Quote:
Originally Posted by 1der
(cue: boy wonder here )
This will require about 60 amps which will be supplied by the OEM dual alternator setup we have.

750W/12V=62.5A (Ohm's law)
...neglecting inverter losses of course.......more like 68A including these
Quote:
Originally Posted by 1der
In theory, the 200w element would use about 18 amps and raise the water temp by around 20 degrees in one hour.

200W/12V=16.67A (Ohm's law)
4 gallons=.53 cu ft
.53ft3*(62.5lbs/ft3)*(1 BTU/(lb*degF)*(20 DegF)=662 BTU's
662 BTU's=194 Watthours (lazy...online conversion
)
(194 Watthours)/200 W=.97 hours
Quote:
Originally Posted by 1der
It takes approx 400 watts (not allowing for some inefficiency/losses) to raise four gals of water by 40 degrees in one hour.

200W/12V=16.67A (Ohm's law)
4 gallons=.53 cu ft
.53ft3*(62.5lbs/ft3)*(1 BTU/(lb*degF)*(40 DegF)=1325 BTU's
1325 BTU's=388 Watthours (lazy...online conversion
)
(388 Watthours)/(1 hour)=388 watts
Quote:
Originally Posted by 1der
200 watts would take about 2 hours to do the same.

It sure would....
Quote:
Originally Posted by 1der
The time/watts/temperature/gals relationships are fairly linear.

Completely linear (assuming adiabatic conditions...ie no heat loss)
Quote:
Originally Posted by 1der
What is NOT linear is the watts fed into a heating element and resultant effective heating . 100 watts (9 amps) into a 200 watt element yields about 50 watts of heating because of resistance losses.

Well, not exactly......it's the amps (current) that makes things nonlinear......due to "I squared R" losses....
P=I^2*R (Ohm's law again)
Calculate the resistance of a 200W element that's rated at 12V operation.
P=V*I (ohm's law again) so I=P/V=200/12=16.67A
rearranging the above equation, solve for R...
R=(P/(I^2))=200/(16.67*16.67)=. 72 ohms
Calculate the power out of 9 amps flowing through .72 ohms:
P=(9*9)*.72=58W
Calculate the voltage required for the above current and resistance:
P=V*I (ohm's law...again!!)
V=P/I=58W/9A=6.5V
...so using a 200W/12V rated element at 6.5V only gives 58W.....so in this case roughly half the voltage yields approx 1/4 the power...
Note: All of the above calcs are simple thermodynamic calcs, and do not account for any heat transfer through the water heater insulation, etc. (Adiabatic calcs) They also do not include the energy/power required to heat up the metal tank, insulation, etc.
Real life will require more power and or time......
...and 1der gets an A for a grade......missed one!