Generator shed

[buickanddeere]

600w 240v in the US is just 2 hots, not actual 240v but 120v, so 120v twice is 240v, 300w a piece. The motor just isn't working to it's potential. My pool pump is the same, I have 12-3 going to it, it's 240v, but that's just 2 120v's. I can run it on 120v just fine, just not as strong.

Wrong, 240V 600W heater on 120V makes 150 watts of heat. The pool pump will make every bit as much power with 120V on the terminals and the main windings in parallel as it will with 240V and the main windings in series.
No insult intended.... but what are your electrical credentials ?

 

[buickanddeere]

If the generator is carrying 5Kw of load, the cooling system will also dump about 5Kw of heat into the generator shack. There is absolutely no way that genes will operate without overheating.

 

[CobyRupert]

What happens if the door isn't opened? Exhaust is already going outside. There should be some sort of ventin just to deal with humidity. I'm at a loss why it's a big deal if it runs for awhile inside the room.

The exhaust is going to be piped out already. I have a 4" dryer vent and 4" 110v fan that will be plugged into the generator. If the generator is running, the fan is on. Even if the fan fails, there's still a 4" vent in each side. I built it much larger than the generator on purpose. Extra area for the heat instead of a box the exact size of the gen. It should be fine running with the door closed.

If the generator is carrying 5Kw of load, the cooling system will also dump about 5Kw of heat into the generator shack. There is absolutely no way that genes will operate without overheating.

Isn't it then a question of how much cfm does the internal combustion engine (air pump) draw into the room (& vent out the exhaust)? I suppose the air flow path between the building's intake vent & engine intake also determines how much heat is removed.
I suspect the cfm an engine moves is a rather large number compared to a 4" fan.

 

[buickanddeere]

Isn't it then a question of how much cfm does the internal combustion engine (air pump) draw into the room (& vent out the exhaust)? I suppose the air flow path between the building's intake vent & engine intake also determines how much heat is removed.
I suspect the cfm an engine moves is a rather large number compared to a 4" fan.

About 765 cubic feet of air used for combustion per hour if making 7HP with a gas engine using 0.6lbs of gasoline per HP per hour. Ain't nowhere near enough airflow for cooling.

 

[CobyRupert]

About 765 cubic feet of air used for combustion per hour if making 7HP with a gas engine using 0.6lbs of gasoline per HP per hour. Ain't nowhere near enough airflow for cooling.

I believe the generator has a 419cc (25.77 cu. in.) motor that runs at 3600rpm.
So doesn't it move about 25.77 cu.in of air every revolution.
That equals 92772 cu. in / min at 3600rpm; =53.69 cu. ft/min; =3221 cu. ft / hr.

If the shed is 3x3x6, that's like 60 air changes an hour. About 1 air change per minute. That will carry a lot of heat away.

Looks like a 4" fan carries about 165 cfm compared to 54 cfm by the engine.
Like I've said, I'm no mechanical engineer, nor do I play one on TV. Maybe someone can confirm or correct.

 

[vsteel]

OK, let me help here with the power and voltage question.

There are 2 formulas we need. Ohm's law which is Voltage=current * resistance or shortened to V=IR. We also need the formula for power, Power = Voltage * current or shortened to P=VI

We know that the power is 600 watts and that we have a voltage of 240 volts. We are going to find the resistance and the current.

First the current draw at 240V. Using our P=VI formula we fill in the values to get 600 watts = 240 volts * I a little algebraic maneuvering gives us 600/240 = I which solves to 2.5 So our circuit running at 240 volts is pulling 2.5 amps of power to make 600 watts of power.

If we take our Ohm's law equation V=IR and rearrange it a little. V/I = R now we fill in our numbers 240 volts / 2.5 amps = R which solves to 96. So our 600W heater has a resistance through the coils of 96 ohms.

We now take our heater and plug it into a 120 volt source. Using our V=IR equation we can get Voltage / resistance = current. Filling in the numbers we get 120 Volts / 96 ohms = I which solves to 1.25 amps. So we now know our little heater is using 120 volts and 1.25 amps to make heat. We just need to figure out how much heat it is making.

Going back to our other formula P=VI we just need to fill in the values. power = 120 volts * 1.25 amps We solve this and come up with 150 watts.

I can see where you would think that if you cut the voltage in half you would cut the power in half but that doesn't work with power. When you reduce the voltage by half you cut the power by a quarter, this is because not only do you not have as much voltage going through the electric coils you have also cut the current in half going through the coils. If you look at our numbers above you can see our current at 120v is half that it was at 240v.

