Suppose I have a 10 hp motor, and then a 20-40 hp motor -- or even a 100 hp
Suppose I am loading any of these motors at exactly 10 hp worth of load.
Which one has the greatest efficiency? I'm interested in both gas and
electric, but not comparing gas with electric.
Just curious about car design, ito total overall hp -- does "over
horsepowering" a car reduce its efficiency, when operating at only a
fraction of that horsepower? I'm sure the answer is "yes", but "how much"
of a yes?
Entirely depends on the design of the motor.
I have hit/miss engines that are rated for 10HP and weigh as much as
I also have a couple 2 strokes that run well over 100 HP and are light
enough to carry.
Both use about the same amount of fuel under load.
That is comparing apples and oranges, though. Someone has to burn
something, in order to boil the water, that would turn the turbine,
that would turn the generator, that would feed the high voltage
transformer, that would send current along a 1,000 mile wire, to a
distribution transformer, and to the low voltage transformer, to run
that "efficient" 10HP motor.
For electric motors, there isn't a whole lot of reason to go a lot
larger than you need, because the 20-HP motor still has the same
internal resistance regardless of load.
IOW it will vary with load, but even loaded at 50% a 20-HP motor will
draw more current than a 10-HP Motor loaded at 100% - BUT there are
those exceptions to consider.
Sometimes having the 20-HP motor lightly loaded is better. Because
you can accidentally or deliberately overload it quite a bit and
nothing bad happens. You try putting 20-HP of load on a 10-HP motor
on something like a rock crusher where it's real easy to overload it
as it chews through a too-large batch, and you'll have a pile of
glowing slag real fast.
Right - but even the most developed 'natural' source, Hydroelectric,
still has to be conserved as if it's a fossil fuel - there isn't
enough dammed-up water to allow rampant waste. And those huge dams
are a huge investment that has to be paid back out of the revenues,
more than dozens of comparable coal or gas fired plants.
That said, by the time you get done with all the processes to get the
fossil fuel out of the ground and ready to use, burn the fuel and run
it through the power plant, then get that electricity to where you
want to use it... Maybe 20% efficient after all is said and done.
You have to figure in all the costs and energy usage like running all
the trucks and heavy equipment at the coal mine, plus the electricity
for running the drag-link shovel. Then transport the coal from the
mine to the power plant, etc. And the energy to reclaim the land
after the coal seam has played out.
Or the energy used to run the drilling rig to drill the oil or gas
well,run the pump jack to get the oil out of the ground or compressors
to clean and transport the gas, build and run the pipeline to the
refinery, the refinery itself, transport the fuel to the power plant.
--<< Bruce >>--
40%? A typical Otto-cycle engine gets about 13% efficiency at its best
operating point, which is near full throttle (to get high cylinder
temperatures and low pumping loss).
The Honda Civic Hybrid variable-cycle semi-Atkinson cycle engine does
much better at a lower power setting. They optimized it so it does
best at constant-speed cruise on a level road. I don't know of any
other production auto engines that work like that.
On Sun, 4 Nov 2012 17:20:29 -0500, "Existential Angst"
With gasoline engines you get maximum efficiency where maximum
torque is produced. So you get best fuel economy in highest gear at
max torque RPM if the aerodynamic drag does not get you first.
My 4.0 Ranger puts out about 220 ft lb at 2400 RPM. I'm playing with
the ScanGuage 2 seeing what gives the best mileage.
I'm finding that short shifting gives me best mileage if I don't care
how long it takes to get up to speed and if I'm willing to loose speed
on upgrades. Poke and glide gets me just about the same if I
accellerate through the gears to 2400 RPM reasonably quickly and then
run in the highest gear I can at the speed. 3.55 gears puts 2400 RPM
at just over 100kph. Haven't had it out on the highway yet to see what
it will do.
1975 Celica got 52MPG at just under 80 MPH sustained on the open road
in 5th gear. Less than 40 at 60MPH. Just happens 80mph was right at
the torque peak in 5th. They changed the gearing in 76, and that
reduced the mileage significantly.
With electric motors, given the same power out, the motor sized
closest to the actual demanded power will be most efficient, given the
same construction on all the motors. It costs energy to move the
rotors after all, bigger motor, bigger rotor, more mass, more energy
needed. Also bearing drag will be more with larger motors. There
could be circumstances where a smaller moter is markedly less
efficient than a good larger motor, less iron, not as good quality
bearings, larger magnetic gaps, etc. Would be like the difference
between an el-cheapo dedicated motor out of a current appliance and a
top quality industrial motor. One is built to a price point, the
other is built to a performance standard. Improved higher temp
insulation has led to a lot of penny-shaving out there in motor
design. They can cut costs by including less iron, but the motors
won't be as efficient as they could be, they lose a lot of energy as
As far as IC engines, you STILL have the same power needed to move a
vehicle down the road at a given speed, the mass of the car isn't
changing, E=MV^2 is still valid. So a small engine still needs to put
out the same power output as a large engine at the same speed. With
Ford, several of their engine families used the same block design for
larger displacement engines as they did for lower displacement ones.
Becuase the higher displacement engines had larger bores, the blocks
actually weighed less than the smaller engines. If you picked the
right RPM the larger engine could be marginally more efficient jsut
because the vehicle mass was slightly less. They usually had
different carbs, though, performance was expected out of the big
bores, so they usually sucked more gas. The answer's going to depend
a lot on engine management, pumping loses, valve train design, amount
of powder just to keep the parts moving and gas quality. So no easy
answer on your second question.
On Thu, 8 Nov 2012 08:45:56 -0800 (PST), Stanley Schaefer
Because of the above, the most efficient engine for the job is the one
that will produce the required torque at the required RPM at about 80%
throttle. This is particularly true for stationary engine use - and
boats and other things that run at a relatively constant load.
Pumping losses are lowest at close to full throttle - particularly if
the intake is sized properly to breath freely at the required speed
As a rough rule of thumb an engine will produce ABOUT 1.2 ft lbs
torque per cubic inch - so if you want an engine to produce 5HP at
1800 RPM you need 14.6 Ft Lbs of torque - which requires about 18
cubic inches displacement. That is about 300cc - apply the 80% and you
get 22.5 cu in or 375cc.
A smaller engine will not be able to produce the required power, and a
larger engine will loose efficiency.
This is not an exact calculation - it depends on the afferage
effective cyl pressure - which depends on intake and exhaust tuning,
compression ratios, etc - but is a close approximation. So a 3.8 liter
or 225 cu inch engine is good for ABOUT 50 HP at 1800 RPM.
The 1.2 ft lbs per cu inch is based on the old flathead low
compression engine if I remember correctly - so a higher compression
automotive engine MAY remove the 80% - or change the 1.2 to closer to
1. or 1.1
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