6 Pumps

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Pumps and Pumping Systems

•Categories and Types

•Performance Characteristics

•Key Design Parameters•Calculation Method

•Specification Data Sheet

•Selection Guidelines•Control Systems

•Typical Operating Problems

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• Positive Displacement

• Reciprocating

• Rotary

• Kinetic or Centrifugal

• Radial• Axial

Categories and Types

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Pumps and Pumping Systems Selection Guidelines

1820m

30.5m

227m3/h2.3m3/h

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Pumps and Pumping Systems Positive Displacement Pumps

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Pumps and Pumping Systems Centrifugal Pumps

1.Casing, volute2. Impeller, vanes, vanetips, backplate,frontplate (shroud),back vanes, pressureequalising passages3. Back cover parallelto Plane of the impellerintake4. Stuffing Box -Gland/mechanical sealhousing,

packing/lantern ring5. Pump shaft6. Pump casing7. Bearing housing8. Bearings9. Bearing seals

8

9

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• Positive displacement Pumps areConstant Volume Pumps

• Centrifugal Pumps areConstant Head Pumps

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Effect of Fluid Density on Pressure with Constant Head

9.8 bar 11.8 bar 7.4 bar

1 0 0 m

WaterSG=1.0

1 0 0 m

BrineSG=1.2

1 0 0 m

GasolineSG=0.75

P = gh: g = 9.81 m/s2

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Pump Head Calculation: Example 1 Calculate the total differential head for the followingpumping system:

3 m

2 0 m

3m of 10” pipe 45m of 8” pipe

8” gate valve

8” check valve

1.8 bar

340 m3/h capacity

Hydrocarbon, sg = 1, μ = 0.9cP

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Pump Head Calculation: Example 1 Discharge head:

Static head = 20mPressure head = 280/(9.81x1) = 28.5 m

Friction: for 8” line, flow = 340 m3/h:

u = 340/3600/((π/4)*0.2032) = 2.9m/s

Re = 1000 x 2.9 x 0.203/0.0009 = 6.5 x 105

ε/d = 0.0018/7.98 = 0.00023

f = 0.016

ΔP/100m = 0.016 x 100 x 1000 x 2.92

/(2 x 0.203) = 33.1 kPa/100m

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Pump Head Calculation: Example 1 Length of 8” pipe = 45m3 x 8” elbows = 3 x 20.2ft = 60.6ft = 18.5m1 x gate valve = 4.5 ft = 1.4 m1 x check valve = 50 ft = 15.2m

Exit loss = 20 ft = 6.1mTotal = = 86.2m

ΔP friction = 86.2/100 x 33.1 = 28.5 kPaΔh friction = 28.5/(9.81 x 1) = 2.9 m, add 20% safety for pumpΔh friction = 1.2 * 2.9 = 3.5 m

Total Discharge head = 3.5 + 20 + 28.5 = 52 m

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Pump Head Calculation: Example 1 Suction head:

Static head = -3mPressure head = 100/(9.81x1) = 10.2 m

Friction: for 10” line, flow = 340 m3/h:

u = 340/3600/((π/4)*0.2552) = 1.85m/s

Re = 1000 x 1.85 x 0.255/0.0009 = 5.2 x 105

ε/d = 0.0018/10.02 = 0.00018

f = 0.015

ΔP/100m = 0.015 x 100 x 1000 x 1.852

/(2 x 0.255) = 10.1 kPa/100m

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Pump Head Calculation: Example 1 Length of 10” pipe = 3m1 x 10” elbows = 26ft = 8mEntrance loss = 15 ft = 4.6mTotal = = 15.6m

ΔP friction = 15.6/100 x 10.1 = 1.6 kPaΔh friction = 1.6/(9.81 x 1) = 0.16 m, add 20% safety for pumpΔh friction = 1.2 * 0.16 = 0.2 m

Total Suction head = -3 + 10.2 – 0.2 = 7 m

Total differential head = 52 – 7 = 45 m

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P d P

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P d P

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For similar conditions of flow (ie the sameefficiency):

– Capacity is directly proportional to speed– Head is directly proportional to the square of

speed &

– Power is directly proportional to the cube of

speed

P d P i

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This translates into the corresponding formulae (for variations inspeed with impeller diameter remaining constant) :

Q2 = Q1 x (N2 ÷ N1)

H2 = H1 x (N2 ÷ N1)2 andP2 = P1 x (N2 ÷ N1)3

However, the above formulae can also be used to determine therelationship between impeller Diameter and flow, Head and Powerby substituting speed (N) with Impeller Diameter (D), keepingSpeed constant. Results in this case are approximate as theseformulae are analogous to the Affinity Laws above.

