1. thermodynamics, ic engine, rac_conv_explan

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1Test 1 (Conv) : Thermodynamics, IC Engine, RAC

ESE-2014 : Test SeriesMECHANICAL ENGINEERING

Test Series-1 (Conv) : Thermodynamics, IC Engines, RAC

Sol.1Sol.1Sol.1Sol.1Sol.1 (a)(a)(a)(a)(a)(i)(i)(i)(i)(i) Cause of irreversibility of a process are:1. Finite potential gradient causing the

irreversibility like temperature,pressure, concentration etc.

2. Presence of dissipative effects likefriction in which macroscopic workdissipates with an increase of internalenergy and then heat.

(ii)(ii)(ii)(ii)(ii) Dissipative effect:Dissipative effect:Dissipative effect:Dissipative effect:Dissipative effect: The transformationof work into molecular internal energyeither of the system or of the reservoirthrough friction, viscosity, inelasticity,electrical resistance and magnetichysteresis. These effects are known asdissipative effect.When work produces heat accompaniedby equivalent increase in kinetic orpotential energy of any system, it is saidto be dissipative.

1.1.1.1.1. (b)(b)(b)(b)(b)

Dwellingat C 20°

T °C∞ = –10

HP

w

Q2 Q1

T1 = 273 + 20 = 293 K

SlSlSlSlSl..... NNNNNooooo .: 060414.: 060414.: 060414.: 060414.: 060414

T2 = 273 – 10 = 263 K

COPmax = T .

T T= =

−1

1 2

293 9 7730

Wmin =max

..×

=6Heat 3 5 10 kJ

COP 9 77 dayIn kWh,

Wmin =.

.× × ×

63 5 10 kWh 19 77 3600 day

= 99.51 kWhday

∴ Operating cost per day = ` 99.51 × 8= ` 796.08

1.1.1.1.1. (c)(c)(c)(c)(c)Average brake power for 3 cylinders:

= NTπ2

60000 =

.× × ×2 3 14 1200 11060000

=13.816 kWAverage ip with 1 cylinder

= 20 – 13.816 = 6.184 kWTotal indicated power, ip

= 6.184 × 4 = 24.736 kW

isfc =bpbsfcip

× = .

×20

36024 736

= 291 grams per kWhFuel consumption

=isfc ip×

×3600 1000

=.

/ −×= ×

×3291 24 736

2 10 kg s3600 1000



2 ESE-2014 : TEST SERIES • MECHANICAL ENGINEERING

ηith =f

ipcv×m�

= .

−× ×3

24 736

2 10 43000

= 0.2876 or 28.76%

1.1.1.1.1. (d)(d)(d)(d)(d)

P

V

2 3

4

1

r =VV

=1

221

Vs = V1 – V2 = 20V2

V3 = 0.05Vs + V2

V3 = (0.05 × 20 + 1)V2

∴ VV

3

2= 2 = rc

η = ( )c

c

rrr

γ

γ −

⎡ ⎤−− ⎢ ⎥γ −⎣ ⎦1

1 11

1

= ( ).

. .

⎡ ⎤−− ⎢ ⎥−⎣ ⎦

1 4

0 4

1 2 11

1 4 2 121

= 0.6536 or 65.36%

1.1.1.1.1. (e)(e)(e)(e)(e)

φ = 50% or v

s

P. ,

P= 0 5

Patm = 101 kPa

Ps = 7.384 kPa

Pv = 0.5 × 7.384 = 3.692 kPa

ω =v

v

. PP P−

0 622 =

. ..

×−

0 622 3 692101 3 692

= 0.23599 kg w.v/kg d.a

∴ ω = 23.6 grams/ kg dry air

2.2.2.2.2. (a)(a)(a)(a)(a)The temperature behaviour of a fluid duringa throttling (h = constant) process isdescribed by the Joule-Thomsoncoefficient, defined as

μJ =h

TP

∂⎡ ⎤⎢ ⎥∂⎣ ⎦

The Joule-Thomson coefficient is a measureof the change in temperature with respectto pressure during a constant enthalpyprocess.

