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CHAPTERSHomeTopicsChapter 1 : General ConceptsChapter 2 : Pelton TurbineChapter 3 : Francis and KaplanTurbineChapter 4 : Centrifugal PumpsChapter 5 : Similarity Relationsand PerformanceCharacteristicsChapter 6 : ReciprocatingPumpsChapter 7 : Hydraulic devicesand Systems

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Chapter 4 : Centrifugal Pumps

Q. 1. Classify hydraulic pumps. Ans. Pumps may bi placed in one of the two generalcategories.(i) Dynamic pressure pumps: centrifugal pump, jetpump, propeller, and turbine.(ii) Positive, displacement pump: Piston plunger, gear,lab, vane, screw etc.

Q. 2. Distinguish between the positive and non-positive displacement pumps. Ans. Positive displacement pump: It causes a fluid tomove by trapping a fixed amount of it then forcing(displacing) that trapped volume into the discharge pipe. E.g. Lobe, gear, screw, vage pump etc. Non-positive displacement pump (rotodynamic pump) - It is pump in which the dynamic motion of a fluid isincreased by pump action. e.g. centrifugal, turbine,propeller etc.

Q. 3. What is the principle of working of aCentrifugal Pump? Ans. It is very clear that the principle used for centrifugalpump is the centrifugal force in the form of dynamicpressure which is generated by rotary motion of one or

more rotating wheels called the impellers.

Q. 4. Sketch layout of a centrifugal pumpinstallation and label it.Ans.

Q. 5. Distinguish between volute casing and vortexcasing in a centrifugal pump. Ans. Volute casing—This is a spiral shaped whose areaof cross-section gradually increases towards the deliverypipe. The velocity of liquid decreases as area increasesalong the path of flow. This arrangement converts K.E.into pressure energy. The efficiency of this casing is lessthan others. Larger amount of energy is lost due toformation of eddies.

Fig. Different types of casings.Vortex casing—When a circular chamber is introducedbetween the impeller and casing, the casing is known asvortex casing. In this case, the liquid from the impellerenters into the vortex chamber and then through thevolute chamber, in this arrangement, the eddy loss is

volute chamber, in this arrangement, the eddy loss isconsiderably reduced and the efficiency of conversionfrom K.E. into pressure energy is increased as comparedwith volute casing.

Q. 6. Explain different types of impellers used incentrifugal pump with neat sketch. Ans. (1) Closed, semi—closed and open impellers inthe closed impellers the vanes are covered with sideplates (shrouds) on both sides. The back shroud ismounted into shaft and front shroud is coupled to theformer by the vanes. The arrangement provides asmooth passage for the liquid; wear is reduced tominimum. This ensures full capacity operation with highefficiency for a prolonged running period.

Fig. (,) Open (ii) Semi-open (iii)

EnclosedSemi-open impeller has a plate only on back side. Thedesign is adopted to industrial pump problems whichrequired a rugged pump to handle liquids containingfibrous material such as paper pulp, sugar molasses andsewage water etc. in open impeller, no shroud or plate isprovided on either side i.e., the vanes are open on bothsides. Such pumps are used where the pump has a veryrough duty to perform i.e. to handle abrasive liquids etc.(ii) Axial, Radial and mixed flow impellers—In the axialflow pumps, the head is developed by the propelling orlift action of the vanes on the liquid which enters theimpeller axially and discharges axially

Fig. Axial radial and mixed flowimpellers.In radial flow impellers, the head is developed by theaction of centrifugal force upon the liquid which entersthe impeller axially at the centre and flows radially to theperiphery. In the mixed flow impeller there is acombination of axial and radial flows. The head isdeveloped partly by the action of centrifugal force andpartly by axial propulsion.

Q. 7. Draw velocity triangles for (i) Curved forwardvanes. (ii) Radial vanes (iii) Curved backwardvanes. How Euler’s head varies with discharge?Also draw variation of power with discharge.

Ans.

Euler’s head variation with discharge: Is negative, so with

increase in flow rate the Euler head rises.

