: 2 : ESE – Offline Test-2017 - ACE Engineering...

: 2 : ESE – Offline Test-2017

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Entropy

Ent

halp

yT

empe

ratu

re

1

2

3

4

4

y

2

x

4 bar

heatexchanger

1 bar c

1212

hhhh

(h3–h4)=(h3–h4)t

(hx–h2)=(h4–hy)

01(a).

Sol: Heat removed from the system

2T

1T

K1002T

K2001T

2V1 dTT042.0dTCQ

=K100

K200

3

3T

042.0

= J1098K200100K/J3042.0 33333

K100

K200

K100

K200

2Vsystem T

dTT042.0

TdT

CS

= K/J630K200100K/J2042.0 2223

K/J100

W1098T

WQS

3

res

1res

0S .E.Hinfluidworking

ressystemuniv SSS

=100

W1098630

3

Since (S)univ 0

0100

W1098630

3

0630100W

980

350100W

W(max) = 35,000 J = 35 kJ

01(b).

Sol: Refer figure. All the state points may be

determined.

We have T1 = 15 + 327 = 288K,

p1= 1 bar, T3 = 650 + 273 = 923K,

rp = 4 or p2 = 4 bar.

Therefore, T2 = T1 (rp)(-1)/

= 288 × (4)0.4/1.4 = 288 × (4)0.286

= 288 × 1.486 = 428K.

Also

12p

12p

12

12c TTC

TTC

hhhh

Therefore, c

1212

TTTT

Q1–W

W

100 KSystem

T1 = 200 K

HE

Reservoir100 K

Q1

: 3 : ME–Conventional Test-13(Solutions)


= 2888.0

140288

8.0288428

= 175 + 288 = 463K

Similarly

K621673.0924/1923r/1TT 286.0/1p34

and t4343 TTTT

or

85.0)621923(923TTTT t4334

85.0302923 = 623 – 256 = 667K

In the ideal heat exchanger, the temperature of

the compressed air gets raised from 4x2 TTtoT

before combustion chamber. And heat supplied

in the combustion chamber is, therefore, given by

xq3 = (h3 – hx) = Cp(T3–Tx)

=1.0.(923–667)=1.0 ×256 = 256 kJ/kg

Work done on the compressor is given by

2884630.1TTCw 12pcompressor

kg/kJ1751750.1

Work done by the turbine is given by

6679230.1TTCW 43pturbine

kg/kJ2562560.1

Therefore, thermal efficiency of the cycle,

%6.31or316.0256

175256ppliedsuHeat

workNetth

And the work ratio is given by rw

= 316.0256

175256workTurbine

workNet

01(c).

Sol: Consider the streamline as shown in the

diagram

Continuity equation between 2 and 3 gives:

A2V2 = A3V3

s/m2410

604

A

VAV

3

223

Applying Bernoulli’s equation between 1

and 3 we get,

f3

233

1

211 hZ

g2

V

g

PZ

g2

V

g

P

)ZZ(hg2

V

g

P31f

233

= m94.3381.92

60025.0

81.92

24 22

The energy head at inlet of blower is,

m58.4081.92

2494.33Z

g2

V

g

P 2

3

233

The energy head at exit of blower is,

m36.29081.92

240Z

g2

V

g

P 2

4

244

The energy head added by blower is,

3

233

4

244

B Zg2

V

g

PZ

g2

V

g

PH

= 29.36 – (–4.58)

= 33.94 m

(1) (2)

(3)

(4)



Power required to drive the blower is,

85.0

94.3346081.92.1gQHP B

B

kW8.112

01(d).

Sol:

04. u = 2xy – x2 = 2(1)(1) – (1)2 = 1 m/s

s/01212x2y2x

u

s/212x2y

u

v = 2xy – y2 = 2(1)(1) – (1)2 = 1 m/s

s/212y2x

v

s/0112y2x2y

v

y

uv

x

uua x

= (1)(0) + (1)(2)

= 2 m/s2

y

vv

x

vua y

= (1) (2) + (1) (0)

= 2 m/s2

2s/mj2i2a

As the flow is inviscid in it is governed by

Euler’s equation

In x direction

xgx

P

Dt

Du

i.e. (1) (2) = 01x

P

Dt

Dua x

m/Pa2x

P

Similarly in y direction

ygy

P

Dt

Dv

101y

P21

m/Pa12y

P

jy

Pi

x

PP

= m/Pa)j12i2(

01(e).

Sol:

At atmospheric pressure, steam condensers

at 100oC

30

25n

3005

70100

15100n

7010015100LMTD m

m = 52.01oC

cicwwmi TTCp.m.AUe

2300.05L52.81= 0.054180(70–15)

Length of the tube L = 12.049m

100oC

15oC

70oC

100oCs/kg05.0mw

Ui = 230W/m2-oC



418005.0

Ld230

Cpm

AU

C

AUNTU

ww

i

min

i

[considering overall heat transfer

coefficient at outer surface = Ui = 230

W/m2K]

412005.0

049.12025.0230

NTU = 1.04141

effectiveness () = 1–e–NTU = 1–e–1.04141

= 0.647

Rate of steam condensation latent heat =

1cceww TTCpm

157018.405.02257m s

s/kg10093.5m 3s

hr/kg33.18m s

02(a).

