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Pharma Calculations
Agitator tip velocity
Tip Velocity : Vt ( m/sec )= π Da N
N = rotational speed in seconds
Da = agitator swip dia. m
ExampleReactor with capacity 5 KL agitator running with 75 rpm agitator diameter is 1200mm Calculate Tip velocity Solution. : GivenDa = 1200 mm = 1.2 mN = 75 rpm = 1.25 rps
Tip velocity Vt =π Da N
Vt = 3.142 X 1.2 X 1.25 = 4.71 m/s
Agitator tip velocity
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
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
Agitator power/Volume ratio
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
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
Agitator pumping capacity
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
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
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
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
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
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
= 79400 ∗Q∗√(T∗G∗Z))/(C ∗ K ∗ P1 ∗ kb ))
= (179400 ∗326∗ 36√(366∗0.78∗1))/(1∗ 315 ∗0.975 ∗1525.7 ))
= 2346 mm2
Reactor vent sizing
Liquid effective agitation depth : dp in m
=(Rvol -Rt )/(ᴫ/4∗DR^2))
(Rvol = reaction mass volume m3
DR = reactor internal dia. m
Rt = Torispherical volume = (ᴫ/24 ∗DR3)
Example
A stainless steel reactor having diameter 1800 mm contain reaction mass volume 1 KL
Calculate liquid effective agitation depth
Solution :
Given
DR= 1800 mm = 1.8 m
Rvol = 1 KL = 1 m3
Rt = ᴫ/24 ∗DR3 = 0.7635 m2
dp = (Rvol -Rt )/(ᴫ/4 ∗DR^2) = (1-0.7635)/2.54) = 0.092 m
Scale up extraction calculation
Settling Test
Procedure : Take ~ 100 ml slurry mass and settle for 3 min.
Obsevation and conclusion :
If settles down less 10 minutes --> Good to filter
If settles down before 30 min & cloudy mass --> Medium Filterability and use preferable centrifuge or high-pressure filter
If settles down in or above 30 min & cloudy mass --> difficult to filter, crystallization conditions need to be altered
Filtration rate test :
Procedure :
Using slurry mass (to be filtered) prepare ~ 5 cm cake thickness over Buckner filter paper.Charge ~100 ml slurry over the 5 cm cake bed and apply vacuum (NLT 650 mm Hg. Measure the time required to filter the mass in seconds.
Obsevation and conclusion :
If ML filters at a rate of > 40 lpm/m2 -->Good to filter
If ML filters at a rate of < 20 lpm/m2 -->Difficult to filter, crystallization conditions need to be altered
Reactor calc
Vent size
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