dumb question on compressor displacement cfm
Grant Erwin
and can run at 900 rpm max.
I am trying to figure the displacement CFM and I'm coming up with an astonishing
and quite unrealistic 45.8 CFM.
I think the actual number is more like 14 to 18. But I can't see what I'm doing
wrong.
Help!
3.14 * 2^2 * 3.5 * 2 / 12^3 * 900 = 45.82-ish
Pete C.
45.82 CFM at 0 PSIG? Like a muffin fan?
I think you need to adjust from raw displacement at 0 PSIG to the
portion of the displacement that remains after the air charge has been
compressed to the output pressure. After all, until the pressure in the
cylinder reaches the pressure in the tank, no air will flow through the
check valves into the tank.
Pete Keillor
Your math for swept volume looks o.k. However, is your rating in CFM
at some pressure? Also, you're not accounting for compression ratio.
The less compression ratio, the less efficiency. The volume you swept
loaded at ~14.7 psia at ambient temp. It exhausts at the tank back
pressure and a higher temperature, and that volume between the piston
and the reed valve at tdc re-expands, limiting the amount of air
sucked in. (Based on that, I guess good design would minimize that
volume by placing the valves very close to the pistons at tdc.)
No, I didn't calculate it. I'd have to dig out thermo texts and I
think I chunked them. Just thinking about it makes my head hurt.
Pete Keillor
Pete C.
If I have this correct, assuming 90 PSIG output pressure you need 7:1
compression to get to output pressure where you actually feed into the
tank. 3.5" stroke / 7 = .5" output stroke * 2 cylinders = 1" effective
output stroke * 12.56 square inches cylinder bore = 12.56 CFM @ 90 PSIG
(with some rounding error and assuming minimal dead air space between
the cylinder and the valves).
Pete C.
Bah! Forgot the conversion from CIM to CFM and RPM in the equation, but
you get the idea. 6.56 CFM @ 90 PSIG I think is the result.
Pete C.
3.14 * 2^2 = 12.56 square inch cylinder bore
15 PSIA to 105 PSIA = 7:1
3.5" stroke / 7 = 0.5" output stroke
0.5" output stroke * 2 cyl = 1" output stroke
1" output stroke * 12.56 square inch bore = 12.56 cubic inches output
per revolution
12.56 cubic inches output per revolution * 900 RPM = 11,304 cubic inches
output per minute
1 cubic foot = 1728 cubic inches
11,304 cubic inches output per minute / 1728 = 6.54 cubic feet output
per minute
Wild_Bill
that result is generally referred to as: free air displacement, meaning
pressure doesn't enter into the calculation, (essentially the same as the
cubic inch or liter, or CC displacement in engines).
If the pump was used to inflate a very thin, empty/collapsed plastic bag
without reaching the point of stretching the bag, the results would be
similar.
--
WB
.........
metalworking projects
www.kwagmire.com/metal_proj.html
"Grant Erwin" <grant...@kirkland.net> wrote in message
news:QPd0m.40479$QS....@newsfe18.iad...
Pete C.
Pete C.
3.14 * 2^2 = 12.56 square inch cylinder bore
Richard J Kinch
> I can't see what I'm doing wrong.
Your displacement CFM calculation looks correct.
Does it really run 900 rpm? That would take like a 10 or 15 HP motor to
deliver 90 psi. Maybe you have something like a 1725 rpm 2 HP motor and
a 10:1 pulley, so you're only running at 172 rpm?
Pete Keillor writes:
> However, is your rating in CFM at some pressure?
No, no, no. Please don't start up this old canard. Compressor CFM is
measured in FREE AIR, not compressed. When a compressor pumps one
"CFM" (cubic foot per minute), that means the intake port inhales one
cubic foot of "free air" (air at atmospheric pressure, which is 0 psig)
every minute.
CFM is a unit of mass flow per time, not volume.
SteveB
Don't worry about it. I've seen ratings on machines at the stores that have
been done by high dollar engineers, and they're farther off than yours.
Steve
Grant Erwin
> Grant Erwin writes:
>
>> I can't see what I'm doing wrong.
>
> Your displacement CFM calculation looks correct.