Q2 = Q1 x (D2 ÷ D1)

H2 = H1 x (D2 ÷ D1)2 and

P2 = P1 x (D2 ÷ D1)3

P d P i S

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Design Parameters

• Temperature• Properties• Capacity• Head (TDH)

• Cavitation (NPSH)• Horsepower• Efficiency

P d P i S

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Liquid Properties

• Specific Gravity• Composition/Quality• Vapour Pressure• Viscosity

P d P i S t

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Pumps and Pumping Systems Effect of Viscosity on Pump Performance

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Pumps andPumping Systems

Viscosity Correction

P d P i S t

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• Cavitation

• NPSH = Net Positive Suction Head• Available• Required

• Calculation

NPSH

P d P i S t

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Cavitation

Formation of vapour bubbles in pump Occurs when pressure NPSHA falls below

vapour pressure

Reduces capacity and efficiency

Eventually flow ceases Causes noise, vibration, erosion and pump

failure if allowed to persist for a prolonged

period of time

P d P i S t

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Avoiding Cavitation Establish accurate range of flow for pump.

If running speed higher than pump curve data multiply

NPSHR at curve speed by speed ratio ( eg. NPSHR = 3.2 x(1800 / 1500) = 3.84 m)

Avoid suction specific speeds of over 6750 (metric) or11,000 (usgpm).

If possible select a pump with rated flow within 10% of

best efficiency point (BEP).

Select for NPSHA 10% above NPSHR.

Require witnessed pump tests over flow range for criticalservice applications.

Don’t accept model tests as a basis for larger pumps.

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Centrifugal Pumps

Net Positive Suction Head Available (NPSHA)= Static Head + Headdeveloped

NPSHA = hps + hs – hvps – hfs

hps Pressure Head

hs Static suction Head (@ min. supply level)

hvps Vapour pressure Head (@ max. pumping temperature)

hfs Friction Head

ps

h s

hf s

hvps

NPSH

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Centrifugal Pumps

NPSH ExampleThe cooling water used for cooling of the reactor during the reaction is

pumped from an open tank. Calculate the NPSHA for the pump given thefollowing conditions:Temperature of Water = 30c, Density (r) @ 30C = 995.7 kg / m3 Vapour Pr @ 30c = 4.241 kPa (abs), Ha = atmospheric pr = 101.325 kPaGravitational acceleration = 9.81 m/s2

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Centrifugal Pumps

NPSH Example 1

Using the formula: NPSHA = Hts – Hvap

Where Hts = total static head= Atmospheric Pr (Ha) – suction lift (Hs) –

friction losses (Hf)= Ha x 1000 – 2.6 – (1 + 0.9 + 0.4 + 0.1)

g x r= 101.325 x 1000 – 2.6 – 2.4

9.81 x 995.7= 10.38 – 2.6 – 2.4= 5.38m

Hvap = 4.241 x 10009.81 x 995.7= 0.43m

NPSHA = 5.38 – 0.43= 4.95m

Add 10% safety factor for value specified to pump supplier:

The NPSH required is thus 4.95/1.1 = 4.5 m

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Centrifugal Pumps

NPSH Example 2Water temperature - 125 C

Density @ 125 C – 939 kg/m3

Vapour pr @ 125 C = 230 kPa (absolute)

Gauge pr = 130 kPag

Gravitational acceleration = g = 9.81 m/s^2

Suction head (Hs) = 4.0m max, 1.0m min

Pipe friction loss = 1.1mValve friction loss = 0.1m

NPSH A = Hts - Hvap

Hs’min = 1m

max

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Centrifugal Pumps

NPSH Example 2

Hts = Ha + Hs - Hf

= (101.325 + 130) x 1000 + 1 – (1.1 + 0.1)

9.81 x 939

= 25.1 + 1 – 1.2

= 24.9 mHvap = 230 x 1000

9.81 x 939

= 25.0 m

NPSH A = 24.9 – 25.0

= -0.1 m

(as NPSH A is negative, the pump will start to cavitate when the tank level drops to

minimum level)

To solve this problem? Increase elevation of tank!

Pumps and Pumpin Systems

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• Define Liquid Properties: SG, VP, Vis 1

• Establish Flow Rates1: Normal & Design• Calculate ‘Net” Suction Pressure • Calculate NPSH ‘Available’ • Calculate ‘Net’ Discharge Pressure at Normal Flow including

Control Valve ΔP 2

• Calculate Differential Pressure & Head

• Select Efficiency: Calculate Power1. At pumping temperature2. Allow ⅓ of total friction ΔP at Normal Flow for control valve.

Verify control valve ΔP at Design Flow minimum 70 kPa

Pump Calculation Method

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Centrifugal Reciprocating RotaryPressure max, bar abs 350 1000 350Temperature max, ºC 500 370 400ΔP max, kPa 200 1400 200Capacity max, m3/h 3600 450 340Viscocity max Pa.s 0.2 400 400Efficiency, % range 50–80 60–90 40-85Capacity Turndown H M MRelative Cost L H M

H = High, M = Medium, L = Low

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• Cavitation / Low NPSH

• Low Suction Pressure• High Discharge Pressure

• High SpGr/Viscocity

• Capacity Loss• Driver Overload

Typical Operating Problems

Date post:	03-Apr-2018
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