IES previous year Thermodynamics solution

Basic Concepts S K Mondals Chapter 1

1

1. Basic Concepts Theory at a Glance (For GATE, IES & PSUs)

Intensive and Extensive Properties Intensive property: Whose value is independent of the size or extent i.e. mass of the system. These are, e.g., pressure p and temperature T. Extensive property: Whose value depends on the size or extent i.e. mass of the system (upper case letters as the symbols). e.g., Volume, Mass (V, M). If mass is increased, the value of extensive property also increases. e.g., volume V, internal energy U, enthalpy H, entropy S, etc. Specific property: It is a special case of an intensive property. It is the value of an extensive property per unit mass of system. (Lower case letters as symbols) eg: specific volume, density (v, ).

Thermodynamic System and Control Volume In our study of thermodynamics, we will choose a small part of the universe to which we will

apply the laws of thermodynamics. We call this subset a SYSTEM.

The thermodynamic system is analogous to the free body diagram to which we apply the laws of mechanics, (i.e. Newtons Laws of Motion).

The system is a macroscopically identifiable collection of matter on which we focus our attention (e.g., the water kettle or the aircraft engine).

System Definition System: A quantity of matter in space which is analyzed during a problem. Surroundings: Everything external to the system. System Boundary: A separation present between system and surrounding.

Classification of the system boundary:- Real solid boundary Imaginary boundary

S

Thas:

Thbei

Ty

C

K Monhe system bou:

Contr Contr

he choice of boing analyzed

ypes of

losed S1. Its afixed i2. Thisreferred3. Therthe syst4. Eneinto or

dals undary may b

rol Mass Systrol Volume Syoundary dep

d.

f Syste

System (a system of dentity. s type of sd to as closre is no mastem boundarrgy transferout of the sy

B

be further cla

tem. ystem. ends on the p

em

(Controfixed mass

system is used system.ss transfer ary. r may take stem.

Basic C

2

assified

problem

ol Masss with

sually . across

place

Conce

System

Fig. A Coor C

epts

m)

ontrol MassClosed Syst

Cha

s System tem

apter 1

S

O

Ex

K Mon

Open Sy1. Its 2. Thi

refea "c

3. Maacro

4. Eneinto

5. A cfixeand

6. Conbouacroma

7. The8. Wh

syst9. The

ma10. Mo

xample: Heat exchacross the syPump - A cenergy from

dals

ystem a system of fis type of serred to as "control voluss transfer oss a control ergy transfeo or out of thontrol volumed region acd energy tranntrol Surfaundary of aoss which thss and energe mass of a co

hen the net intem is fixed ae identity of ss system (clst of the engi

anger - Fluystem boundcontinuous fl

m the surroun

B

(Contrfixed volumsystem is usopen system

ume" can take volume. r may also e system.

me can be seencross which nsfers are stuace Its a control vohe transfer ofgy takes placeontrol volumnflux of masand vice-vers mass in a closed system)ineering devi

id enters andary. low of fluid tndings to the

Basic C

3

rol Volme.

sually m or

place

occur

n as a mass

udied. the olume f both e.

me (open systes across the sa. control volum). ices, in gener

nd leaves the

akes place th system.

Conce

lume S

Fig. A Coor

em) may or mcontrol surfa

me always ch

ral, represen

e system con

hrough the s

epts

System

ontrol Volumr Open Syst

may not be fixace equals ze

hanges unlik

t an open sys

ntinuously w

system with a

Cha

)

me System em

xed. ero then the

ke the case f

stem or contr

with the tran

a transfer of

apter 1

mass of the

for a control

rol volume.

sfer of heat

f mechanical

S

Is1.

2.

3.

QThun

We

A q

K Mon

solatedIt is a sysidentity No interaplace besurroundIn more system isbusy mar

Quasi-Sthe processenrestrained

e need restra

quasi static The dev

equilibriu All states

are equil If we rem

one thedisplace quasista

On the opressurebecause i

In both c Another

On the ot

dals

d Systestem of fixed and fixed eaction of masetween the dings. informal ws like a closrket.

tatic Pres can be

ained process

c process is oviation fromum is infinites of the systelibrium statemove the weig pressure the piston

atic. ther hand if . (This is unit is not in a cases the syste.g., if a perther hand if

B

em d mass withenergy. ss or energy system and

words an isoed shop ami

rocesse restrained

ses in practic

one in which m thermodynesimal. em passes thes. ghts slowly oof the gas gradually.

we remove anrestrained sustained mtems have unson climbs dhe jumps the

Basic C

4

h same

takes d the

olated idst a

d or

e.

namic

hrough

one by s will

It is

all the weighexpansion) banner. ndergone a chdown a laddeen it is not a

Conce

Fig. An

Fig. A q

ts at once thbut we dont

hange of statr from roof t quasistatic p

epts

n Isolated S

quasi static

he piston will t consider th

te. to ground, it process.

