prop rpm question

I believe
there is more to it than just VE. I don't believe that bearing friction
is linear with RPM for example. Also, speed of the flame front becomes
and issue at higher RPM. I believe the drop-off in torque with RPM is a
function of a number of factors.

Yup you're right, there's more than just volumetric efficiency, but
flame front speed in these slow engines is still around 100 feet per
second, while average piston speed won't be much over 40 or 50 fps with
the midpoint travel being somewhat higher. The intake and exhaust
systems present more drag at higher RPMs and start to affect the
performance, and in many modern auto engines four valves per cylinder
are used to ease breathing.
I wonder if the new direct-drive diesel aircraft engines have much
higher torques in the right places?

Dan

wrote

I wonder if the new direct-drive diesel aircraft engines have much
higher torques in the right places?

Torque out the butt, and the torque stays high for a longer period of time
on the power stroke. In short, it will turn the same size prop of an engine
with nearly twice the HP.
--
Jim in NC

wrote:
I believe
there is more to it than just VE. I don't believe that bearing friction
is linear with RPM for example. Also, speed of the flame front becomes
and issue at higher RPM. I believe the drop-off in torque with RPM is a
function of a number of factors.


Yup you're right, there's more than just volumetric efficiency, but
flame front speed in these slow engines is still around 100 feet per
second, while average piston speed won't be much over 40 or 50 fps with
the midpoint travel being somewhat higher. The intake and exhaust
systems present more drag at higher RPMs and start to affect the
performance, and in many modern auto engines four valves per cylinder
are used to ease breathing.
I wonder if the new direct-drive diesel aircraft engines have much
higher torques in the right places?

Dan


It would seem they would. The high compression ratio gives most diesels
outstanding torque at fairly low RPMS, however, many also have a very
narrow torque peak.

Matt

On Tue, 17 Jan 2006 22:48:09 -0800, "skyloon"
wrote:


Kershner states that the maximum thrust force occurs when the plane is
standing still (at a fixed throttle setting, I guess), and decreases as you
go faster. I do not understand this. Is it beacese AOA is largest? I am
trying to see how this relates to power. Power would be force*distance/time
or force*velocity. Maybe the thrust decreases slowly with airspeed, but the
power still goes up as you go faster.

This is just a hand waving argument. Please, anyone who knows more, feel
free to correct this picture.

Dave

I'll pick up on this one. There's a mechanics equation which is
specially straight-forward. It says if you apply a constant force to
an object,
and it moves in the direction of the force, then the work done is the
product of force times distance.

As expressed in the SI system, it's specially simple: F X D = W
gets the units of
F in Newtons times Distance in Meters equals work in joules
Even more interesting: F X V = P
force times speed = power.
In SI units again:
force in Newtons times speed in meters per second = power in Watts

OK that was the engineering/physics.

Now the application:
An airplane with a constant power recip prop engine.
Lets say the engine is putting out 90 HP say a C-152
90 HP = 90 X 746 watts = 67kW

Lets check the numbers at 10 mph, 50 mph and 100 mph
10 mph = 4.5 meters/sec
50 mph = 22.4 meters/sec
100 mph = 44.7 meters/sec

The unknown in the following equation is F
F X V = P or F = P/V

Now force is the same measure as thrust, so now we can
check available thust at these three speeds:

10 mph
F = 67000 W/4.5 M/Sec = 14890 Newtons
A newton, like a small apple weighs a quarter pound about.
So 14890 Newtons = 3340 pound That's a lot of thrust!

Now 50 mph
F = 67000/22.4 = 2990 Newtons or 671 lb.

Now 100 mph
F = 67000/44.7 = 1500 Newtons or 336 lb.

Or the general rule: the faster you go with constant power, the less
the thrust available.
Same applies to boats.

But think about planes with (some) jet engines,
these can be constant THRUST.

That means, the faster they go, the more HP they put out!
(A reason why jets on slow planes is not a great idea)

Brian Whatcott Altus OK

On Sun, 22 Jan 2006 at 20:18:32 in message
, Brian Whatcott
wrote:

But think about planes with (some) jet engines,
these can be constant THRUST.

That means, the faster they go, the more HP they put out!
(A reason why jets on slow planes is not a great idea)

No, is not quite that easy. Another way of looking at force is that it
is rate of change of momentum.

In a simplified way the thrust of a jet engine comes from the change of
momentum from the air captured by the engine to the momentum of the air
that leaves the back of the engine. If you think of a Turbofan engine
then the fan is not so different from a propeller. The internal
efficiency of the two types of engine is somewhat different.

In any case the greatest propulsive efficiency comes from a momentum
change of a large mass of air with a very small velocity change.
--
David CL Francis

On Tue, 24 Jan 2006 22:48:26 GMT, David CL Francis
wrote:

On Sun, 22 Jan 2006 at 20:18:32 in message
, Brian Whatcott
wrote:

But think about planes with (some) jet engines,
these can be constant THRUST.