For radial vanes = 90° and cot = 0, so the headremains constant with variation in discharge.

For curved backward vanes <90° and cot is positive.So with increase in mass flow rate the Euler’s head falls.

Influence of impeller vane exit angle on powercapacity relationship.

Q. 8. Define manometric head of a pump. Ans. The manometric head is defined as the headagainst which a centrifugal pump has to work. It isdenoted by Hm.

Q. 9. Define manometric efficiency. Ans. The ratio of manometric head to the head importedby the impeller to the water is known as manometricefficiency.

Q. 10. Define the following terms:(i) Suction head (ii) Delivery head (iii) Static head(iv) Manometric head (v) Hydraulic (vi)Manometric (viii) Volumetric (viii) Mechanical. Ans. 1.Suction head —It is the vertical height of thecentre line of the centrifugal pump above the watersurface in the tank or pump from which water is to belifted. This height is also called suction lift.

2. Delivery head -The vertical distance between thecentre line of the pump and the water surface in the tankto which water is delivered is known as delivery head.3. Static head-The sum of suction head and deliveryhead is known as static head. This is represented by Hand is written as:

4. Manometric Head -The manometric head isdefined as the head against which a centrifugal pump hasto work. It is denoted by ‘Hm’. EFFICIENCIES OF ACENTRIFUGAL PUMP

1. Manometric efficiency -The ratio of themanometric head to the head imported by the impeller tothe water is known as manometric efficiency.Mathematically, it is written as

2. Mechanical efficiency—Mechanical losses in a pumprepresent the degradation of energy due to mechanicalfriction in packings, glands and bearings etc. “it isdefined as the ratio of the energy transferred to water bythe rotor to the mechanical energy delivered at the shaftcoupling”.

3. Volumetric efficiency—Let Q denote the actualdischarge at the pump outlet and. q represent theinternal leakage, then the total or theoretical volumeflowing through the impeller is (Q + q).‘Volumetric efficiency is defined as the ratio of actual totheoretical discharge’.

4. Hydraulic efficiency—Hydraulic losses due to fluidseparation and energy consumed by friction in the flow.‘Hydraulic efficiency is defined as the ratio of actual headto the theoretical or Euler head’.

Where =No of blades, = Euler head

Q. 11. Define the term Gross head.

Ans. Gross head: The Gross head or Total head is the

difference between the water level at the reservoir (alsoknown as head race) and the level at the tail race.

Q. 12. Justify the statement ‘centrifugal pumpshaving backward curved blades are preferred ascompared to those having radial and forwardcurved blades.’ Ans. An angle of > 90° leads to backward curvedblades, 90° means radial blades and <900 meansforward blades. It is found that with small angle meansalso a large absolute speed. The transformation of thisspeed energy in pressure energy in the peeler isconnected with substantial losses. It is better to choose

90° for getting lower absolute speed c. In addition alarge angle has the disadvantage that it requires withsame delivery head a larger circumferential speed and soit causes larger wheel friction losses. Because if thelarger difference of pressure between entrance and exitof the impeller, larger gap leakages are caused, thesedisadvantages cannot cover crucially better hydraulicefficiency. Therefore in centrifugal pumps only backwardcurved blades with angle> 900 (140—160°) are used.

Q. 13. Discuss the influence of exit blade angle onthe performance and efficiency of a centrifugalpump. Assume radial flow at entrance. Ans. Blade angle affects the efficiency of thecentrifugal pump

When > 90° efficiency - 75%

90° efficiency - 80% to 85%

<90° efficiency -85% to 90%It is clear from the above results that when the exit angle

of the blade is more than 90°the efficiency is very lessi.e. only 75% and as the angle decrease the efficiencyincrease. Efficiency is greater for acutes angles of theexit blades.

When > 90° the speed is slow

When > 90° the speed is slow

= 90° the speed is medium

<90 0 the speed ishighWe know that efficiency depends upon the velocity, asthe velocity increases efficiency also increases

Q. 14. What is ‘slip factor’ of centrifugalcompressor? Ans. Liquid is trapped between the impeller vanes. Sothere is no assurance that the actual fluid will follow theblade shape and leave the impeller in a radial direction.There is usually a sligh slippage of the fluid with respect

to the blade rotation. If is reduced to then (

- ) is called as slip.

