- #1

MattHorbacz

- 18

- 0

I figure that once you get the force acting in the y direction, you can plug into P=Fv=F(ωr) to get the radial velocity of the blade.

And I am not sure how to write "m dot" for mass flow rate, so i will use Ω instead.

**Equations**:

P=½ρA

P=Fv=F(ωr)

Ω=ρA

∑F

∑F

_{swept}V^{3}C_{p}P=Fv=F(ωr)

Ω=ρA

_{blade}V∑F

_{x}=Ω_{in}V_{in}-F_{x,blade}=0∑F

_{y}=-Ω_{out}V_{out}+F_{y,blade}=0**Known values:**

V

_{air}=10 m/s

ρ

_{air}=1.225 kg/m

^{3}

length

_{blade}(swept radius)=.457 m

max chord length=.089 m

A

_{blade}=length

_{blade}*chord length (assuming blade to be a rectangle)

C

_{p}=.45 ( the betz coefficient)

**Attempt at Solution:**

Power is simple to calculate,

Power=.5*1.225kg/m*π*(.457m)

Next, I solve the conservation of momentum equation in the y direction (the x direction doesn't tell me anything useful)

F

Now I plug that into P=Fv and find that v=36.35m/s. assuming the force acts in center of blade, ω=159 rad/s...I still convert to RPM even though this answer is obviously nowhere near correct...

(159 rad/s)(180 degrees/π rad)(1 rev/360 degrees)(60 s/min)=

Power=.5*1.225kg/m*π*(.457m)

^{2}*(10 m/s)^{3}*.45=181 WNext, I solve the conservation of momentum equation in the y direction (the x direction doesn't tell me anything useful)

F

_{y,blade}=ΩV_{out}=ρA_{blade}V^{2}=(1.225 kg/m^{3})*(.457 m)*(.089 m)*(10 m/s)^2=4.98 NNow I plug that into P=Fv and find that v=36.35m/s. assuming the force acts in center of blade, ω=159 rad/s...I still convert to RPM even though this answer is obviously nowhere near correct...

(159 rad/s)(180 degrees/π rad)(1 rev/360 degrees)(60 s/min)=

**1518 RPM**. Our advisor told us that the blade will have and RPM of around 60-120, which I feel is a bit low, but 1518 RPM is wayyyyy to high.I would very much appreciate any advice, whether it be where I messed up, or alternative ways of calculating RPM. Let me know if you would like any clarifications