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: 2 : Mechanical Engg. _ ESE MAINS


01(a).

(i) NPSH (Net Positive Suction Head): NPSH

available is the theoretical amount of head

that could be lost between suction and point

of minimum pressure without causing

cavitation. NPSH refer to the analysis of

cavitation. The minimum value of NPSH

available that is needed to prevent cavitation

in the pump or turbine i.e., the value of

NPSHA that causes pmin to equal pvap. To

avoid cavitation, always operate with

NPSHAvailable ≥ NPSHRequired.

Rvap

Lsabs,atm HSPN

phZ

P

(ii) Slip and (iii) coefficient of discharge (Cd):

Slip of a reciprocating pump is defined as

thedifference of the theoretical discharge and

the actual discharge.

Slip = Theoretical discharge – Actual discharge

= Qth – Qa

Ship can also be expressed in terms of % age

and given by

% slip = 100Q

QQ

th

actth

100Q

Q1

th

act

= (1 – Cd) 100

Slip where Cd is known as co-efficient of

discharge and is defined as the ratio of the

actual discharge to the theoretical discharge.

th

ad Q

QC

Value of Cd when expressed in percentage is

known as volumetric efficiency of the pump.

Its value ranges between 95 to 98%.

Percentage slip of the order of 2% for pumps

in good conditions.

(iv) Air Vessels: Air vessels are a closed

container, in which the bottom part is filled

with water & upper half part is filled with

compressed air. These air vessels installed

very near to the suction valve & delivery

valve to avoid the separation in reciprocating

pumps.

An air vessel usually fitted in the discharge

pipe work to dampen out the pressure

variations during discharge. As the discharge

pressure rises the air is compressed in the

vessel, and as the pressure falls the air

: 3 : Test – 3


expands. The peak pressure energy is thus

stored in the air and returned to the system

when pressure falls.

Purposes of Air vessels:

1) To obtain liquid at uniform discharge.

2) Due to air vessel frictional head and

acceleration head decreases and the

work overcoming friction resistance in

suction and delivery pipe considerably

decreases which results in good amount

of work.

3) Reciprocating pump can run at high

speed without flow separation

(v) Specific speed of pumps:Specific speed is

defined as "the speed of an ideal pump

geometrically similar to the actual pump,

which when running at this speed will raise a

unit of volume in a unit of time through a

unit of head".

In metric units

Specific speed 4/3s H

QNN

N = The speed of the pump in revolutions per

minute (rpm)

Q = The flow rate in liters per second (for

either single or double suction impellers)

H = The total dynamic head in meters

In SI units, Q is in cubic meter per second, N

in rpm and H in meters.

01(b).

Sol: Assumption:-

1–D heat conduction Steady state No heat generation

Given:

2m/W700q T1 = 80oC

L = 0.3m A = 12m2

K = 2.5W/m-K

General heat conduction equation for 3 – D

T1

k

q

z

T

y

T

x

T2

2

2

2

2

2

No heat generation and steady state, 1 – D

0x

T2

2

1cx

T

T = c1x+c2 linear profile

Boundary condition

x = 0, T = 80oC

80 = 0+c2

c2 = 80

at x = L = 0.3m, T =T2

T2 = c10.3+80

T1

T2 = –4oC

x

L=0.3m



According to Fourier law of heat conduction

dx

dTkq [q = heat flux]

L

0

T

T

2

1

kdTqdx

q.L = –k.(T2–T1)

K

L

TTq 21

5.2

3.0

T80700 2

5.2

3.0700T80 2

80–T2 = 84

T2 = –4oC

Comment: This much of temperature

difference, because of very low thermal

conductivity.

: 5 : Test – 3


oil

water

100C

30C

150C

15C

oil

water

75C

T

150C

15C

Before After

Case (I) Case (II)

01. (c)

Sol:

Assumptions :

Radiation effect is neglected

No fouling

Given,

Thi = 150C ,

Tci = 15C

The = 100C case (i)

Tce = 30C

Energy balance for case (i)

)100150(C.m)1530(C.m 0p0pww

100150C.mC.m15 po0pww

powpww Cm50Cm15 ---- (1)

Cmin =0p0 C.m

pwwmax C.mC

For new case:

75150C.m15TC.m 0p0pww

75C.m15TC.m poopww

From equation (1)

poopww Cm15

50Cm

75150Cm15T15

50C.m po0po0

T – 15 =50

1575

T = 37.5C

Exit temperature of water in new case is 37.5C

LMTD for case (i):

30100

15150n

3010015150Im

=

70

85n

7085

=77.25C

Energy balance

U.AI.Im = m0.Cp0(150–100)

5.3770

15150n

5.377515150IIm

=

5.37

85n

5.3785

=2183.0

5.47

m = 58.46C.