Sol: Evaporator temperature = 10–5 = 5C

Condenser temperature = 35 + 5 = 40C

h1 = (hg)5C = 64.20497.148

= 353.61 kJ/kg

h3 = h4 = 238.53 kJ/kg

kg/kJ370h2

Net refrigerating effect = NRE

= 3.51710 = 35.17 KW

Mass flowrate of refrigerant = sec/kgmr

NREkg/kJhhsec/kgm 41r

53.23861.353mr = 35.17

53.23861.353

17.35mr

=08.115

17.35= 0.3056 kg/sec

Efficiency adiabatic = =1

12

12

hh

hh

121

12

hhhh =

9.0

61.353370

21.18hh 12 = 353.61+18.21

= 371.82 kJ/kg

Work of compression (KW)

= kg/kJhhsec/kgm 12r

= 0.3056(18.21)

= 5.565 KW

WC

NRECOP =

565.5

17.35= 6.32

Condenser load (KW)

= kg/kJhhsec/kgm 32r

= 0.3056(371.82–238.53)

= 40.733 KW

Condenser load = NRE +WC

= 35.17 + 5.565

= 40.735 KW

5C

40C3

41

2

22

S

T



02(b).

Sol:

r1 = 0.6m,

r2 = r1 + 0.125 = 0.6 + 0.125 = 0.725m

r3 = r2 + 0.04 = 0.725 + 0.04 = 0.765m

ho = 10 W/m2.K

KA = 0.31 W/m.K

KB = 0.05 W/m.K

th

1

R

TTlossHeat

2765.0210

1725.0765.005.04

725.0765.06.0725.031.04

6.0725.020800

=2157.0

780= 3615.39 watt

Similarly:

Heat loss =

12A

12

21

rrK4rrTT

=

6.0725.031.046.0725.0

T800 2

3615.39 =6.0725.031.04

72800

T2 = 533.31C

Heat loss =

2

3

3o

3

765.02101

20T

Ah1

TT

2

3

765.02101

20T39.3615

T3 = 118.32C

02(c).

Sol: The velocity inside boundary layer is given

by

3

3y

2

1y

2

3Uu

3

2y

2

3

2

3U

dy

du

The wall shear stress at a distance ‘x’ is

given by

0y

w dydu

x

=

x2

U3

=

xRe

x64.4

U

2

3

T1 T2 T3 T

21A

12

rrK4

rr

21B

23

rrK4rr

23r2h

1

r1r2

r33

1

2

AB



b Vbc

a

df

e

= 60120

= 20 r = 17.5 = 48

Vw

Vri

Vro

Vo

Vi

=

xUx1

U64.42

3

i.e. 2/3w U

x64.423

x

------- (1)

The local skin friction coefficient is defined

as

2

fx

U21

xC

=2

2/3

U

2U

x64.42

3

=

xU

64.4/3=

xRe

646.0

The average drag coefficient is defined as

dxCL

1C

L

o

fxD

= dx.xU

646.0

L

1 L

0

=

L

0 x

dx.646.0

UL

1

= L0x2646.0UL

L1

=

U

646.02

=

LU

292.1

=LRe

292.1

03(a).

Sol: Blade speed is given by

sec/m15760

30001

60

DNVb

Heat drop in the fixed blades is given by

10002

VVh

20

2i

fixed

Where,

V0 is the velocity at entry to fixed blades

Vi is the velocity at exit from fixed blades

is the expansion efficiency

is the carry over co-efficient

It may be noted that velocity diagram is

drawn for moving blades and therefore V0

shown there is not the V0 used in the

equation. V0 shown in the diagram is the

absolute velocity of steam leaving the

moving blades and going to second stage, if

any. Thus this V0 need not be equal to 100

m/sec. The values of V0 = 100 and Vi = 250

are fixed blades before the moving blades.

The values are

Vro = 202 m/sec, Vi = 250 m/sec,

Vo = 72 m/sec, Vri = 116 m/sec

Vb = 157 m/sec, Vw = 270 m/sec



Therefore,

9.010002

1009.0250h

22

fixed

Also, hmoving =

10002

VV 2ri

2ro

Where,

Vri is the relative velocity at entry to moving

blades.

Vro is the relative velocity at exit from

moving blades.

The values of Vri is measured from the

velocity triangle bca drawn to scale.

Vri = 116 m/sec

Therefore,

9.010002

1169.0Vh

22ro

moving

Also, degree of reaction is defined as,

fixedmoving

moving

hh

hR

moving

fixed

moving

fixedmoving

h

h1

h

h

R

1

Therefore,

9.010002

1169.0V9.010002

1009.0250

135.0

122

ro

22

22

ro

22

1169.0V

1009.0250186.2

42ro

44

10213.1V

109.01025.686.1

44

2ro 10213.1

86.1

1035.5V

= (2.875 + 1.213) 104

= 4.088 104

Vro = 202 m/sec

Power developed =1000

VVm bw

1000

1572701

= 42.39 kW/kg/sec

fixedmoving

bwstage hh1000

VV

10002

VV 2o

2i

fixed

9.010002

1009.0250 22

9.010002

1035.5 4

= 29.7 kJ/kg

86.1h

h

moving

fixed

Therefore, kg/kJ1686.1

7.29

86.1

hh fixed

moving

(moving + hfixed) = 29.7 + 16 = 45.7 kJ/kg

7.451000

157270stage

= 0.927 or 92.7 %



03(b).

