MACHINE CHEST

Given (from the Material Balance)

MACHINE CHEST

Given (from the Material Balance)

mB1 6911.88ton

day mWHT4 172.693

ton

day

minput mB1 mWHT4 74.387kg

s

ρ 1000kg

m3 μ 0.01poise

τ 5min (Assumption)

(Source: http://www.xyleminc.ca/flygt/canada/mixers/online/AgitPulp.htm)

Required: a. Pulp chest dimensionsb. mixing timec. Power dissipationd. number of impellers

Solution:

a. Pulp Chest Dimensions

Getting the capacity of the chest:

Vcap

minput τ

ρ

Vcap 2.232 104 L

Getting the volume of the chest:

VT 1.2 Vcap

VT 2.678 104 L

Getting the diameter of the chest (since the diameter is equal to the tank height)

DT

4 Vcap

π

1

3

DT 3.051 m

Getting the height of the chest:

HT

4 VT

π DT2

HT 3.662 m

Determination of other pulp chest dimensions:

W 0.1 DT 0.305 m Baffle Width

ZL DT 3.051 m Static Liquid Depth

m 0.25 ZL 0.763 m Baffle Tip Distance from Top to Bottom

DI

DT

31.017 m Impeller Diameter

ZI DI 1.017 m Impeller Distance from Top to Bottom

I 0.25 ZI 0.254 m Impeller blade length

Dd 0.75 DI 0.763 m Impeller Disc Diameter

b 0.2 DI 0.203 m Impeller Blade Width

b. Mixing time

Assume: vp 7m

s (high agitation)

Getting N or impeller revolutions per unit time:

Nvp

π DI1.671

1

s

Getting Froude Number:

NFr

N2 DI

g0.29

Calculating for the mixing time:

tm

5 NFr

1

6DT

DI

2

ZL

DT

0.5

N21.904 s

Checking if there is less mixing time:

tm

τ0.073 less than 0.1 (desired limit)

Therefore, tm 21.904 s

c. Power dissipation:

Getting the impeller Reynolds number:

NRe

ρ N DI2

μ1.729 106

Since the viscosity of the slurry is assumed to be equal to water, impeller types thatcan be used are either propellers or flat-blade turbines.

For a flat-blade turbine, from 4-7, Np 7

Source:https://books.google.com.ph/books?id=mI8v9robL7MC&pg=PR11&lpg=PR11&dq=power+function+vs+impeller+reynolds+number&source=bl&ots=QKnnBM_YF6&sig=6WeyXeg_8-83HIziq13r-1deCj4&hl=fil&sa=X&ei=dURXVbSpCc6B8QX9goHgAg&redir_esc=y#v=onepage&q=power%20function%20vs%20impeller%20reynolds%20number&f=false

Calculating the power dissipation: Vcap 22.316 m3

Pa Np ρ N3 DI5 3.557 104 W

Checking is there will be enough power for the pulp chest:

Pa 3.557 104 W Pa 3.557 104

Vcap 22.316 m3 Vcap 22.316

Pa

Vcap1.594 103 which is greater than 1500 W/m3. This means that ther

will be enough power for the reactor.

d. Number of impellers

Calculating the radius of action:μ 1 10 3 Pa s

RA 9.0 10 2Pa

μ

1

2

536.765

RAH 0.5RA 268.383 greater than the diameter of the pulp chest

RAV 0.2 RA 107.353 greater than the height of the pulp chest

Therefore, only one impeller shall be used.

Summary of the Design Specifications:

Capacity, L 22320Volume of tank, L 26780

Reactor Diameter, m 3.051Reactor height, m 3.662

Static Liquid Depth, m 3.051Baffle Width, m 0.305

Impeller diameter, m 1.017Impeller Disc Diameter, m 0.763Impeller Blade Width, m 0.203

Impeller distance from top to bottom, m 1.017Impeller blade length, m 0.254

baffle tip distance from top to bottom, m 0.763mixing time, s 21.904

Power dissipation, kW 35.57Number of impeller/s 1