The lantern selection and example of primary battery parameters calculation for BEPA series lanterns manufactured by Technomarine
Among all BEPA lanterns manufactured by Technomarine, the red and yellow lanterns have the least luminous intensity. This is related to the technology of the LED production. The further example is given for this situation. The calculation methodology is provided using the on a step-by-step basis. Let us consider the most interesting case – the primary battery parameters calculation for buoys.
1. The first to be determined is the lantern color, guaranteed visibility range, and the operation mode (flashing mode).
For example, we need:
Visibility range at least 5 km;
Flashing mode – group quick flashing (flash + pause = period, seconds): (1 + (0.5) + 1 + (0.5) + 1 + (0.5) + 1 + (6.5) = 12.0) seconds;
Navigation period – 200 days (from May to November);
Average period of darkness – 10 hours.
2. If it's all pretty much clear with the color, what should we do with the visibility range? The visibility range is proportional to the luminous intensity. Luminous intensity is a physical value measured in candelas; it means that we can calculate it.
The required luminous intensity that will guarantee the visibility range that we need is determined using the formula (International Regulations for Preventing Collisions at Sea 1972, Appendix 1, clause 8):
I = 3.43 x106 xT xD2 xK-D,
where: I – luminous intensity under operating conditions, candelas; Т – light stimulus equal to 2 10-7 luxes; D – required visibility range of the lantern, nautical miles (1 m = 1.852 km); K – atmospheric transmissivity. Here is the question, what is the “atmospheric transmissivity”? See the table below:
In the navigation charts, and in the “List of lights” books, this coefficient is determined as 0.8.
In GOST 26600 we find two values: 0.7 and 0.84.
The value to be used is the one most frequently occurring at the buoy location.
Let us calculate some values of the luminous intensity with the most frequently used atmospheric transparency coefficient values, and illustrate them in the form of a table:
3. For our example, let us determine the luminous intensity that guarantees the visibility range when the atmospheric transmissivity K = 0.84 according to the table. We see that the luminous intensity may not be less that about 8 candelas. And we should keep in mind that when the climatic conditions become worse, the atmospheric transparency coefficient also becomes less, so does the visibility range correspondingly. For example, with the slight mist, the visibility range of lantern with luminous intensity 8 candelas will be about 4.2 km.
4. Based on the obtained values of the minimum required (guaranteed) luminous intensity, we choose the signaling light. As for BEPA series lanterns, we choose the globe lantern whose luminous intensity is at least 8 candelas, red color. Among the commercially available BEPA series lanterns, the most optimal option is BEPA-8M, whose luminous intensity is 10 candelas for red color.
5. Based on this selection and technical specifications, we can calculate the parameters of the required power source. The most frequently used power sources for the navigation lanterns are the disposable primary batteries or in common parlance called "batteries".
6. First we need to know our lantern's power consumption characteristics. According to the data sheet of the BEPA-8M product, maximum power consumption is 0.55 W. But you should keep in mind that such power consumption is for the flashing mode, i. e. when the LED is ON as a light source. During the pause between the BEPA-8M flashes, when the LED is OFF, the power consumption is 0.0002 W. Now here is the question: what is the average power consumption rating of BEPA-8M for the flash character period?
To calculate the average power consumption rating of the navigation lantern for the specific flashing mode, we will need such a thing as “duty cycle”.
Duty cycle (in physics, electronics) is one of the generic features of the pulse systems that determines the ratio between the pulse active period to the total period of the cycle. This is a non-dimensional value.
Let us determine the duty cycle for our flashing mode: group quick flashing (flash) + (pause) = period, seconds): 1 + (0.5) + 1 + (0.5) + 1 + (0.5) + 1 + (6.5) = 12.0.
The sum of all flashes for the period equals to 4 seconds.
Now let us divide the period by the sum of all flashes in it:
12 seconds / 4 seconds = 3.
So the duty cycle of this flashing mode equals to 3.
To calculate the average power consumption rating for the period, divide the maximum power consumption by the duty cycle value: 0.55 W / 3 ≈ 0.184 W
7. Now let us calculate the energy capacity that we need based on our data:
Navigation period – 200 days (from May to November),
Average period of darkness – 10 hours per day.
Total active running time for 200 days will be equal to:
200 days × 10 hours = 2000 hours.
