B VENT / Draft Hood / Flue question

[JOSEPH SMOOT]

Hey all,

New homeowner here, been remodeling the whole house. Bought a Rheem 50 gal platinum plus gas (natural venting, only powered flue flap no fan) on clearance. Looked in the manual and it calls for B vent. Ran it out the house and up the side above the roof (per local code).

Issue I have is I discovered it didn't come with the draft hood that stands off from Rheem. The B vent appliance draft hood connector fits fine, but I'm worried about back draft, CO poisoning etc if it's not okay. I had turned it on and lit a match and snuffed it out for a smoke test by the hood and nothing was pulled in, just rose straight up.

The unit has not been turned on since I tried that. Im sure I can get the stand off draft hood locally if I need it, but if I don't that's obviously better. Please let me know if this is a safe way to hook up the vent or do I need to change it out.

Thanks!
Joe
732 485 6943

 

[Phog]

The opening you held the match up to is not the inside of the flue, its just the air gap between the B-vent's inner & outer double walls. So you wouldn't get any draft up it, since it goes nowhere. You do need the draft hood. The purpose of the hood is, first, to allow additional draft air to enter the flue along with the combustion gases, which lowers the temperature of the flue stack. And second to also prevent flame from being pushed downward at the burner if there is a momentary backdraft (such as in high winds).

 

[Sponsor]

 

[JOSEPH SMOOT]

Thanks! I'll be sure to order that then and attach the appliance hood to that.

I really appreciate the advice and explanation. I wasn't sure.

The opening you held the match up to is not the inside of the flue, its just the air gap between the B-vent's inner & outer double walls. So you wouldn't get any draft up it, since it goes nowhere. You do need the draft hood. The purpose of the hood is, first, to allow additional draft air to enter the flue along with the combustion gases, which lowers the temperature of the flue stack. And second to also prevent flame from being pushed downward at the burner if there is a momentary backdraft (such as in high winds).

 

[Dana]

The purpose of the dilution air is to lower the dew point temperature of the exhaust, not the temperature of the stack.

Without the draft hood the exhaust is heading up the flue undiluted and already fairly saturated with water vapor, with a fairly HIGH dew point temperature due to the high concentration of water in the exhaust product of burning methane (natural gas):

CH4 + 2O2 => CO2 + 2H2O <<< the exhaust is mostly water (plus about 15% excess combustion air)

When it's cold outside it is going to result in a LOT of flue condensation running down the inside of the flue, potentially dripping onto the center-flue heat exchanger (even onto the burner itself) and seeping out the seam where the B-vent is hooked up to the flue damper. Any time the stack is cooler than about 125F there will be condensation occurring (assuming natural gas.)

When diluted heavily with indoor air via the vent hood the dew point of the diluted exhaust + air mixture contents plummets, and any initial condensation at the beginning of the burn when the stack is cold quickly re-evaporates. Typical wintertime indoor air has a dew point temperature of about 35-40F even in fairly tight house, lower in an air leaky house. The mixed exhaust + air gases in the flue will have a dew point closer to room temperature, WAY below the 125F of the undiluted exhaust.

 

[jasontaylor7]

"CH4 + 2O2 => CO2 + 2H2O <<< the exhaust is mostly water (plus about 15% excess combustion air)"
Are you sure? It's not about 13%?

To determine the fraction by weight of steam to air in the effluent when methane (CH₄) burns in air, we need to follow these steps:

