- #1

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## 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

Thanks in advance!

Adam

- Thread starter adamjbradley
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- #1

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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

Thanks in advance!

Adam

- #2

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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?

- #3

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

This is a little residential project.

Sincerely,

Adam

- #4

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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?

- #5

Mech_Engineer

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

- #6

brewnog

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

- #7

Mech_Engineer

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

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- #9

brewnog

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

- #10

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der's a srtain seemilarity

- #11

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English lad, English.der's a srtain seemilarity

No wonder you went straight in with a picture.

- #12

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

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?

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

- #13

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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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