J

<⎧⎪μ =⎨⎪>⎩

0 temperature increases0 temperature remains constant0 temperature decreases

The difference in enthalpy between twoneighbouring equilibrium states is

dh = Tds + vdp2nd Tds equation is

Tds = CpdT – p

vT dp

T∂⎡ ⎤

⎢ ⎥∂⎣ ⎦

dh = pp

vC dT T dp vdp

T∂⎡ ⎤− +⎢ ⎥∂⎣ ⎦

dh = pp

vC dT T v dp

T

⎡ ⎤∂⎡ ⎤− −⎢ ⎥⎢ ⎥∂⎣ ⎦⎢ ⎥⎣ ⎦

μJ =h

Tp

⎡ ⎤∂⎢ ⎥∂⎣ ⎦

=p p

vT vC T

⎡ ⎤∂⎡ ⎤ −⎢ ⎥⎢ ⎥∂⎣ ⎦⎢ ⎥⎣ ⎦

1

For an ideal gas, pv = RT

∴p

vT

∂⎡ ⎤⎢ ⎥∂⎣ ⎦

=R vp T

=

μJ =p

vT v

C T⎡ ⎤× − =⎢ ⎥⎣ ⎦

10

2.2.2.2.2. (b)(b)(b)(b)(b)(i)(i)(i)(i)(i) Thermal converters are high-

temperature chambers through whichthe exhaust gas flows. It is fitted inexhaust system of engines to reducethe emission. It promote oxidation ofthe CO and HC which remain in theexhaust.




CO O CO+ →2 212

The temperature must be held above700°C for emission reduction to beeffective. Most thermal converters areessentially an enlarged exhaustmanifold connected to the engineimmediately outside the exhaust ports.Even though HC and CO emissions canbe reduced by oxidation, NOx

emissions can not be reduced usingthermal converter.

2.2.2.2.2. (b)(b)(b)(b)(b)(ii)(ii)(ii)(ii)(ii) Catalytic converters are chambers

mounted in the flow system throughwhich the exhaust gases pass through.These chambers contain catalystmaterial, which promotes the oxidationof the emissions contained in theexhaust flow. Generally, they are calledthree-way converters because they areused to reduce the concentration of CO,HC and NOx in the in the exhaust. Thecatalyst materials most used arePlatinum, Palladium and Rhodium.Palladium and Platinum promote theoxidation of CO and HC. Rhodiumpromotes the reaction of NOx.

2.2.2.2.2. (c)(c)(c)(c)(c)

Δt1

Δt2t

P01

P02

F01, 01tF02, 02t

F0 0t′ ′

P0′

P01 (at 0°C) = 2.928 bar

P′0 (at 10°C) = 4.146 bar

The required pressure difference or initial

compression of the follow-up spring per unit

area of diaphragm is

f14 = P P−′10 0

f14 = 4.146 – 2.928 = 1.218 bar

f14 is the difference in pressure on opposite

sides of the diaphragm or bellows to open

the valve.

The pressure at evaporator inlet, pevp

(at –30°C) = 0.8438 bar

Now the equilibrium pressure of the power

fluid = 1.218 + 0.8438 = 2.0618 bar

Corresponding saturation temperature of

power fluid = – 10°C

Assuming zero pressure drop in evaporator,

Saturation temperature at the evaporator

outlet, t02 (at 0.8438 bar) = –30°C

Effective superheat of the suction gas

= –10 – (–30) = 20°C

3.3.3.3.3. (a)(a)(a)(a)(a)(i) Air standard efficiency vs compression

ratio

r

η

γ = 1.67

γ = 1.4

γ = 1.3

(ii) Relative efficiency vs A/F ratio

1.0

0.8

0.6

0.20.4 0.8 1.2 1.6

A/F ratio

Effi

cien

cy ra

tio




(iii) Break thermal efficiency vs load

BTE

Load

(iv) Volumetric efficiency vs engine speed

η v

Speed

(v) Peak-cycle temperature vs equivalenceratio

T

φ

φ = 1.1

3.3.3.3.3. (b)(b)(b)(b)(b)pv = RT (1 + B ′p + C′p2 + .....)

v =RT RTB RTC p .....p

+ + +′ ′

P

v R dBRB RT

T p dT∂ ′⎡ ⎤ = + +′⎢ ⎥∂⎣ ⎦

dCRTp RpC .....

dT+ + +′

P

v RT dBT RTB RT

T p dT∂ ′⎡ ⎤ = + +′⎢ ⎥∂⎣ ⎦

2

dCRT p RTpC .....

dT′

+ + +′2

P

v dB dCT v RT RT p .....

