How Fast Must Turbine Blades Spin to Store Energy?

[scott.leever]

Angular Velocity Question?

In March 2004, a British company successfully tested a power system to tap the energy of ocean tides. The energy will be stored in an underwater turbine consisting of two metal blades, each 15.0 m long. The movement of the water due to the tides will give kinetic energy to the turbine blades, causing them to spin. In the calculations that follow, ignore any frictional drag due to the seawater.

If we model each of these two blades as a thin uniform steel bar 15.0 m long and 25.0 cm in diameter, at what rate rad/s and rpm must they spin for the turbine to store 1.00 MJ of energy? The density of steel is 7800 kg/m^3 . (Recall that density is equal to an object's mass divided by its volume.)

How do i solve this?? THANKS!

 

[Astronuc]

Here are some equations of motion for both linear and rotational motion
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

Moments of inertia for common geometries.
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi

Rotational kinetic energy
http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html#rke

Select the appropriate geometry.

 

[ogb p]

To solve this problem, we can use the formula for angular velocity, which is given by ω = √(2E/I), where E is the energy stored in the turbine and I is the moment of inertia of the blades.

First, we need to calculate the moment of inertia of the blades. The moment of inertia for a thin uniform steel bar is given by I = (1/12) * m * L^2, where m is the mass of the bar and L is its length.

To calculate the mass of each blade, we can use the density of steel and the volume of the blade. The volume of the blade can be calculated as the volume of a cylinder, given by V = π * r^2 * h, where r is the radius (25 cm = 0.25 m) and h is the length (15 m).

So, the mass of each blade is equal to the density of steel (7800 kg/m^3) multiplied by the volume, which gives us m = 7800 * π * (0.25)^2 * 15 = 731.25 kg.

Now, we can calculate the moment of inertia for each blade: I = (1/12) * 731.25 * (15)^2 = 1539.84 kgm^2.

Plugging this value into the formula for angular velocity, we get ω = √(2 * 1.00 * 10^6 / 1539.84) = 75.49 rad/s.

To convert this to rpm, we can use the conversion factor 1 rpm = 2π rad/s. So, the blades must spin at a rate of 75.49 * (1/2π) = 12.01 rpm to store 1.00 MJ of energy.

In conclusion, the blades of the turbine must spin at a rate of 75.49 rad/s or 12.01 rpm to store 1.00 MJ of energy.