briand2
Mechanical
- Jan 15, 2002
- 180
Apologies for the length of this post, but I'm really trying to find out if my understanding is correct concerning two-phase flow in this particular situation!
In the UK, our typical open vented domestic hot water system in residential properties comprises the following:
atmospheric tank in loft
cold water pipe (20mm (3/4")) drops from bottom connection of loft tank to bottom connection of hot water storage cylinder
hot water storage cylinder (175litres (45USgallon)) in bedroom/kitchen/wherever containing electric immersion heater
open vent pipe (20mm (3/4")) rises from top connection of hot water storage cylinder to terminate over loft tank
Our 'code' requires, in particular, the open vent pipe to be of diameter at least 20mm (3/4").
A fault with the electric immersion heater (of typical output 3kW (10,000btu/hr)) may make it it stay on, in which case I assume the sequence of events is as follows:
To begin with, the pressure in the hot water storage cylinder is just the static head due to the loft tank above. In the hot water storage cylinder, the water temperature rises and, eventually, exceeds the saturation temperature (not just in the immediate vicinity of the electric immersion heater element, but more generally). Due to a combination of vapour generation / boiling / foaming and liquid density decreasing, there is upward, presumably two-phase, flow in the open vent, accompanied by downward flow in the cold water feed pipe. As the steam / water mixture is ejected from the open vent, sufficient flashing off will occur to lower the water temperature to 100C (212F) consistent with saturation temperature at atmospheric pressure. This water falls into the loft tank below, raising its temperature. Left for long enough, and ignoring heat losses from pipe / tank / cylinder, this circulation will result in the temperature of water in the tank reaching and stabilising at 100C (212F). At equilibrium, ignoring energy losses from pipework and ignoring energy required to drive the circulation, the amount of steam flashing off must be equivalent to the 3kW (10,000btu/hr) electrical input. As latent heat of evaporation is about 2200kJ/kg (1000btu/lb), then the amount of steam flashed-off / lost is about 0.0014kg/s (10lb/hr). Also, as in my case the elevation of loft tank above hot water storage cylinder is about 12m (36ft), the static pressure to begin with is about 120kPa(g) (18psi(g)), so the saturation temperature is about 123C (253F). Also, the heat lost due to evaporation must be sufficient to reduce the circulating water temperature from about 123C (253F) to 100C (212F) at the point of expulsion from the open vent. As this heat loss is known (from above, 3kW (10,000btu/hr)), and as the specific heat capacity of water is about 4.2kJ/kgK (1btu/lbR), then then the amount of liquid circulating must be (3/(4.2x23)) or (10,000/(41x1)), i.e. 0.031kg/s (244lb/hr).
From the above, the total mass flow rate is 0.0014kg/s (10lb/hr) vapour plus 0.031kg/s (244lb/hr) liquid making 0.0324kg/s (254lb/hr) total, with the vapour mass flow rate being 4% of the total mass flow rate. The specific volume of vapour is 1.7m3/kg (27ft3/lb) at atmospheric pressure (open end of vent pipe) and 0.8m3/kg (13ft3/lb) at 120kPa(g) (8psig(g)) (inside hot water storage cylinder).
After the above unbearably long preamble, my question is: how do I determine, conservatively, the increase in pressure to which the hot water storage cylinder is subject under the above conditions?
My gut feeling is that I might conservatively allow for all the vapour being at atmospheric pressure, and ignore the relatively very small frictional pressure drop due to liquid flow. Allowing for the vapour to be flowing the entire 12m (36ft) length of the open vent, then this suggests 0.0014kg/s (10lb/hr) vapour at atmospheric pressure, giving me a calculated pressure drop of about 0.3kPa (0.04psi). Alternatively, using imperial units chart and rough extrapolation by square of flow from minimum of 100lb/hr down to 10lb/hr for 3/4" Schedule 40 pipe, a flow of 100lb/hr gives me a pressure drop of 0.8psi per100ft at 100psig, or corrected to about 5.5psi per 100ft at 0psig, so about ((10/100)^2) times this, or about 0.05psi per 100ft, meaning about 0.02psi pressure drop for the 36ft actual length. These metric and imperial answers aren't exactly the same (the imperial figure is about half of the metric answer); however, what they have in common is that they are both very small compared to the original normal operating pressure of 120kPa(g) (18psi(g)) in the hot water cylinder. This tells me that the increase in gauge pressure during the fault condition is less than 1%.
Getting (at last...) to my final query, I'm looking at the above installation where the the vent pipe has been run in 15mm (1/2") rather than 20mm (3/4"). Whilst this certainly contravenes our 'code', I want to comment on whether or not it is likely to lead to a significant decrease in safety due to signicant increase in 'back pressure' in open vent causing rise in pressure in hot water storage cylinder. In relative terms, allowing for pressure drop to be about proportional to diameter^5, then this reduction from 20mm (3/4") to 15mm (1/2") might be expected to lead to back pressure increasing by a factor of about 8. However, in real terms, this would mean the increase in hot water storage cylinder pressure from normal to fault condition being something like 8% as compared to something like 1%. It is this final comparison which leads me to believe that whilst 'looking' like a significant restriction as compared to the 'code' requirement of 20mm (3/4"), this 15mm (1/2") vent pipe is not actually a significantly more dangerous design?
I'm just a HVAC engineer, not a piping specialist, so whilst I've heard of two-phase flow methods like Homogeneous, Lockhart and Martinelli, Baroczy, etc, I'll just be satisfied with obtaining conservative answers rather than 'exact' answers. I'm also not that familiar with imperial units, but I've included them (hopefully accurately) as I know many users of these forums are more comfortable with them!