Slip factor

Q. 15. What is priming of centrifugal pump? Ans. The filling of suction pipe, impeller casing anddelivery pipe upto delivery valve by the liquid fromoutside source before starting the pump is known aspriming.The air is removed and that portion is filled with theliquid to be pumped.

Q. 16. How is the minimum starting speed of acentrifugal pump computed?

Ans. The expression is given by:

Starting speed, N

Where = Manometric efficiency

= whirl velocity

= diameters

Q. 17. What do you understand by characteristiccurves of single stage centrifugal pump?

curves of single stage centrifugal pump? Ans. A centrifugal pump works under its maximumefficiency conditions. However when the pump is run atconditions different from design conditions, it performsdifferently. Therefore to predict the behavior of the pumpunder varying conditions of speeds, heads, discharges orpowers, tests are usually conducted. So characteristiccurves of centrifugal pumps are defined as those curveswhich are plotted from the results of a number of testson the centrifugal pump. Performance characteristics of apump fall into following categories.1. Main characteristic curves2. Operating characteristic curves3. Constant efficiency or muschel curves4. Constant head and constant discharge curves.1. Main characteristic curves—The main characteristiccurves of a centrifugal pump consist of a variation ofhead (Hm), power and discharge with respect to speedas shown in figure (a).

For plotting curves of manometric head versus speed,

For plotting curves of manometric head versus speed,discharge is kept constant. For plotting curves ofdischarge versus speed, manometric head (Hm) is keptconstant and for plotting the curves between power andspeed the manometric head and discharge are keptconstant.2. Operating characteristic curves-The curves plottedfrom the results of a number of tests on a pump whilerunning at its design speed are referred as characteristiccurves. Fig. (b) shows the operating characteristic curvesof a pump.The curves indicate the rotation between efficiency,power and head with respect to discharge. These areplotted for only one speed.Constant efficiency curves or Muschel curves—Theconstant curves are also called the ISO-efficiency curves;depict the performance of a pump over its entire range

Constant efficiency curves of a pumpof operation. Data for plotting these curves is obtainedfrom main characteristic curves i.e. Vs. Q and H Vs. Q.As shown in figure (b) for particular efficiency ahorizontal line is drawn which intersects the curves for

different pump speeds on the Q plot. From the pointsof intersection the corresponding discharge valves areobtained. This information is thus transferred to H Vs. Qcurve for corresponding speeds. Like wise for anothervalue of efficiency, the points are obtained andprotected. Points corresponding to one efficiency, are

protected. Points corresponding to one efficiency, aresmoothly joined to get a constant efficiency curve. A curefor best performance is obtained when peak points tovarious ISO — efficiency curves are joined. Constantefficiency curves help to locate the regions where thepump would operate with maximum efficiency.4. Constant head and constant dischargecurves-The performance of a variable speed pump forwhich the speed constantly varies can be determined bythese curves.

Fig. Curves of a centrifugalpump

Q. 18. Assuming the flow velocity to be constant,prove that manometric head of a centrifugal pumprunning at speed N and having discharge Q can beexpressed as

Where A, B and C are constants. Ans. Applying Bernoull’s equation between inlet andoutlet edges of the impeller for points 1 and 2respectively; Energy at inlet = energy at outlet - work input

Energy at inlet = energy at outlet - work input

Pressure riseFrom the outlet velocity vector.(1)

(2)

Invoking the trigonometrical relationship,

and simplifying

Assuming radial entry of fluid,

Neglecting gravitational effects, any friction losses, andassuming radial entry of water

As derived above,

And

Presuming that velocity of flow remains constant

Where a (1 + k), b =

Also

So Hence proved

Q. 19. Show that the pressure rise in the impellerof a centrifugal pump where friction and otherlosses in the pump are neglected is given by

Provided flow is radial at inlet, velocity of flow isconstant through-out and outlet angle of impellervane is 45°. Solution. Refer to Q. No.

