Inverter to run A.C.
 

"JAXAshby" wrote in message
...
don't understand your question, but will try to explain what _may_ have
been
your question.


I don't understand you explanation.....


An a/c unit will use about 1,400 watts electricity input to remove every
1,000
watts of heat.

The Mermaid Marine a/c in question here is rated at 6500 BTU, and draws 6.4
amps at 120 volts. That is 768 watts of input power.

The a/c is rated at "6500 BTU", which is really 6500 BTU per hour
(manufactuers tend to leave off the "per hour" part of the specification).
This unit has an EER (energy efficiency ratio) of 6500/768= 8.5, which is
comparable to a typical window mounted A/C.

Inverter to run A.C.
 

comments interspersed.

don't understand your question, but will try to explain what _may_ have
been
your question.


I don't understand you explanation.....


An a/c unit will use about 1,400 watts electricity input to remove every
1,000
watts of heat.

The Mermaid Marine a/c in question here is rated at 6500 BTU, and draws 6.4
amps at 120 volts. That is 768 watts of input power.

the 6,500 output is TOTAL output, which = input + heat removed.

Also, I used the figure the HVAC industry uses.


The a/c is rated at "6500 BTU", which is really 6500 BTU per hour
(manufactuers tend to leave off the "per hour" part of the specification).
This unit has an EER (energy efficiency ratio) of 6500/768= 8.5, which is
comparable to a typical window mounted A/C.

see above.

.
6500 BTU/hour = 108 BTU/minute.
1 BTU/minute = 17.58 watts, so 108 BTU/minute * 17.58 = 1899 watts.

My calculations show that 768 watts of input power the unit will remove 1899
watts of heat. How did you get your number????

that's total removed heat, *including* input heat as well.


1,000 watts = 3,012 btu's(according to the figures used by th HVAC
industry [I
have family in the business long term])

1000 watts should be 56.89 BTU/minute, or 3413 BTU/hour.

so I hear, but the 3,012 (sometimes 3,014) is the number used by the HVAC
industry.

A group 27 battery usually has aboout 100 amp-hour capacity, of which
about 50%
is usuable. 2 grp 27's will give about a total of 100 amps before going
dead,
as in unusable

100 amps at 12 volts = 1,200 watts = 3,600 btu's

Now you are mixing amp-hours and watts, not good!

amps used per hour times voltages IS watts per hour.


12 volts at a100 amps will yield 1200 watts, which will provide 4096
BTU/hour. Your estimate of the batteries being able to provide this power
for an hour (fairly reasonable estimate) means that the batteries could
provide 4096 BTU of cooling.

see above to net heat removed, effective.


Also, a group 27 lead-acid battey will provide about 100 amp-hours. The
Optima batteries of the same size are considerably less.

across 1-1/2 hours that makes for about 2,400 btu's per hour INPUT (about
800
watts, about 65 amps).

Which makes for about 1,800 btu's of cooling.

Lost me here. I calculate 4096 BTU of coolig.

total for 1-1/2 hours. reduce to 3,600 btu's for 1-1/2 hours, reduce for HVAC
industry claimed effectiveness and you are down to 1,800 btu's per hour heat
removed from the living area.

btw, 1,800 btu's is about the cooling capacity of 12# of ice melting.

Large airconditioning systems are often rated in "tons". A "ton" is defined
here as the amount of energy required to melt a ton of ice. A ton is 2000
pounds, and requires 12,000 BTU to melt.

144 btu's per pound "change of state" (32* ice to 32* water) times 2,000 pounds
= 288,000 btu's.

"tons of ice" were used originally by a/c (and reefer) manufacturers to give
(commercial) customers a handle to under the "how much cold" the units could
deliver. the term is still used.


2000 lbs ice / 12000 BTU * 1800 BTU = 300 pounds of ice.

I think your math is a little off today.

1,800 btu's per hour/144 (btu's per pound of ice "change of state") = 12.5
pounds of ice times 1-1/2 hours = 18+ pounds of ice.

no way in hell will a marine a/c unit make 300 pounds of ice in a hour.


In other words, for drawing your 2 grp 27 batteries to their knees in an
hour
and a half you will get about the cooling capacity of throwing about 18
pounds
of ice on your cabin sole.

Do you really believe that 18 pounds of ice sitting on the floor will cool
the cabin as well as a 6500 BTU airconditioner????

of course.


