# Appendix B: Determining noise exposure and calculating L_{ex,8}

## Equivalent sound exposure level (L_{eq})

An “equivalent sound exposure level” (L_{eq}) is the continuous sound pressure level (SPL) of noise averaged over time which contains the same amount of energy as that generated by the actual and varying sound levels being measured.

If a worker’s exposure to noise throughout a shift can be characterized by a number of L_{eq} measurements for distinct noise activity periods, the measurements can be combined into a full shift L_{eq} using the equation:

L_{eq}_{, shift} = 10 Log [(1/T) {(t_{i} x 10^{0.1SPLi} ) + ……. .+ (t_{n} x 10^{0.1SPLn})}]

in which

L_{eq}, _{shift} is the full shift equivalent sound pressure level

T is the shift length in hours and equal to sum total of all activities from i=1 to i=n

i is a discrete activity of a worker exposed to a sound level

t_{i} is the duration in hours of i,

n is the total number of discrete activities in the worker’s shift.

SPL_{i} is the L_{eq} for the i^{th} activity in dBA

## Calculation of L_{ex,8}

“L_{ex,8}” is the equivalent sound exposure level (L_{eq}) of noise averaged over 8 hours, the length of a typical work shift. It can be calculated using the equation above, where T is equal to 8, or by the mathematical equation or formula given in the Noise Regulation (O. Reg. 381/15) below.

in which,

L_{ex,8 }is the equivalent sound exposure level in 8 hours,

∑ is the sum of the values in the enclosed expression for all activities from i = 1 to i = n,

i is a discrete activity of a worker exposed to a sound level,

t_{i} is the duration in hours of i,

SPL_{i} is the sound level of i in dBA,

n is is the total number of discrete activities in the worker’s total workday.

For example, if a worker’s exposure consists of two distinct exposure periods, 3 hours at an L_{eq} of 84 dBA and 5 hours at an L_{eq} of 88 dBA, then:

L_{ex,8} = L_{eq}, _{8} = 10 Log [(1/8){(3 x 10^{0.1 x 84}) + (5 x 10^{0.1 x 88})}] = 86.9 dBA

## Determination of L_{ex,8} for non-standard work shifts

### Method 1

For a shift length greater or less than 8 hours, the L_{ex,8} may be calculated using the equation:

L_{ex,8}= L_{eq}_{, T} +10 Log (T/8), where T is the shift duration in hours.

For example, an L_{eq} for a 10 hour shift would be converted to a L_{ex,8} as follows:

L_{ex,8}= L_{eq}_{, 10} + 10 Log (10/8) = L_{eq}_{, 10} + 1 dBA

Therefore, if a worker is exposed to a measured level of 88 dBA L_{eq} for a period of 10 hours, the L_{ex,8} would equal 88 dBA + 1 dBA, or 89 dBA.

### Method 2

The same result can be obtained using the following chart and table, which show the correction in dBA to be made to the L_{eq} for different shift durations.

In this chart the numbers below the line represent the actual shift duration in hours and the numbers above the line are the corresponding correction factor to be added to the shift L_{eq} to convert it to an L_{ex,8}.

Correction (dB) | -3 | -2 | -1.3 | -0.6 | 0 | 0.5 | 1 | 1.4 | 1.7 | 2 | 2.4 | 2.7 | 3 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

Shift duration (hours) | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |

This table is provided for convenience only. It does not provide for the determinations of shifts or exposures for fractions of hours, such as shifts of 7.5 hours.

For a 12-hour shift, both the chart and the table show a 1.7 dBA correction (or more appropriately, a 2 dBA correction since measurement accuracy of better than +/-1 dBA is unlikely). Thus a worker exposed to 83 dBA for 12-hours would have an L_{ex,8} exposure of 83 + 2 = 85 dBA. Another way of looking at this is to consider the 12-hour shift L_{eq} (or L_{ex, 12}) limit to be 83 dBA.

### Method 3

Where a worker’s noise exposure can be easily split into discrete activities during which sound levels are steady, the following table may be used to determine the L_{eq} for the shift.

Steady Sound Level (dBA) | Duration |
---|---|

82 | 16 hours |

85 | 8 hours |

88 | 4 hours |

91 | 2 hours |

94 | 1 hours |

97 | 30 minutes |

100 | 15 minutes |

103 | 7.5 minutes |

106 | 3.75 minutes |

109 | 1.88 minutes |

112 | 0.94 minutes |

115 | 28.12 seconds |

Where a worker is exposed to a steady sound level throughout their shift, the exposure duration must not exceed the value for that sound level in the above Table. Once the worker has reached the permissible duration for that sound level, the worker’s noise exposure has reached 100% of the permissible daily noise dose. For example, if a worker is exposed to a steady sound level of 91 dBA, the maximum permissible daily exposure duration is 2 hours.

Where the daily noise exposure is composed of two or more periods of noise exposure of different levels, the following equation may be used to determine if the overall exposure exceeds the allowable noise exposure limits:

(C1/T1 + C2/T2 + …. C_{n}/T_{n}) x 100 = % of exposure limit

Where

C = total duration of exposure at a specific noise level, and

T = total duration of exposure permitted at that level (Refer to “Table of Equivalent Noise Exposures” above)

The 85 dBA L_{ex,8} exposure limit is exceeded when the dose, calculated using the above formula exceeds 100%.

Example:

A worker is exposed to 85 dBA for 4 hours and 91 dBA for 1.5 hours.

Therefore using above calculation:

(4/8 + 1.5/2) x 100 = 125%

In this case the worker has exceeded his/her allowable noise exposure and therefore measures must be implemented to protect the worker’s hearing. If the exposure duration at 91 dBA were reduced to 1 hour, the combined exposure would be at 100% of the permissible daily noise dose, provided the worker’s exposure is at or below 80 dBA for the rest of his/her shift.