>
> Does it really run 900 rpm? That would take like a 10 or 15 HP motor to
> deliver 90 psi. Maybe you have something like a 1725 rpm 2 HP motor and
> a 10:1 pulley, so you're only running at 172 rpm?
It's run by an 18hp gas engine.
The driving pulley is 5-1/2" and the driven pulley is 19" and the engine
runs between 2200 and 3000 rpm depending on how the variable speed
control is set. So yes, I can drive the pump at 900 rpm, and yes, 900
rpm is the max speed that this air pump (a Quincy model 244) can run.
Grant
Robert Swinney
acting piston there would be only 1/2 active stroke per cyclinder per revolution or 900 total
strokes per minute.
Bob Swinney
"SteveB" <old...@deepends.com> wrote in message news:1eq8h6-...@news.infowest.com...
Pete C.
Robert Swinney wrote:
>
> Swept volume (not compressed volume) would be 1/2 half of Grant's calculation. Assuming a single
> acting piston there would be only 1/2 active stroke per cyclinder per revolution or 900 total
> strokes per minute.
How do you figure that? I think you are confusing a compressor with a 4
cycle engine.
Grant Erwin
Or is it that they both go up each rev but the valves only close on one or the
other?
Actually, a factor of 2 is probably about close. Back in the mid-90s I was told
that this compressor is capable of about 18 CFM IIRC. And half of 45 would be
22.5 which if the ACFM (at some normal condition) were 18 would give a
volumetric efficiency of 80% which isn't an unreasonable number.
Grant
Pete C.
Grant Erwin wrote:
>
> How do you make a crankshaft that only moves the piston up every other rev?
I believe he is confusing a compressor with a 4 cycle engine, where the
four cycle engine requires two crankshaft revolutions to complete the
intake-compress-power-exhaust cycle, unlike a compressor which requires
only one revolution for the intake-output cycle.
Grant Erwin
I don't know about what he's thinking, Pete, but you are correct. I called
my local Quincy service center and talked to the service manager. Each
piston compresses on each stroke.
He also gave me the actual specs for the Quincy Model 244:
770 rpm -> 21.4 CFM @ 100psi
900 rpm -> 25.4 CFM @ 100psi
This also computes, since I remember asking the Quincy guys back in the '90s
if I were to replace the gas engine with an electric motor, what size, and
they told me 7.5hp. Given Richard Kinch's rule of thumb (4 CFM / hp) these
numbers make sense.
What *still* doesn't make sense to me is the free air displacement CFM.
All I know about this compressor is:
2 cylinders
bore: 4"
stroke: 3-1/2"
each cylinder compresses on each rev
max rpm: 900
I'm figuring something like 45 cfm free air displacement at 900 rpm, and
the actual CFM at 100 psi is more like 25.4 cfm. This would give an
absurdly low volumetric efficiency of 62%.
Anyway, it's just a puzzle. Maybe someday I'll figure it out. In the
meantime, like most Quincy pumps, the actual machine just keeps on
running.
Grant
Robert Swinney
aft travel of the piston per revolution. Each piston sucks air in on 1/2 stroke and blows air out
on the other 1/2 stroke. Therefore 900 half strokes of 2 pistons would be the free flowing CFM of
the device. Grant's original math was correct except he multiplied volume by 2. 1/2 + 1/2 =1.
Bob Swinney
"Grant Erwin" <grant...@kirkland.net> wrote in message news:u3t0m.54653$gz5....@newsfe07.iad...
Ignoramus28618
> Probably the statement, "Each piston compresses on each stroke" means that: A stroke is fore and
> aft travel of the piston per revolution. Each piston sucks air in on 1/2 stroke and blows air out
> on the other 1/2 stroke. Therefore 900 half strokes of 2 pistons would be the free flowing CFM of
> the device. Grant's original math was correct except he multiplied volume by 2. 1/2 + 1/2 =1.
Every revolution of crankshaft is two half strokes.
i
Grant Erwin
revolution of the crank and blows air out on the other 1/2. But the piston
travels the full stroke length each time, and of course has the same bore.
And there are two cylinders.
