I understand the creosote-forming danger of burning unseasoned wood, and I understand the performance issues burning unseasoned wood in modern catalytic high-efficiency stoves. But we also talk about the wasted BTUs due to driving off water. Makes sense intuitively, but I did some quick calculations and now I'm questioning how large that effect is.
If we take a model Hearth.com burner running his stove on red oak at 20% moisture content - that cord weighs 3,461 pounds - 2,769 pounds of wood+gas, and 692 pounds of water.
Then we compare him to your "typical" wood burner - cuts that same oak in the spring, burns it in the fall. Let's assume he achieves 30% moisture content. So that same cord would weigh 3,956 pounds - the same 2,769 pounds of wood+gas, plus 1,187 pounds of water. So your typical burner has to drive off an additional 1,187-692 = 495 pounds of water. The energy it takes to do so is wasted heat.
He must raise that water's temperature from room temp (let's say 62F to make the result simpler) to the boiling point, which is 150F away from 212F boiling point at 1 atm pressure. That takes 495 pounds * 150 degrees = 74,250 BTUs.
Then he must evaporate the water as steam. The latent heat of vaporization of water at 1 atm is 970 BTU/lb. So it takes 970 BTU/pound* 495 pounds = 480,150 BTUs to drive off the steam.
So his total wasted heat is 74,250 + 480,150 = 554,400 BTUs. Which sounds significant until you realize that that cord of wood holds 25 MBTU. So the waste heat represents 2.2% of the BTUs in that cord of wood. Hardly anything of practical importance; if I buy a car listing 30 MPG fuel efficiency and I achieve 29.4 MPG I'm a pretty happy camper.
Have I made any errors in my calculations, or is the efficiency improvement of 20% seasoned wood being overstated?
If we take a model Hearth.com burner running his stove on red oak at 20% moisture content - that cord weighs 3,461 pounds - 2,769 pounds of wood+gas, and 692 pounds of water.
Then we compare him to your "typical" wood burner - cuts that same oak in the spring, burns it in the fall. Let's assume he achieves 30% moisture content. So that same cord would weigh 3,956 pounds - the same 2,769 pounds of wood+gas, plus 1,187 pounds of water. So your typical burner has to drive off an additional 1,187-692 = 495 pounds of water. The energy it takes to do so is wasted heat.
He must raise that water's temperature from room temp (let's say 62F to make the result simpler) to the boiling point, which is 150F away from 212F boiling point at 1 atm pressure. That takes 495 pounds * 150 degrees = 74,250 BTUs.
Then he must evaporate the water as steam. The latent heat of vaporization of water at 1 atm is 970 BTU/lb. So it takes 970 BTU/pound* 495 pounds = 480,150 BTUs to drive off the steam.
So his total wasted heat is 74,250 + 480,150 = 554,400 BTUs. Which sounds significant until you realize that that cord of wood holds 25 MBTU. So the waste heat represents 2.2% of the BTUs in that cord of wood. Hardly anything of practical importance; if I buy a car listing 30 MPG fuel efficiency and I achieve 29.4 MPG I'm a pretty happy camper.
Have I made any errors in my calculations, or is the efficiency improvement of 20% seasoned wood being overstated?