Cha

System

process

be kicked uphat the work

is a quasista

apter 1

p by the gas k is done

atic process.

S

La

S

Yo Yo

K Mon

aws of

The ZerothtemperatureThe First Lenergy. The Second matter into The Third Lcrystalline s

ummatFirstly, theconversion owork outputSecondly, mK is unattaiBecause, wescale!! That

ou cant get

ou cant get

dals

Therm

h Law deales. Law deals w

Law of therm work. It alsoLaw of thermsubstance at

tion of 3re isnt a mof heat to wot.

more interestiinable. This ie dont know is why all te something To get work

something To get somebe given.

B

odynam

s with ther

with the cons

modynamics o introduces tmodynamics absolute zero

3 Lawsmeaningful teork. Only at

ingly, there iis precisely th what 0 K lo

emperature s for nothing

k output you m for very litte work output

Basic C

5

mics

rmal equilib

servation of

provides witthe concept odefines the o temperatur

s emperature infinite tem

isnt enough he Third law

ooks like, we cales are at b

g: must give somtle: t there is a m

Conce

brium and

energy and

th the guideliof entropy. absolute zerre is zero.

of the sourcmperature one

work availabw.

havent got best empirica

me thermal e

minimum am

epts

provides a

introduces

ines on the co

ro of entropy

ce from whie can dream

ble to produc

a starting poal.

energy.

ount of therm

Cha

means for

the concept

onversion he

y. The entrop

ich we can g of getting th

ce 0 K. In oth

oint for the t

mal energy th

apter 1

measuring

of internal

eat energy of

py of a pure

get the full he full 1 kW

her words, 0

temperature

hat needs to

S

Yo Vio

Ze

InToprahaTeconof temteminsgivfroIn therad

K Monou cant get

olation of a

eroth LIf two syste(that is A anthermal equ

All tem

nternati provide a sactical consids been refinmperature Snforms with accuracy of

mperature ofmperatures istruments anving the tempom 3.0 to 24. the range frermometers. diation and m

dals every thingHowever mu

all 3 laws: Try to get ev

Law of Tems (say A annd C are in thuilibrium the

mperature m

ional Testandard forderations, thned and exte

Scale of 1990 the thermod measuremenf a number ois accomplisnd values of perature as f5561 K is barom 13.8033 Above 1234measuremen

B

g: uch work you

verything for

Thermond B) are in hermal equilmselves (tha

measuremen

emperar temperatur

he Internationended in se (ITS-90) is

dynamic tempnt obtainabl

of reproducibshed by formthe ITS. In tfunctions of t

ased on measto 1234.93 K

4.9 K the temnts of the inte

Basic C

6

u are willing

r nothing.

odynam thermal equlibrium; B anat is A and B

nts are base

ature Scre measuremnal Temperaveral revisiodefined in superature, thele in 1990. Tble fixed poinmulas that gthe range frothe vapor pr

surements usK, ITS-90 is mperature isensity of visib

Conce

to give 0 K c

mics uilibrium witnd C are in th will be in th

ed on Zeroth

cale ment takingature Scale (Ions, most reuch a way the unit of whiThe ITS90 nts (Table). Ingive the relaom 0.65 to 5ressures of pasing a heliumdefined by ms defined usible-spectrum

epts

cant be reach

th a third syhermal equilermal equilib

h law of the

g into accounITS) was adoecently in 1hat the tempch is the kelis based on

nterpolation ation betwee.0 K, ITS-90articular hel

m constant-vomeans of certing Plancks

m radiation th

Cha

hed.

ystem (say C)librium) thenbrium).

ermodynam

nt both theoopted in 1927990. The In

perature mealvin, to withi the assigne between then readings o is defined bium isotopesolume gas thtain platinum equation fohe absolute t

apter 1

) separately n they are in

mics

oretical and 7. This scale nternational asured on it in the limits ed values of e fixed-point of standard

by equations s. The range hermometer. m resistance r blackbody temperature

S

scaKea Cwa Tim63mejunthe In forthe

K Mon

ale. The absoelvin temperaCarnot enginater arbitrari

me Constan.2% of step ceasured timenction from te thermocoup

general, timr measuremee velocity of t

dals

olute temperature scale ane operating ily assigned t

nts: The timchange in te

e constant arthe welded caple probe gre

me constants fent of liquid. the media.