T dT dT∂ ′ ′⎡ ⎤ = + + +⎢ ⎥∂⎣ ⎦

2 2

∴P

v dB dCT V RT RT p .....

T dT dT∂ ′ ′⎡ ⎤ − = + +⎢ ⎥∂⎣ ⎦

2 2

p P

v dBlim T v RT proved .....( )

T dT→

⎡ ⎤∂ ′⎡ ⎤ − =⎢ ⎥⎢ ⎥∂⎣ ⎦⎣ ⎦2

0i

For Van der Waals gas, to find Boyletemperature TB,

( )T CP

pv

p ==

⎡ ⎤∂=⎢ ⎥∂⎣ ⎦

0

0

pv = B C BRT P P .....

RT RT

⎡ ⎤−+ + +⎢ ⎥⎣ ⎦

22

21

( )pv Bp RT

∂= =

∂0 where B (second virial

coefficient) = a

bRT

−

∴ B ′ =B b a

RT RT R T= −

2 2

⇒dBdT

′=

b a

RT R T

−+

2 2 3

2

From equation (i),

JP o P

RT b alim

C RT R T→

⎛ ⎞μ = − + =⎜ ⎟⎝ ⎠

2

2 2 3

20

b

RT 2 =a

R T2 3

2

∴ Ti =a

bR2

(Inversion temperature)

Ti = 2TB∴ Inversion temperature = 2 × Boyle

temperature




3.3.3.3.3. (c)(c)(c)(c)(c)

1

pv

t

td

0°C

Temp

S

Reference state

Enthalpy of moist air,h = ha + ωhv ...(i)

ha is enthalpy of dry airωhv is enthalpy of water vapour part

ha = cpat = 1.005 t ...(ii)Where specific heat of dry air is 1.005

kJ/kgK and t is the dry bulb temperature

of air in °C.

hv = cpw tdp + (hfg)dp + cpv (t – tdp) kJ/kg

= 4.18 tdp + (hfg)dp + 1.88 (t – tdp)

Here,

h1 = ( )g pvCh c . /

°+ =

01 88 kJ kg

cpw = Specific heat of liquid water = 4.18

kJ/kgK

tdp = Dew point temperature

(hfg)dp = Latent heat of vaporization at DPT.

cpv = Specific heat of superheated vapour

= 1.88 kJ/kgK

hv = 2500 + 1.188t ...(iii)

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

h = 1.005 t + ω(2500 + 1.88t) kJ/kg of dry air

4.4.4.4.4. (a) (i)(a) (i)(a) (i)(a) (i)(a) (i)Tds = dH – vdp

ds∫2

1

= pmc dT mRdpT p

−∫ ∫2 2

1 1

s2 – s1 = p e eT p

mc log mRlogT p

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2

1 1

for 1 kg of air,

p e eT ps s c log RlogT p

⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 22 1

1 1

Change in availability= ψ1 – ψ2 = b1 – b2

= (h1 – T0s1) – (h2 – T0s2)

= ( )p pp Tc T T T R cp T

⎛ ⎞− − −⎜ ⎟⎝ ⎠

2 21 2 0

1 1in in

= ( ). − −1 005 520 300

e. log .⎛ ⎞−⎜ ⎟⎝ ⎠1 573

293 0 287 1 005in5 793

= 1.005 × 220 – 293 (0.3267 – 0.4619)

= 221.1 + 39.6 = 260.7 kJ/kg

(ii)(ii)(ii)(ii)(ii)Wmax = Change in availability = 260.7 kJ/kg

(iii)(iii)(iii)(iii)(iii)Applying S.F.E.E. for turbine,Q + h1 = W + h2

W = (h1 – h2) + Q= 1.005 (520 – 300) – 10= 211.1 kJ/kg

Irreversibility,I = Wmax – W

= 260.7 – 211.1= 49.6 kJ/kg

4.4.4.4.4. (b)(b)(b)(b)(b)T2 = 20°C, (ps)20°C = 2.3385 kPaT1 = 30°C, (ps)30°C = 4.2461 kPa

ω2 = s

s

. p . .p p .