We have

Given And

So

[ w=pg]

Hence proved.

Q.20. Set up the following relation for increase inpressure head across the impeller of a centrifugal

pump;

Where u is peripheral velocity, is the relativevelocity, y is datum height and P is pressure. Thesubscripts 1 and 2 refer to inlet and outletparameters. Solution. The Euller equation for theoretical headgenerated by centrifugal impeller

By Bernoulli’s equation between (suffix 1) and outer(suffix 2) edge of the impeller.

H=

Hence =H-

= increase in piezometric head(hence proved)

Q. 21. Explain mixed flow centrifugal pumps. Ans. Centrifugal pump consists of two or more impellers,the pump is called a multistage centrifugal pump.There are two types of arrangements.1. Multistage centrifugal pump for high heads orimpellers in series.2.Multistage centrifugal pump for high discharge orimpellers in parallel.3.Impellers in series—For developing a high head a

number of impellers are mounted in series or on thesame shaft as shown in figure below.

Fig. Two-stage pumps with impellers in series. The water from suction pipe enters the 1st impellers atinlet and discharged at outlet with increased pressure.The water then from 1st impeller taken to inlet of the2nd impeller with the help of connecting pipe. So atoutlet of 2nd impeller pressure of water will be more.

Total head developed= n × Where n = no. of impellers inseries.(ii) Impellers in parallel—For obtaining high dischargethe pumps can be connected in parallel. Each of thepumps lifts the water from a common pump anddischarges water to a common pipe to which the deliverypipes of each pump is connected.

Fig. Pumps in parallel.

Each of the pump is working against the same head. Total discharge = n Q

1. Suction Head, =

2. Delivery Head, =

3. Total external Head=

4. Monometric. Head, = =

5. NPSH (Net Positive Suction Head) =

6. Torque = = Energy Transfer.

6. Torque = = Energy Transfer.

Problem 1. A centrifugal pump is required to liftwater against total head of 40 m at the rate of50.liters per second. Find the power of the pump, ifthe overall efficiency is 62%. Solution: Given Head, H=40 m Discharge, Q= 50liters/second

=

Overall eff., = 62% = 0.62We know that

Overall efficiency, =

0.62 =

(Take p = 1000 kg/ )

Power = 31612.9 watt = 31 .612 KWAns.

Problem 2. A centrifugal pump having outerdiameter equal to two times the inner diameterand running at 1000 r.p.m. works against a totalhead of 40 m. The velocity of flow through theimpeller is constant and equal to 2.5 m/s. Thevanes are set back at an angel of 40° at outlet. Ifthe outer diameter of the impeller is 50 mm andwidth at the outlet is 50 mm, determine.

1. Vane angle at inlet

2. Work done by impeller on water per second

3. Manometric efficiency.

Solution: Given N = 1000 r.p.m.

Head, =40 m

Velocity of flow, = 2.5m/s

Vane angle at outlet, = 40°

= 500 mm = 0.50 m

Inner dia. of impeller, = =0.25 m

width at outlet = 50 mm = 0.05 m

=13.09 m/s

==26.18 m/s

Discharge is given by, Q = x 0.50 x 0.05 x2.5

(1)Vane angle at inlet

From inlet velocity triangle, tan = = 0.191

0.191 = 48’

(2) Work done by impeller on water per second

But from outlet velocity triangle, we have

26.18- = = =2.979

= 26.18 — 2.979 = 23.2 m/s

Substituting this value of in equation

We get the work done by impeller as

x23.2 X 26.18

= 119227.9Nm/s Ans.

Problem 3. A centrifugal pump impeller hasdiameter of 60 cm and width of 6 cm at the outlet.

The pump runs at 1450 rpm and delivers 0.8 against a head of 80 m. The leakage loss after theimpeller is 4% of discharge, the externalmechanical loss is 10 kW and hydraulic efficiencyis 80%. Determine the blade angle at outlet, thepower required and the overall efficiency of thepump.