B

rature scale lso, the triplbetween resethe value of

e constant isemperature oe sheath walan on an ungeatly influenc

for measuremThe time con

Basic C

7

is also knowe point of waervoirs at tem 273.16 K.

s the amountof a surroundll thickness,

grounded theces the time c

ment of gas cnstant also v

Conce

wn as Kelvinater is taken mperature

t of time reqding media. degree of insermocouple. Iconstant mea

can be estimavaries inverse

epts

n temperatur as the stand and tp, tp

quired for a tSome of thesulation comIn addition, tasurement.

ated to be teely proportio

Cha

re scale. In ddard referencbeing the tri

thermocouplee factors influ

mpaction, andthe velocity o

n times as lonal to the sq

apter 1

defining the ce point. For iple point of

e to indicate uencing the

d distance of of a gas past

ong as those quare root of

S

WWonotthastodonun

FFrandevabec

irr

pdLevolcoomoprep a

K Mon

Work a pork is one of t a point funat the workored in the sne. Therefo

ndergoes a p

ree Expree Expansiod B by a thinacuated. If thecome equal. T

reversible.

dV-wort the gas in tlume V1. Thordinates 1pove out to a essure 2p anand the volu

dals

path funthe basic mo

nction. Therek is possesystem is enere, work is process.

pansionon Let us conn diaphragm. e diaphragm This is known

. Also work

rk or Dithe cylinder

he system is , V1. The pistnew final po

nd volume V2ume V. This m

B

nctionodes of energyefore, work ssed by theergy, but not energy in

n with Znsider an insu Compartmenis punctured,

n as free or

done is zero

splace(Figure showin thermody

ton is the onosition 2, wh

2. At any intemust also be

Basic C

8

y transfer. Th is not a pr system. It t work. A detransit and

Zero Wulated containent A contains the gas in Aunrestrained

o during fre

Free Expa

ment Wwn in below) ynamic equi

nly boundary hich is also aermediate poie an equilibr

Conce

he work doneroperty of th is an interacrease in en

d it can be i

Work Traer (Figure) wh a mass of ga

A will expand expansion. T

ee expansion

ansion

Work be a system librium, the which movea thermodynint in the traium state, si

epts

e by a systemhe system, action acrossergy of the sidentified o

ansfer hich is dividedas, while cominto B until t

The process n.

having initia state of whs due to gas

namic equilibavel of the piince macrosc

Cha

m is a path fuand it cann

s the boundasystem appeaonly when t

d into two commpartment B i

the pressures of free ex

ally the preshich is descr pressure. Lebrium state iston, let the copic propert

apter 1

unction, and not be said ary. What is ars as work the system

mpartments A is completely in A and B

xpansion is

sure 1p and ibed by the

et the piston specified by pressure be ties p and V

S

sig

forthegas

whdra Wham

K Mongnificant onl

r equilibriume piston, the s on the pisto

dWhere dV = adawn at the to

hen the pistomount of work

dals y.

m states. Whe force F actinon.

W= Fdl = infiniteop of it will b

on moves outk W done by t

W

B

en the pistonng on the pist

F dsimal displae explained l

from positiothe system w

1 2W

Basic C

9

n moves an inton F = p.a. a

dl = acement volulater.

on 1 to positiowill be

= V

Conce

nfinitesimal and the infin

padume. The dif

on 2 with the

2

1

V

Vpd

epts

distance dl, itesimal amo

dl = fferential sig

e volume cha

dV

Cha

and if a' beount of work

pdVgn in dW w

anging from V

apter 1

e the area of done by the

V ith the line

V1 to V2, the

S

ThtheSincoosysmumuinfthrTh

on

HHefropat

Th

PRExIn cha

p =So

K Mon

he magnitudee area undernce p is atordinate, all stem as the ust be equiliust be quafinitely slowrough is an e

he integration

a quasi-sta

eat Traeat transfer iom state 1 toth. Therefore

he displaceme

ROBLEMxample 1 a closed systange in inter

2 10VV

= + Wolution:

dals

e of the worr the path 1-t all times the states pa volume chaibrium statesasi-static. T

wly so that equilibrium sn pdV can

atic path.