((pi * (bore/2)^2 * stroke) / 1728) * rpm TIMES NUMBER OF CYLINDERS
i.e. total cyl vol. in cubic inches, converted to cubic feet, times rpm = cfm
of the free air displacement variety
I wish I could understand how you figure a cylinder only pumps half its volume
per crankshaft revolution. If that were true it would make mathemtical sense.
Grant
James Waldby
> Probably the statement, "Each piston compresses on each stroke" means
> that: A stroke is fore and aft travel of the piston per revolution.
> Each piston sucks air in on 1/2 stroke and blows air out on the other
> 1/2 stroke. Therefore 900 half strokes of 2 pistons would be the free
> flowing CFM of the device. Grant's original math was correct except he
> multiplied volume by 2. 1/2 + 1/2 =1.
900 strokes per minutes (which Grant referred to as 900 rpm in
another post) is 900 compressions and 900 intakes per minute.
In Grant's original calculation (which was something like
(4*3.5*pi/12^3)*900*2 CFM), the factor of 2 is for 2 cylinders.
Each compression in one cylinder pumps 4*3.5*pi = 43.98 cubic
inches of air, or about .0255 cubic feet. Hence
4*3.5*pi/1728*900 CFM or 22.9 CFM per cylinder, or about
45.8 CFM for two cylinders.
--
jiw
Pete C.
Grant Erwin wrote:
>
> It seems to me, Robert, that more correctly each piston sucks air in on 1/2 a
> revolution of the crank and blows air out on the other 1/2. But the piston
> travels the full stroke length each time, and of course has the same bore.
> And there are two cylinders.
>
> ((pi * (bore/2)^2 * stroke) / 1728) * rpm TIMES NUMBER OF CYLINDERS
>
> i.e. total cyl vol. in cubic inches, converted to cubic feet, times rpm = cfm
> of the free air displacement variety
>
> I wish I could understand how you figure a cylinder only pumps half its volume
> per crankshaft revolution. If that were true it would make mathemtical sense.
Remember that the compressor may not "pump" anything at all depending on
the input pressure (atmospheric on a single stage compressor) and the
output pressure. If the outlet pressure requirement / back pressure is
sufficiently high, and particularly if there is a moderate amount of
unswept dead space, there will be no output at all from the compressor
and no input either as the compressor will simply keep compressing and
decompressing the same volume of air. This makes more sense when you
consider multi stage compressors and gas boosters.
RoyJ
> Each compression in one cylinder pumps 4*3.5*pi = 43.98 cubic
> inches of air, or about .0255 cubic feet. Hence
> 4*3.5*pi/1728*900 CFM or 22.9 CFM per cylinder, or about
> 45.8 CFM for two cylinders.
>
Correct. And if it uses reed valves, a volumetric efficiency of ~60% is
pretty good.
Richard J Kinch
> It's run by an 18hp gas engine.
Then 45 cfm at 90 psi is entirely within reason; 1 hp being roughly
equivalent to 3 to 4 cfm at 90 psi.
Robert Swinney
"volume" of air each stroke or 1/2 revolution.
Bob Swinney
"Grant Erwin" <grant...@kirkland.net> wrote in message news:LYw0m.28$k_...@newsfe01.iad...
Grant Erwin
Ah, but there is something even you may not know yet, Mr. Kinch. One
horsepower from a gas engine doesn't equal one horsepower from an
electric motor (assuming this is not something sold at Home Depot,
in which case the horsepower rating is largely fictional). The
good people at Quincy a long time ago told me that the electric
equivalent of an 18 hp gas engine is a 7.5hp electric motor.
My compressor's specs are now known:
770 rpm -> 21.4 CFM @ 100psi
900 rpm -> 25.4 CFM @ 100psi
So to get 25.4 CFM you would indeed need most of a 7.5hp electric motor.
Why is a gas engine horsepower different from an electric motor horsepower?
Oh, boy. <ducking>
Grant
Curt Welch
> Grant Erwin writes:
>
> > I can't see what I'm doing wrong.
>
> Your displacement CFM calculation looks correct.