Sol: L = 2.5m, pass = 1,

d0 = 0.025m, di = 0.0225m,

k = 100 W/m-k V = 1.5 m/s,

h0 = 4500 W/m2.k

3060

2060n

30602060LMTD

=

30

40n

3040

= 34.76C

VDR

Le =6109055.0

0225.05.1

LeR 37272.225 > 2000

Flow is turbulent flow

Nu = 4.08.0e Pr.R023.0

= 4.08.0 22.6225.37272023.0

Nu = 216.94

94.216k

D.h i

6078.0

0225.0h i = 216.94

hi = 5860.44 W/m2.k

overall heat transfer coefficient at outer

surface

oo

i

on

iioo A

1

kL2

r

r

Ah

1

AU

1

oo

ii

o

o h

1

k5.22

25n

rh

1

r

r

U

1

=4500

1

100

025.0

5.22

25n

44.5860

1

5.22

25

Uo = 2282.28 W/m2.k

Heat rejected by steam = heat received by

water

m0fg .A.Uhm

23585003600

20000 =2282.28DoLnpm

= 2282.280.0252.5n134.76

n = 841.17

Total no. of tubes , n 842

03(c).

Sol: From 1st law of thermodynamics

Q – W = u

–W = mCv dT

–W = 0.9 0.717 35

(a) W = 22.58 kJ

60C

30C

60C

20C



(b) s2 – s1 =

1

2

1

2v V

VnR

T

TnCm

1

2v12 T

TnCmss

= 0.9 0.717

293

328n

= 0.0728 kJ/K

(c) minimum work = m [(u1 – u2) – T0(s1 – s2)

= 0.9 [(Cv (20 – 55) – 293 (–0.081)]

= 1.2 kJ

(d) Irreversibility = 22.58 – 1.2 = 21.38 kJ

03(d).

Sol: The hydrostatic force on the gate is given by

F = Ahg

Ay

yyg 2

1

= 124219810

= 166.48 kN

The depth of centre of pressure hcp is,

Ah

sinIhh

2G

cp

=

m44.3

124312

45sin241

3

23

Consider the moments produced by self

weight and hydrostatic force as shown in

the diagram

The moment produced by self weight is,

Mw = 150 × 2 = 300 kNm ()

The moment produced by hydrostatic force is,

MF = 2h5F cp

= 166.48 × (5–3.44) × 2

= 367.3 kN-m ()

as MF > Mw the gate will be closed.

04(a).

Sol: No of cylinders = 2

Inlet pressure = P1 = 100kPa

Mass of air = 20 kg/min

Inlet temperature = T1 = 293K

Mass of air / cylinder = 10kg/min

P1V1 = m RT1

100

293287.010

P

RTmV

1

11

= 8.4091m3/min

Pressure ratio = p = 7100

700

P

P

1

2

Work of compression per cylinder

2 m

2 m

hcp

5–hcp45

F

2h5pc



1P

PVP

1n

n n

1n

1

211

1760

10100

13.1

3.1 3.1

13.1

kW22.729

650

TWC= Total work of compression

= 272.22= 144.44kW

Compressor mechanical efficiency

= m = mech = 0.8

Actual work in put to compressor

=m

TWC

kW55.180

8.0

44.144

Work output of engine = Work input to

compressor = 180.55

Brake mean effective pressure (bmep) = 500kPa

N = 2000 rpm

x = No of cylinders = 4

Average volume, N2

Vs (4 stroke engine)

120

xNVbmepPower s

33s m005415.0m

4

02166.0V

Stroke volume of engine cylinder

= 32 m005415.0LD4

1.1D

L

32 m005415.0D1.1D4

m1844.01.1

4005415.0D 3

Stroke length (L) = 1.1D

= 1.1 0.1844 = 0.2028 m

04(b).

Sol: Given:

Pressure on the suction side, PS = 25 cm

of mercury (vacuum)

Pressure on the discharge side,

Pd = 1.4 kgf/cm2

Diameter of suction pipe, ds = 15 cm

Diameter of delivery pipe, dd= 10 cm

Discharge, Q = 30 lit/sec

Efficiency, = 60%

Power required,

mgQH

P

Hm manometric head

= xg2

V

g

P

g2

V

g

P 2ss

2dd

Where x – vertical difference in the levels

of the vacuum and the pressure gauge.

Delivery pressure, Pd = 1.4 kgf/cm2

= 1.4 × 9.81 × 104 N/m2

= 137.34 kPa

m1481.91000

1034.137

g

P 3d

mercuryofcm25g

Ps

= 0.25 13.6



m4.3g

Ps

Discharge, Q = As Vs

s2s

3 Vd4

1030

2

3

s)15.0(4

1030V

Velocity of flow in suction pipe,

Vs = 1.697 m/s

81.92

)697.1(

g2

V 22s

= 0.146 m

2

3

d)1.0(4

1030V

Velocity of flow in delivery pipe, Vd = 3.819 m/s

81.92

)819.3(

g2

V 22d

= 0.7436 m

x = 0.17 + 0.6

x = 0.77 m

Hm = (14 + 0.7436) – (3.4 + 0.146)+0.77

Hm = 11.967m

mgQH

P

6.0

967.11103081.91000P

3

= 5870.107 W

P = 5.87 kW = 7879.3 h.P

04(c).

Sol: Solar pond, also called solar ‘salt pond’, is

an artificially designed pond, filled with

salty water, maintaining a definite

concentration gradient. It combines solar

energy radiation and sensible heat storage,

and as such, it is utilized for collecting and

storing solar energy.