×=

− −0 622 0 622 2 3385

100 2 3385

= 0.0149 kg vap/kg d.a

ω1 =( )pa fg

g f

C T T h

h h

− + ω−

1

2 1 2 2

2

hfg 2 = 2454.1 kJ/kghf 2 = 2538.1 – 2454.1 = 84 kJ/kg




hω1 = hg1 = 2556.2 kJ/kg

ω1 =( ). . .

.− + ×

−1 005 20 30 0 0149 2454 1

2556 2 84

= 0.0107 kg vap/ kg d.a

ω1 =. p

p pω

ω−1

1

0 622

⇒ 0.622 pω1 = (100 – pω1) × 0.0107or 0.622 pω1 = 1.07 – 0.0107 pω1

or 0.6327 pω1 = 1.07or pω1 = 1.691 kPa

Relative humidity,

φ = w

s

p .p .

=1

1

1 6914 2461

= .0 39828 or 39.828%

4.4.4.4.4. (c)(c)(c)(c)(c)Velocity at throat,

C2 = da pp

C C Tp

γ −γ

⎡ ⎤⎛ ⎞⎢ ⎥− ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦

1

21

12 1

C2 = 100 m/s, Cda = 0.84,Cp = 1.005 kJ/kg, T1 = 290 K

. .

.γ − = =

γ1 0 4 0 286

1 4

or.

p. p

⎡ ⎤⎛ ⎞⎛ ⎞ ⎢ ⎥= × × −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

0 28622

1

1002 1005 290 1

0 84

or.

p.

p⎛ ⎞

− =⎜ ⎟⎝ ⎠

0 2862

11 0 0243

or.

p.

p⎛ ⎞

=⎜ ⎟⎝ ⎠

0 2862

10 97568

orp . ,p .p

= =22

10 9175 or 0 9175bar

Δpa = p1 – p2 = 0.0825 barΔpf = 0.85 × 0.0825 = 0.070125 bar

v1 =RT .p

×= = 315

1

287 290 0 8323m10

v2 =/

pv

p

γ⎛ ⎞⎜ ⎟⎝ ⎠

11

12

=.

..

⎛ ⎞⎜ ⎟⎝ ⎠

0 71410 8323

0 9175= 0.885 m3/kg

Throat area,

A2 =am vC

× 2

2

�

=.

. −× = × 4 26 0 8858 85 10 m

60 100

A2 =/.d d

.π ⎛ ⎞= ⎜ ⎟⎝ ⎠

1 222 2

8 85or

4 0 785

⇒ d 2 = 3.3576 cm

fm� = ( ) /f df f fA C pρ Δ 1 22

⇒ ( ) /f

. A . .= × × × ×1 250 4

0 65 2 770 0 070125 1060

Af = 3.121 × 10–6 m2

⇒ df = 1.994 × 10–3 m or 1.994 mm

5.5.5.5.5. (a) (i)(a) (i)(a) (i)(a) (i)(a) (i)

bp =NTπ2

60000

=. . .× × × × ×2 3 14 4500 42 0 54 9 81

60000= 104.8 kW

(ii)(ii)(ii)(ii)(ii)

bmep =bp

LAnK× 60000

=.

. .

×π× × × ×2

104 8 600004500

0 08 0 09 84 2

= 6.87 × 105 Pa




∴ bmep = 6.87 bar

(iii)(iii)(iii)(iii)(iii)

bsfc =

.

. / .

×=

4 4 6010 0 252 kg kWh104 8

(iv)(iv)(iv)(iv)(iv)

bsac = . / .

×=

6 603 435 kg kWh

104 8

(v)(v)(v)(v)(v)

ηbth =f

bpm cv×�

=.

. ××

104 84 4

4400010 60

= 0.3248 or 32.48%

(vi)(vi)(vi)(vi)(vi)

aV� = am RTp

�

= . / × × =

×3

5

6 287 3005 166m min

1 10

Vs = D LnKπ 2

4

= . .π

× × × ×2 45000 09 0 08 8

4 2= 9.16 m3/min

ηv =a

s

V .V .

× = ×5 166

100 1009 16

�

= 0.5639 or 56.39%

(vii)(vii)(vii)(vii)(vii)

A/F = ..