>
> Does it really run 900 rpm? That would take like a 10 or 15 HP motor to
> deliver 90 psi. Maybe you have something like a 1725 rpm 2 HP motor and
> a 10:1 pulley, so you're only running at 172 rpm?
>
> Pete Keillor writes:
>
> > However, is your rating in CFM at some pressure?
>
> No, no, no. Please don't start up this old canard. Compressor CFM is
> measured in FREE AIR, not compressed. When a compressor pumps one
> "CFM" (cubic foot per minute), that means the intake port inhales one
> cubic foot of "free air" (air at atmospheric pressure, which is 0 psig)
> every minute.
Yeah, that's something it took me some time to understand. I assumed
(wrongly) that if it was rated as X CFM @ 90 PSI, they were measuring the
volume at 90 PSI. But that's just not true. It's as you say, the volume of
air sucked in they are measuring. The compressor is rated at different
PSI values only because the CFM drops as the back pressure of the tank
rises.
I'll answer the original question if no one else made it clear in the
thread (I have another 20 posts to read first).
> CFM is a unit of mass flow per time, not volume.
>
> http://www.truetex.com/aircompressors.htm
--
Curt Welch http://CurtWelch.Com/
cu...@kcwc.com http://NewsReader.Com/
Don Young
generally has a throttle and is therefore economically not designed to
deliver its maximum horsepower constantly. They generally have a fairly
constant torque over a wide RPM range so the horsepower is pretty much
proportional to the speed. Onan and probably others have built nearly
identical engines rated at 1800 and at 3600 RPM, with the higher speed
giving about twice the horsepower. The electric motor can deliver its rated
horsepower (or slightly more) constantly but can not deliver a lot more even
for a short time. The gasoline motor is commonly rated for the maximum power
it can deliver, but it can not deliver that amount constantly and have a
satisfactory service life. There are industrial internal combustion engines
which are rated for continuous horsepower.
A useful rating system would give both peak and continous horsepower and
motors are sometimes so rated (and over-rated). Actually, the horsepower to
drive a compressor at a certain speed and pressure is exactly the same
regardless of the type of motor. But a 5HP electric motor can deliver that
5HP constantly with good life but a 5HP gasoline motor would be running
"wide-open" and would not last long.
Lots of other ways to look at it; electric motors can withstand short term
torque overload without stalling better than a gasoline engine, for example.
Just quite different speed, torque, and other characteristics for the two
types of motors.
Don Young
Grant Erwin
My engine (a Wisconsin THD) is an industrial gas engine designed to run at
one speed (at a time). It's horsepower varies somewhat with RPM, from say
14.1 hp @ 2200 rpm up to 18.0 at 3200 rpm. It is fitted with a variable
speed control but it isn't like a gas pedal - it only varies the speed
over a small range. This engine never idles.
I stand by my earlier statement. A hp from a gas engine - ANY gas engine -
is less than half the hp from an electric motor.
GWE
Curt Welch
> Grant Erwin wrote:
> >
> > It seems to me, Robert, that more correctly each piston sucks air in on
> > 1/2 a revolution of the crank and blows air out on the other 1/2. But
> > the piston travels the full stroke length each time, and of course has
> > the same bore. And there are two cylinders.
> >
> > ((pi * (bore/2)^2 * stroke) / 1728) * rpm TIMES NUMBER OF CYLINDERS
> >
> > i.e. total cyl vol. in cubic inches, converted to cubic feet, times rpm
> > = cfm of the free air displacement variety
> >
> > I wish I could understand how you figure a cylinder only pumps half its
> > volume per crankshaft revolution. If that were true it would make
> > mathemtical sense.
You create some of the more fun questions here Grant!
> Remember that the compressor may not "pump" anything at all depending on
> the input pressure (atmospheric on a single stage compressor) and the
> output pressure. If the outlet pressure requirement / back pressure is
> sufficiently high, and particularly if there is a moderate amount of
> unswept dead space, there will be no output at all from the compressor
> and no input either as the compressor will simply keep compressing and
> decompressing the same volume of air. This makes more sense when you
> consider multi stage compressors and gas boosters.
A few people seemed to have touched on the answer, but no one seems to have
made it clear. Let me try.