Solution: Peripheral velocity of impeller at outlet,

= 45.53 m/s

Total quantity of water flowing through the impeller

(Q+q) =0.8+0.04x0.8

=0.832

Flow velocity at impeller outlet

Flow velocity at impeller outlet

= ==7.36 m/s

Fig. Velocity vector diagrams of a centrifugalimpeller

; 0.8= 221.546m

=3.258

Vane angle of impeller outlet, (3.258) = 17.06°

Ideal head = =100 m

Rotor or impeller power = w (Q + q) = (9810 x 0.832

x 100) x

= 816.2 kW

Shaft power = rotor power + mechanical losses

Shaft power = rotor power + mechanical losses

= 816.2 + 10 = 826.2kW

Mechanical efficiency 0.9888 =98.88%

Volumetric efficiency,

=0.961=96.1%

Now, overall efficiency,

= 0.8 x 0.961 x 0.988

= 0.76 or 76%

Problem 4. A three stage centrifugal pump hasimpellers 40 cm in diameter and 2 cm wide atoutlet. The vane angle at out is 45° and reducethe circumferential, area by 10%. If the pumpdelivers 50 liters per second when running at1000 r.p.m., determine

(i) Power of the pump (ii) Manometric head.

Assume manometric efficiency as 90% and overallefficiency as 80%. Solution:1. Number of stages n =32. Dia. of impeller at outlet, =40 cm = 0.40 m3. Width at outlet, = 2 cm = 0.02 m4. Vane angle at outlet, = 45°

5. Reduction in area at outlet = 10% = 0.1

6. Area of flow at outlet = 0.9 x 0.9 x 0.4 x0.02 = 0.02262

Manometric efficiency, = 90% = 0.90 Overall efficiency, = 80% = 0.80 Speed, =N 1000 r.p.m.

Speed, =N 1000 r.p.m. Discharge Q = 50 liters/s = 0.05/s

Velocity of flow at outlet = = =2.21m/sTangential velocity of impeller at outlet

=20.94 m/sFrom velocity triangle at outlet

2.21m/s

2.21 = 20.94 —2.21 = 18.73 m/s

10.90 =

=35.98mFor total head generated by pump

= n x=3x35.98=107.94m

Power outlet of the pump =

= 52.94 kW

We have

So Shaft power ==66.175kW

Problem 5. This diameter of the impeller of acentrifugal pump is 800 mm. The pump deliverswater at the rate of 90 /min. The head raisedby is 72 m and the speed is 900 rpm. Peripheral

by is 72 m and the speed is 900 rpm. Peripheralarea of the outlet is 0.3 The vanes of theimpeller are curved back at 30° to the tangentialdirection. The leakage loss is 2.5%.; the inputhead is reduced by 20% kW. Find the power inputand the overall efficiency of the pump. Solution: Given: D = 800 mm = 0.8m Q =90 /min H =72m N = 900 r.p.m.

Leakage loss = 2.5% Loss in input head = 20% of theoretical head Mechanical losses = 30 kW

Now, Peripheral velocity, = 37.69m/sTotal quantity of water through impeller, (Q + q)

90 + 90× 92.25 /min = 1.53/s

We know = 1.53

Whirl component of velocity, = 37.69-5.1 cot 30° =28.85 m/s

Theoretical head = = 110.84 m ofwaterTotal head against shaft has to work = 72 + 110.84

x = 94.108 m of waterSo, shaft power = Rotor power + mechanical losses = w (Q + q) Hi + mechanicallosses = 9810 (1.53) x 94.168 + 30 x

= 9810 (1.53) x 94.168 + 30 x= 1443.39 kW

Mechanical efficiency,

= 0.9792 =97.92%

Volumetric efficiency, = 0.98039= 98.039%

Hydraulic efficiency, = 0.6495 =64.95%

Overall efficiency, = 0.6495 x 0.98039 x0.9792 = 0.623 = 62.3%.