Let us multiply this value by the lantern's average power consumption rating calculated using the duty cycle:
2000 hours × 0.184 W = 368 Wh.
However, in addition to nighttime the lantern will also function during the daytime, and this daytime operation period will be equal to:
200 days × 24 hours - 2000 hours (active period) = 2800 hours.
Our lantern is “sleeping” for 2800 hours, and during this period, according to the data sheet, it consumes 0.0002 W of power. So the power consumption during the sleep mode in the navigation period will be equal to: 0.0002 W × 2800 hours = 0.56 Wh.
Now we will add the active period power consumption value to the sleep period power consumption value, and round up the result up to the integer: 368 Wh + 0.56 Wh ≈ 369 Wh.
This result is the minimum energy capacity of our power source.
8. Now let us choose the suitable electrochemical power source based on the typical nominal voltage values of the given power sources:
1.2 V, 2.4 V, 2.6 V, 3 V, 4 V, 5 V, 6 V.
When choosing the electrochemical power source, make sure that you use the end voltage value.
Keep in mind that according to the data sheet, the minimum operational voltage of our BEPA-8M lantern is 2 V, and the maximum operational voltage is 6 V.
Example: the primary battery with nominal voltage 1.2 V:
Since 1.2 V of voltage is lower than the minimum operational voltage for our lantern, we will have to connect 2 primary battery is series, so our resulting nominal voltage of this combination will be equal to 2.4 V. In this case, let us consider the example where we use the primary battery with voltage 2.4 V.
Divide the required minimum energy capacity value of our power source by the primary battery nominal voltage to obtain the minimum required electric charge:369 Wh / 2.4 V ≈ 154 Ah.
Self-discharge and temperature factor should also be respected. To avoid digging too much into electrochemistry, we will use the coefficient 1.15 obtained by empiric (practical) way.
154 Ah × 1.15 ≈ 177 Ah.
Now let us look through the commercially available power sources offered by the major manufacturers. We need the primary battery with the following parameters: 2.4 V, 177 Ah: The closest to what we need in terms of parameters is the primary battery manufactured by Energiya OJSC – “Signal 2.6 V 180 Ah”, nominal voltage is 2.6 V, so the minimum charge should be at least:
369 Wh / 2.6 V ≈ 141.92 Ah.
141.92 Ah × 1.15 ≈ 163 Ah.
But the end voltage of this primary battery is 1.8 V, which is lower than the minimum operational voltage of our BEPA-8M lantern.
Let us make approximate calculations of how much of charge this primary battery will give when discharging down to 2 V.
To do so, we will subtract the end voltage of this primary battery from the nominal voltage of the same:
2.6 V - 1.8 V = 0.8 V.
This is the “operating cycle” of the voltage.
Now let us calculate the ratio between this value and the nominal charge of the primary battery.
180 Ah / 0.8 V = 225 Ah per V.
I. e., this primary battery guarantees 225 Ah per each volt of the voltage “operating cycle”.
The final voltage of the primary battery is lower than the minimum operational voltage of our BEPA-8M lantern that we decided to use by 0.2 V.
Let us calculate the amount of charge that will remain in the primary battery when its voltage drops down to 2 V.
0.2 V × 225 Ah = 45 Ah.
We see that our primary battery, before the lantern will go OFF because the voltage is too low (less than 2.0 V) will supply only:
180 Ah (nominal charge) - 45 Ah (residual charge) = 135 Ah.
This value is not suitable, since it is less than what we need – 163 Ah.
And this is the main mistake of the most operators. In practice, they will have to install two primary batteries, which is too much, or replace the primary battery when coming closer to the end of the navigation period. Both of these options lead to unreasonable expenses.
Since we have the negative result, we need to start over the selection of the primary battery.
To get this more diverse, let us assume that the nominal voltage of the primary battery equals to 5 V.
Then, we follow the same procedure as described above:
369 Wh / 5 V ≈ 73.6 Ah.
73.6 Ah × 1.15 ≈ 85 Ah.
Now let us look through the commercially available primary batteries offered by the major manufacturers. We need the primary battery with the following parameters: 5 V, 85 Ah:
The closest to what we need in terms of parameters is the primary battery manufactured by Energiya OJSC – “Signal 5 V 90 Ah”, the end voltage of this primary battery is 3.8 V, which is higher than the minimum operational voltage of our BEPA-8M lantern, it means that the charge of this primary battery will be used completely.
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