Heater Rating Calcultion 3

[NCTHAI]

This looks to be simple for lot of people but I am still learning.

Looking to calculate heater capacity required to heat water from 40 to 150 Deg C at pressure of 6 to 7 Bar. Using imperical formulae Q= Delta T X Cp X V got 183.33 KWH required to heat water. However need to know KW rating of heater.

Need help for following:
- Check if 183.33 KWH is correct energy consumption required.
- Need to know KW size of heater.

Calculation steps would be much appreciated.

Thanks in advance for help!!

NCTHAI

 

[ione]

NCTHAI,

The formula you’ve reported in your post gives as result the energy required to heat up a mass (Q), with a temperature increase (deltaT) and the result is expressed in kJ or equivalent units (kWH).

Now if you want to calculate a thermal power in kW or equivalent units, you need another parameter (i.e. time).

Thermal power = energy/time

To check whether the amount of energy you’ve calculated is correct we need to know the value of Q.

 

[DRWeig]

Yes, as Ione has asked -- and to expand just a bit,

If a static amount of water in a tank, how fast do you want to heat it from 40°C to 150°C?

If a flowing amount of water, what is the mass flow rate?

The answer to either of these questions will allow a power calculation.

Good on ya,

Goober Dave

 

[ione]

Just wanted to add that specific heat of water changes as temperature increases:

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

I suggest you the following approach.

1. Consider just a unit of mass (1 kg)
2. Evaluate specific enthalpy h1 of water at starting conditions (temperature and pressure) [kJ/kg]
3. Evaluate specific enthalpy h2 of water at final (desired) conditions (temperature and pressure) [kJ/kg]
4. Energy required is the difference between E = h2 –h1

The calculator in the attached link could be of some help.

http://www.spiraxsarco.com/resources/steam-tables/sub-saturated-water.asp

Please consider also that this approach does not take into account any heat loss or heating system efficiency.

 

[IRstuff]

While the desired time is, of course, a critical component, it would otherwise mean nothing if you can't actually get the heat to the water in a timely fashion, i.e., if your heater is, for example, outside of the insulation of the tank, and you decided you need to heat up the tank in 1 seconds. 183.33 kWh/1 s = 660 MW. Obviously, you're not going to get that kind of heater, and you will never get that kind of performance even if you had that heater power. While such performance is not impossible, given the right conditions, it's extremely difficult to achieve.

So before you do that calculation, you'll need to look at the heat capacity and heat transfer efficiency of your overall system, including how your heater is installed in the system. Additionally, you'll need to look at how much heat loss there is, and whether that's acceptable, as that will determine your long term energy consumption.

A simple calculation you can do to get a ballpark value is to look at the total thermal mass divided by the average thermal conductance to the center of your tank. The result will be in units of time, and represents a system time constant of a sort. The calculation can be refined further with a more detailed distributed lumped-parameter model of your system.

TTFN

FAQ731-376