Energy balance

U.AII.IIm = m0.Cpo.(150–75) ----------(2)



0.3m

0.3m

l = 10m

Equation (1) (2)

75Cm

50C.m

.A.U

.A.U

0P0

0p0

mII

mI

II

I

75

50

046.58Ld

25.771d

L = 1.9962 m length of new cooler

02(a).

Sol:

Given:

Pr = 0.695

= 1.8510–5 m2/s

k = 0.028 W/m-K

T = 25oC = 298 K

Tw = 65oC = 338 K

2

3L.T..gGr

25

3

1085.1

L2565

2

3382981

81.9

Gr = 3.605109.L3

(Ra)L = Gr. pr

(Ra)L = 2.505109.L3

For upper and lower surfaces,

Characteristic length,

L = 3.0102

3.0104

P

A4 s

= 0.5825m

Ltop = Lbottom = 0.5825m

(RaL)top = (RaL)bottom = 495.257106

For both the vertical sides, characteristic length,

Lvertical = 0.3m

(Ra) L,side = 67.65106

Heat transfer from the top surface:

4/1a )R(54.0uN

L

4/16 )10257.495(54.0 = 80.55

55.80k

L.h top

5825.0

028.055.80h

Km/W8722.3h 2

Qtop = h.AT

= 3.2722100.3(65–25)

Qtop = 464.67W

Heat transfer from bottom surfaces

4/1aL

R27.0uN

4/1610257.49527.0

27.40uN

27.40k

L.h bottom

k27.40h

5825.0

028.0L27.40h bottom

Km/W1937h 2

: 7 : Test – 3


TA.hQ bottom

= 1.937100.3(65–25)

= 232.335W

Heat transfer from vertical sides

4/1aL

R59.0uN

= 0.59(67.656106)1/4

509.53uN

509.53k

h

verticalL

k509.53h

3.0

028.0509.53h

km/W994.4h 2

Heat transfer from both the vertical sides

TAh2Q

= 24.994100.3(65–45)

Qsides = 1198.60W

Total heat transfer from all the sides

Qtotal = Qtop + Qbottom + Qvertical sides

= 464.67+232.335+1198.60

Qtotal = 1895.605W

02(b) (i).

Ans:

(i) In order to predict the behaviour of a water

turbine working under varying conditions of

head, speed, and power, there is need of the

concept of unit quantities. The unit

quantities give the speed, discharge and

power for a particular turbine under a head

of one meter assuming the same efficiency.

The following are the three unit quantities.

(a) Unit speed

(b) Unit power

(c) Unit discharge

Unit speed (Nu) : The speed of the turbine,

working under unit head (say 1m) is known as

unit speed of the turbine.

The tangential velocity is given by,

60

DNu or

D

u60N

If H =1, then HNN u

Where, H = head of water, under which the

turbine is working,

N = speed of turbine under a head,

u = tangential velocity,

Nu = speed of the turbine under a unit head.

Unit discharge (Qu): The discharge of the

turbine working under a unit head (say 1m) is

known as unit discharge.

Q = HKgH2aQ 3

If, H = 1, Then Q = Qu

33u K1KQ

Or HQQ u

Thus,H

QQu



Unit Power (Pu): The power developed by a

turbine, working under a unit head (say 1m) is

known as unit power of the turbine.

Power developed by a turbine is given as,

P = QH and gH2V

HgH2aP

2/32HKP

If H = 1, then P = Pu

Pu = K2 13/2 = K2

P = Pu H3/2

Thus,2/3u H

PP

If a turbine is working under different heads,

the behaviour of the turbine can be easily

known from the unit quantities.

2

2

1

1u

H

N

H

NN

2/32

22/3

1

1u H

P

H

PP

2

2

1

1u

H

Q

H

QQ

Where H1, H2 are the heads under which a

turbine works; N1,N2 are the corresponding

speeds; Q1,Q2 are the discharges, and P1,P2 are

the power developed by the turbine.

02(b) (ii).