ansfer-Ais a path fu state 2 depee dQ is an ine

ent work is g

MS & SO

tem, volume rnal energy g

Where p is in

B

rk done is g-2, as shown a thermod

assed throughanges from Vs, and the p

The piston every state tate.

n be perform

A Path unction, thatends on the iexact differen

2

1 dQgiven by

1 2 =W

OLUTION

changes fromgiven the pres

n kPa and V

Basic C

10

given by n in Fig. dynamic h by the V1 to V2 path 1-2

moves passed

med only

Functiot is, the amointermediatential, and we

1 2=Q Q or

2 2

1 1= = dW

NS

m 1.5m3 to 4.ssure volume

is in m3.

Conce

Fig.

on ount of heat e states throue write

1 2 2 Q Q Q

2 pdV W W

5 m3 and heae relation as

epts

Quasi-Stat

transferred ugh which th

1Q

1W

at addition is

Cha

ic pdV Work

when a systhe system pa

s 2000 kJ. Ca

apter 1

k

tem changes asses, i.e. its

alculate the


11

( )( )( )

[ ]

2 2

1 1

2

3 3 22 1

1

3 3

10.

1 1031 4.54.5 1.5 103 1.51 91.125 3.375 10 33

29.250 10.986 40.236

= = + = + = + = +

= + =

V V

V V

Work done p dV V dVV

VV V lnV

ln

ln

kJ

First Law of Thermodynamics:- Q = W + U 2000 = 40.236 + U U = 2000 40.236 = 1959.764 kJ Example 2. A fluid is contained in a cylinder piston arrangement that has a paddle that imparts work to the fluid. The atmospheric pressure is 760 mm of Hg. The paddle makes 10,000 revolutions during which the piston moves out 0.8m. The fluid exerts a torque of 1.275 N-m one the paddle. What is net work transfer, if the diameter of the piston is 0.6m? Solution: Work done by the stirring device upon the system W1 = 2TN = 2 1.275 10000 N-m = 80kJ This is negative work for the system.

(Fig.)

Work done by the system upon the surroundings. W2 = p.dV = p.(A L) = 101.325

4 (0.6)2 0.80 = 22.9kJ

This is positive work for the system. Hence the net work transfer for the system. W = W1 + W2 = - 80 + 22.9 = - 57.l kJ.


12

ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS)

Previous 20-Years GATE Questions GATE-1. List-I List II [GATE-1998] A. Heat to work 1. Nozzle B. Heat to lift weight 2. Endothermic chemical reaction C. Heat to strain energy 3. Heat engine D. Heat to electromagnetic energy 4. Hot air balloon/evaporation 5. Thermal radiation 6. Bimetallic strips Codes: A B C D A B C D (a) 3 4 6 5 (b) 3 4 5 6 (c) 3 6 4 2 (d) 1 2 3 4

Open and Closed systems GATE-2. An isolated thermodynamic system executes a process, choose the correct

statement(s) form the following [GATE-1999] (a) No heat is transferred (b) No work is done (c) No mass flows across the boundary of the system (d) No chemical reaction takes place within the system GATE-2a. Heat and work are [GATE-2011] (a) intensive properties (b) extensive properties (c) point functions (d) path functions

Quasi-Static Process GATE-3. A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and

0.015 m3. It expands quasi-statically at constant temperature to a final volume of 0.030 m3. The work output (in kJ/kg) during this process will be: [GATE-2009]

(a) 8.32 (b) 12.00 (c) 554.67 (d) 8320.00

Free Expansion with Zero Work Transfer GATE-4. A balloon containing an ideal gas is initially kept in an evacuated and

insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is TRUE at the end of above process?