=6

13 640 44

5.5.5.5.5. (b)(b)(b)(b)(b)Human comfort is defined as the state of mindthat expresses satisfaction with thesurrounding environment. Factors thatdetermine human comfort include: temperatureof the surrounding environment, humidity ofthe air, air motion in addition to air purity.

Factor affecting optimum effectiveFactor affecting optimum effectiveFactor affecting optimum effectiveFactor affecting optimum effectiveFactor affecting optimum effectivetemperaturetemperaturetemperaturetemperaturetemperature1. Climatic and seasonal differences2. Clothing3. Age and sex4. Duration of stay5. Kind of activity6. Density of occupantsPoint considered for cool ing loadPoint considered for cool ing loadPoint considered for cool ing loadPoint considered for cool ing loadPoint considered for cool ing loadestimation and heating load estimationestimation and heating load estimationestimation and heating load estimationestimation and heating load estimationestimation and heating load estimation1. Heat flowing into or going out from the

building by conduction through exteriorwalls, floors, ceilings, doors etc.

2. Heat given off by lights, motor,machinery, cooking operations etc.

3. Heat received from solar radiation.4. Heat liberated by the occupants.5. Heat gain from the fan work.6. Heat gain due to moisture in the outside air.7. Heat gain due to condensation of

moisture from any process such ascooking foods which takes place withinthe conditioned space.

6.6.6.6.6. (a)(a)(a)(a)(a)Pressure loss due to friction for the circularduct:

pfc =a c

c c

f L P QA A

⎛ ⎞ ⎛ ⎞ρ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

2

=( )

a c

c

f L Q P

A

⎡ ⎤ρ ⎢ ⎥⎢ ⎥⎣ ⎦

2

32

Pressure loss due to friction for therectangular duct:

pfR = a R

c R

fL P QA A

⎛ ⎞ ⎛ ⎞ρ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

2

2

=( )

a R

R

fL Q P

A

⎡ ⎤ρ ⎢ ⎥⎢ ⎥⎣ ⎦

2

32

As per given condition,Pfc = PfR

⇒ c

c

P

A3 = R

R

P

A3




⇒D

D

π

π⎛ ⎞⎜ ⎟⎝ ⎠

32

4

= ( )a b

a b

+3 3

2

ora b

D a b

+=

π2 5 3 3

32

or( )a b

Da b

=π +

3 35

2

32

∴ D = /

a b.

a b

⎛ ⎞⎜ ⎟+⎝ ⎠

1 53 3

1 265

6.6.6.6.6. (b)(b)(b)(b)(b)

C.V

Q kJ = 9

1W kJ= 135

2

v1 = 0.37 m3/kg, p1 = 600 kPa

v1 = 16 m/s, z1 = 32 m

v2 = 0.37 m3/kg, p2 = 600 kPa

v2 = 16 m/s, z2 = 0

Writing S.F.E.E. for the above controlvolume:

v dQu p v z gdm

+ + + +21

1 1 1 12 =

zv dWu p v z gdm

+ + + +22

2 2 2 22

∴ u1 – u2 = ( ) v vp v p v

⎛ ⎞−− + +⎜ ⎟⎝ ⎠

2 22 1

2 2 1 1 2

( ) dW dQz z g

dm dm− + −2 1

x

= ( ). .⎛ ⎞−

× − × + ⎜ ⎟×⎝ ⎠

2 2270 16100 0 62 600 0 37

2 1000

+ (0 – 32) × 9.81 + 135 + 9= – 160 + 36.32 – 0.314 + 144= 20.006 kJ/kgSpecific internal energy decreases by20 kJ/kg.

6.6.6.6.6. (c)(c)(c)(c)(c)

T r = 1.3Z = 0.7

Z

0.7

2 Pr

GivenPr = 2, Tr = 1.3, z = 0.7

c

pp

= 2 or p = 2 × 2.73 = 5.46 MPa

c

TT

= . T1 3 or = 1.3 × 44.5 = 57.85 K

pv = zRT

v =. . .

. .

× ×× × 3

0 7 8 3143 57 85

20 183 5 46 10

= 3.05 × 10–3 m3/kg

vr =c

vv

=. .

..

−

−× × =

×

3

2

3 05 10 20 1831 48

4 16 10

1. thermodynamics, ic engine, rac_conv_explan

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