I appreciate your help in providing me a brief equation or rule of thumb to calculate the specific energy consumption as Kwh/Mt/Km length = 5 Km , Flow = 2500 to 3000 Mt/Hr , Route inclination= zero inclination, Width = 1200 mm , trough angle=30 how much electricity will be consumed in electrical motor
Your question cannot be answered until you specify a belt speed at which the material will be carried. 1 watt = 1 newton-meter per second. Once you calculate the required tension (newtons) to pull the belt you must multiply that force by the desired belt speed (m/s). For example, a 5 km long horizontal conveyor carrying 3000mt/hr might have a belt speed of 5 m/s. If belt pull = 222 kilo newtons, then required electric power will be approximately 222 kN x 5 m/s = 1,110 kW.
@@rulmecacorporation4596 here we are with full data I appreciate your help in providing me a brief equation or rule of thumb to calculate the specific energy consumption as Kwh/Mt/Km , belt speed = 2.5 meter per second, trough angle = 30 , surcharge = 25, material = excavated soil (sand) , lump size less than 150 mm , density =1,550 kg/M3 , belt length = 5 Km (multiple flights ) , Flow = 3500 to 4,000 Mt/Hr , Route inclination= zero inclination, Width = 1,200 mm , considering these inputs , how much electricity will be consumed in electrical motor as Kwh per metric ton (Weight) per Km (distance)
You calculated the mechanical power is required, but does that mechanical power equal to the electrical power of the motor which will drive the belt ? in the case of calculation P=0.15KW , so in this case i will use an AC motor with a power at least of 0.15 KW to move all those weights ?
To what inertia are you referring? If you are referring to start-up inertia of a fully-loaded belt, many designers of small conveyors assume that will be covered by the "start-up torque" of the drive motors. Design C motors, such as we supply, have a start-up torque of 200% of running torque. That is usually adequate to do the job for the few seconds it takes to start a fully loaded belt.
Please tell me bellow power and torque relation with all units, power = force x linear distance / time, and power = torque x angular velocity all units and its final value.
ma will always be negligible on short package conveyors which travel at 50 fpm +/-. However, if you need to accelerate very quickly to a high speed with a large load mass, then you should check it.
great video. Thank you so much, sir
Great video
thank you
Please make another video on how to find power of powered roller conveyor .
In which every roller is powered by chain drive.
Try to do it in metric unit .
This was really best and easy calculation to understand power
superb sir its great video
How do we calculate the friction coefficient, do we refer to any table?
I appreciate your help in providing me a brief equation or rule of thumb to calculate the specific energy consumption as Kwh/Mt/Km length = 5 Km , Flow = 2500 to 3000 Mt/Hr , Route inclination= zero inclination, Width = 1200 mm , trough angle=30 how much electricity will be consumed in electrical motor
Your question cannot be answered until you specify a belt speed at which the material will be carried. 1 watt = 1 newton-meter per second. Once you calculate the required tension (newtons) to pull the belt you must multiply that force by the desired belt speed (m/s). For example, a 5 km long horizontal conveyor carrying 3000mt/hr might have a belt speed of 5 m/s. If belt pull = 222 kilo newtons, then required electric power will be approximately 222 kN x 5 m/s = 1,110 kW.
belt speed = 2.5m/s and material is dsand with density 1.55 Mt/M3 @@rulmecacorporation4596
@@rulmecacorporation4596
here we are with full data
I appreciate your help in providing me a brief equation or rule of thumb to calculate the specific energy consumption as Kwh/Mt/Km , belt speed = 2.5 meter per second, trough angle = 30 , surcharge = 25, material = excavated soil (sand) , lump size less than 150 mm , density =1,550 kg/M3 , belt length = 5 Km (multiple flights ) , Flow = 3500 to 4,000 Mt/Hr , Route inclination= zero inclination, Width = 1,200 mm , considering these inputs , how much electricity will be consumed in electrical motor as Kwh per metric ton (Weight) per Km (distance)
Thanks a lot sir..
Most welcome
thanks sir
It is my pleasure.
You calculated the mechanical power is required, but does that mechanical power equal to the electrical power of the motor which will drive the belt ?
in the case of calculation P=0.15KW , so in this case i will use an AC motor with a power at least of 0.15 KW to move all those weights ?
We have some videos explaining the relationship between calculated required mechanical power and suggested electrical power. Did you watch them yet?
@@rulmecacorporation4596 No, i did not , i will search it .
@@wadib3eed7 watch this webinar please m.th-cam.com/video/q5v3M6ZsrrY/w-d-xo.html
@@rulmecacorporation4596 Thank you so much for helping me.
Thx! Why is the inertia not included?
To what inertia are you referring? If you are referring to start-up inertia of a fully-loaded belt, many designers of small conveyors assume that will be covered by the "start-up torque" of the drive motors. Design C motors, such as we supply, have a start-up torque of 200% of running torque. That is usually adequate to do the job for the few seconds it takes to start a fully loaded belt.
Please tell me bellow power and torque relation with all units,
power = force x linear distance / time,
and power = torque x angular velocity
all units and its final value.
No problem. Here is a video which presents what you requested
th-cam.com/video/q5v3M6ZsrrY/w-d-xo.html
Hi, Belt pull required to accelerate = mg x coeff.of friction+ma.But you have not considered ma?Is it wrong?
Please define your terms. Thanks
m=mass of the load(kg/m) +belt weight
g=9.8 m/s2
a=acceleration
ma will always be negligible on short package conveyors which travel at 50 fpm +/-. However, if you need to accelerate very quickly to a high speed with a large load mass, then you should check it.
I think belt weight should be equal 2 times
3 lbs/feet. what parameters is that?
3 lbs/ft = the weight per foot of the belt.
@@mikegawinski3450 how to calculate belt weight per foot ?