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solong [7]
3 years ago
6

A cyclone is operated in a closed circuit with a ball mill. The cyclone is feed from a rod mill with a slurry that has a density

of 1250 kg/m3 while the solid bulk density is 2900 kg/m3. Cyclone overflow goes to flotation circuit consists of a rougher, a scavenger and one cleaner cell. The slury feed to the rougher has 15% solids (overflow) while the feed to ball mill (underflow) from the cyclone has 50% solids with a mass flow rate of 40 t/h.
• Draw a possible flow sheet of the process
• Calculate the mass flow rate of the feed from rod mill to cyclone (hint: you need to calculate % solids in feed and dilution ratios)
Get more help from Chegg

Engineering
1 answer:
pav-90 [236]3 years ago
3 0
Here is the flow sheet. Hope this helps have a great day!!

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I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li
k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted.  For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

6 0
3 years ago
An_______<br> employee is always prompt and on time.
Paha777 [63]
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5 0
4 years ago
A car starts out from rest (zero velocity) at an elevation of 500 m and drives up a hill to reach a final elevation of 2000m and
natta225 [31]

Answer:29,930 kJ

Explanation:

Given

Car starts with an initial elevation of 500 m and drives up a hill to reach a final elevation of 2000 m

Final velocity (V)=20 m/s

Energy of car increases by 100 kJ

mass of car(m)=2000 kg

Total Energy =\Delta PE+\Delta KE+\Delta U

\Delta PE=mg(\Delta h)=2000\times 9.81\times (2000-500)

\Delta PE=29,430 kJ

\Delta KE=m\frac{v_2^2-v_1^2}{2}

\Delta KE=2000\times \frac{20^2-0^2}{2}

\Delta KE=400 kJ

\Delta U=100 kJ

Total Energy=29,430+400+100=29,930 kJ

4 0
4 years ago
A distillation column at 101 kPa is used to separate 350 kmol/h of a bubble point mixture of toluene and benzene into an overhea
AURORKA [14]

Answer:

A)

  D = 158.42 kmol/h

  B =  191.578 kmol/h

B) Rmin = 1.3095

Explanation:

<u>a) Determine the distillate and bottoms flow rates ( D and B ) </u>

F = D + B ----- ( 1 )

<em>Given data :</em>

F = 350 kmol/j

Xf = 0.45 mole

yD ( distillate comp ) = 0.97

yB ( bottom comp ) = 0.02

back to equation 1

350(0.45) = 0.97 D + 0.02 B  ----- ( 2 )

where; B = F - D

Equation 2 becomes

350( 0.45 ) = 0.97 D + 0.02 ( 350 - D )  ------ 3

solving equation 3

D = 158.42 kmol/h

resolving equation 2

B = 191.578 kmol/h

<u>B) Determine the minimum reflux ratio Rmin</u>

The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve

first we calculate the value of the enriching line

Y =( Rm / R + 1 m ) x   +  ( 0.97 / Rm + 1 )

q - line ;  y =  ( 9 / 9-1 ) x -  xf/9-1

therefore ; x = 0.45

Finally Rmin

=  (( 0.97 / (Rm + 1 ))  = 0.42

0.42 ( Rm + 1 ) = 0.97

∴ Rmin = 1.3095

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8 0
3 years ago
Are all transmissions fluids interchangeable
KiRa [710]

Answer:

<u>No</u>.

Explanation:

They are not all the same. Moreover, using a fluid that is not approved by the vehicle manufacturer will void the transmission warranty.

6 0
3 years ago
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