The bottom line is that it will be very difficult to run the airconditioner
off of battery power. If the cabin was really well insulated, and the
difference between your set point and the outside temperature was not too
great then you might get several hours of use from a bank of 4 Optima
batteries. Better have a separate battery for starting the engine though.

nah. no vendor was able to stay in business trying to sell that to the boating
public. Some people bought the story and the product, but all changed out
shortly. There simply was not enough battery capacity available to run the
unit

btw, 1,800 btu's net ain't a hell of a lot of cooling capacity AND to recharge
the batteries you need to produce about 90 amps PER HOUR for each hour run by
the a/c. 12 hours run off the batteries requires 12 hours of engine run time
at 90 amps charging. (that is figuring about 33% charging losses, though some
will say 20% is obtainable)


Rod

Inverter to run A.C.
 

Oh. And I thought the a/c only drew 7 amps @ 8000 btu + 3/4 amp for the
pump. At least that is what the user manual stated.
--
Capt. Frank

__c
\ _ | \_
__\_| oooo \_____
~~~~|______________/ ~~~~~
www.home.earthlink.net/~aartworks
"JAXAshby" wrote in message
...
don't understand your question, but will try to explain what _may_ have
been
your question.

An a/c unit will use about 1,400 watts electricity input to remove every
1,000
watts of heat.

1,000 watts = 3,012 btu's(according to the figures used by th HVAC
industry [I
have family in the business long term])

A group 27 battery usually has aboout 100 amp-hour capacity, of which
about 50%
is usuable. 2 grp 27's will give about a total of 100 amps before going
dead,
as in unusable

100 amps at 12 volts = 1,200 watts = 3,600 btu's

across 1-1/2 hours that makes for about 2,400 btu's per hour INPUT
(about 800
watts, about 65 amps).

Which makes for about 1,800 btu's of cooling.

sorry I used the 1,000 btu figure as I was just doing the numbers
quickly in my
head.

btw, 1,800 btu's is about the cooling capacity of 12# of ice melting.

In other words, for drawing your 2 grp 27 batteries to their knees in an
hour
and a half you will get about the cooling capacity of throwing about 18
pounds
of ice on your cabin sole.

, my 2 group 27 house batteries will run the unit on the
inverter for about one and a half hours.

considering that 2 group 27's will put out about 100 amps in an
hour and
a
half, or about 1,200 watts or about 3,600 btu's, you a/c unit
ain't
putting
out but about 1,000 btu's per hour of cooling.

not a hell of a lot.

Inverter to run A.C.
 

I suppose that is why I chose a generator to run "the stuff" away from the
dock. It just works better.

The inverter is fine for running the microwave to heat up a cup of soup or
somesuch, and even run the tv/dvd. But not much good for hi power
consumption.

Hot water heater on inverter? Forget it.
--
Capt. Frank

__c
\ _ | \_
__\_| oooo \_____
~~~~|______________/ ~~~~~
www.home.earthlink.net/~aartworks
"JAXAshby" wrote in message
...
don't understand your question, but will try to explain what _may_ have
been
your question.

An a/c unit will use about 1,400 watts electricity input to remove every
1,000
watts of heat.

1,000 watts = 3,012 btu's(according to the figures used by th HVAC
industry [I
have family in the business long term])

A group 27 battery usually has aboout 100 amp-hour capacity, of which
about 50%
is usuable. 2 grp 27's will give about a total of 100 amps before going
dead,
as in unusable

100 amps at 12 volts = 1,200 watts = 3,600 btu's

across 1-1/2 hours that makes for about 2,400 btu's per hour INPUT
(about 800
watts, about 65 amps).

Which makes for about 1,800 btu's of cooling.

sorry I used the 1,000 btu figure as I was just doing the numbers
quickly in my
head.

btw, 1,800 btu's is about the cooling capacity of 12# of ice melting.

In other words, for drawing your 2 grp 27 batteries to their knees in an
hour
and a half you will get about the cooling capacity of throwing about 18
pounds
of ice on your cabin sole.

, my 2 group 27 house batteries will run the unit on the
inverter for about one and a half hours.

considering that 2 group 27's will put out about 100 amps in an
hour and
a
half, or about 1,200 watts or about 3,600 btu's, you a/c unit
ain't
putting
out but about 1,000 btu's per hour of cooling.

not a hell of a lot.

Inverter to run A.C.
 

