Calculating % heat loss due to moisture present in fuel to determine the boiler efficiency.
The percentage loss of heat due to moisture present in the fuel is one of the factor to determine the efficiency of the boiler. The moisture enters the boiler along with the fuel leaves the chamber as a superheated vapour. The moisture loss is calculated using the below formula.
Formula:
L3 = ( M x {584 + Cp(Tf - Ta)} / GCV of Fuel ) x 100
Where,
M = Kg of moisture in fuel in 1 kg basis
Cp = Specific heat of superheated steam in kCal/kgoC
Tf = Fuel gas temperature in oC
Ta = Ambient temperature in oC
584 = Latent heat corresponding to partial pressure of water vapour.
GCV 0f Fuel = Gross Calorific Value of Fuel
Let us consider an example, on how to calculate the percentage loss of moisture present in the fuel with the given values.
Step 1: Given that,
Moisture in fuel in 1 kg basis = 0.015 kg
Specific heat of superheated steam = 0.45 kCal/kgoC
Fuel gas in temperature = 250 oC
Ambient temperature =28 oC
GCV of fuel = 13000
Step 2: Applying the values in the formula,
L3 = ( M x {584 + Cp(Tf - Ta)} / GCV of Fuel ) x 100
L3 = ( 0.015 x {584 + 0.45(250 - 28)} / 13000 ) x 100
L3 = 0.0789 %
Result : % Heat Loss Due to Moisture Present in the Fuel = L3 = 0.0789 %