# Motor size and efficiency: both gas and electric

Awl --
Suppose I have a 10 hp motor, and then a 20-40 hp motor -- or even a 100 hp motor..
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?
--
EA

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Existential Angst wrote:

Entirely depends on the design of the motor. I have hit/miss engines that are rated for 10HP and weigh as much as most cars! 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.
--
Steve W.

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100 hp

much"

Electric motors generally convert nearly all of the input energy into work, internal combustion engines very rarely approach even 40%
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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.
i
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Only if you are burning something to generate electricity....
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PrecisionmachinisT expressed precisely :

You have got to burn something. Hydro needs Sunshine (burns hydrogen) Photovoltaic needs sunshine (burns hydrogen) Etc.
--
John G

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Master baiter ?
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PrecisionmachinisT wrote:

No, you're just a piss poor baiter.
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wrote:

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 >>--
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PrecisionmachinisT wrote:

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.
Jon
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best

All of which explains why I said "very rarely approach even 40%"
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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.
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hp

ch"

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 heat.
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.
Stan
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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 and output.
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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