BTW are you in Jacksonville?
--
Capt. Frank

__c
\ _ | \_
__\_| oooo \_____
~~~~|______________/ ~~~~~
www.home.earthlink.net/~aartworks
"JAXAshby" wrote in message
...
don't understand your question, but will try to explain what _may_ have
been
your question.

An a/c unit will use about 1,400 watts electricity input to remove every
1,000
watts of heat.

1,000 watts = 3,012 btu's(according to the figures used by th HVAC
industry [I
have family in the business long term])

A group 27 battery usually has aboout 100 amp-hour capacity, of which
about 50%
is usuable. 2 grp 27's will give about a total of 100 amps before going
dead,
as in unusable

100 amps at 12 volts = 1,200 watts = 3,600 btu's

across 1-1/2 hours that makes for about 2,400 btu's per hour INPUT
(about 800
watts, about 65 amps).

Which makes for about 1,800 btu's of cooling.

sorry I used the 1,000 btu figure as I was just doing the numbers
quickly in my
head.

btw, 1,800 btu's is about the cooling capacity of 12# of ice melting.

In other words, for drawing your 2 grp 27 batteries to their knees in an
hour
and a half you will get about the cooling capacity of throwing about 18
pounds
of ice on your cabin sole.

, my 2 group 27 house batteries will run the unit on the
inverter for about one and a half hours.

considering that 2 group 27's will put out about 100 amps in an
hour and
a
half, or about 1,200 watts or about 3,600 btu's, you a/c unit
ain't
putting
out but about 1,000 btu's per hour of cooling.

not a hell of a lot.

Inverter to run A.C.
 

You are probably not using the battery to run a compressor for the total
cooling. It is probably a heat pump, with the Ocean as the heat sink. So
does not use all the electricity to get the BTU output.

"JAXAshby" wrote in message
...
comments interspersed.

don't understand your question, but will try to explain what _may_ have
been
your question.


I don't understand you explanation.....


An a/c unit will use about 1,400 watts electricity input to remove
every
1,000
watts of heat.

The Mermaid Marine a/c in question here is rated at 6500 BTU, and draws
6.4
amps at 120 volts. That is 768 watts of input power.

the 6,500 output is TOTAL output, which = input + heat removed.

Also, I used the figure the HVAC industry uses.


The a/c is rated at "6500 BTU", which is really 6500 BTU per hour
(manufactuers tend to leave off the "per hour" part of the
specification).
This unit has an EER (energy efficiency ratio) of 6500/768= 8.5, which is
comparable to a typical window mounted A/C.

see above.

.
6500 BTU/hour = 108 BTU/minute.
1 BTU/minute = 17.58 watts, so 108 BTU/minute * 17.58 = 1899 watts.

My calculations show that 768 watts of input power the unit will remove
1899
watts of heat. How did you get your number????

that's total removed heat, *including* input heat as well.


1,000 watts = 3,012 btu's(according to the figures used by th HVAC
industry [I
have family in the business long term])

1000 watts should be 56.89 BTU/minute, or 3413 BTU/hour.

so I hear, but the 3,012 (sometimes 3,014) is the number used by the HVAC
industry.

A group 27 battery usually has aboout 100 amp-hour capacity, of which
about 50%
is usuable. 2 grp 27's will give about a total of 100 amps before
going
dead,
as in unusable

100 amps at 12 volts = 1,200 watts = 3,600 btu's

Now you are mixing amp-hours and watts, not good!

amps used per hour times voltages IS watts per hour.


12 volts at a100 amps will yield 1200 watts, which will provide 4096
BTU/hour. Your estimate of the batteries being able to provide this
power
for an hour (fairly reasonable estimate) means that the batteries could
provide 4096 BTU of cooling.

see above to net heat removed, effective.


Also, a group 27 lead-acid battey will provide about 100 amp-hours. The
Optima batteries of the same size are considerably less.

across 1-1/2 hours that makes for about 2,400 btu's per hour INPUT
(about
800
watts, about 65 amps).

Which makes for about 1,800 btu's of cooling.

Lost me here. I calculate 4096 BTU of coolig.

total for 1-1/2 hours. reduce to 3,600 btu's for 1-1/2 hours, reduce for
HVAC
industry claimed effectiveness and you are down to 1,800 btu's per hour
heat
removed from the living area.

btw, 1,800 btu's is about the cooling capacity of 12# of ice melting.

Large airconditioning systems are often rated in "tons". A "ton" is
defined
here as the amount of energy required to melt a ton of ice. A ton is
2000
pounds, and requires 12,000 BTU to melt.

144 btu's per pound "change of state" (32* ice to 32* water) times 2,000
pounds
= 288,000 btu's.

"tons of ice" were used originally by a/c (and reefer) manufacturers to
give
(commercial) customers a handle to under the "how much cold" the units
could
deliver. the term is still used.


2000 lbs ice / 12000 BTU * 1800 BTU = 300 pounds of ice.

