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# Mechanical Operations MCQ’s – Energy for Size Reduction

The limiting diameter, D_{p} is ____

a) d + r

b) 0.5R + r

c) 0.04R + d

d) R +r

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Answer: c

Explanation: The limiting size D_{p} of particle that can be nipped by the rolls are calculated by

DP = 0.04R + d.

The maximum size of the product is _____

a) 2d

b) 4d

c) 5d

d) 2.5d

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Answer: a

Explanation: The maximum size of the product is equal to 2d, the particle depends on the pacing between the rolls

Roll crushers, are set to a reduction ratio of ____

a) 4:1

b) 3

c) Either 3 or 4:1

d) 1:3

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Answer: c

Explanation: To operate more efficiently the reduction ratio is set to 3 or 4:1, the maximum particle size is 1/3 or ¼.

What is the forces exerted by the rolls?

a) 100 to 500 N/cm

b) 8700 to 70,000 N/cm

c) 500 to 1000 N/m

d) 7 to 300 N/cm

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Answer: b

Explanation: The force exerted by the roll is from 8700 to 70,000 N/cm of the roll width, to allow material to pass without damaging the quality.

Which of the following is defines the theoretical capacity?

a) C = 50 uwd

b) C = 10 uwd

c) C = 5 uwd

d) C = 4.5 uwd

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Answer: a

Explanation: The theoretical capacity is defined as C = 50 uwd, where u is the peripheral speed of rolls in fps, and w is the width.

Which of the following is the capacity Q in tons?

a) 0.0068 ND wds

b) 1.20 ND wds

c) 5 ND wds

d) 10 N D

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Answer: a

Explanation: If N is the speed of the roll in rpm and D roll diameter in inches and thus Q = 0.068 ND wds, which gives results in tons per hour.

The feed size of the roll crusher is _____

a) 2/3 to 3 inches

b) ½ to 3 inches

c) ¼ inches

d) 8 inches

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Answer: b

Explanation: The feed size of the roll crusher is in between ½ to 3 inches and the product ranges between ½ to 20 inches.

Calculate the limiting size, if R is 2.5 cm and the half width is given as 10 cm?

a) 5.1

b) 0.1

c) 0.5

d) 10.1

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Answer: d

Explanation: As D_{p} = 0.04R + d, 0.04*25 + 10 = 10.1 cm.

Calculate capacity, if peripheral speed is 100 fps, width is 10 ft. and d is 5 ft?

a) 250,000

b) 100,000

c) 45,000

d) 50,000

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Answer: a

Explanation: As C= 50 uwd so, C = 50*100*10*5 = 250,000 ft3/hr.

Calculate Q, if Speed is 250 rpm and D is 10 inches and wds factor is given as 500?

a) 11500 T/hr

b) 8500 T/hr

c) 9000 T/hr

d) 5000 T/hr

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Answer: b

Explanation: As Q = 0.068*250*10*500 = 8500 T/hr