(a) The internal energy of the gas decreases from its initial value, but the enthalpy remains constant [GATE-2008]

(b) The internal energy of the gas increases from its initial value, but the enthalpy remains constant

(c) Both internal energy and enthalpy of the gas remain constant (d) Both internal energy and enthalpy of the gas increase


13

GATE-5. Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The Work done in the process is given by:

[GATE-1996, IAS-2000] (a) Pdv (b) +Pdv (c) vdp (d) +vdp

pdV-work or Displacement Work GATE-6. In a steady state steady flow process taking place in a device with a single inlet

and a single outlet, the work done per unit mass flow rate is given by outlet

inlet

vdp = , where v is the specific volume and p is the pressure. The expression for w given above: [GATE-2008]

(a) Is valid only if the process is both reversible and adiabatic (b) Is valid only if the process is both reversible and isothermal (c) Is valid for any reversible process

(d) Is incorrect; it must be outlet

inletvdp =

GATE-7. A gas expands in a frictionless piston-cylinder arrangement. The expansion

process is very slow, and is resisted by an ambient pressure of 100 kPa. During the expansion process, the pressure of the system (gas) remains constant at 300 kPa. The change in volume of the gas is 0.01 m3. The maximum amount of work that could be utilized from the above process is: [GATE-2008]

(a) 0kJ (b) 1kJ (c) 2kJ (d) 3kJ GATE-8. For reversible adiabatic compression in a steady flow process, the work

transfer per unit mass is: [GATE-1996] ( ) ( ) ( ) ( )a pdv b vdp c Tds d sdT

Previous 20-Years IES Questions IES-1. Which of the following are intensive properties? [IES-2005] 1. Kinetic Energy 2. Specific Enthalpy 3. Pressure 4. Entropy Select the correct answer using the code given below: (a) 1 and 3 (b) 2 and 3 (c) 1, 3 and 4 (d) 2 and 4 IES-2. Consider the following properties: [IES-2009] 1. Temperature 2. Viscosity 3. Specific entropy 4. Thermal conductivity Which of the above properties of a system is/are intensive? (a) 1 only (b) 2 and 3 only (c) 2, 3 and 4 only (d) 1, 2, 3 and 4 IES-2a. Consider the following: [IES-2007, 2010]

1. Kinetic energy 2. Entropy


14

3. Thermal conductivity 4. Pressure Which of these are intensive properties? (a) 1, 2 and 3 only (b) 2 and 4 only (c) 3 and 4 only (d) 1, 2, 3 and 4

IES-3. Which one of the following is the extensive property of a thermodynamic

system? [IES-1999] (a) Volume (b) Pressure (c) Temperature (d) Density IES-4. Consider the following properties: [IES-2009] 1. Entropy 2. Viscosity 3. Temperature 4. Specific heat at constant volume Which of the above properties of a system is/are extensive? (a) 1 only (b) 1 and 2 only (c) 2, 3 and 4 (d) 1, 2 and 4 IES-4a Consider the following: [IES-2010]

1. Temperature 2. Viscosity 3. Internal energy 4. Entropy

Which of these are extensive properties? (a) 1, 2, 3 and 4 (b) 2 and 4 only (c) 2 and 3 only (d) 3 and 4 only.

Thermodynamic System and Control Volume IES-5. Assertion (A): A thermodynamic system may be considered as a quantity of

working substance with which interactions of heat and work are studied. Reason (R): Energy in the form of work and heat are mutually convertible. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false [IES-2000] (d) A is false but R is true IES-5a A control volume is [IES-2010] (a) An isolated system (b) A closed system but heat and work can cross the boundary (c) A specific amount of mass in space (d) A fixed region in space where mass, heat and work can cross the boundary of that

region

Open and Closed systems IES-6. A closed thermodynamic system is one in which [IES-1999, 2010, 2011] (a) There is no energy or mass transfer across the boundary (b) There is no mass transfer, but energy transfer exists (c) There is no energy transfer, but mass transfer exists (d) Both energy and mass transfer take place across the boundary, but the mass transfer

is controlled by valves IES-7 Isothermal compression of air in a Stirling engine is an example of


15

(a) Open system [IES-2010] (b) Steady flow diabatic system (c) Closed system with a movable boundary (d) Closed system with fixed boundary IES-8. Which of the following is/are reversible process(es)? [IES-2005] 1. Isentropic expansion 2. Slow heating of water from a hot source 3. Constant pressure heating of an ideal gas from a constant temperature

source 4. Evaporation of a liquid at constant temperature Select the correct answer using the code given below: (a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 4 IES-9. Assertion (A): In thermodynamic analysis, the concept of reversibility is that, a

reversible process is the most efficient process. [IES-2001] Reason (R): The energy transfer as heat and work during the forward process

as always identically equal to the energy transfer is heat and work during the reversal or the process.