1. Write the balanced chemical equation for the combustion of methane:
   CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}CH4+2O2→CO2+2H2O
2. Determine the molar masses of the reactants and products:
   • Molar mass of CH₄: 12+4×1=16 g/mol12 + 4 \times 1 = 16 \text{ g/mol}12+4×1=16 g/mol
   • Molar mass of O₂: 2×16=32 g/mol2 \times 16 = 32 \text{ g/mol}2×16=32 g/mol
   • Molar mass of CO₂: 12+2×16=44 g/mol12 + 2 \times 16 = 44 \text{ g/mol}12+2×16=44 g/mol
   • Molar mass of H₂O: 2×1+16=18 g/mol2 \times 1 + 16 = 18 \text{ g/mol}2×1+16=18 g/mol
3. Calculate the mass of products from 1 mole of CH₄:
   • 1 mole of CH₄ produces 1 mole of CO₂ (44 g) and 2 moles of H₂O (2 × 18 = 36 g).
4. Calculate the mass of air required for the combustion:
   • Air is about 21% O₂ and 79% N₂ by volume (and approximately by mole, since the molar masses of O₂ and N₂ are relatively close).
   • For 2 moles of O₂ (64 g), we need 2 moles of O20.21≈9.52 moles of air\frac{2 \text{ moles of O}_2}{0.21} \approx 9.52 \text{ moles of air}0.212 moles of O2≈9.52 moles of air.
     Therefore, the mass of air needed is approximately:
     • Molar mass of air (approximate): 0.21×32+0.79×28≈29 g/mol0.21 \times 32 + 0.79 \times 28 \approx 29 \text{ g/mol}0.21×32+0.79×28≈29 g/mol
     • Mass of air: 9.52×29≈276 g9.52 \times 29 \approx 276 \text{ g}9.52×29≈276 g
5. Calculate the fraction by weight of steam to air in the effluent:
   • Mass of steam (H₂O) produced: 36 g
   • Mass of air used: 276 g
   • Thus, the fraction by weight of steam to air in the effluent is:
   Mass of H2OMass of air=36 g276 g≈0.1304\frac{\text{Mass of H}_2\text{O}}{\text{Mass of air}} = \frac{36 \text{ g}}{276 \text{ g}} \approx 0.1304Mass of airMass of H2O=276 g36 g≈0.1304

Therefore, the fraction by weight of steam to air in the effluent is approximately 0.1304 or 13.04%. Since 13%<50%, I question the use of most.

 

[Fitter30]

"CH4 + 2O2 => CO2 + 2H2O <<< the exhaust is mostly water (plus about 15% excess combustion air)"
Are you sure? It's not about 13%?

To determine the fraction by weight of steam to air in the effluent when methane (CH₄) burns in air, we need to follow these steps:

1. Write the balanced chemical equation for the combustion of methane:
   CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}CH4+2O2→CO2+2H2O
2. Determine the molar masses of the reactants and products:
   • Molar mass of CH₄: 12+4×1=16 g/mol12 + 4 \times 1 = 16 \text{ g/mol}12+4×1=16 g/mol
   • Molar mass of O₂: 2×16=32 g/mol2 \times 16 = 32 \text{ g/mol}2×16=32 g/mol
   • Molar mass of CO₂: 12+2×16=44 g/mol12 + 2 \times 16 = 44 \text{ g/mol}12+2×16=44 g/mol
   • Molar mass of H₂O: 2×1+16=18 g/mol2 \times 1 + 16 = 18 \text{ g/mol}2×1+16=18 g/mol
3. Calculate the mass of products from 1 mole of CH₄:
   • 1 mole of CH₄ produces 1 mole of CO₂ (44 g) and 2 moles of H₂O (2 × 18 = 36 g).
4. Calculate the mass of air required for the combustion:
   • Air is about 21% O₂ and 79% N₂ by volume (and approximately by mole, since the molar masses of O₂ and N₂ are relatively close).
   • For 2 moles of O₂ (64 g), we need 2 moles of O20.21≈9.52 moles of air\frac{2 \text{ moles of O}_2}{0.21} \approx 9.52 \text{ moles of air}0.212 moles of O2≈9.52 moles of air.
     Therefore, the mass of air needed is approximately:
     • Molar mass of air (approximate): 0.21×32+0.79×28≈29 g/mol0.21 \times 32 + 0.79 \times 28 \approx 29 \text{ g/mol}0.21×32+0.79×28≈29 g/mol
     • Mass of air: 9.52×29≈276 g9.52 \times 29 \approx 276 \text{ g}9.52×29≈276 g
5. Calculate the fraction by weight of steam to air in the effluent:
   • Mass of steam (H₂O) produced: 36 g
   • Mass of air used: 276 g
   • Thus, the fraction by weight of steam to air in the effluent is:
   Mass of H2OMass of air=36 g276 g≈0.1304\frac{\text{Mass of H}_2\text{O}}{\text{Mass of air}} = \frac{36 \text{ g}}{276 \text{ g}} \approx 0.1304Mass of airMass of H2O=276 g36 g≈0.1304

Therefore, the fraction by weight of steam to air in the effluent is approximately 0.1304 or 13.04%. Since 13%<50%, I question the use of most.

Your looking at at a post from 11/20 2019
First if there wasn't a fan it would need a draft diverter. First place to go check the install manual. You take all your math I'd put a draft gauge in the flue didn't like the reading. My first call with all the numbers would be tech at the factory. Next a flue analyzer would give me all the readings i need. Co would be very high.