I think your math is a little off today.

1,800 btu's per hour/144 (btu's per pound of ice "change of state") = 12.5
pounds of ice times 1-1/2 hours = 18+ pounds of ice.

no way in hell will a marine a/c unit make 300 pounds of ice in a hour.


In other words, for drawing your 2 grp 27 batteries to their knees in
an
hour
and a half you will get about the cooling capacity of throwing about 18
pounds
of ice on your cabin sole.

Do you really believe that 18 pounds of ice sitting on the floor will
cool
the cabin as well as a 6500 BTU airconditioner????

of course.


The bottom line is that it will be very difficult to run the
airconditioner
off of battery power. If the cabin was really well insulated, and the
difference between your set point and the outside temperature was not too
great then you might get several hours of use from a bank of 4 Optima
batteries. Better have a separate battery for starting the engine
though.

nah. no vendor was able to stay in business trying to sell that to the
boating
public. Some people bought the story and the product, but all changed out
shortly. There simply was not enough battery capacity available to run
the
unit

btw, 1,800 btu's net ain't a hell of a lot of cooling capacity AND to
recharge
the batteries you need to produce about 90 amps PER HOUR for each hour run
by
the a/c. 12 hours run off the batteries requires 12 hours of engine run
time
at 90 amps charging. (that is figuring about 33% charging losses, though
some
will say 20% is obtainable)


Rod

Inverter to run A.C.
 

"JAXAshby" wrote in message
...


the 6,500 output is TOTAL output, which = input + heat removed.

No. The ratings are based on how much heat transfer the unit will do, not
what the output is. What a useless number the heat output would be! When a
contractor needs to select a A/C unit he doesn't care how much heat is
blowing out the coils on the outside of the house, he cares about how much
heat he has to get out of the room.

The EER rating is where the input power is brought into the equation.



Also, I used the figure the HVAC industry uses.

What do you mean by "the figure" ??


1000 watts should be 56.89 BTU/minute, or 3413 BTU/hour.

so I hear, but the 3,012 (sometimes 3,014) is the number used by the HVAC
industry.

So you are derating on some "rule of thumb" basis.


amps used per hour times voltages IS watts per hour.


Yes, but you left off the "per hour" part.

In your first post, you said:
quote:
"An a/c unit will use about 1,400 watts electricity input to remove every
1,000
watts of heat."
unquote.

Will you please back those numbers up?

Your 1400 Watts of input power is equivalent to 80 BTU/minute (1400 watts /
17.58 watts per btu-minute) or 4800 BTU/hour. The 1000 watts of heat
removed = 3412 BTU / hour.

I maintain that the BTU rating of an A/C unit is a measure of its ability to
remove heat from the room, not how much heat it shoves outside. Thus, the
EER of your example would be 3412 BTUh/1400 watts = 2.44. That would be an
awful number!

Even using your explanation of the rating being the output power, the EER
would be (4800 BTU input + 3412 BTU removed) / 1400 watts = 5.8. That is
still a bad number.

Typical EER ratings are from 8 to 10. Why are you assuming an EER so much
lower?

Rod

Inverter to run A.C.
 

comments interspersed.

the 6,500 output is TOTAL output, which = input + heat removed.

No. The ratings are based on how much heat transfer the unit will do, not
what the output is.

ah, check it again.

When a
contractor needs to select a A/C unit

a general constractor uses an HVAC contractor which understands the issues
involved and sizes accordingly. I am family in the business long-term

Also, I used the figure the HVAC industry uses.

What do you mean by "the figure" ??


1,000 watts = 3,012 btu's.

So you are derating on some "rule of thumb" basis.

it's the number used by the industry for decades.

In your first post, you said:
quote:
"An a/c unit will use about 1,400 watts electricity input to remove every
1,000
watts of heat."
unquote.

Will you please back those numbers up?

I have family in the business long term, and that is the number used by the
HVAC industry. I also sold large scale computer systems for over 20 years and
every installation I sold, the HVAC contractor asked how many kva input my
machine required and then sized the a/c such that the a/c input was about 140%
of the computer input.

I maintain that the BTU rating of an A/C unit is a measure of its ability to
remove heat from the room, not how much heat it shoves outside.

maintain that if you want, but size the a/c as if the heat rating was a total
of input plus heat removed.

Why are you assuming an EER so much
lower?

because that's what the HVAC industry says they can do under the very best of
conditions.