Problem 6. The outer diameter of an impeller of acentrifugal pump is 400 mm and outlet width 50mm. The pump is running at 800 r.p.m. and isworking against a total head of 15 m. The vanesangle at outlet is and manometric efficiency is75%. Determine1. Velocity of flow at outlet2. Velocity of water leaving this vane3. Angle made by the absolute velocity at outletwith the direction of motion at outlet 4. Discharge Solution:

Outer diameter, = 400 mm = 0.4 m

Width at outlet, =50 mm = 0.05 mSpeed, N = 800 r.p.m.

Speed, N = 800 r.p.m.

Head, =15mVane angle at outlet, = 40°

Manometric efficiency, = 75% = 0.75Tangential velocity of impeller at outlet

=16.75 m/s

0.75

=11.71m/sFrom the outlet velocity triangle, we have

(1) (2) Velocity of water leaving the vane

=12.45 m/s(3) Angle made by absolute velocity at outlet

= 0.36

0.36= 19.80°or 19° 48’ Ans.(4) Discharge through pump is given by

0.4x0.05x4.23

Problem 7. Find the rise in pressure in theimpeller of a centrifugal pump through whichwater is flowing at the rate of 0.01 the

internal and external diameters of. The impellersare 15 cm and 30 cm respectively. The widths ofthe impellers at inlet and outlet are 1.2 cm and0.6 cm. The pump is running at 1500 r.p.m. Thewater enters the impeller radially at inlet and

impeller vane angle at outlet, is Neglect lossesthrough the impeller. Solution. Given, Discharge Q =10 , Internal dia. = 0.15 m

External dia. = 0.30 m, width at inlet

= 0.012 m, Width at outlet = 0.006m, speed N =1500 r.p.m. Vane angle at inlet, = 450

Velocity of flow at inlet,

=1.768m/s

Velocity of flow at outlet, =1.768. m/sTangential velocity of impeller at outlet

=23.56 m/s

Pressure rise

Pressure rise = =28.13 m

Problem 8. A centrifugal pump with 1.2 m

diameter runs at 200 r.p.m. and discharge 1880litters/s the average lift being 6m. The anglewhich the vanes make at exit with the tangent tothe impeller is 26° and radial velocity of flow is2.5 m/s. Determine the manometric efficiency andthe least speed to start pumping against a head of6 m, the inner diameter of the impeller being 0.6m. Ans. (a) Peripheral or tangential velocity of impeller atoutlet,

= 12.56m/s

Whirl component =12.56 - 2.5 cot 25°= 7.43 m/s

Manometric efficiency =0.63 or 63%.

(b) Minimum speed, H,

Now 2 and so

=6;

=6.26 m/s

Also 6.26

Minimum rotational speed, N = =199.36 r.p.m.

Problem 9. The impeller of a centrifugal pump hasa diameter of 10cm and breadth 3.5 cm at theinner periphery; the corresponding dimensions at

the outer periphery are 20 cm and 1.7 cmrespectively. The pump runs at 1500 rpm, has 7vanes with vane angle at entry and exit equal to16° and 30° respectively. Calculate: (i) thetheoretical discharge for shockless entrance, (ii)the theoretical head developed, (iii) the actual

the theoretical head developed, (iii) the actualhead produced, the losses and the power requiredto drive the pump.Neglect the effect of vane thickness and presumethat the hydraulic efficiency is 85% and theoverall efficiency is 75%. Solution. For the nomenclature and velocity vectordiagrams,The peripheral speeds are

=7.85 m/s.

= 15.70 m/s.

(i) From the inlet velocity triangle, = 7.85tan 160 = 2.25m/s

x 0.1 x 0.035 x 2.25= 0.0247

(ii) =2.31 m/s

= 15.7 —2.31cot 30° = 11.69 m/s

For shockless entry, the theoretical or Euler head isgiven by]

= 18.71 m(iii) The slip factor can be found from the relation

1.34

Slip factor, = 1/1.34 = 0.746The actual head produced is

H =18.71 x0746 x 0.85 = 11.86 m

The hydraulic losses are = = 18.71 x 0.746 x (1— 0.85) = 2.09 mPower required to drive the pump, P

P

= 3831.68 W = 3.832 kW.