A solar pond reduces the convective and

evaporative heat losses by reversing the

temperature gradient with the help of non-

uniform vertical concentration of salts.

Figure illustrates the principle of solar pond.

The vertical configuration of “salt gradient

solar pond” normally consist of the

following three zones :

1. “Surface (homogeneous) convective zone

(SCZ)”. It is adjacent to the surface and

serves as a buffer zone between

environmental fluctuations at the surface

and conductive heat transport from the layer

below. It is about 10 to 20 cm thick with a



low uniform concentration at nearly the

ambient air temperature.

2. “Lower connective zone (LCG) : It is at the

bottom of the pond, and this is the layer

with highest salt concentration, where high

temperatures are built up.

3. “Concentration / Intermediate gradient zone

(CGZ) : This zone keeps the two convective

zones (SCG and LCG) apart and gives the

solar pond its unique thermal performance.

It provides excellent insulation for the

storage layer, while transmitting the solar

radiation. To maintain a solar pond in this

non-equilibrium stationary state it is

necessary to replace the amount of salt that

is transported by molecular diffusion from

the LCG to SCZ. This means that salt must

be added to the LCG and fresh water to the

SCG whilst brine is removed. The brine can

be recycled, divided into water and salt (by

solar distillation) and returned to the pond.

The major heat loss occurs from the surface

of the solar pond. This heat loss can be

prevented by spreading a plastic grid over

the pond’s surface to prevent disturbance by

the wind. Disturbed water tends to lose heat

transfer faster than when calm.

Due to the excessively high salt

concentration of the LCZ, a plastic liner or

impermeable soil must be used to prevent

infiltration into the nearby ground water or

soil. The liner is a factor that increases the

cost of a solar pond. A site where the soil is

naturally impermeable, such as the base of a

natural pond or lake, or can be made

impermeable by compaction or other means,

will allow considerably lower power costs.

The optical transmission properties and

related collection efficiency vary greatly

and depend on the following factors :

Salt concentration

The quantity of suspended dust or other

particles

Surface impurities like leaves or debris,

biological material like bacteria and algae.

The type of salt.

It becomes obvious that much higher

efficiencies and storage can be achieved

through the utilization of refined or pure salt

whenever possible, as this maximizes

optical transmission.

The solar pond is an effective collector of

diffuse, as well as direct radiation, and will

gather useful heat even on cloudy or

overcast days. Under ideal conditions, the

pond’s absorption efficiency can reach 50%

of incoming solar radiation, although actual

efficiencies average about 20% due to heat

losses. Once the lower layer of the pond

reaches over 60C the heat generated can be

drawn off through a heat exchanger and

used to drive a low temperature organic



0.1 m

L=

0.5

mrankine cycle (ORC) turbine. This harnesses

the pressure differentials created when a

low boiling point organic fluid (or gas) is

boiled by heat from the pond via a heat

exchanger and cooled by a condenser to

drive a turbine to generate electricity. The

conversion efficiency of an organic rankine

cycle turbine driving an electric generator is

5 – 8 %. (which mean 1 – 3% from

insolation to electricity output).

Applications of solar ponds :

Power generation

Space heating and cooling

Crop drying

Desalination

Process heat

Limitation :

Sunny climate

Need for large land area

Availability of salt

Availability of water.

04(d).

Sol:

K333

2322032100

Tmean

K/3331

T1

mean

Grashoff’s number of air

26

3

2

3

109.18

5.0201003331

81.9L.T..gGr

Gr = 818634629.6

Rayleigh number (Ra) = Gr.Pr

= 818634629.6 × 0.696

Ra = 569769702.2

from the given correlation

Nu = 0.1 (Gr.Pr)1/3

= 0.1 (569769702.2)1/3 = 82.902

Nu = 82.902

902.82K

L,h

5.002896.0902.82

LK902.82

h

h = 4.8 W/m2-K

Energy balance – Heat transfer by natural

convention = decrease of internal energy of

plate

plateas

Tc.m)TT(A.h

plate

T3831201001.05.08.4

S/C0501.0T

Rate of cooling of plate min/C3T



v

p

1

5

2

34

pvr=c

(a)

S

T

1

5

2

3

4

v = c

(b)

v = c

when 10 cm side is vertical

Gr L3

Nu (Gr)1/3

Nu (L3)1/3

Nu L

LKhL

h = CK (where C = Constant)

∵ h is in dependent on length of the plate.

Whatever the length of vertical side, heat

transfer will be constant.

Result will not change.

05(a).

Sol:

T1 = 273 + 50 = 323 K

4.01

2

1

1

2 16vv

TT

T2 = 979 K

p2 = bar5.48160.1v

vp 4.1

2

11

K14135.48

70979

pp

.TT2

323

9791413718.0TTcQ 23v32

kg/kJ312

Now 34p4332 TTcQQ

K17231413005.1

312T4

22.114131723

TT

vv

3

4

3

4

1.1322.1

16vv

vv

vv

4

3

2

1

4

5

K6151.13

11723

vv

TT 4.0

1

5

445

bar9.1323615

0.1TT

pp1

515

34p23v

15v

1

2cycle TTcTTc

TTc1

QQ

1

=

312312323615718.0

1

= %5.66or665.0624

292718.01

221

11

m/kN10

K323K.kJkg287.0

p

RTv



QQr

Qa (1)

(2)

1→ plate2→ surrounding

kg/m927.0 3

11

121 v1615

16v

vvv

Wnet = Q1 × cycle = 0.665 × 624 kJ/kg

kg/m927.0

1615

kg/kJ624665.0vv

W.p.e.m

221

net

= 476 kN/m2

= 4.76 bar

05(b).

