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How do I calculate efficiency on wood burning appliances? |
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You can calculate the efficiency of wood burning appliances by knowing the "K" values of the fuel, or you could just buy a KANE425 with the wood pellet fuel table built in.
The calculation below shows how the "K" values can be obtained:
CALCULATION OF FUEL DATA
For any fuel not specified by Kane International Ltd. The net calorific value, gross calorific value and composition should be obtained from the fuel supplier.
The following fuel data has been calculated with reference to the efficiency claculation.
Example 1:
Chemical composition : C 25%
H2 3%
H20 50%
Qnet 8.35 MJ/kg
Qg 9.3 MJ/kg *
Max CO2 20.4%
K1n = (255 x % carbon in fuel) / Qnet (Kj/Kg)
= (255 x 25) / 8350 = 0.763
K1g = ( 255 x % carbon in fuel) / Qg (Kj/Kg)
= (255 x 25) / 9300 = 0.685
K2 = Max % CO2 = 20.40
K3 = Wet Loss = [(9 x % H2 + % H2O) / 9300] x 2425
= [(9 x 3 +50) / 9300] x 2425
= (77 / 9300) x 2425 = 20.08
K4 = 65 (an approximation for wood) *
The fuel values to program into the Analyser are as follows:
NATURAL GAS:
K1g : 0.763 K1n : 0.685
K_2 : 20.4 K_3 : 20.08
K_4 : 65 O2r : 8.0
The table below shows the "K" values for common fuels:
Fuel |
K1g |
K1n |
K2 |
K3 |
K4 |
O2r |
Natural gas |
0.35 |
0.39 |
11.89 |
9.83 |
32 |
3 |
Town gas |
0.467 |
0.519 |
12.9 |
14.7 |
32 |
3 |
Light Oil |
0.48 |
0.51 |
15.51 |
6.1 |
48 |
3 |
Heavy Oil |
0.51 |
0.54 |
15.8 |
5.7 |
48 |
3 |
Coal |
0.62 |
0.65 |
18.39 |
2.2 |
63 |
8 |
Coke |
0.75 |
0.76 |
20.61 |
0.3 |
47 |
8 |
Anthracite |
0.67 |
0.69 |
19.1 |
1.8 |
65 |
8 |
Propane |
0.42 |
0.45 |
13.79 |
7.7 |
32 |
3 |
Butane |
0.43 |
0.46 |
14.11 |
7.4 |
32 |
3 |
Gascor |
0.453 |
0.503 |
13.46 |
13 |
32 |
3 |
Kinsale gas |
0.366 |
0.406 |
11.68 |
10.4 |
32 |
3 |
Wood Pellets |
0.629 |
0.67 |
20.1 |
6.97 |
65 |
3 |
Calculated data: Tnet = Net Temperature
% CO2 content in flue gas
% Dry Flue Gas losses
% Wet losses
% Unburned carbon loss
% Efficiency
%CO2 = (20.9 - %O2m) x K2 / 20.9
Tnet = Flue Temperature - Inlet Temperature
Dry flue gas loss = 20.9 x K1n x (Tnet) / K2 x (20.9 - %O2m)
Wet loss = 9 x % H2 + %H2O / Qgr x [2488 + 2.1Tf - 4.2 Ti]
Simplified = [(9 x %H2 + %H2O) / Qgr] x 2425 x [1 + 0.001 Tnet]
Wet loss = K3 (1 + 0.001 x Tnet)
Where K3 = [(9 x %H2 + % H2O) / Qgr] x 2425
Net efficiency = 100% - dry flue gas losses
= 100% - 20.9 x K1n x (Tnet) / K2 x (20.9 - % O2m)
Gross efficiency = 100% - (dry flue gas losses + wet losses)
= 100% - [20.9 x K1g x (Tnet) / K2 x (20/9 - %)2m)] + [K3 x (1 + 0.001 x Tnet)]
Excess Air = [(20.9% / (20.9% - O2m%) - 1] x 100%
CO2% = [(20.9% - O2m%) x K2% / 20.9%]
Unburned fuel loss = K4 x CO% / (CO% + CO2%)
Where K4 = 70 fo coke
= 65 for anthracite
= 63 for Bituminous coal
= 62 for coal tar fuel
= 48 for liquid petroleum fuel
= 32 for natural gas
The formula for K4 is based on the gross calorific value Qgr. To obtain the loss based on net calorific value multiply by Qgr/Qnet. Since this loss is usually small this conversion has been ignored.
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