That means, the faster they go, the more HP they put out!
(A reason why jets on slow planes is not a great idea)

No, is not quite that easy. Another way of looking at force is that it
is rate of change of momentum.

In a simplified way the thrust of a jet engine comes from the change of
momentum from the air captured by the engine to the momentum of the air
that leaves the back of the engine. If you think of a Turbofan engine
then the fan is not so different from a propeller. The internal
efficiency of the two types of engine is somewhat different.

In any case the greatest propulsive efficiency comes from a momentum
change of a large mass of air with a very small velocity change.


Hi David,

let's forget about jets and recips. Let's imagine a vehicle that is
provided with a constant thrust device.
Then, the faster it goes, the more hose power it provides.
You can take it to the bank

Brian

On Thu, 26 Jan 2006 at 19:29:21 in message
, Brian Whatcott
wrote:

Hi Brian,

Hi David,

let's forget about jets and recips. Let's imagine a vehicle that is
provided with a constant thrust device.
Then, the faster it goes, the more hose power it provides.
You can take it to the bank


No don't let us forget the basic ideas of propulsion.

A constant thrust device is doing one of two things;

1. Accelerating. In which case it is adding to its kinetic energy
and its power is going into that or

2. It reaches a constant speed against a constant drag and a steady
state occurs..

I presume you are not claiming that a constant thrust motor can generate
infinite power? In that case you would be right!

What do you have in mind as a constant thrust device? Newton's laws are
pretty good and I am not aware of any means of getting around them. They
only need adjustments at velocities and masses far beyond normal
terrestrial transport activities.

It is of course true that the efficiency of propulsion devices does vary
with speed and many other conditions.

--
David CL Francis

On Mon, 30 Jan 2006 00:26:39 GMT, David CL Francis
wrote:

[Brian]
let's forget about jets and recips. Let's imagine a vehicle that is
provided with a constant thrust device.
Then, the faster it goes, the more hose power it provides.
You can take it to the bank


[David]
No don't let us forget the basic ideas of propulsion.

A constant thrust device is doing one of two things;

1. Accelerating. In which case it is adding to its kinetic energy
and its power is going into that or

...Good

2. It reaches a constant speed against a constant drag and a steady
state occurs..


....Good

I presume you are not claiming that a constant thrust motor can generate
infinite power? In that case you would be right!

I am claiming that engineers are familiar with two Newtonian equations
1) force times distance (in the direction of the force) = work
2) force times velocity in the direction of the force = power

I understand that it is non-intuitive to non-engineers that the
arrangement of
eqn 2) as
3) force = power / velocity is always true until large fractions of
c.

Though I cannot offer any further debate with you on this topic,
(unless you wish to pay me) it is helpful for you to know that thrust
is an equivalent term to force, and so for a vehicle with thrust
5 units and velocity 10 units, its power is 50 units.
5 = 50 / 10

Moreover, I am pretty sure you can work out the missing term in this
question
What is the power of a vehicle with thrust 5 units, and velocity 20
units? 5 = P / 20





(power = 100 units)

I urge you to contemplate the great simplicity of Newton's laws - and
their interesting practical applications.

Respectfully

Brian Whatcott Altus OK

On Thu, 2 Feb 2006 at 17:59:33 in message
, Brian Whatcott
wrote:

I am claiming that engineers are familiar with two Newtonian equations
1) force times distance (in the direction of the force) = work
2) force times velocity in the direction of the force = power


The above statements are true but not the whole story.

Are you considering a situation on a vehicle passing through an
atmosphere or a rocket in empty space? If there is no drag then the body
will accelerate as long as the thrust is present. Rockets can produce a
constant thrust but by their very nature the mass of the body reduces as
fuel is used up and the acceleration increases. What is the measure of
the power developed? The momentum change taking place as the fuel and
working fluids are expelled through the jet at their inherent velocity
or the distance the rocket moves? In space there is no resistance and
therefore the distance moved is irrelevant to power. The power is
transformed into a velocity change and kinetic energy. In an atmosphere
speed will increase to a steady state when thrust and drag are equal. Of
course they are both forces.

I understand that it is non-intuitive to non-engineers that the
arrangement of
eqn 2) as
3) force = power / velocity is always true until large fractions of
c.

Though I cannot offer any further debate with you on this topic,
(unless you wish to pay me) it is helpful for you to know that thrust
is an equivalent term to force, and so for a vehicle with thrust
5 units and velocity 10 units, its power is 50 units.
5 = 50 / 10

I do not feel inclined to continue this against such patronising
elementary statements.
--
David CL Francis