# How to Calculate Heat Loss Due to Radiation and Convection?

Calculating percentage heat loss due to radiation and convection to determine boiler efficiency

## % Loss Due to Radiation and Convection

The loss of heat by radiation and convection is determined from the boiler casting into the surrounding boiler house. This percentage loss of heat is one of the factor to determine the efficiency of the boiler. The surface loss and other unaccounted percentage loss is assumed to have certain values based on the type and size of the boiler.
→ For industrial fire tube / packaged boiler = 1.5 to 2.5%
→ For industrial watertube boiler = 2 to 3%
→ For power station boiler = 0.4 to 1% Below given the formula to calculate the % loss of heat due to radiation and convection.
Formula:
Total Radiation and Convection Loss L6 = 0.584 x [(Ts / 55.55)4 - (Ta / 55.55)4] + 1.975 x (Ts - Ta)1.2x 2âˆš[(196.85Vm+ 68.9) / 68.9]
% Loss by Radiation and Convection = (L6 x 100) / (GCV of Fuel x Fuel Firing Rate)

Where,
L6 = Radiation loss in W/m2
Vm = Wind Velocity in m/s
Ts = Surface Temperature (K)
Ta = Ambient Temperature (K) Let us consider an example on how to calculate the percentage heat loss caused by radiation and convection with the given values.
Step 1: Given that,
Vm = Wind Velocity = 22 m/s
Ts = Surface Temperature = 180 K
Ta = Ambient Temperature = 33 K
GCV of fuel = 3501 kJ/kg
Fuel firing rate = 5599.17 % Step 2: Applying the values in the formula,
L6 = 0.584 x [(Ts / 55.55)4 - (Ta / 55.55)4] + 1.975 x (Ts - Ta)1.25 x 2âˆš[(196.85Vm + 68.9) / 68.9]
L6 = 0.584 x [(180 / 55.55)4 - (33 / 55.55)4] + 1.975 x (180 - 33)1.25 x2âˆš[(196.85 x 22+ 68.9) / 68.9]
L6 = 8142.436 w/m2 Step 3: Finding % Heat loss
To convert to kCal/m2 = 8142.436 x 0.86 = 7002.495
To convert to kCal = 7002.495 x 90 = 630224.62 kCal
% Heat Loss by Radiation and Convection = (L6 x 100) / (GCV of Fuel x Fuel Firing Rate)
= (630224.62 x 100) / (3501 x 5599.17)
= 3.334% Result : % Heat Loss by Radiation and Convection = 3.334% 