Sol: Given data:

Power of the prototype (larger turbine) turbine

(Pp) =7500KW

Head on the prototype Kaplan turbine (Hp)

=10m

Speed of the prototype turbine (Np) =100rpm

Scale Ratio= 1:10

Head on the model turbine (Hm) =4m

Efficiency of both turbine equal= 0.85

Nm, Qm, Pm, Nsm =?

Head coefficient = N.D H

Discharge coefficient = Q D2. H

Power coefficient = P D2. 2

3

H

p

m

pp

mm

H

H

D.N

D.N

10

4

10

1

100

Nm

Nm = 632.46rpm (Speed of the model

turbine)

2

32

.

m

p

m

p

m

p

H

H

D

D

P

P

5.126

4

10

1

10105.7

pm

Pm =18.97 kW

model =mm H.gQ

Pm

0.85 = 4Q81.91000

970,18

m

Qm = 0.569m3/sec

Nsm =4/5)4(

97.1846.632 =486.96

: 9 : Test – 3




03(a) (i).

Sol:

SLNo

Aspects ImpulseTurbine

Reactionturbine

1 Conversionof fluidenergy

The availablefluid energyis convertedinto K.E by anozzle.

The energyof the fluid ispartlytransformedinto K.Ebefore it(fluid) entersthe runner ofthe turbine.

2 Changes inpressure andvelocity

The pressureremains same(atmospheric)throughoutthe action ofwater on therunner

Afterentering therunner withan excesspressure,waterundergoeschanges bothin velocityand pressurewhilepassingthrough therunner.

3 Admittanceof waterover thewheel

Water may beallowed toenter a part orwhole of thewheelcircumference.

Water isadmittedover thecircumference of thewheel

4 Water-tightcasing

Not Required Necessary

5 Extent towhich thewater fillsthe

Thewheel/turbinedoes not runfull and air

Watercompletelyfills all thepassages

wheel/turbine

has a freeaccess to thebuckets.

between theblades andwhileflowingbetween inletand outletsections doeswork on thevanes.

6 Installationof Unit

Alwaysinstalledabove the tailrace. “Nodraft tube isused”.

Unit may beinstalledabove orbelow thetail race-“useof a drafttube isnecessary”

7 Relativevelocity ofwater

Eitherremainingconstant orreducesslightly dueto friction.

Due tocontinuousdrop inpressureduring flowthrough theblade, therelativevelocityincreases.

8 Flowregulation

By meansof a needlevalve fittedinto thenozzle. Possible

withoutlosses

By meansof a guide-vaneassemble. Always

accompanied by loss

: 11 : Test – 3




03(a) (ii).

Sol: Given Data:

H = 250m

= 0.45

Cv = 0.98

m =18

o = 0.76

f = 50Hz

Pair of poles = 6

D = ?

d = ?

Po = ?

Synchronous speed (N) = 60f/P

= 60 50/6 = 500rpm

Speed ratio ():

=gH2

U

0.45 =25081.9260

500

D

D = 1.2 m

Jet Ratio (m) =D/d=18

d= D/18=1.2/18=0.06667m=66.67mm

Discharge through the nozzle

(Q) = Cv A nozzle.VJet

= 0.98 22/7/ 4 (0.06667)2 25081.92

= 16.8m3/sec

Output power of the pelton turbine

= o.gQH

= 0.76.1000. 9.81.16.8.250=31.31MW

03(b).

Sol: d1 = 2cm = 0.02m, 1 = 0.4,

T1 = 600C = 873 K

d2 = 0.06m, 2 = 0.2, T2 = ?, T3 = 300 K

F1–2 = 1, F2–3 = 1

Thermal circuit :

Net radiation exchange between (1) and (2) =

Net radiation exchange between (2) and (3)

22

2

12111

1

42

41

A

1

FA

1

A

1TT.

=

23222

2

43

42

FA

1

A

1TT

11

A

A1

ATT.

22

1

1

142

41 =

2222

43

42

A

1

A

1

A

1TT

11

d

d1

A.TT

22

1

1

142

41 = 22

43

42 ATT

(1)(2)

(3)

Eb1

11

1

A

1

22

2

A

1

22

2

A

1

232FA

1

122FA

1

Eb2 Eb3

: 13 : Test – 3


xc

L

11

d

d1

Ld.TT

22

1

1

142

41 = LdT.T 22

43

42

12.0

1

6

2

4.0

12T873 4

24

= 62.0300T 442

8734 – 42T = 44

2 300T3.2

8734– 42T = 2.3 4

2T –2.33004

3.3 42T = 5.99471011

T2 = 652.85 K

04(a).

