Pat_the_engineer
Civil/Environmental
Just wanted to know how to find out how much water is lost between a shower head and the drain and how to determine the drop in temperature.
Thanks,
Pat
Thanks,
Pat
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Shower Head Problem 1
- Thread starter Pat_the_engineer
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What do you mean by "lost?" And why would you care about the drop in temperature?
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I assume you mean the amount lost through vaporisation causing cooling?
A couple of variables to think about.
Droplet size will have a significant effect.Droplet size is really a product of shower head design and operating pressure. Smaller droplets will likely evaporate faster and have more surface area per unit volume of water.
The humidity will have an impact on the amount of vaporisation occuring. Inside a shower stall with reasonably warm water running will create an environment of very high humidity. So it is likely that there will be little evaporation occurring.
The ambient temperature and the temperature of the water will impact on the temperature drop.
So who can give any reasonable answer to your question unless you define the variables.
Regards
Ashtree
"Any water can be made potable if you filter it through enough money"
A couple of variables to think about.
Droplet size will have a significant effect.Droplet size is really a product of shower head design and operating pressure. Smaller droplets will likely evaporate faster and have more surface area per unit volume of water.
The humidity will have an impact on the amount of vaporisation occuring. Inside a shower stall with reasonably warm water running will create an environment of very high humidity. So it is likely that there will be little evaporation occurring.
The ambient temperature and the temperature of the water will impact on the temperature drop.
So who can give any reasonable answer to your question unless you define the variables.
Regards
Ashtree
"Any water can be made potable if you filter it through enough money"
LittleInch
Petroleum
Assuming what ashtree says,
My estimate is that any evaporation water loss for a standard domestic shower is <5% and probably <2% once the shower room is more like a steam room.
That's asusming you don't count the water seeping out of the shower...
Temperature? ~Start temp is probably about 40C for a "hot" shower.
Air temp at start about 20C then rising as shower continues.
So an estimate is around 35C allowing for heat loss, heating up the shower surfaces, drain pipe etc etc.
Is that good enough?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
My estimate is that any evaporation water loss for a standard domestic shower is <5% and probably <2% once the shower room is more like a steam room.
That's asusming you don't count the water seeping out of the shower...
Temperature? ~Start temp is probably about 40C for a "hot" shower.
Air temp at start about 20C then rising as shower continues.
So an estimate is around 35C allowing for heat loss, heating up the shower surfaces, drain pipe etc etc.
Is that good enough?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Lots of help for what looks like homework
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
SuperSalad
Chemical
gehardl pretty much nailed it as far as I can tell.
The amount of variables is ridiculous for something like this. Spray pattern, droplet size, mineralization on the shower head, bath/shower surface area, drain rate, room temperature, room humidity, bath/shower surface temperature, shower head height, shower head angle, water pressure, room airflow, curtain size, curtain material, bath/shower surface material, bath/shower surface thickness, body size, hair length, hair thickness, hair quantity, skin saturation, evaporation rate, temperature, how one showers (ie. if you fling water around while being wetted or stand there unmoving), etc., etc., the list goes on and on...
If you actually need to know this information for some reason, directly measure your flow rate & temperature in and directly measure your flow rate & temperature out.
Andrew H.
The amount of variables is ridiculous for something like this. Spray pattern, droplet size, mineralization on the shower head, bath/shower surface area, drain rate, room temperature, room humidity, bath/shower surface temperature, shower head height, shower head angle, water pressure, room airflow, curtain size, curtain material, bath/shower surface material, bath/shower surface thickness, body size, hair length, hair thickness, hair quantity, skin saturation, evaporation rate, temperature, how one showers (ie. if you fling water around while being wetted or stand there unmoving), etc., etc., the list goes on and on...
If you actually need to know this information for some reason, directly measure your flow rate & temperature in and directly measure your flow rate & temperature out.
Andrew H.
This is a control volume problem, Qin = Qout - Qlost
Qlost = Q(into the air) + Qothers (hair, floor, shower walls ect)
I would ignore Qothers since they do not lower the temperature of the water (the water is still liquid)
Q(into the air) is an Basic HVAC question, use the psychometric chart and the volume of the room plus any volume exchange due to the exhaust fan. For those who polish the cannon ball add the condensation on the mirror. The psychometric chart will give values needed, essentially the room goes from ~50% to saturated and that will be the Qlost,
but for long showers
Once the room is saturated without an exhaust fan those losses stop.
Once the temperature of the mirror is at the dew point those losses stop.
Once the shower walls, curtain, hair are fully wetted those losses stop.
So then the steady state loss is how much air is the fan moving and how much is spilling onto the floor
Hydrae
Qlost = Q(into the air) + Qothers (hair, floor, shower walls ect)
I would ignore Qothers since they do not lower the temperature of the water (the water is still liquid)
Q(into the air) is an Basic HVAC question, use the psychometric chart and the volume of the room plus any volume exchange due to the exhaust fan. For those who polish the cannon ball add the condensation on the mirror. The psychometric chart will give values needed, essentially the room goes from ~50% to saturated and that will be the Qlost,
but for long showers
Once the room is saturated without an exhaust fan those losses stop.
Once the temperature of the mirror is at the dew point those losses stop.
Once the shower walls, curtain, hair are fully wetted those losses stop.
So then the steady state loss is how much air is the fan moving and how much is spilling onto the floor
Hydrae
One can run the problem backwards; measure the RT before, turn the shower one for 15 mins, measure RT after. The amount of condensation is relative small, compared to the typical 1 gpm of modern shower heads. The change in heat of the air of the room can be equated to the loss of heat of the water.
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I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
I agree with the theory but in terms of actual temperature change, this will only be true if the room has perfect insulation and there are no other sources of hot or cold in the room
Regards
Ashtree
"Any water can be made potable if you filter it through enough money"
Regards
Ashtree
"Any water can be made potable if you filter it through enough money"
LittleInch
Petroleum
Well dear old Pat has posted only this question and logged back in once.
So clearly not high in his or her list of priorities. ...
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
So clearly not high in his or her list of priorities. ...
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
It's a number, not necessarily any better or worse than any other number
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
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1
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Have we been nerd snipped?
Hydrae
Hydrae
Where is?….. eh Pat_the_engineer ?
Guess they don't care about their answer or problem, or it was homework, after all.
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
- Thread starter
- #18
Pat_the_engineer
Civil/Environmental
Hi all thank you for your responses they were truly appreciated and hope you all had a great easter break. Happy easter and thank you once again.
LittleInch
Petroleum
Now Pat, the way it works here is that you actually engage with the people who bothered to reply and respond to the questions and close out the issue.
E.g. did it answer your question
Why did you ask the question?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
E.g. did it answer your question
Why did you ask the question?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
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