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I am trying to work out how much energy (ideally electrical heating) I need to put into a heater to raise the temperature of water from say 10 degrees C to 40 degrees C, if the water is flowing at 10 litres per hour.

Assume that the heater is 80% efficient.

I know that the specific heat of water comes into this at 4.18 J/g degree C but how do I work in the flow rate?

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4  
Electrical heaters are basically 100% efficient. Some of the heat of the overall system may be lost to the environment, but all of the electrical power going into a resistor is going to be converted to heat. – Olin Lathrop Mar 11 at 13:10
    
Yes agreed, but the efficiency in transfer of that electrical energy to the water is where these are losses. This will heat will be lost in the wires going to the heater and as the temperature of the resistor goes up, it's resistance will increase too. – Richard Mar 11 at 15:14
2  
1: Electrical energy isn't transferred to the water, only the heat from the resistor is. 2: Increasing resistance doesn't change the efficiency. For a fixed voltage source, there will be less power into the resistor. However, all this power is still converted to heat with essentially 100% efficiency. – Olin Lathrop Mar 11 at 15:19
    
Efficiency depends on where you measure. If you measure the energy at the input to the heating elements, it's probably 100% or very close to it. If you measure at the receptacle and are powering the heater via a 4 or 5 extension cords, it will be less. – DLS3141 Mar 11 at 15:56
    
Energy is transfered to the water, the electrical energy is converted to heat in the resistor and some of this heat is transfered to the water, some heat is also conducted out along the wires. these wires are short and thick so any losses there should be small. – Richard Mar 11 at 17:17

You just need to convert the volumetric flow rate to a mass flow rate by multiplying by its density. This is easy for water:

$$10\ \mathrm{l/hr} \cdot 1\ \mathrm{kg/l} = 10\ \mathrm{kg/hr} = 10^4\ \mathrm{g/hr} = \frac{10^4}{3600}\ \mathrm{g/sec} = 2.78\ \mathrm{g/sec}$$

Now you can multiply by the specific heat and the temperature rise to get the power required.

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$$P = \dot m \cdot C_p \cdot \Delta T $$

Where $P$ is the power required in Watts ($Joules/sec$)

$\dot m$ is the mass flow rate ($kg/sec$) (you'll have to convert the volumetric flow given to a mass flow rate by using $\dot m = Flow_{Vol} \cdot \rho$ where $\rho$ is the density of the fluid)

$C_p$ is the specific heat of the fluid ($ \frac{Joules}{kilogram \cdot K}$)

and $ \Delta T$ is the temperature difference in $K$

This will give you the power required to heat the water as it flows. The amount of energy will, of course depend on how long the water flows.

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  • 10 litres per hour
  • 10 kg per hour
  • 10,000 g per hour
  • 166.66 g per minute
  • 2.77 g per second

Every second, you need to raise the temperature of 2.77 g of water by 30 °C.

  • 2.77 g of water
  • 4.18 J/g specific heat
  • 11.6 J every second for 1 degree
  • 348 J every second for 30 degrees

Assuming 80% efficient.

  • 348 J / 80% = 435 J

You need 435 J/s.

Joules per second is a Watt

  • 435 W
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4  
You forgot about the 30C temperature rise required. – Dave Tweed Mar 11 at 15:31
1  
OK, so 14.51 W is the power required to raise by 1 degree C? To get a 30 degree C rise, I'd need 435.3 Watts? That sounds more reasonable. – Richard Mar 11 at 17:13
    
Actually, more like 350W. As Olin said, losing 20% of the heat to the environment would be an extremely unusual situation. – Dave Tweed Mar 11 at 17:17
    
OK, so assuming losses of 5%, 366 Watts. In my test, I put in 144 watts (DC 12v @ 12A) and got an increase of 20 degrees, which appears to be a bit under half the required power? Ignoring losses, a 20 degree rise needs 11.6*20=232 Watts. I should only have seen a rise of 12.4 degrees... – Richard Mar 11 at 17:27
    
It's probably just combined measurement errors, as I don't have an accurate flow meter, so a drop in flow would give me hotter water for less power in. I trust the temp gauge, it's a digital thermocouple and I trust the PSU display of current and voltage, but the flow measurement was just how long it took to get 100ml of water... – Richard Mar 11 at 17:32

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