Sol:

Tw = 150C, L = 2m, T = 30C,

U = 12 m/s,

= 0.962 kg/m3, = 2.13110–5 Pa-s,

k = 0.031 W/m-k, Cp = 1010 J/kg-k.

Assumptions :

Steady state condition.

Properties of fluid are taken at average

temperature.

Schematic:

Re =510131.2

212962.0L.U.

= 1083435.007 > 5105

The flow is turbulent flow.

Critical Reynolds no. = 5105

5c 105xu

55

c 10510131.2

x12962.0

xc = 0.9229m

L

0

x dxhL

1h =

dx.hdx.h

L

1 L

x

turb

x

0

lam

c

c

Nux = 0.332 3

15.0

x Pr.Re

3

15.0

Pr.xu.P

.332.0K

x.h

3

15.0

5.0

Pr.x.U.

332.0h

hx = 0.332.C.x–0.5laminar

Similarly, hx = 0.0296 C.x–0.5Turbulent

=

8.0

x.C.0296.0K

5.0

xc.332.0K

L

1L

x8.0x

0 c

c

=

2

18.0

x8.0

L3

15.0

x PrReRe037.0Pr.Re664.02

Kcc

= 7729.10167517.4152

K

= 52.14322

031.0

Km/W204.22h 2



LaminarTurbulenth

24.22h

0.92 2x m

w/m2-k

L = 2m

xc = 0.92

U, T

Comment: Average heat transfer coefficient

is less in case of forced connection because

the thermal conductivity of fluid is too small.

The largest local heat transfer coefficient

occur at loading edge of flat plate, where

laminar thermal boundary layer is extremely

thin. And just downstream of xc, where

turbulent boundary layer is thinnest.

04(b).

Sol: Refer to study materials

05(a) (i).

Sol: Distinguish between base load power

plant and a Peak load power plant:

Base load: It is the minimum level of

electricity demand required over a period of

24 hours. It is needed to provide power to

components that keep running at all times

(also referred as continuous load).

Refrigerator and HVAC (Heating ventilation

systems) are examples of base demand.

Peak load: It is the time of high demand.

These peaking demands are often for only

shorter durations. In mathematical terms,

peak demand could be understood as the

difference between the base demand and the

highest demand.

Examples of household loads: microwave

oven, toaster and television are examples of

peak demand,

Power plants are also categorised as base

load and peak load power plants.

Base Load Power plants:

Plants that are running continuously over

extended periods of time are said to be base

load power plant. The power from these

plants is used to cater the base demand of the

grid. A power plant may run as a base load

power plant due to various factors (long

starting time requirement, fuel requirements,

etc.).

Examples of base load power plants are:

1. Nuclear power plant

2. Coal (Thermal)power plant

3. Hydroelectric plant

4. Geothermal plant

5. Biogas plant

6. Biomass plant

7. Solar thermal with storage

8. Ocean thermal energy conversion

: 15 : Test – 3


Peak Load Power plants:

To cater the demand peaks, peak load power

plants are used. They are started up

whenever there is a spike in demand and

stopped when the demand recedes.

Examples of peak load power plants are:

1. Gas plant

2. Solar power plants

3. Wind turbines

4. Diesel generators

Distinguish between Firm power and

Secondary power:

Firm power – The net amount of power

which is continuously available from a plant

without any break on firm or guaranteed

basis.

Secondary power – The excess power

available over the firm power during the off

peak hours or during monsoon season.

Firm (primary) power: Minimum Power that

can be produced by a plant with no risk. For

a single hydroelectric plant, it corresponds to

the minimum availability of storage based.

Firm energy is marketed with high price.

Secondary (Surplus) power: All the power

available in excess of firm power. Secondary

power cannot be relied upon. Its rate is

usually less than that of firm power. It can

be generated ~9 to 14 hours/day.

Firm (primary) power: It depends upon

whether storage is available or not for the

plant since a plant without storage like run-

of-river plants would be produced power as

per the minimum stream flow. For a plant

with storage, the minimum power produced

is likely to be more since some of the stored

water would also be used for power

generation when there is low flow or no flow

in the river.