The problem lies with the dead space at the top of the cylinder. You can't
calculate the CFM unless you also have an good measure of the effective
volume of that dead space (and I have no clue what is typical).
The problem Grant, is what Pete says here. Not all the air in the stroke
actually gets pushed out into the tank. The worst case is that none of it
gets pushed out. But even with the normal case, not all of the air which
was in the stroke volume, will get pushed out.
The calculations that Pete posted yesterday I believe were invalid because
they seemed to be based on the assumption that CFM was a measure of
compressed volume (which is a good guess - but just not true).
Lets do an example with simple numbers. Lets say we have a stroke volume
of 90 cubic inches, and 10 cubic inches of dead space (not trying to be
realistic here). This means that the 100 ci of the cylinder gets
compressed down to 10 ci at TDC. So the pressure will go from 15 PSI, to
150 PSI in this compressor. That means that the max it can do is 150 PSI.
At 150 PSI, no air will go into the tank per the exmaple Pete gave.
At 140 PSI of tank pressure, the air won't start to flow until the cylinder
pressure reaches 140 PSI. 140/15 is 9.3e so that's a 1 to 9.3 compression
ratio we need. But because the piston must compress both the stroke volume
AND the dead space volume, we need more than 1:9.3e of stroke compression.
We need enough piston movement to create a 1:9.3e compression on that 110
ci of air before air will start to flow to the tank. The cylinder volume
needs to be reduced from 110 to 110/9.33 or 11.79 ci before air flows.
From that point, the stroke is only 1.79 ci away from TDC. So as it pumps
the air out, the pressure effectively stays at 140 PSI, but only 1.79 ci of
air gets pumped out. The rest stays in the dead space and doesn't go
into the tank.
1.79 ci of air at 140 PSI, is the same as 1.79 * 9.33 = 16.7 ci of air at
15 PSI.
So where the stroke of this example was 100 ci per stroke, the air moved
per stroke was only 16.7 ci when the pressure reached 140 PSI. So the
efficiency of the compressor drops as the pressure rises. And it's all
because of the dead space at the top of the cylinder. The larger the dead
space, the faster the CFM numbers drop as tank pressure goes up.
Now I assume for air compressors, they attempt to minimize that dead space.
But with a reed valve popping open, there must be some "blow back" as air
in the reed valve area pushes back into the cylinder as the piston starts
to fall and the valve closes. Or in other words, the area created by the
opening of the valve, adds a little to the dead space.
At the same time, I wonder if too little dead space might be a safety
concern? The compressor can only take so much pressure before blowing up
or breaking a connecting rod, and one simple sure way to limit the max
pressure is by intentionally including a little dead space. Like the
numbers I used above the system was limited to 150 PSI. Maybe for a 120
PSI compressor they might limit the pressure to 200 PSI by adding a little
dead space for safety reasons? They way, even if the pressure cutoff
switch malfunctioned, the pressure could not go above 200 PSI even with the
motor constantly running?
None the less, the only problem with Grant's numbers that I can see, is
that he didn't include the effect of the dead space at the top of the
cylinder. With the tank pressure at 0, his calculation should be close to
correct. But as the tank pressure rises, the CFM numbers will drop.
Without knowing the effective volume of the dead space, you won't be able
to calculate how much the CFM drops.
But since he has a number for the CFM, we should be able to calculate the
dead space.
Richard J Kinch
> I stand by my earlier statement. A hp from a gas engine - ANY gas
> engine - is less than half the hp from an electric motor.
Horsepower is a physical measure of power and as such it has NO reference
or variation to the source. It is the same unit for waterwheels, horses,
electric motors, bicyclists, turbines, IC engines, rockets, etc.
Of course we know that the word "horsepower" is applied to every
fantastical characterization of power with no fixed relation to the
physical unit. Like vacuum cleaners, air compressors, hi-fi amplifiers,
etc.
Jim Wilkins
> Grant Erwin writes:
> > I stand by my earlier statement. A hp from a gas engine - ANY gas
> > engine - is less than half the hp from an electric motor.
>
> Horsepower is a physical measure of power and as such it has NO reference
> or variation to the source. It is the same unit for waterwheels, horses,
> electric motors, bicyclists, turbines, IC engines, rockets, etc.
True, but the rated HP is measured at the optimum speed and load. Less
well matched real-world loads shift the operating point along the
speed / torque curve.
jsw
Bruce In Bangkok
<ki...@truetex.com> wrote:
>Grant Erwin writes:
>
>> I stand by my earlier statement. A hp from a gas engine - ANY gas
>> engine - is less than half the hp from an electric motor.
>
>Horsepower is a physical measure of power and as such it has NO reference
>or variation to the source. It is the same unit for waterwheels, horses,
>electric motors, bicyclists, turbines, IC engines, rockets, etc.
Err, not really. for example:
# One mechanical horsepower of 550 foot-pounds per second is
equivalent to 745.7 watts
# A metric horsepower of 75 kgf-m per second is equivalent to 735.499
watts
# A boiler horsepower is used for rating steam boilers and is
equivalent to 34.5 pounds of water evaporated per hour at 212 degrees
Fahrenheit, or 9809.5 watts
# One horsepower for rating electric motors is equal to 746 watts
Cheers,
Bruce
(bruceinbangkokatgmaildotcom)
Wild_Bill
type of power source.. 60 cycle AC vs. 4 cycle IC.
A 5 HP 4 cycle gas engine adequately replaces a 2 HP electric motor.
--
WB
.........
metalworking projects
www.kwagmire.com/metal_proj.html
"Grant Erwin" <grant...@kirkland.net> wrote in message
news:3OD0m.18871$t_7....@newsfe20.iad...
>
Jim Wilkins
> It's often said that the rated HP power requirement difference is in the
> type of power source.. 60 cycle AC vs. 4 cycle IC.
>
> A 5 HP 4 cycle gas engine adequately replaces a 2 HP electric motor.
>
> --
> WB
That 2 HP is (or should be) the continuous rating limited by
temperature rise, and not nearly the short-time maximum.
jsw
Robert Swinney
Dead space. It too is filled with the CFM flow. We call that "clearance in steam engines.
Bob Swinney
"Curt Welch" <cu...@kcwc.com> wrote in message news:20090626003919.328$h...@newsreader.com...
RoyJ
But:
A gas engine has a max torque that is purely dependent on the burning
fuel in the cylinder. What's there is there. If you overload it, it just
kills. A big flywheel helps but that means the speed sag is quite long.
In contrast, an electric motor is quite capable of putting out much more
**PEAK** HP (The Sears HP number) for a short period. The instantaneous
peak can be anywhere from 3x to 10x rated HP depending on motor type and
construction. The motor simply draws more current than rated for the
short period. This matches nicely with real world loads that often have
short period high loads.
The net is that for high peak load situations, you can indeed use a much
smaller rated electric motor.
If you want to see where some of this falls apart, think about running a
large water pump (sewage lift station) out at the end of a LONG power
line. When you overload the electric motor on a hard start, the extra
current drops the voltage down so low that the motor doesn't really get
the extra power. In some cases you can get a full motor stall, the unit
will never start. It lets the smoke out of a rather expensive piece of
equipment in a few seconds.
Jim Wilkins
The short version is that if the tank pressure is high enough, the
compressed air in the dead space won't expand to below atmospheric
pressure when the piston is down, and no more air will be sucked in.
jsw
Richard J Kinch
> # One mechanical horsepower of 550 foot-pounds per second is
> equivalent to 745.7 watts
> ...
> # One horsepower for rating electric motors is equal to 746 watts
Rounding errors do not constitute a variation in the fundamental units of
physics.
Richard J Kinch
> In contrast, an electric motor is quite capable of putting out much more
> **PEAK** HP (The Sears HP number) for a short period.
At the age of 53 I am still myself capable of putting out more than 1 HP
for brief periods (and they seem to become briefer with each birthday, soon
to fall below zero). But surely even the strongest man alive still
qualifies for only fractional "rated" horsepower. Prefixing "rated" to a
specific physical unit makes it meaningless, and any argument in support of
such flimsy talk is a definitional retreat.
Grant Erwin
Well, *something's* going on here. My Quincy compressor requires an 18hp
gas engine to run it. But it can be run by a 7.5hp electric motor.
