briand2
Mechanical
- Jan 15, 2002
- 180
Apologies for asking a non-chemical question here, but as my HVAC application is apparently very similar to a 'reboiler' (as far as I can tell!), I thought this is the best place to ask!
I have a hot water storage cylinder heated by a 3kW electric immersion heater. Cold water supply is from an atmospheric tank about 12 metres above the hot water storage cylinder. Off the top of the hot water storage cylinder, an open vent pipe runs upwards and terminates at an open end in the water tank, about 1 metre above the water level in the tank.
Before turning on the electric immersion heater, with all water at the same temperature, the level in the open vent pipe is at the same height as the level in the tank, and the pressure in the hot water storage cylinder is about 1.2bar(g) (due simply to static head from tank / open vent pipe). Under normal operation, the electric immersion heater is thermostatically controlled to maintain water temperature at about 60degC. Assuming no heat loss from the pipes, etc, as the cold water is supplied at about 20degC then the the difference in density between the water in the cold feed and that in the open vent would be about 2%, so the level in the open vent would rise by about 2% (i.e. 0.2 metres) for equilibrium. The pressure in the hot water storage vessel would remain unchanged, at 1.2bar(g).
What I'm trying to understand is what happens to the pressure in the hot water storage vessel if the thermostatic controller fails leaving the 3kW electric immersion heater permanently on.
Assuming, to begin with, that the water temperature in the open vent is the same as that in the hot water storage cylinder, then when that temperature has risen high enough for density to drop by 10%, water will start to dribble out of the open vent into the tank below. However, that would require the water temperature to rise to about 158degC, which is saturation temperature for a pressure of 6bar(a), so it won't happen because before that, when the water temperature has risen to about 123degC (saturation temperature for a pressure of 1.2bar(g)), the water will boil. I assume that once boiling has started, as the vapour bubbles rise from the electric immersion heater up through the body of water in the hot water storage cylinder and then on into the open vent pipe, their size is increasing due to decreasing static pressure (due to increasing elevation) and so the fluid in the open vent becomes a mixture of vapour and liquid. This mixed state will presumably significantly decrease the 'average' density of fluid in the open vent, and so it will rise up in the tube and spurt out of the open vent, with the vapour leaving the system (taking heat energy with it) and the remaining water, now at 100degC, dropping back into the tank below.
Leaving this situation until equilibrium is reached, I assume the water in the tank and 'cold' supply pipe will be 100degC and the steam / water mixture at the top of the hot water cylinder / entering the open vent pipe will be about 123degC, dropping along / up the open vent due to flashing off until it is at 100degC at the outlet of the open vent. I further assume that (ignoring energy requirement for circulation / friction losses), the amount of water evaporating will be equivalent to 3kW (about 0.0014kg/s at latent heat of 2200kJ/kg) and that, as this results in cooling of remaining water from 123degC to 100degC, then the amount of liquid circulating must be about (3/(4.2x23)), i.e. 0.031kg/s. From this, the vapour proportion by mass is about 4%.
Now, under these conditions, as the total flow downwards in the 'cold' water supply pipe is about 0.0324kg/s, the friction loss is about 9Pa/m, making a total friction loss of about 108Pa (say 0.001bar) over the 12m metre length of the 20mm pipe. This is pretty insignificant compared to the original hot water storage cylinder pressure of about 1.2bar(g) so, if this were the case, the hot water storage cylinder pressure would remain, essentially, constant. Now at the 0.0014kg/s vapour in the open vent pipe (which is also 20mm), assuming it to be at an average pressure of 0.6bar(g), with specific volume about 1.1m3/kg, the pressure drop due it flowing up the open vent (ignoring any liquid reducing the cross sectional area of the open vent?) would only be about 0.2kPa (0.00002bar). This is where I'm becoming more confused; if the pressure at the top open end of the open vent is atmospheric, and I need only a differential pressure / driving force of 0.00002bar to move the steam up it, where do I 'lose' the approximately 1.2bar(g) in the hot water storage vessel? Could it be that there is still 'back pressure' due to a column of liquid remaining in the open vent, and the vapour is just pushing its way up through that?
Also, if under original static conditions the pressure in the hot water storage cylinder was 1.2bar(g) and yet, under thermosyphon flowing conditions as described, there must be a friction pressure drop (however small) in the 'cold' supply pipe, then the pressure in the hot water cylinder would appear to be lower under 'fault' conditions (electric immersion heater remaining on irrespective of water temperature), which is a bit counter-intuitive?
Apologies once again for the length of this post, and thank you to anyone able to correct any misunderstanding I have of the process involved here. (Apologies also for 'almost' cross-posting; I put a very similar post in the Pipelines, Piping and Fluid Mechanics engineering Forum but got no replies.)