Problem 10. A centrifugal pump has to discharge225 liter of water per second against a head of25 m when the impeller rotates at a speed of1,500 r. p.m. Determined (a) the impellerdiameter, and (b) the vane angle at the outletedge of the impeller Assume that 0.75 ; theloss of head in pump in meters due to fluidresistance is 0.03 ; where m/s is theabsolute velocity of water leaving the impeller, t1area of the impeller outlet surface is (1.2. ) ,

where is the impeller diameter m, and waterenters the impeller without whirl. Solution: = 33.33 mThe manometric head is losses of headin the pump.Losses in the pump = (33.33 — 25) = 8.33 mThus 0.03 =8.33

=16.66 m/s

Substituting the above value of in (i)

Now from the outlet velocity triangle of Fig. 24.5

Or

Or

= 0.2534 m or 253.4 mm.

Thus =2.93 m/s; = 19.90 m/s

And = 16.42 m/s = [2.93/ (19.90 — 16.42)] = 0.8420

Problem 11. The internal and external diameter ofthe impeller of a centrifugal pump which isrunning at 1000 rpm are 200 mm and 400mmrespectively. The discharge through the pump is0.04 and velocity of flow is constant and equalto 2.0 m/s. The diameters of the suction anddelivery pipes are 150 mm and 100 mmrespectively and suction and delivery heads are6m (abs.) and 30 m (abs.) of water respectively.If the outlet vane angle is 45° and the power

required to derive the pump is 16.186 kWdetermine:(a) Vane angle of the impeller at inlet.(b) The overall efficiency of the pump and(c) Manometric efficiency of the pump. Ans. Given:

Speed, N = 1000 r.p.m.

Internal dia., = 200 mm = 0.2 m

External dia., = 400 mm 0.4 m

Discharge, Q = 0.04

Velocity of flow,

Dia. of suction pipe, = 150 mm = 0.15 m

Dia. of suction pipe, = 150 mm = 0.15 m

Dia. of delivery pipe, = 100 mm 0.10 m

Suction head, = 6 m (abs.)

Delivery head, 30m (abs.) Outlet vane angle, = 45°Power required to drive the pump, P = 16.186/kW(a) Vane angle of the impeller at inlet

From inlet velocity, we have tan

Where

=10.47 m/s

= 0.191 or.191 = 10°48’ Ans.

(b) Overall efficiency of the pump

We know Where S.P. = Power required to derive the pumpand equal to P here.

… (1)

= 0.02424

Now is given by equation as

… (2)

Where = Pressure head at outlet of pump

= 30 m

= Velocity head at outlet of pump =

= Pressure head at inlet of pump ==6 m

= Velocity head at inlet of pump =

= Vertical height of outlet and inlet of pumpfrom datum line.

If then equation (ii) becomesas

… (3)

Now =5.9m/s

And =2.26 m/sSubstituting these values in equation (iii), we get

= (30+ 132) — (6 + .26) = 31.32— 6.2625.06 m.

Substituting the value of in equation (i), we get= .02424 x 25.06 = 0.6074 60.74%. Ans.

Problem 12. The internal and external diametersof the impeller of a centrifugal pump are 200 mmand 400 mm respectively. The pump is running at1200 rpm. The vane angles of the impeller at inletand outlet are 20° and 30° respectively. Thewater enters the impeller radially and velocity offlow is constant. Determine the work done by theimpeller per unit weight of water. Solution.

Internal diameter, = 200 mm = 0.2 mExternal diameter, = 400 mm = 0.4 m Speed, N = 1200 rpmVane angle at inlet, = 20°

Vane angle at outlet, = 30°

Water enters rodially means, a 90°, =0

Velocity of flow,

=12.56m/s

= 25.13 m/s

From inlet triangle,

=12.56 tan = 12.56 x tan 20 4.57m/s

= 4.57 m/s

25.13 - = 7.915

= 25.13 - 7.915 = 17.215 m/sThe work done per kg per second

= 44.1 Nm/s