Rays53hatt
Veteran Member
I posted about needing to replace my existing water heater and I am considering an on demand 110 electric water heater. Here are my thoughts on this. please share where my thinking is incorrect. I am open to redirection. I used Chat GPT for help with this, so if I'm wrong, I blame technology.
Now for the Geeks who want to debate energy suck, here is some data.
We did some measuring, and we will need 11 gallons of hot water a day and that makes it easy for the calculations. As the replacement would be an 11-gallon whale.
Assuming I am at anchor and not running the motors (no heat exchanger) or the genset, just operating on solar and batteries.
And assuming I will use the hot water at 3 or 4 different times during the day.
And assuming I replace the heater with one like the original 11-gallon tank
To calculate the energy needed to heat all eleven gallons in the Whale water heater:
First, let's calculate the energy required to raise the temperature of all eleven gallons by forty degrees Fahrenheit. (40 degrees seems adequate if we are starting with 80-degree water)
The mass of one gallon of water is approximately 8.34 pounds.
The specific heat of water is 1 BTU per pound per degree Fahrenheit.
Bu using the formula:
Energy = Mass X Specific Heat X Temperature Change
For eleven gallons:
The mass of eleven gallons of water is approximately 8.34 pounds per gallon, so 91.74 pounds.
The energy required to raise the temperature by forty degrees Fahrenheit is:
91.74 pounds X 1 BTU/pound/F X 40 = 3669.6 BTU’s
Converting BTUs to kilowatt-hours, knowing that one kWh equals 3,412 BTU’s
Energy in kWh = 3669.6 BTUs / (3412 BTUs / kWh)~.098 kWh
So, it would take approximately 1.075 kilowatt-hours to heat eleven gallons of water by forty degrees Fahrenheit. Assuming zero loss in efficiency. The above DOES NOT TAKE IN TO ACCOUNT The Whale has a typical efficiency rating of 90% and the on demand is 99% efficient.
With that data, if I wash the dishes and need one gallon of hot water. With the original heater I will have to heat all 11 gallons and use one, then later in the day I will want a shower from all that work washing those two cups and two plates and I will need to heat all 11 gallons again. How is this possibly more efficient?
Additionally, if I am keeping the water warm all day I use additional energy and that is calculated below.
The energy required to maintain the temperature depends on factors like insulation, ambient temperature, and how often the heater cycles to maintain the temperature.
Let's assume a reasonably well-insulated tank and typical conditions. The standby loss for an electric water heater might be around 0.1 to 0.2 kWh per hour. If we assume the heater cycles periodically throughout the day (let's estimate this at 0.15 kWh per hour):
Standby energy per day = 0.15 kWh/hour X 24 hours = 3.6 kWh/day
So, to keep the water warm all day long, the water heater might use an additional 3.6 kWh.
- To heat the eleven gallons will take 1.075 kWh
- To keep it warm all day will take approximately 3.6 kWh
For a total of 4.675 kWh a day
I am new to this solar and battery energy calculating, so please be kind in the smack down you are about to give me.
Now for the Geeks who want to debate energy suck, here is some data.
We did some measuring, and we will need 11 gallons of hot water a day and that makes it easy for the calculations. As the replacement would be an 11-gallon whale.
Assuming I am at anchor and not running the motors (no heat exchanger) or the genset, just operating on solar and batteries.
And assuming I will use the hot water at 3 or 4 different times during the day.
And assuming I replace the heater with one like the original 11-gallon tank
To calculate the energy needed to heat all eleven gallons in the Whale water heater:
First, let's calculate the energy required to raise the temperature of all eleven gallons by forty degrees Fahrenheit. (40 degrees seems adequate if we are starting with 80-degree water)
The mass of one gallon of water is approximately 8.34 pounds.
The specific heat of water is 1 BTU per pound per degree Fahrenheit.
Bu using the formula:
Energy = Mass X Specific Heat X Temperature Change
For eleven gallons:
The mass of eleven gallons of water is approximately 8.34 pounds per gallon, so 91.74 pounds.
The energy required to raise the temperature by forty degrees Fahrenheit is:
91.74 pounds X 1 BTU/pound/F X 40 = 3669.6 BTU’s
Converting BTUs to kilowatt-hours, knowing that one kWh equals 3,412 BTU’s
Energy in kWh = 3669.6 BTUs / (3412 BTUs / kWh)~.098 kWh
So, it would take approximately 1.075 kilowatt-hours to heat eleven gallons of water by forty degrees Fahrenheit. Assuming zero loss in efficiency. The above DOES NOT TAKE IN TO ACCOUNT The Whale has a typical efficiency rating of 90% and the on demand is 99% efficient.
With that data, if I wash the dishes and need one gallon of hot water. With the original heater I will have to heat all 11 gallons and use one, then later in the day I will want a shower from all that work washing those two cups and two plates and I will need to heat all 11 gallons again. How is this possibly more efficient?
Additionally, if I am keeping the water warm all day I use additional energy and that is calculated below.
The energy required to maintain the temperature depends on factors like insulation, ambient temperature, and how often the heater cycles to maintain the temperature.
Let's assume a reasonably well-insulated tank and typical conditions. The standby loss for an electric water heater might be around 0.1 to 0.2 kWh per hour. If we assume the heater cycles periodically throughout the day (let's estimate this at 0.15 kWh per hour):
Standby energy per day = 0.15 kWh/hour X 24 hours = 3.6 kWh/day
So, to keep the water warm all day long, the water heater might use an additional 3.6 kWh.
- To heat the eleven gallons will take 1.075 kWh
- To keep it warm all day will take approximately 3.6 kWh
For a total of 4.675 kWh a day
I am new to this solar and battery energy calculating, so please be kind in the smack down you are about to give me.