \cdot 9.95 \mathrm{MW} / \mathrm{h}}{29.117 \cdot 10^3}} \\ & =1568.57 \mathrm{rad} / \mathrm{s} \end{aligned}