A bucket conveyer is used to move grain from the ground to the top of a storage
ID: 1717455 • Letter: A
Question
A bucket conveyer is used to move grain from the ground to the top of a storage bin 100 ft high. The overall length of the bucket conveyer is 150 ft. The buckets are 1ft x 4in x 4in and spaced at 1 foot intervals. The bucket system is operated at a constant speed of 15 ft/s.
a. Determine the power required to operate the system if friction is neglected.
b. Determine the power required to accelerate the bucket system from rest in 10 seconds to operational speed (consider the work to accelerate and work to lift loaded buckets). Assume the density of the grain to be 721 kg/m^3.
Explanation / Answer
>> Firstly, let's find total mass of grains which will be on the conveyor.
Let total mass =M and massof one bucket, full of grains is "m"
>> Now, as dimensions of bucket are:- 1ft X 4 in X 4 in = 12in X 4in X 4in
=> Volme of 1 bucket = 12*4*4 = 192 in3 = 192*0.02543 = 0.00315 m3
As, density of the grain = 721 Kg/m3
>> mass, m = 721*10.00315 = 2.268 Kg ....mass of 1 bucket.
>> Now, total length of conveyor = 150 ft
and, buckets are spaced at 1 foot intervals, and length of 1 bucket = 1 ft
=> Total no. of buckets on conveyor at a time are = 150/(1 + 1) = 75
=> Total mass on conveyor,M = 75*m = 170.14 Kg
>> Part (1).
=> Power required to operate the system = Energy per secrequired to take all buckets to an elevation of 100 ft
= M*g*h = 170.14*9.81*100*12*0.0254 = 508.73 W .....ANSWER..
>> Part (2).
>> For accelerating the bucket system from rest to 15 ft/s in 10 sec, let work done = W
As, W = F*L
and, F = M*a
and, a = (15 - 0)/10 = 1.5 ft/s2 = 1.5*12*0.0254 m/s2 = 0.457 m/s2
=> W = 170.14*0.457*150*12*0.0254 = 3556.468 Joules
>> As, power required to accelerate the bucket system from rest in 10 seconds to operational speed = work to accelerate and work to lift loaded buckets
= 3556.468 + 508.73 = 4065.198 W
=> Power required to accelerate the bucket system from rest in 10 seconds to operational speed = 4.1 KW .....ANSWER....
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