If it didn't take an 18hp gas engine to run it, why would they have built
this machine with one on it?
Your own rule of thumb (4 cfm / hp) bears out 25.4 cfm from 7.5hp electric.
I know that theoretically a horse is a horse.
So how do *you* explain it?
I'm not trying to diss you, Richard. I very much enjoyed your CO2 page and
made a similar setup with which I taught my children to make their own
soda pop. The result of that was that after they saw cup after cup of
sugar going in, they stopped drinking pop. I also learned really a lot
from you in the big air compressor thread from a few years ago.
Grant Erwin
RoyJ
peak) hp that a given motor can put out. If you try to make your 53 year
old body put out the same hp as a 19 year old body for an extended
period you will DIE just as certainly as a 1 hp motor putting out 1.5 hp
for an extended period.
Jim Wilkins
Nevertheless that's how electric motor HP is given, and the time
dependence of power output that electric motors have but gas ones
don't is part of the reason for the non-equivalence. Another may be
because small gas engines have been found to rarely meet spec. If you
want to compare gas HP to Sears HP they may be equal. Watt's original
definition of the 24/7 horse power rating of his steam engines was
extremely indefinite; at any instant several of the horses might be
sleeping.
I'm 62 and found out last weekend that I can't quite manage 1 HP any
more, while unloading a pallet of cement bags. Lifting an 80 Lb bag 4
feet in 1/2 second is a bit over 1 HP.
jsw
Richard J Kinch
> I'm 62 and found out last weekend that I can't quite manage 1 HP any
> more, while unloading a pallet of cement bags. Lifting an 80 Lb bag 4
> feet in 1/2 second is a bit over 1 HP.
Quite so. Climbing stairs is a better "impedance match" to the power
capabilities of your blood-meat-bone engine. Since 1 HP = 550 ft-lb/sec,
if you (ahem) happen to weigh 225 lbs, you can climb 2 ft/sec to calibrate
1 HP output. Given a typical stair riser height of 8 inches, that's 3
steps/sec, or 180 beats per minute on your metronome.
Now, to calibrate your internal metronome for standardized stair climbing,
you synthesize a waltz rhythm in your head to 60 measures/minute, which you
can check on your wristwatch. ONE(tick) two three, ONE(tick) two three,
etc. See how far you can climb 8-inch steps at that rate. Tabulate over a
lifetime and chart your personal mortality. Whew!
Richard J Kinch
> If it didn't take an 18hp gas engine to run it, why would they have built
> this machine with one on it?
Of course. My point is simply that the unit of horsepower is a specific
and well-defined physical quantity, and that the word "horsepower" on a
gasoline engine is not that unit, and there is no conversion between the
two because the latter usage can mean anything the manufacturer whimsically
chooses. Contrast this to an automobile they are honest in specifying
"brake horsepower" to reflect useful rpm x torque delivered to the wheels.
In the case of gasoline-powered air compressors, the situation is difficult
to characterize in terms of true horsepower units, because the crankshaft
power delivered by the engine is a complicated function of many factors,
and the crankshaft sink at the piston compressor is likewise varying
through factors like the cycle hysteresis.
Richard J Kinch
> The term 'rated hp' in this context is the amount of continuous (or
> peak) hp that a given motor can put out.
Not really, because those quantities are themselves are arbitrary and
ill-defined. Engines are "derated" for all sorts of things like
longevity. So the "rating" (which is after all a word implying that
things should reduce to some kind of ratio) is quite meaningless except
perhaps as a comparison between two like machines "rated" by the same
arbitrary method.
At one time Sears vacuum cleaners were "rated" by the largest observed
stall speed instantaneous current draw of the motor, times 120 VAC
divided by 746 watts/HP. Which current could be induced by peak AC
cycle voltage of 170 volts by properly selecting an instantaneous
measurement at the AC line peak. Which of course had nothing to do with
the motor shaft power (zero at zero rpm), even less the thermodynamic
turbine potential output delivered to a vacuum cleaner nozzle on the end
of a hose.