Sol: K30327330C30T o

Km/W4.17h 2v

lengthwavesolarat9.0

lengthwavelongat1.0

)etemperaturroomat(

K32327350C50T op

428b Km/W1067.5 ,

Qa = 0.9Q

= HT by convection+ HT by radiation

0.9Q = h T +

22

2

21111

1

42

41

A

1

FA

1

A

1TT

= 17.4 (50 – 30) +

2

2

2

1

211

1

42

411

1

A

A

F

11

TTA

∵ A1 <<<A2

0A

A

2

1

F1-2 = 1

The above equation became

0.9 Q =

0111

TTA204.17

1

42

411

= 42

4111 TTA204.17

= 348 + 0.9 5.67 10-8 1 (3234–3034)

0.9 Q = 473.31 Q = 525.9 W/m2

05(c).

Sol: Let Q be the discharge through pump and q

be the discharge through the bypass pipe.

The pump increases the pressure of fluid

therefore flow through the bypass valve

must be in backward direction. (Because

inside valve fluid must flow from high

pressure to low pressure)

By continuity equation,

0.2 + q = Q

q = (Q – 0.2) m3/s ……….. (1)

As the pump and bypass valve are parallel

to each other, head increased by pump and

head loss inside pipe must have same

magnitude.

HP = hL



i.e.100 – 25 Q2 =

g2

A/q2.0

g2

VK

22

L

=81.92

q

1.04

12.0

2

22

i.e. 100 – 25 Q2 = 165.25 q2

100 – 25 (q + 0.2)2 = 165.25 q2

i.e 100 – 25 (q2 + 0.49 + 0.04) = 165.25 q2

i.e 190.25 q2 + 10q – 99 = 0

q = 0.6955 m3/s

and discharge through the pump is

Q = q + 0.2 = 0.8955 m3/s

05(d).

Sol: Yes, According to the correlation

31

t

Pr026.1

For liquid metal

<<t

Pr<<1

Nux = 0.546(Rex.Pr)1/2, Pr0.04, Rex100

For oil

Pr > 1 [ for liquid except liquid metal]

> t

05(e).

Sol: For no heat transfer and negligible exit

velocity, we have (mi – mf)h1 = miui – mfuf

we have, ffiif mmmorm21

m

also gK323Kkg/kNm287.0

m1m/kN2000

RT

Vpm

32

i

iii

k5.21

kg75.102

5.21mmm ffi

2

T323005.175.10 f

fT71.075.1032371.05.21

[∵ Temperature inside the tank changes so

problem can be solved with accuracy by

taking average temperature]

1744.0 + 5.402Tf = 4930.6 – 7.6325Tf

or

K4.2446325.7402.5

6.17446.4930Tf

Thus

kPa7541

4.244287.075.10VRTm

pf

fff

TBL

HBL

x

HBL

TBL

t

x



06(a).

Sol: The velocity at inlet (V1) and exit (V2) is

given by

s/m989.14.0

4

25.0

A

QV

211

s/m958.72.0

4

25.0

A

QV

222

The pressure at exit is obtained by

Bernoulli’s equation as

2

222

1

211 Z

g2

V

g

PZ

g2

V

g

P

i.e., 2122

2112 zzgVV

2PP

5.09810958.7989.12

100010150 223

= 125.22 kPa

Consider control volume as shown in the diagram

The linear momentum equation to the

control volume is

CVinout Vmt

VmVmF

Here last term is zero because flow is steady

In x direction the momentum equation is

Fx = Q (V2x -V1x)

i.e., Rx + P1A1 + P2 A2 = Q (–V2 – V1)

Rx = – P1 A1 – P2A2 - Q (V2 + V1)

989.1958.725.01000

2.04

22.1254.04

10150 223

= –25.27 kN

Similarly the momentum equation is y

direction gives,

Fy = Q (V2y – V1y) = Q (0 – 0) =0

i.e, Ry – Wnet = 0

Ry = Wnet = Wpipe + Wwater

= 500 + 1000 0.1 9.81 = 1481 N

In order to find moment reaction at ‘O’,

angular momentum equation is applied the

control volume as

CVinout Vmrt

VmrVmrE

The moments are taken about ‘O’ and

moment in z’ direction (anticlockwise) is

considered positive.

1122

wnet222111Z

VrQVrQ

rWrAPrAPM

(1) C.G

Ry

RxO

Mz

V2(2)

V1

P1 = 150 kPa

P2 = 125.22 kPa

Wnet



2211wnet222111Z VrVrQrWrAPrAPM

958.73.0989.12.0

25.010002.014812.04

3.01022.1252.04.04

10150

2

323

= 2388.6 Nm

06(b).

Sol:

(i) Supercharging of IC engines

The method of increasing the inlet air

density, called supercharging, is usually

employed to increase the power output of

the engine. This is done by supplying air at

a pressure higher than the pressure at which

the engine naturally aspirates air from the

atmosphere by using a pressure boosting

device called a supercharger.

(ii) Morse Test:

The Morse test consists of obtaining

indicated power of the engine without any

elaborate equipment. The test consists of

making inoperative, in turn, each cylinder of

the engine and noting the reduction in brake

power developed.

This test is applicable only to multi cylinder

engines.