Secondary (surplus) Power: This is the

power produced by a hydropower plant over

and above the firm power.



05(a)(ii).

Sol: Given data:

Discharge (Q) = 11m3/sec

Head (H) = 16m

Efficiency of the plant () =0.75

Peak load operation time = 5 hours per day

Firm capacity of the plant =?

Case-I:

Withoutpondage (peak load operation)

Output Power of the plant = Efficiency

Hydro power

= 0.75 1000 9.81 11 16

= 1.3 MW

Firm capacity of the plant (Energy generated

for specific duration)

= 1.3 5 = 6.5 MWH per day

= 6.5 365 24 =56,940 MWH per annum

Case-II: With pondage (peak load operation)

Output Power of the plant =

Different efficiencies (Plant efficiency

Evaporation loss factor) Hydro power

= 0.75 0.9 1000 9.81 11 16

= 1.17 MW

Firm capacity of the plant (Energy generated

for specific duration)

= 1.17 5 = 5.85 MWH per day

= 5.85 365 24 = 51,246 MWH per annum

05(b).

Sol:

T0 = 100oC

T = 20oC

kA = 60W/mK

kB = ?

hair = 5W/m2-K

c

c

0 LmcoshxLmcosh

TTTT

Lc = corrected length =4

12.01

4

dL

Lc = 1.003m

(TA)at x = 15cm = (TB)at x = 7.5cm

c

c

c

c

mLcosh

xLmcosh

mLcosh

xLmcosh

d.k

h.4

kA

phm

c

012.060

54m

(m)A = 5.27

BA003.1m.cosh

075.0003.1mcosh003.127.5cosh

15.0003.127.5.cosh

m003.1cosh

m920.0cosh

7595.92

8212.44

A100oC

B

0.012m

1m

: 17 : Test – 3


2034.2

m928.0cosh

m003.1cosh

cosh(1.003m)–2.2034cosh(0.928m) = 0

Using Newton’s Rapson’s method

f = cosh(1.003m)–2.2034cosh(0.928m)

f = 1.003m sinh1.003m – 2.2034 0.928

sinh0.928m

0

00i xf

)x(fxx

After solving, we get

533.10mB

533.10d.k

h4

951.110d.k

h.4

951.110012.0

54k B

kB = 15.02W/m-K

Bcc

Acc

B

A

mLtanhT.PhkA

mLtanh.T.PhkA

QQ

BcB

AcA

mLtanh.k

mLtanh.k

003.1533.10tanh.02.15

003.127.5tanh.60

999.0

9999.09907.1

2Q

Q

B

A

06(a).

Sol: Assumption:

No radiation

Steady state

properties are constants.

D = 10mm = 0.01m

u = 10m/s

T = 23oC

Ti = 75oC

T = 35C

= 8933kg/m3

cp = 388J/kg-K

K = 350W/m-K

Pr = 0.708

= 15.5310–6m2

Kair = 0.025W/m-K

)sphereFor(6

D

3

R

A

VS

D.u

Rde

61053.15

01.010

15.6438Rde

4.03/2e

2/1dd Pr.R06.0R4.02Nu

d

Nud = 48.0445

sphere

u = 10m/s

T = 23oC



70

T

3523

70.48

0445.48k

D.h

fluid

01.0

025.00045.48h

h = 120.11W/m2-K

Biot number

635001.011.120

kS.h

Bisolid

Bi = 0.00057<0.01

Biot number is less than 0.1, lumped

capacity analysis can be applied

According to lumped capacity analysis

T

mC)TT(hA

.mC

hA

TT

T

0 p

T

Ti

.

C.V

hATTn

p

T

Ti

.

VC

hA

i

peTT

TT

.

VC

hA

i peTTTT

38801.08933

611.120

e2335

2375

4.33 = e(0.02079)

= 70.486second

06(b).

