Agitator Tip Velocity

agitator tip velocity is an important factor to consider. It refers to the speed at which the agitator blades move through the liquid in the reactor. This velocity can affect the mixing efficiency and the heat transfer rate in the reactor.

Tip velocity of agitator means rotation speed of agiatator. It is expressed as m/sec. Mixing intensity depends on rotation of agitator.

It means mixing intensity of different agitator is same if tip velocity is same, Answer is no.

Tip Velocity (V) = π x D x N

Where,
V = Velocity in m/s
D = Diameter of agitator in m
N= Rotation per second

Example :
Reactor with capacity 5 KL agitator running with 75 rpm. Agitator diameter is 1200 mm.
Calculate : Tip velocity in m/sec.

Sol:
Given : rpm = 75, rps = 1.25
Agitator Dia. = 1200 mm = 1.2 m

Tip velocity = π.D.N = 3.14 x 1.2 x 1.25 ( m/sec)
= 4.71 m/sec

Agitator power calculation

Agitatator power calculation important for ensuring suitability of use as well as mapping equipment during technolgy transfer for same volume or increased volume.

Pharmaceutical applications :

Agitator power should be same for identical reactor and same volume of handling.
Reaction volume required power should be maximum 75 % of rated capacity provided by reactor designer.

Formula :

P = n x Np x ρ x N^3 x D^5

Where,

P = Agitator Power watt ( W)

1 KW = 1000 W

1 HP = 746 W

n = No. agitator stages

Np = Power No.

N = Rotation per sec

D = Dia. Of Agitator in m

ρ = Density in kg/m3

Example :

Reactor Anchor running with 75 rpm. Agitator diameter is 1200 mm, liquid viscosity is 1.2 kg/m.sec and density 1000 kg/m3.

Calculate : Agitator Power

Sol:

Given : rpm = 75/60 = 1.25 RPS, n = 1,

Np = 0.6, Density = 1000 kg/m3, Dia. = 1200 = 1.2 m

Agitator Power = n x Np x ρ x N^3 x D^5 =1 x 0.6 x 1000 x 1.25^3 x 1.2^5 = 2916 W= 2.91 KW= 3.9 HP

Considering Motor efficiency & transmission losses, Agitator Power = 3.9 / 0.8 = 4.87 i.e. 5 HP

Tip Velocity : Vt ( m/sec )

= π Da N

N = rotational speed in seconds

Da = agitator swip dia. m

Example

Reactor with capacity 5 KL agitator running with 75 rpm agitator diameter is 1200mm

Calculate Tip velocity

Solution. :

Given

Da = 1200 mm = 1.2 m

N = 75 rpm = 1.25 rps

Vt =π Da N

Vt = 3.142 X 1.2 X 1.25 = 4.71 m/s

Reynold's Number : NRE

= (ρ∗N∗Da2 )/µ

Where,

ρ = density of liquid kg/m3

N = rotational speed in seconds

Da = agitator swip dia. m

µ = viscosity of liquid kg/m sec

Example

Reactor agitator running with 75 rpm. Agitator diameter is 1200 mm,

Liquid viscosity is 1200 cp and density is 1800 kg/m3

calculate

Reynolds number NRE

solution :

Given viscosity = 1200 cp = 1.2 kg/m.s

RPM= 75 = 1.25 rps

Density = 1800 kg / m3

NRE = (ρ∗N∗Da2 ) = 2700

Agitator Power Watt : P

= Np ρ * N3 * Da5

ρ= density of liquid kg/m3

N = rotational speed in seconds

Da = agitator swip dia. m

Np= power number

Example

Reactor running with 75 rpm . Agitator diameter is 1200mm, Liquid viscosity is 1.2 kg/m.s and density 1000 kg/m3 .

Calculate Agitator power

Solution :

Given ρ = 1000 kg/m3 N= 1.25 rps Da = 1.2 m Np = 0.6

Agitator power = Np* ρ *N^3* Da^5

= 0.6 x 1000 x 1.25^3 * 1.25 ^5

= 2916 W = 2.916 KW = 3.9 HP

Power / Volume (W/lt)

= (Np ∗ρ ∗ N3 ∗Da5 )/Vmass

ρ = density of liquid kg/m3

N = rotational speed in seconds

Da = agitator swip dia. m

Np= power number

Vmass = Liquid volume in lt.

Example

Reactor running with 75 rpm . Agitator diameter is 1200mm, Liquid viscosity is 1.2 kg/m.s and density 1000 kg/m3 .

Reaction mass volume 1500 lt.

Calculate : Agitator power / volume

Solution :

Given

ρ = 1000 kg/m3 N= 1.25 rps Da = 1.2 m , Np = 0.6 V = 1500 lt

Agitator power = (Np ∗ ρ ∗ N3 ∗ Da5)/(Vmass )

= (0.6 ∗ 1000∗ 1.25^3 ∗ 1.25^5 )/1500

= 1.944 W/ lt

Agitator Pumping Capacity : Q in m3/s

= NQ N Da3

N = rotational speed in seconds

Da = agitator swip dia. in m

NQ = flow number

Example

Reactor running with 75 rpm . Agitator diameter is 600mm, reactor diameter is 1100 mm and density 1000 kg/m3

Calculate Agitator pumping capacity

Solution :

