1. Welcome to
Project Presentation on
“DESIGN OF CENTRIFUGAL PUMP”
Prepared By
Nishant R Shah
8th
Semester
Y760038
Carried out at
LUPIN LIMITED
TARAPUR

2. COMPANY PROFILE
One of the world’s largest Manufacturers of Anti-TB Drugs
and Leading Global Player of Cardio-vascular Drugs
It has a notable presence in the areas of diabetes and
respiratory therapy
One of the Topmost Pharmaceutical Company in the country
 A public limited company established by Dr.Deshbandhu
Gupta in the year 1968 and has annual turnover of Rs.6000
crores
 Various Plants in the Company
 Fermentation Plant
 Chemical Synthesis Plants
 Multi-Purpose Plants
 Effluent Treatment Plant
 Reverse Osmosis Plant
 Multiple Effect Evaporator Plant
 Solvent Recovery Plants

3. INTRODUCTION
PUMP
 Pump is a device which converts mechanical energy into
hydraulic energy by any of various technologies, typically by
suction orcompression, used to move water, airorotherfluids
into through orout of a system.

4. CENTRIFUGAL PUMP
Introduction
A kinetic device which adds energy to
the pumped liquid by increasing its velocity
 It transforms mechanical energy from a
rotating impellerinto kinetic and potential
energy
 A simple machine which consists of a set
of rotating vanes enclosed with a housing or
casing

5. TYPES OF CENTRIFUGAL PUMP
HORIZONTALPUMP VERTICALPUMP

6. COMPONENTS
Stuffing Box Mechanical Seal Coupling
Reduces Leakage Forms a seal between Connects pump to
rotary & stationary the pump driver
part of the pump

7. COMPONENTS
Impeller
Open Impeller Semi Open Impeller Closed Impeller

8. TERMINOLOGY
1. AIRBINDING:
If airenters the impeller, the pump is not capable to drive the liquid
through the delivery pipe. This is called as airbinding of the pump.
2. PRIMING:
If pump is initially full of airto avoid airbinding, the centrifugal pump
needs
priming. The removal of airfrom the suction line &pump casing is known
as
priming. The priming should be done before the pump is started.
3. NETPOSITIVESUCTION HEAD(NPSH):
The NPSHof the pump is the amount by which the pressure
available at the suction point is in excess orgreaterthan the vapour

9. TERMINOLOGY
4. CAVITATION:
If the pressure in the suction line is less than the vapour
pressure of
the liquid then some of the liquid flashes into the vapourwhich
leads
to the formation of airbubbles that damages the impeller. This
phenomenon is Cavitation. Cavitation results in no pumping of
liquids. The occurrence of cavitation can be noted by a marked
increase in noise &vibration as the vapourbubbles collapse &
also a
loss of head &efficiency.
 Two important terms :
 Net Positive Suction Head Available (NPSHA)
 Net Positive Suction Head Required (NPSHR)

10. PUMP CALCULATIONS
 As perProcess Flow Diagram, Data forPump selection are as below:
 Materials of Construction = SS-316
 Pumping Fluid = Methanol
 Flow rate = 10 m³ /hr
 Density = 790 kg/m
3
 Sp.gravity = 0.79
 Viscosity = 0.6Cp
 Vapourpressure = 400mmHg = 5.44mWc
 Pressure in suction tank =1.033 kg/cm² = 10.33 mWc
 Pressure in discharge reactor=1.033 kg/cm² = 10.33 mWc
 Suction static head (hs)= 3m
 Discharge static head (hd)= 19 m
 Suction side velocity (vs) = 0.55 m /s
 Discharge side velocity (vd) = 1.41 m /s
 Methanol Temperature in Storage Tank = 35° c
 Temperature of Chilled Waterin Heat Exchanger= 10
0
C
 Methanol Temperature in Reactor= 15
0
C

12. PUMP CALCULATIONS
 Determination of Discharge Pipe Diameter
Flow rate(m
3
/s) = π*D
2
*2m/s
4
0.0027 = π*D
2
*2
4
.˙. D = 0.042m = 42mm ~ 50mm
.˙. Discharge pipe size = 50mm ~ 50NB ~ 2
’’
.˙. Suction pipe size = 80mm ~ 80NB ~ 3
’’
SUCTION SIDE
 Nre = D*v*ρ = 57933
µ
 Friction Factor
ƒ = 0.078 = 0.005
(Nre)
 Pressure Drop(ΔP)
ΔP = 4*ƒ*L*v2
*ρ
2*D*g

13. PUMP CALCULATIONS
= 4*0.005*9.82*0.55
2
*790
2*0.08*9.81
= 0.003 kg/cm2
= 0.037mLc
 Suction Pressure
Suction Pressure = Pressure inside the storage tank + Static height - ΔP
= 13.07 + 3 - 0.037
= 16.03mLc = 1.22 kg/cm
2
 NPSH Available
NPSH Available = Suction Pressure – Vapour Pressure
= 16.03 – 6.53
= 9.49mLc
DISCHARGE SIDE
 Nre = D*v*ρ = 92825
µ
 ƒ = 0.078 = 0.004
(Nre)

14. PUMP CALCULATIONS
 Pressure Drop(ΔP)
ΔP = 4*ƒ*L*v
2
*ρ
2*D*g
= 4*0.004*39.69*1.41
2
*790
2*0.05*9.81
= 0.11 kg/cm2
= 1.43mLc
 System Losses
System Losses = (hd - hs) + (∆Ps + ∆Pd) + vs
2
+ vd
2
2*g
= 19-3 + 0.037 + 1.43 + 1.14
2
+ 0.55
2
2*9.81
= 17.54mLc = 1.34 kg/cm
2
ΔP across heat exchanger = 0.33 kg/cm
2
= 4.31mLc
 Discharge Pressure
Discharge Pressure = Pressure inside the reactor + Static height
+ DischargeΔP + ΔP due to heat exchanger

15. PUMP CALCULATIONS
Power Requirement
Assuming Pump Efficiency(n) = 55% = 0.55
Power Requirement = ( Q*System Head(hd- hs)*ρ*g)
n
= ( 0.0027*16*790*9.81)
0.55
= 626 W
= 0.62 kW
= 0.83 HP ~ 1HP
.˙. Power Required in Motor = 1HP
 Energy Requirement
Assuming pump runs for 4 hrs
Energy required = 0.62 kW*4 hrs
= 2.48 kW/day
 Cost Requirement
Pumping cost = 2.48 kW/day*5.5 Rs/kW
= Rs. 13.64/day
.˙. Pumping cost = Rs. 13.64/day

17. PUMP CURVES

19. Advantages
 Requires Less floorspace &simple foundation
 Steady discharge
 The pump does not get damaged even if the
discharge
gets blocked
 It can handle Liquids containing high proportions of
suspended solids
 Lowermaintenance cost
 Cheaper
 Operated at high speed
 Used forlarge capacity &low heads