In this test the engine is first run at the

required speed by adjusting the throttle in SI

engine or the pump rack in CI engine and

the output is measured. The throttle rack is

locked in this position.

Then, one cylinder is cut out by short

circuiting the spark plug in the SI engine or

by disconnecting the injector in the CI

engine. Under this condition all the other

cylinders will motor the cut out cylinder and

the speed and output drop. The engine speed

is brought to its original value by reducing

the load.

If there are k cylinders, then

ip1 + ip2 + ip3 + ip4 + ... + ipk =k

1kbp + fpk

………..(1)

where, ip, bp and fp are respectively

indicated, brake and frictional power and

the suffix k stands for the cylinder number.

If the first cylinder is cut-off, it will not

produce any power but it will have friction,

then

ip2 + ip3 + ip4 + …..+ ipk =k

2

bpk + fpk

…...(2)

Subtracting Eqn (2) from Eqn (1)

1

k

2k

k

k1 bpbpip

Similarly we can find the indicated power

of the cylinders, viz., ip2, ip3, ip4,…, ipk

The total indicated power developed by the

engine, ipk, is given by



k

1kk ipip …..(3)

When all the k cylinders are working, it is

possible to find the brake power, bpk, of the

engine.

The frictional power of the engine is given

by

fpk = ipk – bpk …..(4)

(iii) Simple carburetor:

The process of preparation of combustible

mixture by mixing the proper amount of

fuel with air before admission to the engine

cylinder is called carburation. A device

which atomises the fuel and mixes it with

air and prepares the charge for Otto cycle

engine is called carburetor.

Simple Carburetor:

The carburetors are used on most SI engines

for preparation of combustible air – fuel

mixture charge. The figure shows a simple

carburetor, which provides an air – fuel

mixture for normal range at a single speed.

It consists of a float chamber, fuel discharge

nozzle and metering orifice, a venturi, a

throttle valve and a chock valve.

Air-fuel mixture

Idling adjustment

Fuel meteringorifice

Float chamber

Float pivot

Fuel supply valveFuel in

Strainer

Vent to atmosphereIdling airrestriction

Idling air bleed

Air

A AChokevalve

Fueldischarge nozzle

z

B B2

Thottlevalve

Venturi

Fig: The Simple Carburetor

Float



How it works: The basic carburetor as

shown in fig. is built around a hollow tube

called a throat. The downward motion of the

piston creates a partial vacuum inside the

cylinder that draws air into the carburetor’s

throat and past a nozzle that sprays fuel.

The mixture of air and fuel produced inside

the carburetor is delivered to the cylinder

for combustion.

06(c).

Sol: Wein displacement law:- Black body

spectral distribution has a maximum and

that the corresponding wave length max

depends on temperature.

mT = Constant = 2898m.K

From the equation, the maximum

temperature of the body occur at a shorter

wavelength.

Given, T = 5780K

For maximum spectral emissive power

m.T = 2898

m5780 = 2898

m = 0.5013m

Shorter wavelength because of very high

temperature.

06(d).

Sol: Difference between heat pump and

refrigeration cycle

A refrigerator is a device which, operating

in a cycle, maintains a body at a

temperature lower than the temperature of

surroundings.

Performance parameter in a refrigerator

cycle called the coefficient of performance,

abbreviated to COP, which is defined as

COP =W

Q

inputwork

effectDesired 2

∴21

2

QQ

Qref]COP[

…… (1)

A heat pump is a device which, operating in

a cycle, maintains a body, say B at a

temperature higher than the temperature of

the surroundings. By virtue of the

temperature difference, there will be heat

leakage Q1 from the body to the

surroundings. The body will be maintained

at the constant temperature t1, if heat is

discharged into the body at the same rate at

which heat leaks out of the body.

The COP is defined as

COP =W

Q1

∴ [ COP]H.P =21

1

QQ

Q

….. (2)

Relation between COPhp and COPR

From Equations (1) and (2), it is found that

[COP]H.P = [COP]ref +1



The COP of heat pump is greater than the

COP is a refrigerator by unity.

Differences between Kelvin Planck and

Clausius Statements of 2nd Law

The Kelvin-Planck statement of the second

law states: It is impossible for a heat engine

to produce net work in a complete cycle if it

exchange heat only with bodies at a single

fixed temperature.

Clausius’ statement of the second law gives:

It is impossible to construct a device which,

operating in a cycle, will produce no effect

other than the transfer of heat from a cooler

to a hotter body.

Heat cannot flow of itself from a body at a

lower temperature to a body at a higher

temperature. Some work must be expended

to achieve this.

07(a).

Sol:

Heat generation is only in wall A.

Heat conduction equation 1-D

kq

dxTd2

2

Integrate with respect to x

1Cx.k

q

dx

dT

Integrate with respect to x

T = 21

2

CxC2

x.

k

q

----------(1)

Boundary condition, at x = 0, T = TB

TB = 0 + 0 + C2

C2 = TB

At the interface AB:

BB

AA dx

dTk

dx

dTk

12.0

100T230k.Cx.q B

A1

[at interface AB, x = 0]

12.0

100T23024C B

1

C1 = 79.86(TB–100) ----------(2)

At x = 0.1m, T = TC

TC = B1

2

A

T1.0C2

1.0.

k

q

BB

25

C T100T86.791.02

1.0

24

105.2T

TC = 850.68+8.983TB ----------(3)