I tried to write this at a level that everyone could follow and not make any assumptions on what everyone knew and that is why I didn't use equations which can solve things directly. I have a degree in electrical engineering so I purposely left out the shortcuts and made is a straight forward as I could though longer than it needed to be.

 

[CobyRupert]

Why power is not cut 1/2 when voltage is cut 1/2 (condensed):
P=V x I,
but I=V/R, because R is how it's built and can't change;
so P= V x (V/R), or
P= V^2/R.
QED.

 

[buickanddeere]

I believe the generator has a 419cc (25.77 cu. in.) motor that runs at 3600rpm.
So doesn't it move about 25.77 cu.in of air every revolution.
That equals 92772 cu. in / min at 3600rpm; =53.69 cu. ft/min; =3221 cu. ft / hr.

If the shed is 3x3x6, that's like 60 air changes an hour. About 1 air change per minute. That will carry a lot of heat away.

Looks like a 4" fan carries about 165 cfm compared to 54 cfm by the engine.
Like I've said, I'm no mechanical engineer, nor do I play one on TV. Maybe someone can confirm or correct.

First of all you are thinking two stroke instead of four stroke, so half that projected airflow. 2nd that engine does not have 100 % volumetric efficiency even at wide open throttle so figure 75% pumping efficiency. 3rd the gasser has a throttle plate restricting airflow at partload. A 419cc is an approx 14HP motor , working at half load to make 7HP. Let's run a little extra , slightly lean for more at 105%. Figurered an air fuel ratio of 14.7 to 1.
25.77 X 3600 X 0.5 X.75 X 0.5 /1728 X60 = 604 CFPH for a mechanical based calculation. In hind sight if the volumetric efficiency was 85% . The mechanical based airflow calculation would be 685 CFPH.
i had ran the other numbers figuring 7HP, using 0.6 lbs of gasoline per HP per hour. Air weights 0.0807 lbs to obtain 765 CFPH. close enough for a back of the napkin after some Crown Royal.
To be more accurate the military's fuel consumption of 0.63 lbs of gasoline per HP per hour for a Honda GX 35 at 60% power could be used . So leaned out and with a little more fuel consumption and leaning her a little . The airflow into the engine making 7HP would be approx 843 CFPH. http://dtic.mil/dtic/tr/fulltext/u2/a540184.pdf
I have portrayed in stage plays Bert Healy in Orphan Annie, the farmer in Anne of Green Gables, the School Master in Oliver and "Big Chief" in Peter Pan. In an upcoming B movie " Red Spring" I have about a minute, centre screen as a vampire.

 

[buickanddeere]

OK, let me help here with the power and voltage question.

There are 2 formulas we need. Ohm's law which is Voltage=current * resistance or shortened to V=IR. We also need the formula for power, Power = Voltage * current or shortened to P=VI

We know that the power is 600 watts and that we have a voltage of 240 volts. We are going to find the resistance and the current.

First the current draw at 240V. Using our P=VI formula we fill in the values to get 600 watts = 240 volts * I a little algebraic maneuvering gives us 600/240 = I which solves to 2.5 So our circuit running at 240 volts is pulling 2.5 amps of power to make 600 watts of power.

If we take our Ohm's law equation V=IR and rearrange it a little. V/I = R now we fill in our numbers 240 volts / 2.5 amps = R which solves to 96. So our 600W heater has a resistance through the coils of 96 ohms.

We now take our heater and plug it into a 120 volt source. Using our V=IR equation we can get Voltage / resistance = current. Filling in the numbers we get 120 Volts / 96 ohms = I which solves to 1.25 amps. So we now know our little heater is using 120 volts and 1.25 amps to make heat. We just need to figure out how much heat it is making.

Going back to our other formula P=VI we just need to fill in the values. power = 120 volts * 1.25 amps We solve this and come up with 150 watts.

I can see where you would think that if you cut the voltage in half you would cut the power in half but that doesn't work with power. When you reduce the voltage by half you cut the power by a quarter, this is because not only do you not have as much voltage going through the electric coils you have also cut the current in half going through the coils. If you look at our numbers above you can see our current at 120v is half that it was at 240v.

I tried to write this at a level that everyone could follow and not make any assumptions on what everyone knew and that is why I didn't use equations which can solve things directly. I have a degree in electrical engineering so I purposely left out the shortcuts and made is a straight forward as I could though longer than it needed to be.

Looks good to me.

 

[fatjay]

I underestimated this project, but I usually do that with things. Apparently stucco is a pain in the ***. We'll see how that goes. In the mean while I got the roof done. Didn't get final shots, but I think it looks good.