The CEMA Horsepower Equation
The CEMA Horsepower Equation
To transform the C and W_{RIRR} into a new friction factor we simply divide the resistance by the applied load and multiply by C_{hr}:
The question then becomes, “which W_{RL} should I use to select a W_{RIRR}?” The LSIT Section in CEMA7 recommends selecting the W_{RIRR} which corresponds to the average load on the belt cross section:
However, since the relation between W_{RL} and W_{RIRR} is nonlinear, we need to correct this friction to account for the nonlinear pressure distribution. The small sample method in CEMA6 and CEMA7 has the same problem. OCC did not provide CEMA with a pressure distribution and did not describe how to perform the integration discussed in the previous section. Instead, users of the small sample method in CEMA7 apply the following correction factor to the small sample method results:
Where bulk density has units of lbf/in^{3}, angles are in degrees, belt width has units of inches, and idler spacing has units of inches. The author of this paper has no idea how this formula was derived, but to maintain consistency the writers of CEMA7 LSIT also recommend using the same formula to scale LSIT results. Accordingly, we can define a new friction factor:
The indentation loss is then calculated using the following formula:
For the example conveyor in the previous section, W_{RL_AVG} = 2.69 N/mm, K_{LSIT} = 0.00913, C_{wd} = 1.113, K_{y1} = 0.0103
To calculate the drag on a single idler set we set L = S_{i}. Thus, using this friction factor based method, we predict that the drag is:
Comparison with Field Measurements
While commissioning the Dahej Overland Conveyor in India [21], the author used strain gauges to measure the drive torque during the exact scenario described in the previous two sections. The author also measured the torque on the motor shafts while the belt ran empty. Fig. 8 shows the torque on the shaft and speed of the belt during an empty drift stop. 26% motor torque is equivalent to a drag of 35,468 N. As the conveyor slows, there is less friction retarding conveyor motion and this is reflected in the deceleration curve shown in Fig. 9. Using these curves and Newton’s 2nd law we compute that mass of the empty conveyor is 196,546 kg.
Fig. 8: Dahej conveyor, empty stop.
Fig. 9: Belt deceleration.
Although the weigh scale was not operational while CDI was present at site, we repeated the drift stop when the belt had enough load to require 63.2% of motor torque. This test showed that 4860 t/h require 63.2% motor torque which is equivalent to 85,885 N of drag. There are 817 idler sets on the carry side of the Dahej conveyor and thus the additional drag created by loading material on the belt is (85,885 N – 35,468 N) / 817 sets = 61.7 N/set.
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