Regards,
Brian
I have a hot water storage cylinder heated by a 3kW electric immersion heater. Cold water supply is from an atmospheric tank about 12 metres above the hot water storage cylinder. Off the top of the hot water storage cylinder, an open vent pipe runs upwards and terminates at an open end in the water tank, about 1 metre above the water level in the tank.
Before turning on the electric immersion heater, with all water at the same temperature, the level in the open vent pipe is at the same height as the level in the tank, and the pressure in the hot water storage cylinder is about 1.2bar(g) (due simply to static head from tank / open vent pipe). Under normal operation, the electric immersion heater is thermostatically controlled to maintain water temperature at about 60degC. Assuming no heat loss from the pipes, etc, as the cold water is supplied at about 20degC then the the difference in density between the water in the cold feed and that in the open vent would be about 2%, so the level in the open vent would rise by about 2% (i.e. 0.2 metres) for equilibrium. The pressure in the hot water storage vessel would remain unchanged, at 1.2bar(g).
What I'm trying to understand is what happens to the pressure in the hot water storage vessel if the thermostatic controller fails leaving the 3kW electric immersion heater permanently on.
Assuming, to begin with, that the water temperature in the open vent is the same as that in the hot water storage cylinder, then when that temperature has risen high enough for density to drop by 10%, water will start to dribble out of the open vent into the tank below. However, that would require the water temperature to rise to about 158degC, which is saturation temperature for a pressure of 6bar(a), so it won't happen because before that, when the water temperature has risen to about 123degC (saturation temperature for a pressure of 1.2bar(g)), the water will boil. I assume that once boiling has started, as the vapour bubbles rise from the electric immersion heater up through the body of water in the hot water storage cylinder and then on into the open vent pipe, their size is increasing due to decreasing static pressure (due to increasing elevation) and so the fluid in the open vent becomes a mixture of vapour and liquid. This mixed state will presumably significantly decrease the 'average' density of fluid in the open vent, and so it will rise up in the tube and spurt out of the open vent, with the vapour leaving the system (taking heat energy with it) and the remaining water, now at 100degC, dropping back into the tank below.
Leaving this situation until equilibrium is reached, I assume the water in the tank and 'cold' supply pipe will be 100degC and the steam / water mixture at the top of the hot water cylinder / entering the open vent pipe will be about 123degC, dropping along / up the open vent due to flashing off until it is at 100degC at the outlet of the open vent. I further assume that (ignoring energy requirement for circulation / friction losses), the amount of water evaporating will be equivalent to 3kW (about 0.0014kg/s at latent heat of 2200kJ/kg) and that, as this results in cooling of remaining water from 123degC to 100degC, then the amount of liquid circulating must be about (3/(4.2x23)), i.e. 0.031kg/s. From this, the vapour proportion by mass is about 4%.
Now, under these conditions, as the total flow downwards in the 'cold' water supply pipe is about 0.0324kg/s, the friction loss is about 9Pa/m, making a total friction loss of about 108Pa (say 0.001bar) over the 12m metre length of the 20mm pipe. This is pretty insignificant compared to the original hot water storage cylinder pressure of about 1.2bar(g) so, if this were the case, the hot water storage cylinder pressure would remain, essentially, constant. Now at the 0.0014kg/s vapour in the open vent pipe (which is also 20mm), assuming it to be at an average pressure of 0.6bar(g), with specific volume about 1.1m3/kg, the pressure drop due it flowing up the open vent (ignoring any liquid reducing the cross sectional area of the open vent?) would only be about 0.2kPa (0.00002bar). This is where I'm becoming more confused; if the pressure at the top open end of the open vent is atmospheric, and I need only a differential pressure / driving force of 0.00002bar to move the steam up it, where do I 'lose' the approximately 1.2bar(g) in the hot water storage vessel? Could it be that there is still 'back pressure' due to a column of liquid remaining in the open vent, and the vapour is just pushing its way up through that?
Also, if under original static conditions the pressure in the hot water storage cylinder was 1.2bar(g) and yet, under thermosyphon flowing conditions as described, there must be a friction pressure drop (however small) in the 'cold' supply pipe, then the pressure in the hot water cylinder would appear to be lower under 'fault' conditions (electric immersion heater remaining on irrespective of water temperature), which is a bit counter-intuitive?
Apologies once again for the length of this post, and thank you to anyone able to correct any misunderstanding I have of the process involved here. (Apologies also for 'almost' cross-posting; I put a very similar post in the Pipelines, Piping and Fluid Mechanics engineering Forum but got no replies.)
Regards,
Brian
