All categories

3 years ago

Answer the following questions briefly 1.Write the definitions of Comminution and Calcination

Engineering

1 answer:

VikaD [51]3 years ago

8 0

Answer:

Comminution is defined as the reduction of solid materials from one average particle size to a smaller average particle size through various methods which include crushing, grinding, cutting, vibrating etc,

Calcination is also defined as a thermal treatment process in the absence or limited supply of air or oxygen applied to ores and other solid materials to bring about a thermal(heat) decomposition.

You might be interested in

Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the tempe

kow [346]

Answer:

a. 81 kj/kg

b. 420.625K

c. 101.24kj/kg

Explanation:

$\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }$

t1 = 360

p1 = 0.4mpa

p2 = 1.20

y = 1.13

substitute these values into the equation

$\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }$

$\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347$

when we cross multiply

t2 = 360 * 1.1347

= 408.5

a. the work required in the firs compressor

w=c(t2-t1)

c=1.67x10³

t1 = 360

t2 = 408.5

w = 1670(408.5-360)

= 1670*48.5

= 80995 J

= 81KJ/kg

b. $n=\frac{t2-t1}{t'2-t1}$

n = 80%

t2 = 408.5

t1 = 360

0.80 = 408.5-360 ÷ t'2-360

$0.80 =\frac{48.5}{t'2-360}$

cross multiply to get the value of t'2

0.80(t'2-360) = 48.5

0.80t'2 - 288 = 48.5

0.8t'2 = 48.5+288

0.8t'2 = 336.5

t'2 = 336.5/0.8

= 420.625

this is the temperature at the exit of the first compressor

c. cooling requirement

w = c(t2-t1)

= 1.67x10³(420.625-360)

= 1670*60.625

= 101243.75

= 101.24kj/kg

8 0

3 years ago

If the same amount of force were applied to all four balls in the picture, which would experience the greatest change in motion?

nydimaria [60]

There is no image connected to this question. Next time attach something to your question to get a more successful answer.

6 0

4 years ago

Read 2 more answers

Which of the following shot pellets has the smallest diameter?

vitfil [10]

I believe i got this correct when i had an assignment with this question. It should be 7 1/2. B!

5 0

3 years ago

Air leakage tests are conducted by using a large fan to blow air into a structure.

lord [1]

TRUE
An airtightness test it’s a whole building test that measures how easy it is for air to leak through a building enclosure or “skin” one common method is to use a large fan or “blower” to extract air from or supply air to the building.

5 0

4 years ago

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft