mechanical engineering - paper i...mechanical engineering - paper i f-126, katwaria sarai, new delhi...

MECHANICAL ENGINEERING - Paper I

F-126, Katwaria Sarai, New Delhi - 110 016 011-41013406, Mobile: 8130909220, 9711853908Ph:

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Mains Exam Solution

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ASTER

SECTION–A

Q.1: (a) A main pipe divides into two parallel pipes which again form as one pipe. The length anddiameter of the first parallel pipe are 1000 m and 0.8 m respectively, while the length anddiameter of the second parallel pipe are 1000 m and 0.6 m respectively. Find the rate of flowin each parallel pipe, if total flow in the main is 2.5 m3/sec. The coefficient of friction foreach parallel pipe is same and equal to 0.005. [12 Marks]

Sol:

1000m

1000m

2

3

11

L2 = L3 = 1000 m,

d2 = 0.8m, d3 = 0.6m

Q1 = Q2 + Q3 = 2.5 m3/s ...(i)

Coefficient of friction, Cf = 0.005

Friction factor, f = Cf × 4 = 0.02For parallel pipes 2 and 3,

‘hf’ will be same.

hf2 = hf3

2

2 2

2

fL V2gD

=2

3 3

3

fL V2gD

22

22 2

QA D

=23

23 3

QA D

Q A V

2252

QD =

2353

QD

22

5Q

0.8=

23

5Q

0.6

Q2 = 2.053 Q3

From (i)

2.05 3Q3 +Q3 = 2.5

Q3 = 0.819m3/s

Q2 = 1.681 m3/s

Q.1: (b) A reversible engine works between three thermal reservoirs, A, B and C. The engineabsorbs an equal amount of heat from the thermal reservoirs A and B kept at temperaturesTA and TB respectively, and rejects heat to the thermal reservoir C kept at temperature TC.The efficiency of the engine is times the efficiency of the reversible engine, which worksbetween the two reservoirs A and C.




Mains Exam Solution

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ASTER

Prove that: A A

B C

T T( 2 1) 2(1 )T T

[12 Marks]

Sol: TA TB

TC

QC

E1 W

QA Q =B QATA

TC

QC

E2 W

QA

1 = A C C2

A A

T T T1T T

...(i)

1 =A C C

A A A

2Q Q QW 12Q 2Q 2Q

...(ii)

CA B

A B C

QQ QT T T

= 0

C

C

QT =

A BA

A B

T TQ

T T

...(iii)

C

A

QQ =

A BC

A B

T TT

T T

From (i) (ii) and (iii)C

A

T1T

= C

A B

1 1T1T T2

C

A

TT

= C C

A B

T T12T 2T

C

A

T2 2T

=C C A

A A B

T T T2T T T

A

C

T2 2T

=A A

C B

T T2 1T T

A

B

TT = A

C

T22 1 1 T

Q.1: (c) With the help of neat sketch, explain the working of a thermostatic expansion valve. Howdoes it cope up with the variable load? [12 Marks]

Sol: An expansion device in refrigeration system reduces the pressure of refrierant liquid from condenserpressure to evaporator pressure. In refrigeration system, these devices have two purposes one ispressure reduction and other is control function i.e. supply of liquid refrigerant to evaporator at




Mains Exam Solution

IES M

ASTER

evaporation rate. Point to be noted here is that capillary tube does not perform second functioni.e. control and perform only first one. The thermostatic expansion valve (TEV) serve both function.So details of TEV are–

Capillary tube for power fluid.

Bellows

FPFP

Fe

Fe

Pe

Fs

Pc

Screw

Adjustablespring

Needle fororifice

High Pressureliquid fromcondenser

EvaporatorSuction tube/line

Superheatedrefrigerant tocompressor

(Feeler Bulb)

Bulb containingpower fluid.

TEV maintains a constant degree of superheat at exit of evaporator. TEV is the most versatile and most commonly used expansion valve in refrigeration system. By ensuring a constant degree of superheat at evaporator exit, it protects the compressor from

liquid entry. The TEV has a feeler bulb at exit tube of evaporator so that it senses the temperature at exit of

evaporator. The feeler bulb is connected to top of bellow by a narrow tube. The feeler bulb and narrow tube contain some fluid called as power fluid. This power fluid may

be same as refrigerant or some other. Under normal conditions, the pressure of power fluid Pp is the saturation pressure corresponding

to temperature at exit of evaporator.

Force on bellows due to power fluid,

Fp = p bP A

Fs – Spring force that can be adjusted according to need/requirement

Fe = evaporator force

Fe = e bP A

In steady state, Fp = Fs + Fe

Fs = (Fp – Fe)

In given situation or for a particular setting, Fs is constant.

Fp – Fe = constant

Again Pp – Pe is also constant.

Since (Pp – Pe) is constant, this means (Tp – Te) i.e. degree of superheat is constant.

Hence for given setting of spring the TEV maintains a constant pressure difference between evaporatorand power fluid in power line i.e. (Pp – Pe) - constant. So constant degree of superheat.

It degree of superheat is required to be varied, it varied by adjusting spring force by screw i.e.

superheatT 1 eP P

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Mains Exam Solution

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ASTER

Response of TEV to Load Variation: (Control function)

Load variation means how the TEV regulate the flow rate of refrigerant as load on evaporator increaseor decrease.

(a) Increase in Refrigeration load: As load of refrigeration increases, the rate of evaporation in-creases, the degree of superheat try to increase.

Since degree of superheat sT is proportional to (Pp – Pe), So pressure of power fluid in feelerbulb try to increase. This increase force Fp on bellows and orifice needle moves down and morearea for flow of refrigerant is available at orifice, so mass flow rate of refrigerant increases.

This increased mass flow rate restores the pressure difference (Pp – Pe) so the degree of super-heat.

(b) Decrease in Refrigeration load: Due to reduced load on evaporator, less refrigerant boils whichreduces the temperature in suction line i.e. degree of superheat reduces. This reduction in degreeof superheat results in reduced pressure in power fluid ‘Fp’. The situation is

Fp < Fs + Fe

Due to this reduction in Fp a new balance point at higher level of orifice needle is achieved. Thehigher level of orifice needle reduces area of flow. This reduction in flow area reduces the massflow rate of refrigerant.

So TEV always makes a balance point in each case of load i.e. increased or decreased load.

The refrigerant flow rate is always proportional to load. So we can say that TEV is highly effectiveand behaves like close loop control.

Q.1: (d) The fuel rod of a nuclear reactor is lagged with a tight fitting cladding material to preventoxidation of the surface of the fuel rod by direct contact with the coolant. The heat generation

occurs only in the fuel rod according to the following relation: qg = 2

0 2rq 1R

. Under

steady state conditions, heat generated in the fuel rod is conducted through the claddingmaterial and then dissipated to the coolant flowing around the cladding by convection.

Assuming that there is no contact resistance between the fuel rod and cladding, derive anexpression for the heat flux through the fuel rod and cladding material. [12 Marks]

Sol: The heat generated per unit volume in the nuclear fuel rod is given by

CladdingNuclear fuel rod

tw

R

RclO

Temperatureprofile

Nuclear cylindrical fuel rodwith cladding

tfr

tcl




Mains Exam Solution

IES M

ASTER

qg =2

0rq 1R

where, qg = Heat generation rate at radius r,

q0 = Heat generation rate at centre of the rod (r =0), and

R = Outer radius of fuel rod

Rcl = Outer radius of cladding,

qcl = Heat generation rate in the cladding,

qfr = Heat generation rate in the fuel rod,

kfr = Thermal conductivity of fuel rod material, and

kcl = Thermal conductivity of cladding material.

One-dimensional steady state heat conduction in radial direction is given by

gqd dt rrdr kdr

= 0 ...(i)

Also, heat flow rate per unti area, QqA

is given by

q = dt dt qk ordr dr k

Substitutingdtdr

=q in eqn. (i) we havek

gqd rq rdr kk

= 0 (Assuming k to be constant)

or, drq

dr= gq r ...(ii)

or frd r qdr

= gq r ...for fuel rod

or, frd r qdr

=2

0rq r1R

...(iii)

Since there is no internal heat generation in cladding, therefore

cld r qdr

= 0 ...(iv)

Integrating eqn (iii), we obtain

frr q =2 4

0 12r rq C2 4R

...(v)

or, qfr =3

10 2

Cr rqr2 4R

...(vi)

Now, integrating eqn. (iv), we have

clr q = C2 ...(vii)

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Mains Exam Solution

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ASTER

or, qcl = 2Cr

...(viii)

In order to evaluate C1 and C2 (integration constants), using the following boundary conditions, we have

At r = 0, qfr = finite 1C 0 ...[From equation (v)]

At r = Rfr qfr = qcl

C2 = frq R ...[From eqn. (vii)][Substituting the value of qfr from equation (vi)]

=3

0 2R Rq R2 4R

or, C2 = 20qR

4

The rate of heat flow through fuel rod, qfr = 3

frfr 0 2

dt r rk qdr 2 4R

...(ix)

and the rate of heat flow through the cladding, qcl = 2cl 0cl

dt qk R

dr 4r ...(x)

Q.1: (e) Compare compression ignition engine with spark ignition engine so far as the followingpoints are concerned:

(i) Working cycle

(ii) Method of ignition

(iii) Method of fuel supply [12 Marks]

Sol: (i) Working cycle:

3

4

1

2

P

V

4

32

1V

P

(a) The ideal cycle for the SI engine is the Otto cycle.

(b) Heat addition at constant volume.

( c) Compression ratio is low i.e. 6 to 10.

(d) Complete cycle on PV coordinate.

(a) It works on diesel or dual combustion cycle.

(b) Heat addition takes place at constant pressure in diesel cycle and In dual cycle at constant pressure and constant volume.

( c) Compression ratio is high i.e. 12 to 16.

(d) Diesel cycle on PV co-ordinate.

SI CI

(ii) Method of ignitionSI engine: Air fuel mixture is ignited at the end of compression stroke by means of an electricspark because SI engine fuel have high ignition temperature. So, a spark plug is necessary.




Mains Exam Solution

IES M

ASTER

CI engine: CI engine works with high compression ratio and ignition temperature is low comparedto SI fuels. So, in CI engine self ignition take place at the end of compression stroke.

(iii) Method of fuel supplySI engine: In SI engine a combustible fuel air mixture is prepared outside the engine cylinder. Thisprocess is known as carburation. During the suction stroke air fuel mixture is supplied by carburator.

CI engine: Only air is drawn into the cylinder during suction stroke and compressed to highpressure. Fuel is injected into the cylinder at the end of compression stroke so a fuel injector isrequired for injection of fuel and carburator is replaced by the injection system.

Q.2: (a) A jet of water is discharging at 25 kg/sec from a nozzle of 25 mm diameter. The jet fromthe nozzle is directed towards a window of a building at a height of 30 m from the ground.Assuming the nozle discharge to be at a height of 2 m from the ground, determine thegreatest distance from the building where the fireman can stand, so that the jet can reachthe window. [20 Marks]

Sol: Velocity of water at jet,

v = 2 2Q 4 m 4 25 50.9m/sA 1000d 0.025

x

2m

y = 28m

v

v = v cos x v = v sin y

In x-direction

Sx = 2x x

1v t a t2

x = v cos t 0

cos =xvt

sin =2x1

vt

In y-direction

h = 2y y

1v t a t2

, where ay = –g m/s2

28 = 21v sin t gt2

28 =2

21xvt 1 gt2vt




Mains Exam Solution

IES M

ASTER

28 = 2 2 21(vt) x gt2

x2 = 2

2 2128 gtvt2

x =2

2 21(vt) 28 gt2

For maximum x,dxdt

= 0

2 21v t gt28 gt2

= 0

v2 = 2 2128g g t2

t2 = 2

2

2v 28gg

t =2

2(50.93) 28 9.81) 2 6.942

9.81

The greatest distnace from the building,

x =2

2 2 2150.93 6.942 234.74 m28 9.81 6.9422

Q.2: (b) Two rigid tanks shown in Figure 2(b) each contain 10 kg of N2 gas at 1000 K, 500 kPa. Theyare now thermally connected to a reversible heat pump, which heats one and cools theother with no heat transfer to the surroundings. When one tank in heated to 1500 K, theprocess stops. Find the final (P, T) in both tanks and the work input to the heat pump,assuming constant heat capacities. [20 Marks]

W

AN2

BN2

QA QBH.P

Figure 2(b)

Sol:

QA QBW

QA10 kg/W2

B10 kg/W2

T = 1000KA1

T = ?A2

T = 1000KB1

P = 500KT = 1500K

B1

B2

Assumptions: (1) Reversible heat pump

(2) 2' ' 1.4 for N




Mains Exam Solution

IES M

ASTER

For tank ‘B’

TdS = dU + pdV

TdS = dU Rigid tank

BS =B2

B1

Tv T

dTmCT = v

1500mC ln1000

Similarly AS = A2v

TmC ln1000

Now univverseS 0

Reversible heat pump is there

universeS = 0

mCvln A2v

T 1500mC ln10001000

= 0

A2v 2

T 1500mC ln1000

= 0

A22

T 15001000

= 1

TA2 = 666.67 K

Pressure in both tanks

Since volume is constant

P T

So,A1

A1

PT =

A2

A2

PT

PA2 =A1 A2

A1

P T 500 666.67T 1000

PA2 = 333.335 kPa

Similarly PB2 =B1 B2

B1

P T 750kPaT

A 666.67K 333.335 kPaB 1500K 750 kPa

Tank Final Temperature Final Pressure

RN2 =R 8.314 0.297 kJ/kg-KM 28

Now Cv =R 0.297

1 0.4

= 0.7425 kg/kg.K

So, Work input = QB – QA

= mCv (TB2 – TB1) – mCv (TA1 – TA2)

= 10 × 0.7425 [(1500 – 1000) – (1000 – 666.67)]

= 1237.52 kJ

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Mains Exam Solution

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ASTER

Q.2: (c) Water is flowing steadily over a smooth flat plate with a velocity of 2 m/sec. The length ofthe plate is 30 cm. Calculate(i) The thickness of the boundary layer 10 cm from the leading edge of the plate.(ii) The rate of growth of the boundary layer at 10 cm from the leading edge; and(iii) The drag coefficient on one side of the plate.Assume parabolic velocity profile.Kinematic viscosity of water = 1.02 × 10–6 m2/sec.Derive the expressions used in the calculation. [20 Marks]

Sol: Given: V = 2m/s, L = 30 cm

Reynolds number,

Re = 5 56

VL 2 0.3 5.88 10 5 101.02 10

So Laminar boundary layer

The parabolic velocity profile is 2u y y2

U

Now momentum thickness, = 0

u u dy1U U

=2 2

0y y y y dy2 1 2

=2 3 2 3 4

0y y y y y y dy2 4 2 2

=2 3 4

0y y y y dy2 5 4

=3 52 4

2 3 40

2 5 4 1y yy y3 52 4

=5 1 23 5 15

From Von Karman momentum equation

0 =2

2 d 2 U dUdx 15 dx

...(i)

From Newton’s law of viscosity,

0 =y 0

uy

=22

y 0

d y y2U Udy

=y 0

2Uy2U




Mains Exam Solution

IES M

ASTER

=2 U

...(ii)

From equation (i) and (ii), we get

22 dU15 dx

=

2 U

d =15 dx

U

2

2

= 15 x CU

Boundary condition: = 0 at x = 0, we get C = 0

Boundary layer growth,x

=x

30µ 5.48ρUx Re

Boundary layer thickness, =ex

5.48xR ...(iii)

From equation (ii) and (iii), we get

0 = ex2 U R5.48x

Cfx =2

exU RU 0.36492 x

Cfx =ex

x

R0.7299

U

or Cfx =ex

0.7299R

The average drag coefficient,

Cf =L L 1/20 0

ex

1 0.7294 0.7299 x dxL L VR

=eL

1.46R

(i) Now thickness of boundary layer at 10 cm from leading edge of plate

= 3

ex6

5.4x 5.4 0.1 1.22 10 mR 2 0.1

1.02 10

(ii) Rate of growth of boundary layer at 10 cm from leading edge of plate ddx

=1/2

x

5.4x 5.4 x 5.4 xRe ux v

ddx

=

1 12

x

15.4 x 2.7 2.72v vx Re




Mains Exam Solution

IES M

ASTER

at x = 10 cm = 0.1 m

=3

6

2.7 6.097 10 m / m2 0.1

1.02 10

(iii) Drag coefficient, Cf =3

5eL

1.46 1.46 1.9 10R 5.88 10

Q.3: (a) A four-stroke cycle gasoline engine has six single-acting cylinders of 8 cm bore and 10 cmstroke. The engine is coupled to a brake having a torque radius of 40 cm. At 320 rpm, withall cylinders operating, the net brake load is 350 N. When each cylinder in turn is renderedinoperative, the average net brake load produced at the same speed by the remaining 5cylinders is 250 N. Estimate the indicated mean effective pressure of the engine. With allcylinders operating, the fuel consumption is 0.33 kg/min; calorific value of fuel is 43 MJ/kg; the cooling water flow rate and temperature rise is 70 kg/min and 10°C respectively. Ontest, the engine is enclosed in a thermally and acoustically insulated box through whichthe output drive, water, fuel, air and exhaust connections pass. Ventilating air blown upthrough the box at the rate of 15 kg/min enters at 17°C and leaves at 62°C. Draw up a heatbalance of the engine stating the items as a percentage of the heat input. [20 Marks]

Sol: 4 Stroke gasoline engineK = 6 (No. of cylinders)

D = 0.08m (Bore)

L = 0.1 m (Stroke)

B.P = T = F

=2 3200350 0.4

60

= 46.914 kW

Net brake load produced by remaining 5 cylinders,

B.P = T = 2 N250 0.460

= 33.51 kW

I.P = 6 BP 6 B.P = 80.424 kW

Now I.P =N KmeP L A2 60

( For strokes are there in one cycle)

80.424 × 1000 = 2 3200meP 0.1 60.084 120

80.424 × 1000 = meP × 0.0804

meP =80.424 1000

0.0804

meP = 10.003 bar

Now, Heat input = fm calorific value

= 30.33 13 1060

= 236.5 kW

Heat lost to coding water = w pm C T

=70 4.187 1060

= 48.848 kW

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Mains Exam Solution

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ASTER

Other losses (Exhaust connection, etc)

= a,2 a,1a p,air T Tm C

= 15 1.005 62 1760

= 11.306 kW

Heat balance sheet

( )Heat input 236.5 100

Heat equivalent to B.P 46.914 19.837Heat lost to cooling water 48.848 20.655

Exhaust Losses 11.306 4.78Unaccounted Loss 128.932 54.51

Head kW Percentage of heat

Q.3: (b) A simple saturation refrigeration cycle uses R134a as refrigerant. The refrigeration systemoperates at 40°C condenser temperature and –16°C evaporation temperature respectively.If a liquid vapour heat exchanger is installed in the above simple saturation refrigerationcycle, find the COP and power per ton of refrigeration. The outlet vapour of heat exchangeris 15°C temperature. [20 marks]

Evaporator

Heat Exchanger

Figure 3(b)

ExpansionValve (EV)

Compressor

Condenser

Sol: A simple saturation vapor cycle

Condenser5 3

2

6

7 1

H.E

Evaporator

Note: Superheatingof vapor is not taking

place in the evaporator.

T

40°C

15°C

–16°C

6 5 4

7 1

2

3

s

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Mains Exam Solution

IES M

ASTER

h1 = 389.02 kJ/kg

h4 = 419.43 kJ/kg, h5 = 256.41 kJ/kg

h2 – h1 = C1(T2 – T1)

= 0.831 × (15 + 16)

h2 = h1 + 0.831 × 31

h2 = 414.781 kJ/kg

Now, S3 = S2

3

4 p4

TS C lnT

=2

1 p1

TS C lnT

3T1.7111 1.145 ln313

=273 151.7379 0.831 ln273 16

T3 = 348.02 K

h3 = h4 + Cp (T3 – T4)

= 419.43 + 1.145 × (348.02 – 313)

= 459.53 kJ/k

Now, h5 – h6 = h2 – h1

256.41 – h6 = 0.831 × 31

h6 = 230.649 kJ/kg

Refrigeration effect, Q0 = h1 – h7 = h1 – h6

= 309.02 – 230.649

= 158.371 kJ/kg

Work input, W = h3 – h2 = 459.53 – 414.781

= 44.749 kJ/kg

COP = 0Q 3.539W

Power per ton of refrigeration

P/TR =0

W 0.283Q

or Horse power per ton of refrigeration

HP/TR = 1.345

Q.3: (c) Moist air at 28°C DBT and 20.6 WBT and 101.325 kPa barometric pressure flows over acooling coil and leaves it at a state of 10°C DBT and with specific humidity 7.046 gm/kgof dry air.(i) If the air is required to offset a sensible heat gain of 2.35 kW and a latent heat gain

of 0.31 kW in a space to be air-conditioned, calculate the mass of dry air which mustbe supplied to the room in order to maintain a DBT of 21°C in the room.

(ii) What will be the relative humidity in the room?(iii) If a sensible heat gain diminishes by 1.175 kW but latent heat gain remains unchanged,

at what temperature and moisture content must the air be supplied to the room?Take specific capacity of air as 1.012 kJ/kg, K, latent enthalpy of water at 21°C is 2454 kJ/Kg. Show the processes on the psychrometric chart. [20 Marks]




Mains Exam Solution

IES M

ASTER

Sol:

Roomi

DBT = 21°C

1 2

DBT = 10°C = 7.046 gm/Kg of dry air

DBT = 28°CWBT = 20.6°C

(i) RSHF = RSH 2.35 0.883

RSH RLH 2.35 0.31

From graph;

RSH RLH

1

2(0.883)

h1 = 40 kJ/Kg of d.a.

h2 = 28 kJ/Kg of d.a.

Now, a 1 2m (h h ) = 2.35 + 0.31

am = 0.222 Kg/s

(ii) From graph RH in room = 49%.

(iii) When RSHF diminishes by 1.175 kW

New, RSHF =2.35 1.175 0.791

(2.35 1.175) 0.31

Room conditions are also same.i.e. 21°C, 49% RH.

New supply conditions are 16°C DBT, 60% RH.

2

1

(0.791)

RLH

(Sam

e)

(i)




Mains Exam Solution

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Q.4: (a) A geothermal source provides 10 kg/s of hot water at 500 kPa, 150°C flowing into a flashevaporator that separates vapour and liquid at 200 kPa. FInd the three fluxes of availability(inlet and two outlets) and the irreversibility rate. Take ambient temperature as 25°C.

[20 Marks]

2

3

Vapour

Liquid

1

Sol: At 200 kPa for vapour,

hg = (hg)at 200 kPa = 2706.63 kJ/kg

s2 = (sg)at 200 kPa = 7.1271 kJ/kgK

At 200 kPa for liquid,

h3 = (hf)at 200 kPa = 504.68 kJ/kg

s3 = (sf)at 200 kPa = 1.53 kJ/kgK

At 500 kPa, 150°C (Tsat = 151.86°C)

h1 = (hf)500 kPa – 4.1868 × (151.86 – 150)

= 640.21 – 4.1868 × 1.86 = 632.42 kJ/kg

s1 = (sf)at 500 kPa – 4.1868 × ln 425.01123.15

= 1.842 kJ/kgK

Tsat = 151.86 + 273.15 = 425.01K

T1 = 150 + 273.15 = 423.15 K

Availability at inlet

1 = (h1 – T0s1) – (h0 – T0s0)

= (h1 – h0) – T0 (s1 – s0)

At ambient condition of 25°C, T0 = 273.15 + 25 = 298.15 K

Assuming enthalpy and entropy as 0 at 0.01°C (triple point)

h0 = 4.1868 × (298.15 – 273.16)

= 104.628 kJ/kg

s0 = 298.154.1868 ln 0.3665 kJ/kgK273.16

1 = (632.42 – 104.628) – 298.15 × [1.842 – 0.3665]

= 527.79 – 439.92 = 87.87 kJ/kg

Availability of outlet vapour,

2 = (h2 – h0) – T0(s2 – s0)

= (2706.63 – 104.628) – 298.15 × (7.1271 – 0.3665)




Mains Exam Solution

IES M

ASTER

= 2602 – 298.15 × 6.7606 = 586.33 kJ/kg

Availbaility of outlet liquid

3 = (h3 – h0) – T0 (s3 – s0)

= (504.68 – 104.628) – 298.15 × (1.53 – 0.3665)

= 400.05 – 298.15 × 1.1635

= 53.15 kJ/kg

1m = 10 kg/s

1m = 2 3m m 10

and 1 1m h = 2 2 3 3m h m h

10 × 632.42 = 2 3m 2706.63 m 504.68

2 2m 2706.63 504.6810 m = 6324.2

2m 5046.82706.63 504.68 = 6324.2

2m =6324.2 5046.8 1277.4 0.58kg/s

2706.63 504.68 2201.95

3m = 210 m 9.42 kg/s

Flux of availability at inlet = 1 1m = 10 × 87.87 = 878.7 kW

Flux of availability of outlet liquid = 3 3m

= 9.42 × 53.15 = 500.67 kW

Irreversibility rate I = 1 1 2 2 3 3m m m

= 878.7 – 340.07 – 500.67 = 37.96 kW

Q.4: (b) Air at a mean velocity of 20 m/sec flows through a 2 cm diameter tube whose surface ismaintained at 200°C. The temperature of air as it enters the tube at inlet is 20°C and leavesthe tube at 180°C. Determine(i) The length of the tube required to heat the water from 20°C to 180°C, and(ii) The pumping power required to maintain the flow.Assume f = 0.3164/(ReD)1/4.

Properties of air at the mean film temperature fT : = density = 0.8345 kg/m3; specific heat = cp = 1015 J/kg K; dynamic viscosity, = 2.3825 ×

10–5 kg/m.s; Pr = 0.703;

thermal conductivity, k = 0.034425 W/mK. [20 Marks]

Sol: Given: Friction factor, f = 0.3164/(Re)1/4

V = 20m/s Air

T = 20°C = 293Ki2

2cm

T = 200°C = 473 Kw

T = 180°C = 453Ke

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Mains Exam Solution

IES M

ASTER

Properties of air at mass film temperature, Tf = i e2T T2

= 20 180 100 C are:

2

P = 0.8345 kg/m3, Cp = 1015 J/kgK, 52.3825 10 kg/ms , Pr = 0.703, K = 0.034425 W/mK

Reynolds number, Re = 5VD 0.8345 20 0.02 14010 2000

2.3825 10

So, turbulent flow

Friction factor, f = 1/4 1/4

e

0.3164 0.3164 0.0290814010R

For turbulent flow through pipe,Nu = 0.023 Re0.8 Pr0.4

h 0.020.034425

= 0.8 0.40.023 (14010.49) (0.703)

h = 71.37 W/m2 K

Mass flow rate of air, m = AV

= 20.8345 0.02 204

= 5.24 × 10–3 kg/s

Heat recieved by air, Q = p e imC T T

= 5.24 × 10-3 × 1015 × (180 –20)

= 851.51 W

where mT = i e

i

T T 200 20 200 180T 200 20nn

200 180T

= 72.82°K

Also Q = mhA T

or 851.51 = 71.37 0.02 L 72.82

L = 2.607 m

Pumping power, W = fP Q g h Q

=2

ff lVmg h mg2gD

=3 25.24 10 0.02908 2.607 20 3.972 W

0.02 2

Q.4: (c) A single-cylinder, single-acting reciprocating compressor using R12 as refrigerant has a bore 80 mmand stroke 60 mm. The compressor runs at 1450 rpm. If the condensing temperature is 40°C, findthe performance characteristics of the compressor when the suction temperature is –10°C. Specificheat of vapour at 40°C is 0.759 kJ/kg K.

Assume the simple cycle of operation and no clearance. [20 Marks]




Mains Exam Solution

IES M

ASTER

Sol: Bore, D = 80mm = 0.08 m

Stroke, L = 60 mm = 0.06 m

N = 1450 rpm,

v1 = 0.0767 m3/kg

h1 = 183.2 kJ/kg

s1 = 0.7020 kJ/kgK

2h = 203.2 kJ/kg

2s = 0.6825 kJ/kgK

s2 =2

2 pv 2 12

Ts C ln , But s sT

T

3

4 –10°C

h

2

1

40°C

2

0.7020 = 2T0.6825 0.759 ln313.15

2Tln313.15

=0.7020 0.6825 0.02569

0.759

2T313.15

= 0.02569e 1.026

T2 = 321.29 K

h2 = 2 pv 2 2h C T T

= 203.2 + 0.759 × (321.29 – 313.15) = 209.38 kJ/kg

Mass rate of refrigerant, m =

2

1

ND L4 60

v

=

20.08 0.06 14504 0.095 kg/s

60 0.0767

Volume flow rate at compressor inlet

=

22

0.08 0.06 1450N 4D L4 60 60

= 7.2885 × 10-3 m3/s = 0.4373 m3/min

Refrigerating effect = (h1 – h4)

h4 = h3 = (hf)40°C = 74.6 kJ/kg

Refrigerating effect = 183.2 – 74.6 = 108.6 kJ/kg




Mains Exam Solution

IES M

ASTER

Total refrigerating capacity, o(Q ) = 0.095 × 108.6

= 10.317 kW

Power required to drive the compressor, (W) = 2 1m h h

= 0.095 × (209.38 – 183.2) = 2.487 kW

COP =Total refrigerating capacity

Power required

=10.312 4.152.487

p = pko

W 0

Design point

PeakPower

maxW

p = po,mina

m 0W 0

po

pk = constant

oQ

Fig. Performance characteristics of a reciprocating compressor

From fig. generally refrigeration systems operate on the left-hand side of this curve. But just afterstarting, the compressor passes through the peak power. The compressor motors are, therefore, oversizedto enable them to take the peak load during pull-down. The starting current is more than the runningcurrent.

SECTION–BQ.5: (a) A single-cylinder, single-acting, square reciprocating pump has piston diameter and stroke

length of 300 mm. The pump is placed such that the vertical distance between the center-line of the pump and sump level is 5 m. The water is being delivered at a height of 22 mabove the centerline of the pump. The suction and delivery pipes are 8 m and 28 m longrespectively, and diameter of both the pipes is 150 mm. If the pump is running at 30 rpmand coefficient of friction for suction adn delivery pipes is 0.005, estimate the theoreticalpower required to drive the pump (kW). [12 Marks]

Sol: Given:

Piston dia (D) = 300 mm

Stroke length (L) = 300 mm

Hs = 5 m




Mains Exam Solution

IES M

ASTER

Hd = 22 m

Ls = 8 m

Ld = 28 m

ds = dd = 150 mm

N = 30 rpm

fs = fd = 0.005

P = Q H

H = s df f s d2 (h h ) H H3

sfh =

2s s s

s

4f l V2g d

where Vs = 2A 300 2 30, wr 0.15

a 150 60

sV 1.88 m/s

=24 0.005 8 (1.88)

2 9.81 0.15

sfh 0.192 m

and dfh =

24 0.005 28 (1.88)2 9.81 0.15

dfh 0.672 m

H = s df f s d2 (h h ) H H3

=2 (0.192 0.672) 5 223

H 27.567 m

Hence, P = 9.81 × Q × (27.567 m)

where, Q = 2

3ALN (0.3) 0.3 30 0.0106 m /s60 4 60

P = 9.81 × 0.0106 × 27.567

P 2.8648 kW




Mains Exam Solution

IES M

ASTER

Q.5: (b) Show that the diagram work per unit mass of steam for maximum blading efficiency of a50% reaction stage is Vb

2, where Vb is the mean blade velocity.[12 Marks]

Sol: The Velocity Diagram of reaction turbine with usual notations is,

A

Vr12

EV2

C BVb

V

Rise in velocity in moving blade

D

1

V1 Vr2

Assuming no friction aV = 0 i.e. no change in axial componentof velocity.

For impulse turbine Vr2 = kbVr1 = CE

Due to reaction in moving blades, the velocity in moving bladesincreases by ED.

For 50% reaction and similar geometry.

= 2 1 1 r2 r1 2; and V V , V V (AB = CD) (AC = BD)

Since 1 2 . So blades are unsymmetrical.

Since aV = 0 So no axial thrust due to axial velocity component. But the thrust will be present dueto pressure difference across each rotor disc.

So whirl velocity or power velocity component across the rotor—

V = 1 2 r1 1 r2 2V cos V cos V cos V cos

= 1 b 1 2 1 r2V cos V V cos ( V V )

= 1 b 1 2V cos V V cos ( )

= 1 b2V cos V

So, Diagram Work (WD), WD = s bm V V

WD = s b 1 bm V 2V cos V

So diagram work per unit flow D

s

Wm = b 1 bV 2V cos V

So energy input to blades per kg of steam—

=

2 22 r2 r11

V V1 V2 2

=

2 22 1 r11 1 r2

V V1 V ( V V )2 2

= 12

r21

VV

2From ABC using cosine rule, 2

r1V = 2 21 b 1 bV V 2V V cos

Input energy per unit flow

Input energy = 2 2 21 1 b 1 b

1V V V 2V V cos2

= 2 2

1 b 1 bV V 2V V cos2

So diagram efficiency D =

b

2 21 b 1 b

2 V .VV V 2V V cos

D =

b 1 b

2 21 b 1 b

2V 2V cos V

V V 2V V cos




Mains Exam Solution

IES M

ASTER

For maximum efficiency the blade speed, Vb = V1 cos

The expression for maximum efficiency,

=

2

22cos

1 cosVelocity Diagram for maximum efficiency condition & R = 50%

Vr1

1

Vr2

2

V1V2

Vb

V

Vb = V1 cos

V = bV

WD = 2bV

Hence in this condition, the blade inlet and exit angles are right angle. Whirl component of velocity isequal to blade speed.

Q.5: (c) Derive an expression for efficiency of a combined cycle where two thermodynamic cyclesare coupled in series. The expression should be derived in terms of efficiencies of thecoupled cycles. Coventional notations may be used. [12 Marks]

Sol:

Topping cycle

Pump

CB Q2

Bottoming cycle

Q3

Pump

WT2WT1

Q1

T

a

S

Q1

Q2

Q2

Q3

Two vapor power cycle in series.

I = 322

1 2

QQ1 and 1Q Q

Q2 = 11 Q1 and Q3 = Q2 21

Now = total 1 3

input 1

W Q QQ Q

= 3

1

Q1Q

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Mains Exam Solution

IES M

ASTER

= 2 2

1

Q 11

Q

= 1 1 2

1

Q 1 11

Q

= 1 21 1 1

= 1 2 1 21 1

= 1 2 1 2

Q.5: (d) Explain with neat sketch hwo solar absorption refrigeration system works for space cooling.[12 Marks]

Sol: SOLAR COOLING AND REFRIGERATION• Absorption refrigeration cycle is used for space cooling which requires energy input as heat which

in turn provided by Sun. There is also seasonal matching between the energy needs of the spacecooling system and availability of solar radiation.

• Commonly used refrigerant absorbant combinations are ammonia-water and water-lithium bromide.

A schematic of solar operated absorption system is shown below:

Flat-platecollector array

Hotwater

1. Generator2. Condenser3. Expansion valve4. Evaporator5. Absorber6. Liquid pump7. Heat exchanger

2

Coolingwater

1

7

45

6

Cooled space

Coolingwater

3

Water heated in a flat plate collector array is passed through a heat exchanger called the generator,where it transfer heat to a solution mixture of absorbant and refrigerant, which is rich in refrigerant.Refrigerant vapour is boiled off at a high pressure and goes to condenser where it is condensed intoa high pressure liquid. The high pressure liquid is throttled to a low pressure and temperature in anexpansion value and passes through evaporator coil. Here, the refrigerant vapour absorber heat andcooling is therefore obtained in the space surrounding the coil.

The refrigerant vapour is now absorbed into a solution mixture withdrawn from the generator, which isweak in refrigerant concentration. This yields a rich solution which is pumped back to the generator,thereby completing the cycle. The rich solution flowing from the absorbant to generator is usually heatedin a heat exchanger by the weak solution withdrawn from the generator. This will help in improvingperformance of cycle.




Mains Exam Solution

IES M

ASTER

Q5: (e) How do fuel cells work? Explain the principle with the help of a sketch.[12 Marks]

Sol: Principle of Operation of Fuel Cells:A fuel cell is an electro-chemical device in which the chemical energy of fuel is continuously convertedinto electric energy. This conversion of energy takes place at constant pressure and temperature.

The main components of a hydrogen (H2) – oxygen (O2) fuel cell are:(1) A fuel electrode (anode)(2) An oxidant electrode (cathode)(3) An electrolyte (a solution of H2SO4 for acidic fuel cell and KOH for alkalil fuel cells)(4) Additional components are container, separators, sealings, fuel and oxidant supply etc.

The basic feature of a fuel cell is that the fuel and the oxidant are combined in the form of ionsthan in form of neutral molecules.

Fig. represents the schematic diagram of a fuel cell using hydrogen as fuel and oxygen as oxidantand alkaline solution of KOH as electrolyte.

Heat

– +

Electrolytesolution

40%KOH

Permeable NiElectrode (anode)

Permeable NiElectrode (cathode)

SpentO2

O2in

H2

SpentH and H O

vapour2 2

A hydrox (H – O ) fuel cell22

It is called alkaline fuel cell (AFC). It consists of two permeable Nickel electrodes immersed inan electrolyte of good conductivity. An electrolyte may be an alkaline solution of KOH as shownor acidic solution of H2SO4 called respectively as alkali fuel cells and acidic fuel cells.The porous fuel electrode is anode (negative pole) and the other porous oxidant electrode iscathode (positive pole). These electrodes are separated by a porous gas barrier called separator(not shown in Fig.)The anode is supplied H2 gas as fuel at a certain pressure and the cathode is supplied O2 asoxidant at a pressure. These gases pass through the respective electrodes and bubble aroundthrough the electrolyte solution. The pores provide an opportunity to gases, electrodes and electrolyteto come in contact for their electrochemical reactions. The electrodes are connected through anexternal circuit as shown.

The electro-chemical reactions are generally slow and a catalyst is required in the electrodes toaccelerate the reaction. Platinum is the best catalyst but costly. In general, less expensive catalystslike nickel and silver are used according to application and design.




Mains Exam Solution

IES M

ASTER

Chemical reactions with alkaline H2–O2 fuel cell:The hydrogen gas is ionized at anode and it produces a free electron and H+ ions. Everyhydrogen molecules brought to electrode surface is dissociated into two H atoms by catalyticproperty of electrode. These enter into electrolyte solution as hydrogen ions leaving behind twoelectrons which pass through the external circuit to the cathode (positive electrode) The reactionat anode is as follows:

Anode: 2H 2H 2e ...(i)

The oxygen supplied to cathode (positive electrode) reacts with water of electrolyte and theelectrons transmitted to it to produce hydroxyl (OH¯) ions. Thus,

Cathode : 2 21 O H O 2e 2OH¯2

...(ii)

These hydroxyl ions migrate from cathode to anode through electrolyte. The hydrogen and hydroxylions then combine in the electrolyte to produce water i.e.

22H 2OH 2H O ...(iii)

Above equation shows that OH ions produced at one electrode (cathode) are involved in areaction at the other electrode (anode). By adding the above three equations, the overall processis chemical reaction of H2 and O2 gases to form water i.e.

2 2 21H O H O2

...(iv)

Thus the net reaction of a fuel cell in which hydrogen and oxygen supplied is the produce water,electrical energy and heat.

Chemical reaction with acidic electrolyte (H2SO4):

Anode : 2H 2H 2e ...(v)

Cathode : 2 21 O 2H 2e H O2

...(vi)

Overall reaction : 2 2 21H O H O2

...(vii)

Q.6: (a) A centrifugal pump has an impeller diameter at outlet as 1 m and delivers 1.5 m3/s of wateragainst a head of 100 m. The impeller is running at 1000 rpm. The width of the impeller is85 mm. If the manometric efficiency is 85%, determine the type of impeller (forward, radialor backward curved), and the blade angle at outlet. Draw velocity triangle at outlet.

[20 Marks]

Sol:

Vw2

u2

2 2

Vf2Vr2V2




Mains Exam Solution

IES M

ASTER

D2 = 1.0m, N = 1000 rpm, Hm = 100m

b = 85mm, Q = 1.5m3/sec

u2 = 2D N 1 100060 60

=

50 m/sec 52.35 m/sec3

Assuming given width of impeller, b = 85mm is at exit,

Q = 2 2 f 2D b V

Flow velocity, Vf2 =2 2

Q 1.5 5.617 m/secD b 1 0.85

Since manometric efficiency, (assuming radial entry at impeller inlet)

m =m

w2 2

gHV u

Whirl velocity at exit of impeller,

Vw2 =m

2 m

gH g 100 22.04 m/secu 52.35 0.85

Blade angle at exit,

2tan = f 2

2 w2

V 5.617 0.1853u V 52.35 22.04

2 = 10.50°

Type of impeller is backward curved.

Q.6: (b) Consider the combined gas steam power cycle shown in the figure. The topping cycle isa gas turbine cycle that has a pressure ratio of 8. Air enters the compressor at 300 K andthe turbine at 1300 K. The isentropic efficiency of the compressor is 80 percent, and thatof the gas turbine is 85 percent. The bottoming cycle is a sample ideal Rankine cycleoperating between the pressure limits of 7 MPa and 5 kPa. Steam is heated in a heatexchanger by the exhaust gases to a temperature of 500°C. The exhaust gases leave theheat exchanger at 450 K. Determine(i) The ratio of the mass flow rates of the steam and the combustion gases.(ii) The thermal efficiency of the combined cycle.

T

3500°C7 MPa

7 MPa

5 kPa

52

141

2

4

450

300

31300

SAssume specific heat of gas as 1.005 kJ/kg K. [20 Marks]




Mains Exam Solution

IES M

ASTER

Sol: Assumption: The heat rejected by the exhaust gases of gas turbine is the only heat source for Rankinecycle.

T

3500°C7 MPa

7 MPa

5 kPa

52

141

2

4

450

300

31300

S

2

4

Gas turbine:

1T 300 K , 3T 1300 K , rp = 8

2

1

TT =

1

p(r )

2T =0.41.4300 8 543.43 K

comp. 0.80 =2 1

2 1

T TT T

2T = 604.29 K

3

4

TT =

1

3

4

PP

4

1300T =

0.41.4(8)

4T = 717.66 K

turbine 0.85 =3 4

5 4

T TT T

4T = 805.01 K

Assuming air gm m

Wnet = p 3 4 p 2 1C T T C (T T )

= 1.005 (1300 805.01) (604.29 360)

= 191.65 kJ/Kg of air




Mains Exam Solution

IES M

ASTER

Rankine cycle

Let mass flow rate of gases in Gas turbine is gm Kg / s

Then a 4 5 3 2 steamm 1.005 T T (h h ) m

h3 = 3411.4 kJ/Kg

h2 = h1 + vdP

= 137.75 + 0.001005 × (7 × 103 – 5)

= 144.78 kJ/Kg

steam

g

mm =

1.005 (805.01 450)3411.4 144.78

(i)steam

g

m 0.1092m

Wnet, steam = steam 3 4m (h h )

Now, S3 = S4

5.8148 = 0.4762 + x × 7.9176

x = 0.674

h4 = h1 + x × hfg

= 137.75 + 0.674 × 2423.0

= 1770.852 kJ/Kg

Wnet,steam = g0.1092 m (3411.4 1770.852)

net,steam

g

Wm = 179.18 kJ/Kg of air

So, Wnet,total = 191.65 + 179.18

= 370.83 kJ/Kg of air

combined =net, total

input

WQ

=370.83

1.005 (1300 604.24)

= 53%

Q.6: (c) What is Betz limit for wind turbines ? Derive an expression for Betz limit for wind turbines.[20 Marks]

Sol: Energy from wind stream is extracted by a wind turbine, by converting the kinetic energy (K.E.) of thewind to rotational motion required to operate an electric generate.

In order to compute the mathematical relationships, let us make the following assumptions:(i) The flow of wind is ‘incompressible’, and hence the air stream diverges as it passes through the

turbines.

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Mains Exam Solution

IES M

ASTER

(ii) The mass flow rate of wind is ‘constant’ at far upstream, at the rotor and at far down stream.Let, p = Atmospheric wind pressure,

pus = Pressure on upstream of wind turbine,pds = Pressure on downstream of wind turbine,Uw = Atmospheric wind velocity,

(Uw)us = Velocity of wind upstream of wind turbine,(Uw)bl = Velocity of wind at blades,(Uw)ds = Velocity of wind downstream of wind turbine before the wind front reforms and

regains the atmospheric level.Abl = Area of blades,

m = Mass flow rate of wind, and

= Density of air.

The kinetic energy of wind stream passing through the turbine rotor is given by:

K.E. = 2w bl

1 m(U )2

And, m = bl w blA (U )

K.E. = 2bl w bl w bl

1 A (U ) (U )2

= 3bl w bl

1 A (U )2

The force on the rotor disc, F is given as:F = (pus – pds)Abl ...(i)

Also, F = w wus dsm U U ...(ii)

Applying Bernoulli’s equation to upstream and downstream sides, we get:

2w us

1p (U )2

= 2us w bl

1p (U )2

and 2ds w bl

1p (U )2

= 2w ds

1p (U )2

Solving the above equations, we obtain:

pus – pds =

2 2w us w ds

1 (U ) (U )2

...(iii)

Equating eq. (i) & (ii), we get:

(pus – pds)Abl = w wus dsm U U = bl w w wbl us dsA U U U ...(iv)

Solving eq. (iii) & (iv), we get:

2 2

w wus ds1 U U2

= w bl w us w ds(U ) (U ) (U )

or, (Uw)bl = w wus dsU U

2

In a wind turbine system “speed flow work”, W, is equal to the difference in kinetic energy between

upstream and downstream of turbine for unit mass flow, m = 1.




Mains Exam Solution

IES M

ASTER

Therefore, W = (K.E.)us – (K.E.)ds

or, W = 2 2

w us w ds1 (U ) (U )2

The power output ‘P’ of wind turbine (rate of doing work) is given as:

P = 2 2

m wus ds1 m U U2

=

2 2

w wus dsU Um

2

=

2 2w us w ds w us w ds

bl(U ) (U ) (U ) (U )

A2 2

w us w dsbl w bl bl

(U ) (U )m A (U ) A

2

or P = 2 2

bl w w ds w wus us ds1 A U (U ) U U4

...(v)

The get Pmax (maximum turbine output), differentiating eq. (5) w.r.t. (Uw)ds and equating to zero, we get:

w ds

dPd(U ) = 2 2

w ds w us w ds w us3(U ) 2(U ) (U ) (U ) 0

The above quadratic equation has the following two solutions.

(Uw)ds = w us1 U3 and (Uw)ds = (Uw)us

For power generation (Uw)ds < (Uw)us' so we can have only

(Uw)ds = w us1 U3

We get: Pmax =

22

bl w us w us w us w us1 1 1A (U ) (U ) (U ) (U )4 3 3

=

2bl w us w us

1 4 8PA (U ) (U )4 3 9

or, Pmax = 2bl w us

8 A (U )27

= 3

bl w us16 1 A U27 2

= 3

bl w us10.593 A U2

Total power in the wind, stream, Ptotal = 3bl w us

1 A (U )2

Pmax = 0.593 Ptotal

Now, “coefficient of power”, Cp = maxPP

= 0.593

The factor 0.593 is known as Blitz limit.




Mains Exam Solution

IES M

ASTER

Q.7: (a) A Pelton turbine with a wheel diameter of 1.5 m, operating with four nozzles, produces 16MW of power. The turbine is running at 400 rpm and operating under a gross head of 300m. Water is supplied through penstock of length 2 km. The coefficient of friction in penstockis 0.004. There is 10% of head loss taking place in the penstock. If the velocity coefficientis 0.97, blade velocity coefficient is 0.9, overall efficiency is 0.84 and Pelton bucket deflectsthe jet by 165°, determine(i) Discharge through the turbine (m3/s)(ii) Penstock diameter (m)(iii) Jet diameter (m)(iv) Hydraulic efficiency of the turbineDraw velocity triangles. [20 Marks]

Sol: Velocity triangles

V1

V2

2

V1rV

2rVVF

2 = 180 – 165 = 15°

= Deflection angle = 165°

Blade velocity (v) =DN 1.5 400 31.415 m/s60 60

(i) Discharge through turbine (Qtotal)

overall =P (shaft)

Water power at Penstock end

0.84 =6

net total

16 10 Wg H Q

Hgrass = 300 m

Hnet = Hpenstock = Hgross – 0.1 × Hgross = 270 m

0.84 =3

total

16 101000 9.81 270 Q

3totalQ 7.1913 m / sec

(ii) Penstock diameter (dp)

Head loss in penstock = 300 – 270 = 30 m

(hf) =2

2 58flQ30

d g




Mains Exam Solution

IES M

ASTER

where, f = 4 × coefficient of friction

= 4 × 0.004 = 0.016

30 =3 2

2 5p

8 0.016 2 10 (7.1913)(d ) 9.81

dp = 1.354 m

(iii) Jet diameter (dj)

Qper jet = totalQ 7.1913 1.797825Number of nozzle 4

Qper jet = Ajet × Vnozzle

Vnozzle = V1 = v netheadC 2g H

V1 = 70.599 m/s

1.797825 =2jd 70.599

4

dj = 0.18 m

(iv) Hydraulic efficiency

hhydro. =1 2w w

nethead

V(V V )g H

where 1w 1V V 70.599 m/s

1r 1V V V 70.599 31.415

1rV 39.184 m/sec

2rV KVn where K = 0.9 (Given)

2rV = 0.9 × 39.184 = 35.2656

2 2w r 2V V cos V

= 35.2656 × cos 15 – 31.415 = 2.64855 m/sec

hhydro =31.415 (70.599 2.64895)

9.81 270

hydroh 0.8687 86.87%

Q.7: (b) What do you mean by compounding in steam and gas turbines ? What are the variousmethods of compounding in steam and gas turbines? Explain all the methods with neatsketch. [20 Marks]

Sol: Compounding is a method for reducing the rotational speed of the impulse turbine to practical limit. Ifsteam at high pressure is allowed to flow through one row of moving blades, it produces a rotor speedof about 30000 rpm which is too high for practical use. Not only this, the leaving loss is also very high.So to reduce rotational speed and losses, the transfer of energy from steam to turbine is done in steps.

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Mains Exam Solution

IES M

ASTER

This is possible by making use of more than one set of nozzles, blades, rotors in series, keyed to acommon shaft, so that either the steam pressure or the jet velocity is absorbed by the turbines in stages.The leaving losses also will be less. This process is called compounding of steam turbines. Thecompounding is done in three ways. These are done by three types:

(a) Pressure Compounded Impulse Turbine Or Rateau StagingIn this type of turbine, the compounding is done for pressure of steam only i.e. to reduce the highrotational speed of turbine the whole expansion of steam is arranged in a number of steps by employinga number of simple turbine in a series keyed on the same shaft. Each of these simple impulse turbineconsisting of one set of nozzles and one row of moving blades is known as a stage of the turbine andthus this turbine consists of several stages. The exhaust from each row of moving blades enters thesucceeding set of nozzles. Thus we can say that this arrangement is nothing but splitting up the wholepressure drop.

From the steam chest pressure to thecondenser pressure into a series of smallerpressure drop across several stages of impulseturbine and hence this turbine is called,pressure-compound impulse turbine. Thevariation of pressure and velocity at variousstages is shown in figure below:

Stea

m p

ress

ure

ches

t

Velo

city

Con

dens

er p

ress

ure

N M N M N M N M N M

(b) Velocity-Compounded Impulse Turbineor Curtis StagingIn this type of turbine, the compounding is donefor velocity of steam only i.e. velocity drop isdone in small drops through many moving rowsof blades instead of a single row of movingblades. It consists of a set of nozzles and rowsof moving blades attached to the rotor and rowsof fixed blades attached to casing.

The fixed blades are guide blades whichguide the steam to succeeding rows ofmoving blades, suitably arranged betweenthe moving blades and set in a reversedmanner.

In this turbine, three rows or rings ofmoving blades are fixed on a single rotorand this type of rotor is termed as thethree row rotor.

Stea

m p

ress

ure

ches

t

Velo

city

Con

dens

er p

ress

ure

N M G M G MM-moving bladesG-guide blades

The whole expansion of steam from thesteam chest pressure to the exhaustpressure takes place in the nozzles only.

There is no pressure drop in either in the moving blades or the fixed i.e. the pressure remainsconstant in the blades as in the simple impulse turbine.

The steam velocity from the exit of the nozzle is very high as in the simple impulse turbine. Steam with high velocity enters the first row of moving blades and on passing through these

blades, the Velocity slightly reduces i.e. the steam gives up a part of its kinetic energy and existsfrom this row of blades with a fairly high velocity.

It then enters the first row of guide blades which directs the steam to the second row of movingblades.




Mains Exam Solution

IES M

ASTER

Inface, there is a small drop in velocity in the fixed or guide blades due to friction. On passingthrough the second row of moving blades some drop in velocity again occurs i.e. steam gives upanother portion of its kinetic energy to the rotor.

It is again redirected by the second row of guide lades to the third row of moving blades whereagain some drop in velocity occurs and finally the steam leaves the rotor with small velocity inaxial direction and it is about 2 percent of initial total available energy of steam.

Hence this arrangement is nothing but splitting up the velocity gained from the exit of the nozzlesinto many drops through several rows of moving blades and hence the name velocity compounded.

Q.7: (c) A reaction steam turbine having diameter of 1400 mm is rotating at 3000 rpm. The turbinestages are designed in such a fashion that the enthalpy drop in both, rotor and stator, issame in each stage. if the speed ratio is 0.7 and blade angle at outlet is 20°, draw velocitytriangles and determine degree of reaction, blade angle at inlet and diagram efficiency.

[20 Marks]

Sol:D 1.4 m DNu 219.9 220 m/s

N 3000 rpm 60

F.Bh = M.Bh

Degree of reaction =M.D

M.B F.B

hh h

1Degree of reaction2

Also,1

UV = 0.7

V1 =U DN 220 314.29 m/s

0.7 60 0.7 0.7

Since the given turbine is a 50% reaction turbine.

Outlet angle of blade, 2 1 20

V1 = 2rV 314.16 m/s

V2 = 1rV

Velocity triangle: Taking ‘V’ as base

Taking scale as 1 cm = 44 m/s

Then U = 5 cm

V1 = 7.14 cm = 2rV

1 = 20°

1 2 1 2

1fV 2fV1rV

2rVV1

U

From the above velocity diagram blade angle at inlet 1 59 .

Diagram efficiency, d =2 1

w2 22r r1

V U

V VV2 2 2




Mains Exam Solution

IES M

ASTER

=1

w2

2 r1

V U

VV

2

From velocity diagram, wV = 8.2 × 44 = 360.8 m/s

d = 22

360.8 220 88.71%(136.4)(314.29)

2

Q.8: (a) A single-stage air compressor delivers air at 6 bar. The pressure and temperature at theend of suction are 1 bar and 27°C. It delivers 1.5 m3 of free air per minute when thecompressor is running at 350 rpm. The clearance volume is 5% of stroke volume. The freeair conditions are 1.013 bar and 15°C. The index of compression and expansion is 1.3. Find(i) The volumetric efficiency,(ii) Bore and stroke of cylinder if both are equal,(iii) The power required if the mechanical efficiency is 80%. [20 Marks]

Sol: Single stage air compressor.

3 2

4 1

VVC

P

P2 = 6 bar

T1 = 300 K, P1 = 1 bar

FAD = 1.5 m3/minute

m =2

air

air

P FAD 1.013 10 1.5R T 0.287 288 60

m = 0.0306 Kg/s

Also, m = 1 1 4

1

P (V V )R T

1 4V V =0.0306 0.287 300

100

1 4V V = 0.026 m3/s

Now, v =1/n

2

1

P1 C C

P

=1

1.361 0.05 0.051




Mains Exam Solution

IES M

ASTER

(i) v 85.16%

(ii) D = L

v = 1 42

V VND L

4 60

0.8516 = 30.026 4 60

D 350

D = 0.1882 m

D 18.82 cm L

(iii) Power required =

n 1n2

1 1 41

Pn P (V V ) 1n 1 P

=

0.31.31.3 100 (0.026) 6 1

0.3

= 5.769 kW

Actual power required =5.769

0.8

= 7.21 kW

Q.8: (b) Consider an ideal steam regenerative cycle in which steam enters the turbine at 3 MPa,300°C and exhausts to the condenser at 10 kPa. Steam is extracted from the turbine at 0.8MPa and supplied to an open feed water heater. The feed water leaves the heater assaturated liquid. The appropriate pumps are used for the water leaving the condenser andfeed water heater. If the mass flow rate through the boiler is 1 kg/s, determine the amountof steam extracted (kg/s), the total pump work (kW) and total turbine work (kW). Draw theschematic of this set-up. [20 Marks]

Sol:

WT

WP2 WP1

Boiler

1kg/s1

2 3

4

56

7

(1 m)kg/smkg/s

FWH

Condenser

Schematic of Regenerative Cycle




Mains Exam Solution

IES M

ASTER

T

65

1

2

s

7

43

3MPa

10kPa

(1 m)0.8 MPa m

T =300°C=573K1

At P1 = 3MPa; Tsat = 233.9°C

Since T1 = 300°C > Tsat , so superheated steam

s1 = 6.5389 kJ/kgK

h1 = 2993.48 kJ/kg

Isentropic compression, so s1 = s2 = 6.5389 kJ/kgK

At 0.8 MPa, sf = 2.0461 kJ/kgK and sg = 6.6627 kJ/kgK

hf = 721.1 kJ/kg, hfg = 2048.04 kJ/kg

since sf < s2 < sg so wet steam vf = 0.001115 m3/kg

s2 = sf + x2 sfg

6.5389 = 2.0461 + x2 (6.6627 – 2.0461)

x2 = 0.97

h2 = hf + x2 hfg = 721.1 + 0.97 × 2048.04 = 2708 kJ/kg

Now, s2 = s3 = 6.5389

At 10 kPa; Tsat = 45.81°C kJ/kgK

sf = 0.6492 kJ/kgK, sg = 8.1501 kJ/kgK

hf = 191.81 kJ/kg, hg = 2584.63 kJ/kg, vf = 0.001010m3/kg

Since sf < s3 < sg, so wet steam

s3 = sf + x3 (sg – sf)

6.5389 = 0.6492 + x3 (8.1501 – 0.6492)

x3 = 0.785

h3 = hf + x3 (hg – hf)

= 191.81 + 0.785 (2584.63 – 191.81)

= 2070.17 kJ/kg

h4 = hf = 191.81 kJ/kg

h5 = h4 + vdP = 191.81 + 0.001010 (800 – 10)

= 192.61 kJ/kg

h6 = hf at 0.8 MPa = 721.1 kJ/kg

Now, heat balance for open feed heater

5 2h mh1 m = h6

192.61 m 27081 m = 721.1

2515.39m = 528.49




Mains Exam Solution

IES M

ASTER

(i) Amount of steam extracted,

m = 0.21 kg/s

Now

Pump work for feed water heater, Wp2 = Vfdp

= 0.00115 × (3000 – 800)

= 2.453 kJ/kg

Now h7 = h6 + vdP = 721.1 + 2.453 = 723.55 kJ/kg

(ii) Total pump work, Wp = Wp1 + Wp2

= (1 – m ) vdP + 1 vdP

= (1 – 0.21) 0.00100 (800 – 10) + 1 × 2.453

= 3.08 kW

(iii) Total turbine work, WT = 1 × WT1 + (1 – m ) WT2

= 1(h1 – h2) + (1– 0.21) (h2 – h3)

= 1 (2993.48 – 2708) + (1 – 0.21) (2708 – 2070.17)

= 789.37 kW

Q.8: (c) A Brayton cycle works between 1 bar, 300 K and 5 bar 1250 K. There are two stages ofcompression with perfect inter-cooling and two stages of expansion. The work out of firstexpansion stage is being used to drive the two compressors. Their air from the first stageturbine is again heated to 1250 K and expanded. Calculate the power output of free powerturbine and cycle efficiency without and with a perfect heat exchanger and compare them.Also calculate the percentage improvement in the efficiency because of the addition ofheat exchangers. [20 Marks]

Sol: Assumptions:(i) Ideal processes of compression and expansion.

(ii) a gm m = 1 Kg/s and Cpaa = Cpg

5 7

6810

T

s

3 1

4 29

P = 1 bar1

P = 5 bar2

Pi

Given T1 = T3 = 300 K, P1 = 1 bar

T5 = T7 = 1250 K, P2 = 5 bar

Perfect intercooling, so T1 = T3 and T2 = T4

Intercooling pressure, Pi = 1 2PP 1 5 2.236

Now2

1

TT =

1

i

1

PP

1.4 11.42T 2.236

300 1

T2 = TT4 = 377.55 K




Mains Exam Solution

IES M

ASTER

Work input per compressor, Wc = Cp (T2 – T1) = 1.005 (377.55 – 300) = 77.94 kW

HP turbine is required to drive both compressor, so

Cp (T5 – T6) = 2 × Wc

1.005 (1250 – T6) = 2 × 77.94

T6 = 1094.9 K

Now, 6

5

PP =

1.41 1.4 16 6

65

T P 1094.9 P 3.14barT 5 1250

= P7

Also8

7

TT =

1

8

7

PP

1.4 11.48

8T 1 T 901.43K

1250 3.14

(i) Without perfect heat exchanger

Power output of power turbine,

WT = Cp (T7 – T8) = 1.005 (1250 – 901.43)

= 350.31 kW

Cycle efficiency, 1 =work outputheat input

= T

p 5 4 p 7 6

WC T T C T T

= 350.31

1.005 1250 377.55 1250 1094.9

= 33.92%

(iii) With perfect heat exchanger; T9 = T4 and T10 = T8

Power output remains some, so WT = 350.31 kW/kg of air

Cycle efficiency, 2 =work outputheat input

= T

p 5 10 p 7 6

WC T T C T T

= 350.31

1.005 1250 901.43 1250 1094.9

= 69.21%

Improvement in efficiency after addition of heat exchanger

=2 1

1

69.2 33.92 104.39%33.92

mechanical engineering - paper i...mechanical engineering - paper i f-126, katwaria sarai, new delhi...

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