Delta IED-S Series, Руководство пользователя

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Chapter 10 Auto-tuning Process | IED-S
10-20
10-6-4 Examples of Installing Upward and Downward Forced Deceleration Sensors
Assume that the distance between the sub-top/sub-bottom floor and top/bottom floor is 3.3 m, the
effective length of the forced deceleration sensor CAM is 2.2 m, Pr.04-08 = 0.5 m/s 2 , and Lm =
1.485 m [0.45 x 3.3].
1. If rated speed = 1 m/s, then Lx = (1 × 0.75) 2 ÷ (2 × 0.5) = 0.563 m.
When Lx – Lm [= 0.563 – 1.485 = -0.922] is a negative value, install the first one forced
deceleration sensor
L1 = 0.563 m
2. If rated speed = 1.5 m/s, then Lx = (1.5 × 0.75) 2 ÷ (2 × 0.5) = 1.266 m.
When Lx – Lm [= 1.266 – 1.485 = -0.219] is a negative value, install the first one forced
deceleration sensor.
L1 = 1.266 m
3. If rated speed = 1.75 m/s, then Lx = (1.75 × 0.75) 2 ÷ (2 × 0.5) = 1.723 m.
When Lx – Lm [= 1.723 – 1.485 = 0.238] is a positive value and
≤
CAM, install the second
one forced deceleration sensor.
L2 = 1.723 m
L1 = 1.485 m
4. If rated speed = 2 m/s, then Lx = (2 × 0.75) 2 ÷ (2 × 0.5) = 2.25 m.
When Lx – Lm [= 2.25 – 1.485 = 0.765] is a positive value and
≤
CAM, install the second one
forced deceleration sensor
L2 = 2.25 m
L1 = 1.485 m
5. If rated speed = 2.5 m/s, then Lx = (2.5 × 0.75) 2 ÷ (2 × 0.5) = 3.516 m.
When Lx – Lm [= 3.516 – 1.485 = 2.031] is a positive value and
≤
CAM, install the second
one forced deceleration sensor.
L2 = 3.516 m
L1 = 1.485 m
6. If rated speed = 3 m/s, then Lx = (3 × 0.75)
2
÷ (2 × 0.5) = 5.063 m.
When Lx – Lm [= 5.063 – 1.485 = 3.587] is a positive value, and
≤
two times of CAM but >
one times of CAM, install the third one forced deceleration sensor.
L1 = 1.485 m
L3 = 5.063 m
L1 + CAM
≥
L2
≥
L3 - CAM
1.485 + 2.2
≥
L2
≥
5.063 - 2.2
3.685 m
≥
L2
≥
2.863 m
7. If rated speed = 4 m/s, then Lx = (4 × 0.75)
2
÷ (2 × 0.5) = 9 m.
When Lx – Lm [= 9 – 1.485 = 7.515] is a positive value and
≤
four time of CAM but > three
times of CAM, install the fourth one forced deceleration auxiliary sensor.
L1 = 1.485 m
L2 = L1 + 2.2 m = 1.485 + 2.2 = 3.685 m
L3 = L2 + 2.2 m = 3.685 + 2.2 = 5.885 m
L4 = Lx = 9 m
L3 + CAM
≥
L4 auxiliary
≥
L4 – CAM
5.885 + 2.2
≥
L4 auxiliary
≥
9 – 2.2
8.085 m
≥
L4 auxiliary
≥
6.8 m