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APPENDIX B ERROR ESTIMATION
Table of Contents
- 1. GENERAL
- 2. ERROR PROPAGATION FORMULAE
- 2.1 Addition/Subtraction of Two Variables
- 2.2 Multiplication/Division of Two Variables
- 3. EXAMPLE OF ERROR CALCULATION FOR DIRECT MEASUREMENT
- 4. EXAMPLE OF ERROR CALCULATION FOR MASS BALANCE
1. General
The error of an individual measurement can be estimated by the using
the standard deviation for the measurement device, for example, a scale
may be accurate to ± 100 kg or a laboratory device may measure
concentration ± 2%.
Two measured variables (e.g., flow F and concentration C) with uncertainties
(∆F and ∆C) might then be used to calculate a new value,
Releases (R). The error of the release estimate (∆R) can be calculated
using traditional statistical techniques.
This appendix provides basic formulae with examples as well as examples
of error estimation for direct measurement and mass balance release calculations.
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2. Error Propagation Formulae
Two measured variables (e.g., x and y) with uncertainties (∆x
and ∆y) might be used to calculate a new value, z. The following
two subsections provide formulae which are used to calculate the uncertainty, ∆z
in z for addition/subtraction and multiplication/division.
2.1 Addition/Subtraction
of Two Variables
The following formula is typically used to determine the variance of
the sum of two values, the variances of which are known:
∆z = SQRT [(∆x)2 +
(∆y)2]
where:
- ∆z is the error for the sum
- ∆x is the error for the first variable
- ∆y is the error for the second variable
For example, if
x = 2.0 ± 0.2
cm
y = 3.0 ± 0.6 cm
and,
z = x + y = 2.0 cm + 3.0 cm = 5.0 cm
then,
∆z = SQRT [(0.2 cm)2 + (0.6 cm)2]
= 0.6 cm
therefore,
z = 5.0 ± 0.6 cm
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2.2 Multiplication/Division of
Two Variables
The following formula is typically used to determine the variance of
the product of two values, the variances of which are known:
∆z/z = SQRT [(∆x/x)2 +
(∆y/y)2]
For example, if
x = 2.0 ± 0.2 cm
y = 3.0 ± 0.6 cm
and,
z = x * y = 2.0 cm * 3.0 cm = 6.0 cm2
then,
∆z/z = SQRT [0.2/2.0)2 + (0.6/3.0)2]
= 0.22
∆z = 0.22 * 6.0 cm2= 1.3 cm2
therefore,
z = 6.0 ± 1.3 cm2
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3. Example of Error Calculation for Direct Measurement
If a facility was calculating particulate matter emissions based on
a measurement of flow in a particular stack and the particulate matter
concentration, the following information might be available:
Flow: 2,000 m3/min with an estimated error
of 100 m3/min
Concentration: 15 mg/m3 with
an estimated error of 1.5 mg/m3
Operating hours:
16 hours a day, 340 days in the year.
The releases of particulate matter (PM) is calculated as follows:
PM = 15 mg/m3 * 2,000m3/min
* 60 min/hour * 1X10-6kg/mg * 16 hours/day
* 340 days/year * 0.001 tonnes/kg = 9.8 tonnes/year
The error is calculated using the error estimated for each variable
and the formula provided in the preceding section.
∆PM/PM = SQRT [(100/2,000)2 + (1.5/15)2]
= 0.11
This error can then be converted into an absolute value by multiplying
the error by the release estimate.
∆PM = 9.8 tonnes/year * 0.11 = 1.1 tonnes per year
The particulate
matter, for this facility is therefore, 9.8 ± 1.1 tonnes.
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4. Example of Error Calculation for Mass Balance
If a facility is calculating its sulphur dioxide release using mass
balance, the following information might be available:
Table 4-1: Mass and Concentration of Sulphur in Each Stream - Mass
Balance Example with Error Estimates
Stream |
Mass of Stream
(tonnes) |
Error Estimate
(%) |
Concentration
of Sulphur
(%) |
Error Estimate
(%) |
Concentrate |
30,000 |
1% |
30 |
0.3% |
Recyclables |
3,000 |
2% |
5 |
5% |
Other |
1,500 |
5% |
5 |
5% |
Product |
7,500 |
0.5% |
15 |
1% |
Slag |
25,000 |
10% |
1 |
10% |
By-product (sulphuric acid) |
20,000 |
0.5% |
32 |
0.5% |
The mass of sulphur in each stream and the calculated error is provided
in the table below:
Table 4-2: Mass of Sulphur in Each Stream - Mass Balance Example
Stream |
Mass of Sulphur
in Stream
(tonnes) |
∆S/S
(square root of sum of errors squared) |
Standard Error
(tonnes) |
Concentrate |
9,000 |
0.0104 |
94.0 |
Recyclables |
150 |
0.0539 |
8.1 |
Other |
75 |
0.0707 |
5.3 |
Total Input |
9,225 |
|
|
Product |
1,125 |
0.0112 |
12.6 |
Slag |
250 |
0.1414 |
35.4 |
By-product (sulphuric acid) |
6,400 |
0.0071 |
45.3 |
Total Output |
7,775 |
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The error on each of the input and output streams is equal to the square
root of the sum of the errors squared as shown in the tables below.
Table 4-3: Error Estimation on Input Streams
Stream |
Standard Error
(tonnes) |
Standard Error
Squared |
Concentrate |
94.0 |
8,829 |
Recyclables |
8.1 |
65 |
Other |
5.3 |
28 |
Sum of Standard Errors Squared |
|
8,922 |
Standard Error of Sum
(tonnes) sulphur |
|
94 |
Table 4-4: Error Estimation on the Output Streams
Stream |
Standard Error
(tonnes) |
Standard Error
Squared |
Product |
12.6 |
158 |
Slag |
35.4 |
1,250 |
Sulphuric Acid |
45.3 |
2,048 |
Sum of Standard Errors Squared |
|
3,456 |
Standard Error of Sum
(tonnes) sulphur |
|
59 |
The error for the emission calculation is calculated as shown in Table
4-5.
Table 4-5: Error Estimation on the Annual Release
|
Sulphur
(tonnes) |
Standard Error
(tonnes) |
Standard Error
Squared |
Total Inputs |
9,225 |
94 |
8,922 |
Total Outputs |
7,775 |
59 |
3,456 |
Releases |
1,450 |
|
|
Sum of standard errors squared |
|
|
12,379 |
Standard Error of Sums
(tonnes sulphur) |
|
|
111 |
Standard Error (%) |
|
|
7.7 |
If this mass balance calculation estimates the monthly emissions, an
additional calculation is required to determine standard error on an
annual basis. The error is divided by the square root of 12 (i.e., the
number of data points) to yield an error on an annual basis of 2.2%
The calculated standard error can then be multiplied by 2 to yield a
95.5% confidence limit. The annual error at 95.5% confidence would be ± 4.4%.
In order to yield an absolute error value, this percentage is multiplied
by the annual release.
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