Total heat generation inside the wall A = Heat

conducted through wall B + Heat conducted

through wall C

2.5 105 0.1 A

kB=230 W/mKKA=24

W/mK KC=200W/mK

(B) (A) (C)

100C

150C

TC

TB

x=00.12m 0.1m 0.15m



=

12.0100T230A B

15.0

A200150TC

A = perpendicular area in the direction of heat

flow

34.5TB + 24TC = 7500 ------(4)

From equation number (3) and (4)

TB = 111.624C

TC = 152.04C

at maximum temperature, 0dxdT

0Cx.k

q1

and from equation (2)

C1 = 79.28(TB–100)

C1 = 928.292 C/m

qkC

x A1

5105.224292.928

x

x = 0.089116 m

at x = 0.089116m, T=Tmax

from equ. (1)

21

2

Amax CxC

2x

.k

qT

624.1110891.029.9282

0891.0

24

105.2T

25

max

Tmax =152.9868C

% of total heat conducted by wall B =

AwallingenerationheatTotal

BwallthroughconductionHeat

=

A5

B

BB

LA105.2

L

100TA.k

=

1.0105.212.0

100624.111230

5

=25000

33.22279 = 89.11%

Hence, 89.11% of the total heat transferred

through wall B and 10.89% of the total heat

transferred through wall C.



07(b).

Sol:

Make point (1) on psychrometric chart at

DBT = 5C, WBT = 2.5CFrom chart:

v1 = 0.792 m3/kg dry air

1 = 0.0032 kg.vap/kg dry air

h1 = 14 kJ/kg dry air

volume flow rate of air = V = 90 m3/min

mass flow rate of air = mv

V

1

792.0

1

60

90m = 1.894 kg/sec

Amount of steam pick up = hr/kg40ms

Latent heat of steam = 2691.3 kJ/kg

Sensible heat load pick up =

kg/kJLHsec/kgmQ sL

= 3.26913600

40 = 29.903 KW

Total heat load pick up = LsL QQQ

= LS13a QQhhm

= 1.894(h3–h4) = 40.7 + 29.903

h3 = 14+894.1

903.297.40

= 14 + 37.28 = 51.28 kJ/kg

s13a mwwm

a

s13 m

mww

=894.13600

400032.0

= 0.009067 kg/vap/kg dry air

h3 = 51.28 kJ/kg and w3

= a.d.kg/vap.kg009067.0

mark this point (3) on psychrometric chart

from chart

DBT = 27.5C; WBT = 18.5C

3

(WBT)3=?

(DBT)3=?

21

WBT=25

DBE 5C



07(c) (i).

Sol: Given data: QActual = 0.018 m3/sec

N = 60 rpm; L = 0.5 m

D = 0.22 m; H = 15 m

(i) Theoretical discharge of the pump:

60

LND4

60

ALNQ

2

Theritical

60

605.022.04

2

= 0.019 m3/sec

(ii)Slip (S) =The

ActualThe

Q

QQ

019.0

018.01 = 0.0526

Percentage of slip of the pump = 0.0526 100

= 5.26%

07(c) (ii).

Sol: Given:

Head at the base of Nozzle, H = 32 m

Discharge, Q = 0.18 m3/s

Area of jet, Aj = 7500 mm2

Power available at shaft, Pshaft = 44 kW

Mechanical efficiency, m = 94%

To find:

(i) Power lost in Nozzle

(ii) Power lost in Runner

(iii) Power lost in mechanical friction

(i) Power lost in nozzle:

Power supplied at inlet of turbine

kW1000

gQH

1000

3218.081.91000

Water Power =56.505 kW

Kinetic energy of jet per second,

2Vm2

E.K = 2VAV

2

1 2VQ

2

1

2V18.010002

1E.K

Discharge, Q = Area of jet × Velocity of jet

6107500

18.0

A

QV

V = 24 m/s

22418.010002

1E.K

kW84.51s/E.K

Power cost in Nozzle = Water power (K.E/s)

= 56.505 51.84

(Plost)Nozzle = 4.665 KW

(ii) Power cost in Runner:

Mechanical efficiency,

powerRunner

turbineofshaftatPowerm

P.R

P.Sm

Where,

S.P Shaft power;



R.P Runner power

P.R

4494.0

Runner power94.0

44

= 46.808 kW

Power lost in runner =

K.E/s Runner power

= 51.84 46.808 = 5.031 kW

(iii) Power lost in mechanical friction:

Power lost in mechanical friction = Runner

power –shaft power

= 46.808 44

= 2.808 kW

08(a).

Sol:

km/W475U 20

Latent heat = 600kJ/kg – k

d = 25mm = 0.025m

L = 4.87m

V = 2m/s, Cpw = 4.18kJ/kg-K

w = 1000kg/m3

Heat loss by oil = Heat gain by water

500600 = 36004.18(T–286)

T = 305.93K

93.305355286355

93.305355286355LMTD m

063.49

69n

063.4969

m = 50.466oC

286TCmAUwpwms0

466.50pnL.d475

202693.305418060

3600

n.p = 470.56

mass flow rate of water AVmw

Vd4

nm 2w

Vd4

n60 2

21000025.04

n60 2

n = 61.115

n62 tubes

n.p = 470.56

p = 7.699 pass

p 8 pass

355k

286k

min/kg3600mw

min/kg500mO



10 m50 m

500 m

g

60

08(b).

