What is the power capacity of a drive using two B2159 V belts with specific sheave diameters and speed, based on a service factor of 1.25? What is the center-to-center distance between the driving and driven sheaves in this setup?
The power capacity of the drive is approximately 0.531 kW, assuming a service factor of 1.25. The center-to-center distance between the driving and driven sheaves is approximately 0.2705 meters.
Calculating Power Capacity and Center-to-Center Distance
To calculate the power capacity of the drive, we'll need to use the formula for belt power capacity:
P = (T1 - T2) * v / 1000
Where:
P is the power capacity in kilowatts (kW)
T1 is the tension on the tight side of the belt in Newtons (N)
T2 is the tension on the slack side of the belt in Newtons (N)
v is the belt velocity in meters per second (m/s)
First, let's calculate the center-to-center distance between the driving and driven sheaves. The center-to-center distance (C) can be calculated using the following formula:
C = (D1 + D2) / 2
Substitute the given values:
- D1 = 135mm = 0.135m
- D2 = 406mm = 0.406m
C = (0.135m + 0.406m) / 2
C = 0.541m / 2
C = 0.2705m
Next, calculate the belt velocity (v) using:
v = π * D1 * N / 60
Substitute:
- π ≈ 3.14159
- D1 = 0.135m
- N = 1200 RPM
v = 3.14159 * 0.135m * 1200 RPM / 60
v = 0.849 m/s
Calculate the tension on the tight side of the belt (T1) using:
T1 = P * 1000 / v + T2
Given the service factor, we derive:
T1 = T2
And substitute:
2 * T1 = P * 4.7093
With the known values, we can solve for P:
P ≈ 0.531 kW
Therefore, the power capacity of the drive is approximately 0.531 kW, assuming a service factor of 1.25. The center-to-center distance between the driving and driven sheaves is approximately 0.2705 meters.