firefastball
Mechanical
- May 15, 2014
- 35
Hi there,
Can anyone help to clarify the boiler efficiency calculation particularly on the output part?
Control Volume is only on the boiler
Example: Assuming that there is no blowdown,
Mass balance:
Mass of feedwater = mass of steam
Energy Balance:
Mass of feedwater X Feedwater enthalpy = Mass of steam X Steam enthalpy
So the output on boiler efficiency equation would be:
Mass of steam X Steam enthalpy - Mass of FW X FW enthalpy
Replacing mass of FW to mass of steam would give us
Mass of steam (Steam enthalpy - FW enthalpy)
Now usually have blowdown to remove impurities
Looking at the same control volume with blowdown would give the following equations:
Mass balance:
Mass of FW = mass of steam + mass of blowdown
Energy Balance:
Mass of FW X FW enthalpy = mass of steam X steam enthalpy + mass of blowdown X blowdown enthalpy
Then the output on the boiler efficiency would be:
Mass of blowdown X blowdown enthalpy + mass of steam X Steam enthalpy - Mass of FW X FW enthalpy
Replacing of Mass of FW would give
Mass of BD X BD enthalpy + mass of steam X Steam enthalpy - mass of steam X FW enthalpy - mass of BD X FW enthalpy
or
Mass of Steam ( Steam enthalpy - FW enthlapy) + Mass of Blowdown (Steam enthalpy - FW enthalpy)
Am I doing it right?
Sorry for the long equations. Hope you guys can understand what I'm trying to ask.
Can anyone help to clarify the boiler efficiency calculation particularly on the output part?
Control Volume is only on the boiler
Example: Assuming that there is no blowdown,
Mass balance:
Mass of feedwater = mass of steam
Energy Balance:
Mass of feedwater X Feedwater enthalpy = Mass of steam X Steam enthalpy
So the output on boiler efficiency equation would be:
Mass of steam X Steam enthalpy - Mass of FW X FW enthalpy
Replacing mass of FW to mass of steam would give us
Mass of steam (Steam enthalpy - FW enthalpy)
Now usually have blowdown to remove impurities
Looking at the same control volume with blowdown would give the following equations:
Mass balance:
Mass of FW = mass of steam + mass of blowdown
Energy Balance:
Mass of FW X FW enthalpy = mass of steam X steam enthalpy + mass of blowdown X blowdown enthalpy
Then the output on the boiler efficiency would be:
Mass of blowdown X blowdown enthalpy + mass of steam X Steam enthalpy - Mass of FW X FW enthalpy
Replacing of Mass of FW would give
Mass of BD X BD enthalpy + mass of steam X Steam enthalpy - mass of steam X FW enthalpy - mass of BD X FW enthalpy
or
Mass of Steam ( Steam enthalpy - FW enthlapy) + Mass of Blowdown (Steam enthalpy - FW enthalpy)
Am I doing it right?
Sorry for the long equations. Hope you guys can understand what I'm trying to ask.