Thank you to anyone who's managed to read all the above without falling asleep, and I'd be really grateful for any comments telling me where I've gone wrong in my simplistic analysis.
Brian
In the UK, our typical open vented domestic hot water system in residential properties comprises the following:
atmospheric tank in loft
cold water pipe (20mm (3/4")) drops from bottom connection of loft tank to bottom connection of hot water storage cylinder
hot water storage cylinder (175litres (45USgallon)) in bedroom/kitchen/wherever containing electric immersion heater
open vent pipe (20mm (3/4")) rises from top connection of hot water storage cylinder to terminate over loft tank
Our 'code' requires, in particular, the open vent pipe to be of diameter at least 20mm (3/4").
A fault with the electric immersion heater (of typical output 3kW (10,000btu/hr)) may make it it stay on, in which case I assume the sequence of events is as follows:
To begin with, the pressure in the hot water storage cylinder is just the static head due to the loft tank above. In the hot water storage cylinder, the water temperature rises and, eventually, exceeds the saturation temperature (not just in the immediate vicinity of the electric immersion heater element, but more generally). Due to a combination of vapour generation / boiling / foaming and liquid density decreasing, there is upward, presumably two-phase, flow in the open vent, accompanied by downward flow in the cold water feed pipe. As the steam / water mixture is ejected from the open vent, sufficient flashing off will occur to lower the water temperature to 100C (212F) consistent with saturation temperature at atmospheric pressure. This water falls into the loft tank below, raising its temperature. Left for long enough, and ignoring heat losses from pipe / tank / cylinder, this circulation will result in the temperature of water in the tank reaching and stabilising at 100C (212F). At equilibrium, ignoring energy losses from pipework and ignoring energy required to drive the circulation, the amount of steam flashing off must be equivalent to the 3kW (10,000btu/hr) electrical input. As latent heat of evaporation is about 2200kJ/kg (1000btu/lb), then the amount of steam flashed-off / lost is about 0.0014kg/s (10lb/hr). Also, as in my case the elevation of loft tank above hot water storage cylinder is about 12m (36ft), the static pressure to begin with is about 120kPa(g) (18psi(g)), so the saturation temperature is about 123C (253F). Also, the heat lost due to evaporation must be sufficient to reduce the circulating water temperature from about 123C (253F) to 100C (212F) at the point of expulsion from the open vent. As this heat loss is known (from above, 3kW (10,000btu/hr)), and as the specific heat capacity of water is about 4.2kJ/kgK (1btu/lbR), then then the amount of liquid circulating must be (3/(4.2x23)) or (10,000/(41x1)), i.e. 0.031kg/s (244lb/hr).
From the above, the total mass flow rate is 0.0014kg/s (10lb/hr) vapour plus 0.031kg/s (244lb/hr) liquid making 0.0324kg/s (254lb/hr) total, with the vapour mass flow rate being 4% of the total mass flow rate. The specific volume of vapour is 1.7m3/kg (27ft3/lb) at atmospheric pressure (open end of vent pipe) and 0.8m3/kg (13ft3/lb) at 120kPa(g) (8psig(g)) (inside hot water storage cylinder).
After the above unbearably long preamble, my question is: how do I determine, conservatively, the increase in pressure to which the hot water storage cylinder is subject under the above conditions?
My gut feeling is that I might conservatively allow for all the vapour being at atmospheric pressure, and ignore the relatively very small frictional pressure drop due to liquid flow. Allowing for the vapour to be flowing the entire 12m (36ft) length of the open vent, then this suggests 0.0014kg/s (10lb/hr) vapour at atmospheric pressure, giving me a calculated pressure drop of about 0.3kPa (0.04psi). Alternatively, using imperial units chart and rough extrapolation by square of flow from minimum of 100lb/hr down to 10lb/hr for 3/4" Schedule 40 pipe, a flow of 100lb/hr gives me a pressure drop of 0.8psi per100ft at 100psig, or corrected to about 5.5psi per 100ft at 0psig, so about ((10/100)^2) times this, or about 0.05psi per 100ft, meaning about 0.02psi pressure drop for the 36ft actual length. These metric and imperial answers aren't exactly the same (the imperial figure is about half of the metric answer); however, what they have in common is that they are both very small compared to the original normal operating pressure of 120kPa(g) (18psi(g)) in the hot water cylinder. This tells me that the increase in gauge pressure during the fault condition is less than 1%.
Getting (at last...) to my final query, I'm looking at the above installation where the the vent pipe has been run in 15mm (1/2") rather than 20mm (3/4"). Whilst this certainly contravenes our 'code', I want to comment on whether or not it is likely to lead to a significant decrease in safety due to signicant increase in 'back pressure' in open vent causing rise in pressure in hot water storage cylinder. In relative terms, allowing for pressure drop to be about proportional to diameter^5, then this reduction from 20mm (3/4") to 15mm (1/2") might be expected to lead to back pressure increasing by a factor of about 8. However, in real terms, this would mean the increase in hot water storage cylinder pressure from normal to fault condition being something like 8% as compared to something like 1%. It is this final comparison which leads me to believe that whilst 'looking' like a significant restriction as compared to the 'code' requirement of 20mm (3/4"), this 15mm (1/2") vent pipe is not actually a significantly more dangerous design?
I'm just a HVAC engineer, not a piping specialist, so whilst I've heard of two-phase flow methods like Homogeneous, Lockhart and Martinelli, Baroczy, etc, I'll just be satisfied with obtaining conservative answers rather than 'exact' answers. I'm also not that familiar with imperial units, but I've included them (hopefully accurately) as I know many users of these forums are more comfortable with them!
Thank you to anyone who's managed to read all the above without falling asleep, and I'd be really grateful for any comments telling me where I've gone wrong in my simplistic analysis.
Brian
