How Much Power Does a 30m Diameter Windmill Generate?

AI Thread Summary
A discussion on the power generation of a 30m diameter windmill highlights the importance of renewable energy amidst rising energy costs. The windmill entrains air at a speed of 14 m/s, with an outlet speed of 5 m/s, and the air density is given as 1.2 kg/m3. The relevant equation for calculating power involves mass flow rate and shaft work, with no changes in pressure or elevation affecting the outcome. Participants clarify that the initial velocity and area can be used to determine mass flow rate, emphasizing that shaft work equates to power. The conversation aims to guide the calculation of the windmill's power output in megawatts.
desi01
Messages
1
Reaction score
0

Homework Statement


In this era of high energy prices, there is a focus on renewable sources of energy such as wind energy. Consider a large windmill 30m in diameter. On a windy day, suppose that the windmill entrains a stream of air 25 m in diameter at a speed of 14 m/s. Downstream of the windmill, the entrained stream exits over a large diameter at a speed of 5 m/s. The pressure equals atmospheric pressure at both inlet and outlet. Density of air: ρ = 1.2 kg/m3

Find the power (in megawatts) generated by the windmill.

25r2ona.jpg


Homework Equations



1/2 Δ(v^2) + gΔz + (ΔPressure)/density + lost work + shaft work/mass flow rate = 0

The Attempt at a Solution



So I know that that is no change in Pressure and no lost work so those go to 0. I don't think that you would need the gΔz part either?? So basically I'm thinking plug in the velocities given and that would only leave shaft work which I'm solving for and mass flow rate. Now I don't know what velocity and area I would use for that either so that's where I'm stuck?? Am I headed in the right direction?
Thanks.
 
Physics news on Phys.org
Welcome to Physics Forums. You're correct that gΔz is not needed (it is zero).

You can use the initial velocity and area, and density, to get the mass flow rate.

Also, note that "shaft work" would actually be power here.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top