Q: How would I go about working out how I can make at least 180 F and 3 gallon/min

[adamjbradley]

Main Question or Discussion Point

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

 

[viscousflow]

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

 

[adamjbradley]

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.

Sincerely,
Adam

 

[JaredJames]

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?

 

[Mech_Engineer]

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

 

[brewnog]

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

 

[Mech_Engineer]

...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.

 

[kandelabr]

 

[brewnog]

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

 

[kandelabr]

der's a srtain seemilarity

 

[JaredJames]

der's a srtain seemilarity

English lad, English.

No wonder you went straight in with a picture.

 

[adamjbradley]

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.

Sincerely,
Adam

 

[adamjbradley]

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