Sol: A = 15 cm2

Tw = 200C, T = 30C

(i). Smaller side is vertical

Average temperature =2

30200 = 115C

Taking the property from table at 115C

= 0.91 kg/m3,

Cp = 1009 J/kg-K

= 22.6510–6 Pa-s,

K = 0.0331 w/m-k

= 24.8910–6 m2/s

meanT

1

l

b

: 19 : Test – 3


=

227330273200

1

= 2.5577 10–3/K

2

3L.T..gGr

(here, L = b)

=

26

33

1089.24

b3020010577.281.9

K

C.Pr p =

0331.0

10091065.22 6

= 0.69

Gr.Pr = 4.787109 b3

Nu = 0.59(Gr.Pr)0.25

Nu = 155.193b3/4

4/3b193.155KL.h

h =b

0331.0b193.155 4/3

h = 5.1369b–1/4

Heat transfer

Q = h.A.T

= 5.1369b–1/41510–4(200–30)

Q = 1.31b–1/4 watt

(ii) Longer side is vertical

similarly

Gr = 6.938109 l3 (here, L = l )

Pr = 0.69

Gr.Pr = 4.78109 l3

Nu = 155.193 l3/4

h = 5.1369 l–1/4

Q = 1.31 l–1/4

(Q)smaller side vertical = 1.14(Q)longer side vertical

5.1369b–1/4 = 1.145.1369l–1/4

4/14/1

14.1

b

1

14.1b 4/14/1

6889.1b

= 1.6889b

Area = 15cm2

lb = 15

1.6889.bb = 15

b2 = 8.8812

b = 2.98 cm

l = 5.033 cm

heat transfer when smaller side is vertical.

Q = 1.31b–1/4

= 1.31(0.0298)–0.25

Q = 3.152 watt

Heat transfer when longer side is vertical

14.1

152.3Q

765.2Q Watt



: 21 : Test – 3


07(a).Sol.

Given:

d = 5mm

L = 10cm = 0.1m

(Tb)inlet = 27oC

(Tb)exit = 77oC

Pr = 0.703

Kair = 0.028W/m-K

= 18.2210–6m2/s

= 1.1774kg/m3

cp = 1006J/kg-K

6e 1022.18005.03d.u

R

27.823Rde

For developing flow:-

10

d

z

Rde

d

z

10

703.027.823

10

703.027.823

d

z

z<0.2893m

z<28.93cm

upto 28.93cm, the flow is developing flow,

then the new z = 10cm = length of pipe.

3/1

rez

d

z

P.R3.1Nu d

3/1

5

100703.027.823

3.1

Nuz = 3.991

991.3k

d.h

005.0

028.0991.3h

h = 22.35W/m2K

(i) Total rate of heat removed

= )TT(C.minletexit bbp

2777.Cud4

20 p2

)50(100631774.1005.04

z20 2

= 69.77W

(ii) Heat removed by single conduit

W4885.320

77.69

3.4885 = hAT

3.4885 = 22.350.0050.1T

T = 99.366oC

The exit conduit wall temperature

= 77 + 99.366

= 176.367oC

10cm

u = 3m/s



H

Q

P Inputpower

Max eff

Head

Qoptimu

m

(N = constant)

(iii) Local heat transfer coefficient at the exit

of each conduit

h = 22.35W/m2-K

07(b) (i).

Sol: These curves are obtained by running the

centrifugal pump at the design speed, which

is also the driving speed of the motor. The

design discharge, power and head are

obtained from the corresponding curves,

where the efficiency is maximum as shown

below figure.

07(b) (ii).

Sol: Given Data:

D2 = 2D1

N = 1200 rpm

H = 75m

Vf1 = Vf2 = 3 m/s

D2 = 0.6m,

Therefore

D1 = D2/2 = 0.6/2 = 0.3m

B2 = 0.05m

= ?

W.D/sec = ?

mano = ?

Tangential Velocity of the pump wheel

(Impeller)at inlet (U1):

601

1

NDU

60

12003.0

= 18.85 m/s

Outlet Diagram

1

1

u

Vftan 159.0

85.18

0.3

TW

Ta

x

T= 99.366oC

d

h

22.35x

T

L= 10cm

: 23 : Test – 3


= 9.04 (inlet vane angle)

U2 =60

12006.0

602

ND

= 37.7 m/sec

Assumption: It seems that the impeller vane

bend backward exit angle typing or print

error, in order to get the solution, it may be

assumed exit vane angle as 300

tan30o = 2ww2

f

V7.37

0.3

VU

V

2

2

Vw2 = 32.5m/s

Discharge through the impeller (Q):

Q=D2.B2.2fV

= .(0.6).(0.05).(3.0) = 0.283m3/sec

Work done per second by the impeller

= Q.Vw2.U2

= 1000 0.283 32.5 37.7

= 346.75 KN-m/sec

2wmano uV

gHmn

2

= (9.81 75) / (32.5 37.7)

= 0.6 = 60%



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