It is all meaningless. The US House yesterday passed the ration-and-tax
coercion of fuel, so all our engines may have to be derated. Socialists
just wanted to take half our stuff for their own benefit but let us keep
the rest; these guys decree that we'll fight over dividing half the
stuff to begin with. Talk about ratios.
Jim Wilkins
> Jim Wilkins writes:
> ...
> Quite so. Climbing stairs is a better "impedance match" to the power
> capabilities of your blood-meat-bone engine. Since 1 HP = 550 ft-lb/sec,
> if you (ahem) happen to weigh 225 lbs, you can climb 2 ft/sec to calibrate
> 1 HP output. Given a typical stair riser height of 8 inches, that's 3
> steps/sec, or 180 beats per minute on your metronome.
>
> Now, to calibrate your internal metronome for standardized stair climbing,
> you synthesize a waltz rhythm in your head to 60 measures/minute, which you
> can check on your wristwatch. ONE(tick) two three, ONE(tick) two three,
> etc. See how far you can climb 8-inch steps at that rate. Tabulate over a
> lifetime and chart your personal mortality. Whew!
In college I could run up the fire stairs in an 8 floor building on
three deep breaths, and then back down on one. Now I'm exhausted from
even remembering it.
jsw
RoyJ
Richard J Kinch wrote:
>
> Of course. My point is simply that the unit of horsepower is a specific
> and well-defined physical quantity,
correct
and that the word "horsepower" on a
> gasoline engine is not that unit,
Blather
and there is no conversion between the
> two because the latter usage can mean anything the manufacturer whimsically
> chooses.
As long as they choose it but that does not change the definition
Contrast this to an automobile they are honest in specifying
> "brake horsepower" to reflect useful rpm x torque delivered to the wheels.
REAL blather. There is NO vehicle manufacturer that talks about "brake
hp to the rear wheels.
>
> In the case of gasoline-powered air compressors, the situation is difficult
> to characterize in terms of true horsepower units, because the crankshaft
> power delivered by the engine is
torque at engine RPM
a complicated function of many factors,
> and the crankshaft sink at the piston compressor is likewise varying
> through factors like the cycle hysteresis.
Complete blather.
The real point is that manufacturer's either publish a spec sheet and
meet those specs. If they don't they are guilty of misrepresentation or
outright fraud.
But you have to realize that a gasoline engine has NO OVERLOAD capacity,
unlike an electric motor.
dca...@krl.org
> I think the actual number is more like 14 to 18. But I can't see what I'm doing
> wrong.
At 900 rpm when the piston is at bottom dead center, the pressure
inside the cylinder is not 1 atmosphere. It is somewhat less because
of the drop in pressure due to the filter and intake manifold.
When the piston is at top dead center, there is still a little volume
inside the cylinder. The compression ratio on compressors is quite
high because they use reed valves. But it is still not infinite.
Otherwise with just a slight change due to manufacturing tolerances or
temperature changes, the piston would be hitting the head.
And then those reed valves on both the intake and the exhaust, need
some pressure across them to cause them to open.
And then there is some leakage by the piston rings.
None of these are biggies but they all affect how much air actually
gets pumped.
So you calculated the swept volume correctly, but the efficiency can
only be determined by an actual test.
Dan
Private
"Richard J Kinch" <ki...@truetex.com> wrote in message
news:Xns9C37AEEA7BE...@216.196.97.131...
> Of course. My point is simply that the unit of horsepower is a specific
> and well-defined physical quantity, and that the word "horsepower" on a
> gasoline engine is not that unit, and there is no conversion between the
> two because the latter usage can mean anything the manufacturer
> whimsically
> chooses. Contrast this to an automobile they are honest in specifying
> "brake horsepower" to reflect useful rpm x torque delivered to the wheels.
>
> In the case of gasoline-powered air compressors, the situation is
> difficult
> to characterize in terms of true horsepower units, because the crankshaft
> power delivered by the engine is a complicated function of many factors,
> and the crankshaft sink at the piston compressor is likewise varying
> through factors like the cycle hysteresis.
Just to throw another (largely irrelevant) definition of HP into this
discussion,
http://en.wikipedia.org/wiki/Tax_horsepower