Calculating Flow Rate of 180°F and 3 GPM

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SUMMARY

This discussion focuses on calculating the energy required to heat water to 180°F (80°C) at a flow rate of 3 gallons per minute (11 liters per minute). The key formula used is q = h / (cp ρ dt), where h represents heat flow rate, cp is specific heat capacity, ρ is density, and dt is the temperature difference. The calculated energy input needed for heating water from 20°C to 80°C at this flow rate is 42 kW. The context suggests a residential solar water heating project.

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adamjbradley
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I'll elaborate a little. I trying to work out how to calculate and how to make at least 180 F (80 C) and 3 gallons per minute (11 liters/min)?

Thanks in advance!
Adam
 
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Elaborate a bit more please. Would you like to know how to make a flow rate of 3gal/min at 180F?
 


viscousflow said:
Elaborate a bit more please. Would you like to know how to make a flow rate of 3gal/min at 180F?

I'd like to know how to calculate how many J of energy I need to provide in order to satisfy that requirement and some suggestions on how (preferably solar thermal, ideally with storage!)

This is a little residential project.


Adam
 


Do you mean you want to raise the temperature to 180 and then pump at 3 gallons a minute (provide energy for both)?

Or

Do you mean you are pumping at 3 gallons a minute and you need the energy input required to heat that flow rate to 180?
 


You'll also need the temperature of the input water being heated.
 


...and that's assuming he means water.
 


brewnog said:
...and that's assuming he means water.

True. Based on his posts it sounds like he's making (or trying to make) a solar water heater.
 


93mb.png
 


Erm, what? Is that the most constructive thing you can post? Don't bother.
 
  • #10


der's a srtain seemilarity
 
  • #11


kandelabr said:
der's a srtain seemilarity

English lad, English.

No wonder you went straight in with a picture.
 
  • #12


jarednjames said:
Do you mean you want to raise the temperature to 180 and then pump at 3 gallons a minute (provide energy for both)?

Or

Do you mean you are pumping at 3 gallons a minute and you need the energy input required to heat that flow rate to 180?

Sorry I dropped off the radar, been caught up in the floods in Brisbane Australia.

The latter, I'm pumping water at 11 litres/minute and need to know the energy input required to heat that flow rate to 80C (from 20C).

Thanks for your patience.


Adam
 
  • #13


I think the following applies for heating water 11L/minute (0.183 kg/s) at 80C (dt is 55C)

q = h / ( cp ρ dt )
where
q = volumetric flow rate
h = heat flow rate
cp = specific heat capacity
ρ = density
dt = temperature difference

h = q ( cp ρ dt ) and
h = 0.183 ( 4.2 ) ( 55)
h = 42 kW
 

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