Given

NQ = 0.6

N = 75 rpm = 1.25 rps ,

Da = 0.6 m

Q = NQ N Da3

= 0.6 X 1.25 X 0.6^3 = 0.16 m3/s

Mixing time : tm in sec

= 5∗ Vmass)/NQ ∗ N ∗ Da^3

N = rotational speed in seconds

Da = agitator swip dia. m

NQ = flow number `

Vmass = Liquid volume m3

Example

Reactor with capacity 5 KL contain 1.5 KL of liquid , running with 90 rpm. Agitator diameter is 1200 mm

Calculate : Mixing time

Solution :

Given

V mass = 1.5 m3

NQ= 0.5 ( for anchor ) ,

N = 90 rpm ( 1.5 rps ),

Da = 1200 mm = 1.2 m

tm = 5∗ Vmass)/NQ ∗ N ∗ Da^3 =5 x 1.5 / 0.6∗ 1.5 ∗ 1.2^3 = 4.82 s

Shear rate

= [P/(v∗ µ)) ] ^0.5

r = average shear rate (S-1 )

P = Agitator power in W

µ = Liquid viscosity (Pa . S) or kg/m s or poise

1 Pa. s = 10 Poise

V = liquid volume in m3

Example

A Reactor contain 2000 L of Liquid , Agitator power is 2916 W, viscosity of liquid is 1.2 pa. s

Calculate shear rate

Solution :

Given

V = 2 m3 P = 2916 W µ = 1.2 pa.s

r = [P/(V ∗ µ)) ] ^ 0.5 = [2916/(2∗1.2 )) ] ^0.5 = 34.85 S-1

Just Suspension speed (NJS) (Sec)

= S v^0.1 [g (ρS – ρL)/ρL ]0.45 X^0.13 dp^0.2 D^–0.85

NJS= just suspension speed, rps

S= geometry factor

dp= Particle diameter, m

g = gravitational acceleration, m/s2

ρL = Liquid density , kg/m3

ρs = Solid density, kg/m3

D = impeller diameter, m

X= Wt. of solid / wt. of liquid x 100, %

v= Kinematic viscosity, m2/s

v = µ / ρ

µ = Dynamic viscosity in kg/m.s or pa.s or Poise

1 kg/ms or pa. s = 10 poise

Example

Calculate the just suspension impeller speed for AICI. The impeller is 45° a pitched blade with D/T value of 1/3 ànd blade width of D/4 located.

Calculate : NJS in rps

Solution :

Given

S= 3.7

dp= 5000 µm ( 0.005 m )

g = 9.81 m/s2

ρL = 1326 kg/m3

ρs = 2440 kg/m3

D = 0.724 m

X= 0.4

µ = 1 cp = 0.001 kg /s

v = µ / ρ

v = 7.54 * 10-7 m2/s

Njs = = S v^0.1 [g (ρS – ρL)/ρL ]0.45 X^0.13 dp^0.2 D^–0.85

= 3.7* (7.54 10^-7)^0.1 [9.81 x (2440 – 1326) /326 ]^0.45 0.4^ 0.13 0.005 ^0.2 0.724^–0.85

= 0.95 rps =57 rpm

Batch reactor heat transfer area :

Cylindrical section + Torispherical section

πDL + [ (ᴫ/24 )∗ ((1.147 ∗ D)^2) ]

Where,

D=Diameter of reactor cylindrical section in m

L =Height of cylindrical section in m

Reactor condenser sizing m2

Ac = (UR ∗AR ∗LMTD rea )/(Uc ∗LMTD cond)

UR = reactor jacket overall HTC, kcal/hr.m2. °C

AR = Jacket heating area, m2

Uc = Condenser overall HTC, kcal/hr.m2. °C

Example

A stainless steel reactor having capacity 4 KL. Reactor subjected to maximum temperature 25 °C

( LMTD = 20 °C ) jacket side . Condenser side ( LMTD = 25 °C ).

Calculate reactor condenser sizing in m2

Solution :

Given

V = 4 KL

UR = 300 kcal/hr m2 °C

Uc = 350 kcal / hr.m2. °C

LMTD rea = 20 °C

LMTD cond = 25 °C

Reactor surface area = 3.8 V0.68 = 3.8 40.68 = 9.75 m2

Ac = (UR ∗AR ∗LMTD rea )/(Uc ∗LMTD cond) =(300∗9.75∗20)/(350∗25)) = 6.68 m2

Reactor vent sizing mm2

= 179400 ∗Q∗√(T∗G∗Z))/(C ∗ K ∗ P1 ∗ kb ))

Q=Required relieving capacity, m3/min

C = coeffient For sp. Heats ( use value 315 )

T= temp in ℃

G=sp. Gravity of gas

Z=Compressibility factor ( used value 1 )

K=coefficient of discharge ( used value 0.975 )

P1=Relieving pressure, abs+guage+21% excess in kPa

Kb= capacity correction factor ( use value 1 )

Example

Rate of vaporization of ethylene gas vapour through the reactor vent at flow rate Q is 326 m3/min T is 93 ℃ and P is 1170 Kpa. Calculate reactor vent size

Given

Q= 326 m3 / min

P 405.3 Kpa

T =93 ℃ =366 k

G= 0.78

Z= 1

K = 0.975

P1 = 1170 + 245.7 + 110 = 1525.7 kpa

Kb = 1

C = 315

Solution :

Reactor vent size