# Thread: Extra Power Due to Hopper

1. ## Extra power due to hopper

I'm estimating power for a belt conveyor which includes a feed hopper to be fed by dump trucks. The hopper top area is 4x4 m2 and bottom's (in contact with conveyor's skirtplates) is 2x0.65 m2. the Height of the hopper is 3.5m. I considered no gate between the hopper and conveyor (direct feed), while a vertical slide gate will be installed at the outlet of skirtplates area to regulate the capacity of conveyor. I need to know how to calculate extra power due to hopper to be able to size conveyor driving motor (my calculations are based on ISO).

Regards,

2. Dear Ehab A. Hassan,

Please inform the belt width, belt speed, capacity, material name, material bulk density, material size and moisture content to give you some suggestions. It is also assumed that you will not be creating material extra height (pile) above hopper opening.

The material will be falling from truck (dumper). The material fall from truck to hopper mouth will be at least 2 to 3m. So, worst fall height can be around 6.5 m to belt line. This needs that hopper outlet should have spout (chute segment) which should not allow any direct fall on belt.

In general one needs to use feeder below hopper, however, if your belt speed is very low than possibly you can avoid feeder, but it also depends upon lump size etc. Proper layout at hopper outlet will have great influenced on the performance and life of the equipment underneath.

Calculation of drag force due to material in hopper is not difficult issue, which can be informed on knowing the details mentioned above.

Regards,
Ishwar G Mulani.
Author of Book : Engineering Science and Application Design for Belt Conveyor.
Email : parimul@pn2.vsnl.net.in
Tel.: 0091 (0)20 5882916

3. Dear Mr. Mulani:

1. The belt width is 800mm flat at the feed hopper area, and 30 deg. troughed after.

2. Belt speed is not finally decided, but I suggest 0.5- 0.7 m/s which will result in a siutable filling at the required capacity.

3. Capacity is 100 t/h

4. Material to be transfered is Pyrites / 2.25 t/m3

5. Material size is 10% (-20 +5mm) and 90% (-5mm)

6. Material is dry

Also it sould be noted that this conveyor is secondary (to be used max. 1hr/day).
I'm interested in the method of calculations rather than the final figure.

Regards,

4. Dear Ehab A. Hassan,

You have decided to design feeder cum conveyor, using 800mm wide belt. In this context, you can investigate for following solution:
1) 800 mm wide, 30 degree troughing from tail end to head end
2) Belt speed 0.24 mps
3) Skirt board width 520 mm
4) Sr. no. 1, 2 & 3 will result into material layer depth of 100 mm at skirt board, which will permit some degree of regulation by control gate in skirt board. This also implies material cross section 52000 sq. mm
5) Maximum allowable cross section beyond skirt board is 64520 sq. mm for the said belt. Hence, conveying is possible.
6) Material drag will be corresponding to column height 2.5 times the skirt board width i.e. 1.3 m height
7) Material column weight is 0.52 x 2 x 1.3 x 2.25 x 1000 = 3042 kg.
8) Material shear factor / drag factor can be considered approximately 0.5 (I am not going in details about this issue which has relevance to material repose angle, surcharge angle, etc.). So drag force is 1521 kgf (14921 N ). The drag force fluctuates from static to dynamic (continuos flow condition). The flow tends to reduce the pressure and drag force, but you have to go for the worst situation. You may also come across someone using shear factor 0.7
9) Above corresponds to 3.6 kW at belt line, due to aforesaid drag only.

I hope the above information will be of use to you in relation to your chosen arrangement.

Regards,
Ishwar G Mulani.
Author of Book : Engineering Science and Application Design for Belt Conveyor.
Email : parimul@pn2.vsnl.net.in
Tel.: 0091 (0)20 5882916

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6. Dear Mr. Mulani:

Thank you for your reply. As per your calculations you consider a head of material (Box shaped volume) bounded by area of opening under hopper and height equals to 2.5-3 width of opening (skirt width). Another method uses the same boundaries, but considering conical volume concluding in 1/3 the power required in your method. Do you think this method would result in under estimating the power?

Regards,

7. Dear Mr. Ehab,

I / we have always used the method as described by me, for sizing drive and other components of belt conveyor or belt feeder. This gives safe and dependable results.

I have also mentioned that drag force during dynamic condition (flow condition) would be less, but we have to go by the worst situation such as starting etc.

The method mentioned by you also has logic i.e. as if material above the opening has ‘caved in’, so one can say caved-in volume is something like spherical or conical in shape (like arch in monument gates, turned by 360 degrees on vertical axis). However, if one is taking conical shape, he might be taking more height and result may not be very far off.

These are all calculation models to derive engineering solution. Actual flow pattern is somewhat different, wherein more material flow out near exit end and less at rear end. People have done enough research on this, so opinion / method can differ.

Regards,
Ishwar G Mulani.
Author of Book : Engineering Science and Application Design for Belt Conveyor.
Email : parimul@pn2.vsnl.net.in
Tel.: 0091 (0)20 5882916

8. Dear Ehab,

There is an established, old, conservative, method that significantly departs from Mr. Mulani's description. It goes as follows:

A. Create a horizontal shear plane that resists material withdrawl where material mass times material coefficient of friction acts in resistance to forward motion. Given slot width 0.65; len = 2.00 m

1. w1 = opening at front leaving hopper - plan view = 0.650 m
2. w2 = opening at back leaving hopper - plan view = 0.650 m
3. h1 = height of material at front of hopper above w1
4. h2 = height of material at back of hopper above w2
5. len = length of slot at shear plane = 2.0 m
6. phi = arching angle @ base of hopper at shear plane = 70 deg.
7. f = material internal friction coef. = tan(37 deg.) = 0.75
8. shear resistance = mass over shear plane * material int. coef.
9. mass over shear plane = triangle at front & back vertical plane where height is defined by the material arch angle above half width or h1 and h2 (given below) * the lenght of slot* mat. den.

We know:
10. w1 = width of skirt = 2/3 belt width (already given at 0.650)
11. w2 = typically width at skirt beginning and tapering 3 degrees in plan view to w1 front (given at 0.650m)
12. h1 = (0.5*w1)*tan(phi) = (0.5*0.65)*(tan(70))= 0.893 m
13. h2 = (0.5*w2)*tan(phi) = 0.893 m

then,
14. mass = area *len*density = ((0.893 * 0.65 /2)*2.00)*2250kg/m = 1306 kg
15. shear resist. = 1306 * 0.753 = 984 kg or 9650 N
16. shear resistance power = 9.650 * .24 / (eff = 0.9) = 2.6kw

B. To this you must add the skirt resistance plus belt, idler, pulleys and scraper resistances. Skirt resistance is take as two parts: a) Rankine hydrostatic head pressure on skirt walls * frict. coeff. of material rubbing on wall plus b) material resistance beyond hopper pressure. Rankine pressure is calculated from h1 and h2 above plus skirt height for region under hopper opening.

Some other points:
1. width cannot be 650 mm and flat transitioning to 30 deg trough because you do not have enogh belt width to hold material after leaving hopper discharge point. I calculate the vertical opening leaving hopper to be 79.1 mm for 650 mm width given at 0.24 m/s and 2250 kg/m^3
2. any length beyond end of hopper region will add to the power consumption and should be as short as possible.

3. You can purchase BELTSTAT from our website and it will do it all for you:
www.conveyor-dynamics.com

Lawrence Nordell
Conveyor Dynamics, Inc.

9. Added drag must be included for rubber skirts if used

LKN