Power To Move And Lift
Andy you say-
You can try for a simple-rough formula:
kW = 0.090 x tonnage (metric) x kilometers (1000m) - conservative
xxx = 0.09 x 3000 t/h x 3 km = 810 kW/0.95. (select 900 kW)
Lifting 3000 t/h up 804 m lift will take altogether about 9000 kW and a super belt. The above multiplier (0.090) will not be applicable for such a heavy belt. I also improved standard idler configurations.
You do not need a 1800mm wide belt to handle 3000 t/h. How about 1050 mm going fast but not too fast, or a conservative 1200 mm for the high lift?
Me thinks you need someone in the know. ■
Re: Power Demand For A Trough Belt Conveyor
Thank U, Mr Nordell for the early reply.
Yeah I am a Port Planner with very limited knowledge on the technicality of the Conveyor Systems.
But since we have to arrive at the overall power requirement of the entire Port (wherein conveyors contribute a significant share) we are to assess a very approximate power demand for the conveyor system.
in my current situation, coal has to be unloads by two Ship unloaders (each of 1500tph capacity) to a single common conveyor system which takes to the stackyard about 3km from the jetty.
we do not requie the 824m lift. it is not 15deg slope from the starting point to the end. We need the slope to negotiate road crossings at certain intermediate stretches. the maximum lift may be 7m only. Is there any significant power difference for a flat conveyor and an inclined one? ■
Re: Power Demand For A Trough Belt Conveyor
Originally Posted by andy7512
Can anyone provide the approximate Power requirement (connected load) per meter run of the conveyor system.
What will be the increase in connected load for an an inclination of 15deg conveyor system.
you have proposed an open ended question as you have not even investigated your desired operating voltage and then the amperage loading for that width of belt which is readily available apparently.
1. it all depends on how much high voltage power you have available to you
2. what your desired operating voltage is
3. friction and resistance of belting, splices, idlers, weight of belt per foot, weight of ore per foot of loading.
4. a conveyor requires a a continuos emergency stop cable system on both sides
of the belt
5. if a belt brake is being employed- I hope
There are much more efficient methods of transporting ores than belt conveyors, rakes of coal cars, barges, lorrys for overland transport of coal and they are not affected by grade increases.
If you would like to, please contact me off the board to learn more about the Pneumatic Capsule Pipeline method of mining mass transportation of ores and overburden.
lzaharis@lightlink.com ■
Power Demand for a Trough Belt Conveyor
As we are in the process of calculating the approximate power demand (connected load) for a Dry bulk Handling system for coal unloading. we require the power requirement for a[B] conventional Trough type Belt conveyor of about 3000TPH capacity and 1.8 to 2m wide belt carrying coal from the Jetty to stackyard approximately3km.( Assumed that the conveyor is flat along the length)
Can anyone provide the approximate Power requirement (connected load) per meter run of the conveyor system.
What will be the increase in connected load for an an inclination of 15deg conveyor system. ■