Ducted Fan Force and nozzle optimisation

[Digimagic]

I have a model aircraft ducted fan with exit velocity of 22.4m/s and area of 6221sq mm. Is there an equation I can use to estimate static thrust? Also, if I added a nozzle with exit area of 4418sq mm, what effect would that have on static thrust and is there an equation I can use to optimise the nozzle area ratio?

Any help would be much appreciated,
Andrew

 

[zekeman]

If you just have fan in a duct, the answer is no, since the fan power would determine it and you would need fan curves for that.

Adding a nozzle won't help.

 

[moon161]

<Disclaimer> The blue dress was in the news when I took propulsion </Disclaimer>

Static Thrust= mass flow * average exit velocity.
Mass flow ~ Air density * exit area * average exit velocity

Ts = rho * Ae * Ve^2.

Thrust is zero when your your plane moves as fast forward as your exhaust is backwards.

So if you constrict the exit duct w/ a convergent nozzle, you reduce mass flow because the fan is working against a restriction, but you increase average exit velocity. This gives you less static thrust, but when properly done, a higher maximum airspeed.

 

[zekeman]

Just noticed that the fan is airborne within the duct.
The thrust is, without external air friction is

p1A1-p2A2+mass flow rate*(V2-V1)

p1=static pressure of air entering duct
p2= static pressure at exhaust

A1 entering duct area
A2 exhaust duct area

p1 and p2 are found from fan curves total pressure vs flow rate and the fluid flow equation (energy equation)for the duct.
V1,V2 velocity at entrance, exhaust

 

[rb1957]

the thrust depends on the fan blade ... but you know the exhaust velocity ... then maybe you can use bernoulli ... 1/2*rho*V^2*A.

intuitively a nozzle would create a faster airflow, and more losses ... the thrust is developed by the fan blade ... the power from the engine is applied to the airflow by the blade, increasing the airflow's speed. making the airflow increase it's speed downstream (by a nozzle) doesn't sound like you've added energy to the airflow, but caused some to be wasted.

i disagree with the above post ... the exhaust velocity is the delta V imparted by the van. if the forward velocity is equal to this, i think the fan will still be producing thrust, albeit reduced, becuase the fan blade efficiency will be reduced, but not zero.

 

[btrueblood]

"Thrust is zero when your your plane moves as fast forward as your exhaust is backwards."

Um, no. Quick thought experiment: typical spacecraft launch vehicles reach velocities of some 7,000 meters/sec or more, yet have typical exhaust velocities of some 3,000 to 4,500 m/sec. Propulsive efficiency may decrease, but thrust pretty much stays constant, or even increases with forward speed for typical airbreathing engines (due to higher ram pressure on engine intakes).

 

[CastMetal]

btrueblood, I don't think a rocket engine is a good comparison for a 5-10lb propeller driven model. Generic equations for propeller thrust.
Thrust=.5*rho*A*(Ve^2-Vo^2)
rho= air density
Ve= entry velocity
Vo= exit velocity

.5*rho*(Ve^2-Vo^2)=pressure differential

With that I agree with moon161 given the circumstances that the exit velocity will limit your top speed.

A rocket creates a large LOCAL pressure differential and thrust velocity vs vehicle velocity is marginalized

 

[rb1957]

umm, i don't think so ...

stationary the outflow velocity is 22.3 m/s because of the energy the blade is adding to the airflow, a deltaV of 22.3 m/s.

if the plane is moving forward the deltaV will still be there, slightly reduced from the stationary value 'cause the efficiency of the blade is slightly reduced 'cuase the AoA has increased slightly. At some speed the blade will stall and thrust produced will drop significantly but i don't think it ever becomes zero.

thought exercise ... is the slipstream velocity of a spitfire something like 400 mph, ie it's max speed ? ... again, i don't think so.

 

[CastMetal]

The max speed of a spitfire not in a dive is said to be 380mph.....

Not trying to say thrust is not produced once you reach an airspeed equal to the stationary outflow only that it is a limiting factor and unlikely you can surpass it by much if at all in a propeller driven aircraft.

 

[rb1957]

actually that was my point ... i don't think the slipstream of a spit' was 380 mph, yet the plane managed to go that fast.

thus i don't think the plane's speed is limited to 22.4 m/s 'cause that's the fan's deltaV.

it's max speed is going to be limited to it's drag (and not the fan's deltaV).

 

[CastMetal]

Slipstream velocity is relative to the prop. When slipstream equals airspeed zero thrust is produced and you've reached a max speed for those conditions.

Apologize for a lot of over simplifications in my previous posts.
but drag alone will not determine a max airspeed

Drag, then Prop efficiency, then power curve of the engine will all be important factors in determining max airspeed.

Stationary delta V is related to the prop and engine power curves but is not directly correlated to max airspeed

 

[btrueblood]

What rb said. Given CastMetal's theory, the propulsive thrust of a propellor would be a maximum with the plane at its lowest speed, or even at zero speed, and then fall off as the plane's speed increased. In reality, all the plots of prop thrust I've seen show a somewhat nonlinear increase in thrust as speed increases, up to about Mach 0.5 or 0.6, then it starts to fall off. Free propellors limit out at medium-high Mach, because the prop tips start to break mach 1, locally, limiting their effectiveness at high speeds. Ducted fans are a way to get around that limit, by diffusing and slowing incoming air, limiting prop tip speeds to ones that are closer to design speed of the fan. Modern turbofan engines on jet airliners get the majority of their thrust from the fan section, not the turbine core.

The basic thrust equations for rockets work for ducted fans too. Thrust = (mdot)*Vexit, plus some pressure expansion terms.

 

[rb1957]

i'd prefer to say slipstream velocity is relative to the plane (rather than the prop). when you're doing lift calcs for a wing there's a correction for the wing-in-slipstream, where the dynamic pressure is higher.

slipstream velocity = vehicle velocity + deltaV
the OP is telling us with vehicle velocity = 0, slipstream = deltaV = 22.4 m/s

the vehicle can quite possibly travel faster than 22.4 m/s, depending on the drag of the vehicle and the efficiency of the fan producing thrust.

 

[zekeman]

I reconsidered this problem and if for the moment we ignore internal friction and look at the overall picture, I see that the power required to move the syatem is simply

U*T

U = Air speed
T= thrust

The maximum thrust you can get would coincide with the maximum fan power that can be delivered. If you look at the fan curve, there is a point (assuming constant fan speed) that the power is maximized, at a combination of
m' and pL
m'= mass flow rate lb/hr
pL= total pressure of duct airflow=p+rhoV^2/2g
To get the maximum you have to adjust the exit area to get m' and the corresponding pL that maximizes fan power

Absent friction, I did the Bernouli equation and got

Uv1/g+v1^2/2g+pl/rho+pL/rho+pl/rho=Uv2/g+v2^2/2g+ p2/rho

Now, the left side is known= K and the RHS is a function of v2 and p2

The airframe thrust equation is
p2A2-p1A1+m'(v2-v1)

p2A1*v1/v2-p1A1+m'(v2-v1)

I maximzed this to get

-(U+v2)/g*A1*v1/v2+A1*v1/v2(K-Uv2/g+v2^2/2g)+m'/g=0

solve for v2 and from that you get A2 and v2