Energy required to raise temperature of water flowing through a heater

Question

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?

Answer 1

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.

Answer 2

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

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.

Answer 3

• 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