Sol:

For rectangular channel the diameter is

taken as hydraulic diameter (Dh) as

m67.1610502

10504

P

A4Dh

The velocity of the flow is,

s/m52001050

106.2

A

QV

6

= 5.2 × 10-3 m/s

The Reynolds number of the flow is,

3

63h

10t8.0

1067.16102.51000VDRe

= 0.108

Flow is laminar

For laminar flow through inclined pipe

pressure drop is given by

gzpPwhereD

VL32P *

2*

21221 D

VL32ZZgPP

i.e. 12221 ZZgD

VL32PP

60cos105009810

1067.16

10500102.5108.032

6

26

633

= 239.5 + 2.45

= 241.95 Pa

(% of pressure drop due to gravity)

= %01.110095.241

45.2

for laminar flow friction factor is function

of Reynolds number only and is given by

6.592108.0

64

Re

64f

08(c).

Sol:

(i). Pyranometer : A pyranometer is a device

used to measure the “total hemispherical

solar radiation”. The total solar radiation

arriving at the outer edge of the atmosphere

is called the ‘solar constant’.

The working principle of this instrument is

that sensitive surface is exposed to total

(beam, diffuse and reflected from the earth

and surrounding) radiations. ‘

The description of pyranometer is given

below.



Construction: It consists of a “black

surface” which receives the beam as well

diffuse radiations which rises heat. A “glass

dome” prevents the loss of radiation

received by the black surface. A

“thermopile” is a temperature sensor, and

consists of a number of thermocouples

connected in series to increase the

sensitivity. The “supporting stand” keeps

the black surface in a proper position.

Working: When the pyranometer is

exposed to sun, it starts receiving the

radiations. As a result, the surface

temperature starts rising due to absorption

of the radiation. The increase in the

temperature of the absorbing surface is

detected by the thermopile. The thermopile

generates a thermo emf which is

proportional to the radiations absorbed, this

thermo emf is calibrated in terms of the

received radiations. This will measure the

global solar radiations.

(ii) Pyrheliometer : A pyrheliometer is a

device used to measure “beam or direct

radiations”. It collimates the radiation to

determine the beam intensity as a function

of incident angle.

This instrument uses a collimated director

for measuring solar radiation from the sun

and from a small portion of the sky around

the sun at normal incidence.

The description of a thermoelectric type

pyrheliometer is given below:

Construction: In this instrument, two

identical blackened manganin strips A and

B are arranged in such a way that either can

be exposed to radiation at the base of

collimatimg tubes by moving a reversible

shutter.



Working : One strip is placed in radiation

and a current is passed through the shaded

strip to heat it to the same temperature as

the exposed strip. When there is no

difference in temperature, the electrical

energy supplied to shaded strip must equal

the solar radiation absorbed by the exposed

strip. Solar radiation is then determined by

equating the electrical energy to the product

of incident solar radiation, strip area and

absorptance.

(iii) Sunshine Recorder : A Sunshine recorder

is a device used to measure the “hours of

bright sunshine in a day”.

The description of a sunshine recorder is

given below :

Construction: It consist of a ‘glass-sphere’

installed in a section of “spherical metal

bowl” having grooves for holding a recorder

card strip and the glass sphere.

Working : The glass-sphere, which acts as

a convex lens, focuses the sun’s rays/ beam

to a point on the card strip held in a groove

in the spherical bowl mounted

concentrically with the sphere.

Whenever there is a bright sunshine, the

image formed is intense enough to burn a

spot on the card strip. Through the day, the

sun moves across the sky, the image moves

along the strip. Thus a burnt space whose

length is proportional to the duration of

sunshine is obtained on the strip.

08(d).

Sol: Given:

Head, H = 30 m

Discharge, Q = 10 m3/s

Speed of runner, N = 300 rpm

Peripheral velocity of wheel at inlet,

gH29.0u1

Radial Velocity of flow, gH23.0Vf

Overall efficiency, 0 = 80%

Hydraulic efficiency, h = 90%

To Find:

(i) Power developed

(ii) Diameter and width of runner at inlet

(iii) Guide blade angle at inlet

(iv) Inlet angle of runner vane

(v) Diameter of runner at outlet



(i) Power developed:

Overall efficiency =gQH

P

P = o × gQH

= 0.8 × 1000 × 9.81 × 10 × 30

P = 2354.4 KW

(ii) Diameter and width of runner at inlet:

Given, gH29.0u1

gH29.060

ND1

gH29.0N

60D1

3081.929.0300

60D1

D1 = 1.39 m

Discharge,

Q = Peripheral Area × Velocity of flow

Given velocity of flow at inlet,

gH23.0V1f

3081.923.0V1f

1fV = 7.278 m/s

Q = D1B1 ×1f

V

10 = × 1.39 × B1 × 7.278

B1 = 0.3146 m = 31.46 cm

(iii) Guide Blade angle at inlet:

2 inlet angle of runner vane.

Hydraulic efficiency of turbine is given by

gH

UVUV 2w1wh

21

Given that the discharge is radial at outlet

Whirl velocity,2wV at exit is zero,

0V2w

gH

UV 1wh

1

1

hw u

gHV

1

3081.929.0u1

u1 = 21.834 m/s

834.21

3081.99.0V

1w

s/m131.12V1w

Since1wV < u1, angle 1 is acute

1

1

w1

f1 Vu

Vtan

131.12834.21

278.7

1 = 36.87

tan1

1

1

w

f

V

V

131.12

278.7

1 = 30.96

1 1

1rVV1

1fV

u1

1wV

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