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-9a Which one of the following represents open thermodynamic system?

(a) Manual ice cream freezer (b) Centrifugal pump [IES-2011] (c) Pressure cooker (d) Bomb calorimeter

IES-10. Ice kept in a well insulated thermo flask is an example of which system? (a) Closed system (b) Isolated systems [IES-2009] (c) Open system (d) Non-flow adiabatic system IES-10a Hot coffee stored in a well insulated thermos flask is an example of (a) Isolated system (b) Closed system [IES-2010] (c) Open system (d) Non-flow diabatic system IES10b A thermodynamic system is considered to be an isolated one if [IES-2011]

(a) Mass transfer and entropy change are zero (b) Entropy change and energy transfer are zero (c) Energy transfer and mass transfer are zero (d) Mass transfer and volume change are zero

IES-10c. Match List I with List II and select the correct answer using the code given

below the lists: [IES-2011] List I

A. Interchange of matter is not possible in a B. Any processes in which the system returns to its original condition or state is called C. Interchange of matter is possible in a

List II 1. Open system 2. System


16

D. The quantity of matter under consideration in thermodynamics is called

3. Closed system 4. Cycle

Code: A B C D A B C D (a) 2 1 4 3 (b) 3 1 4 2 (c) 2 4 1 3 (d) 3 4 1 2

Zeroth Law of Thermodynamics IES-11. Measurement of temperature is based on which law of thermodynamics?

[IES-2009] (a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics IES-12. Consider the following statements: [IES-2003]

1. Zeroth law of thermodynamics is related to temperature 2. Entropy is related to first law of thermodynamics 3. Internal energy of an ideal gas is a function of temperature and pressure 4. Van der Waals' equation is related to an ideal gas

Which of the above statements is/are correct? (a) 1 only (b) 2, 3 and 4 (c) 1 and 3 (d) 2 and 4 IES-13. Zeroth Law of thermodynamics states that [IES-1996, 2010] (a) Two thermodynamic systems are always in thermal equilibrium with each other. (b) If two systems are in thermal equilibrium, then the third system will also be in

thermal equilibrium with each other. (c) Two systems not in thermal equilibrium with a third system are also not in thermal

equilibrium with each other. (d) When two systems are in thermal equilibrium with a third system, they are in

thermal equilibrium with each other.

International Temperature Scale IES-14. Which one of the following correctly defines 1 K, as per the internationally

accepted definition of temperature scale? [IES-2004] (a) 1/100th of the difference between normal boiling point and normal freezing point of

water (b) 1/273.15th of the normal freezing point of water (c) 100 times the difference between the triple point of water and the normal freezing

point of water (d) 1/273.15th of the triple point of water IES-15. In a new temperature scale say , the boiling and freezing points of water at

one atmosphere are 100 and 300 respectively. Correlate this scale with the Centigrade scale. The reading of 0 on the Centigrade scale is: [IES-2001]

(a) 0C (b) 50C (c) 100C (d) 150C


17

IES-16. Assertion (a): If an alcohol and a mercury thermometer read exactly 0C at the ice point and 100C at the steam point and the distance between the two points is divided into 100 equal parts in both thermometers, the two thermometers will give exactly the same reading at 50C. [IES-1995]

Reason (R): Temperature scales are arbitrary. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-17. Match List-I (Type of Thermometer) with List-II (Thermometric Property) and

select the correct answer using the code given below the [IES 2007] List-I List-II A. Mercury-in-glass 1. Pressure B. Thermocouple 2. Electrical resistant C. Thermistor 3. Volume D. Constant volume gas 4. Induced electric voltage Codes: A B C D A B C D (a) 1 4 2 3 (b) 3 2 4 1 (c) 1 2 4 3 (d) 3 4 2 1 IES-18. Pressure reaches a value of absolute zero [IES-2002] (a) At a temperature of 273 K (b) Under vacuum condition (c) At the earth's centre (d) When molecular momentum of system becomes zero IES-19. The time constant of a thermocouple is the time taken to attain: (a) The final value to he measured [IES-1997, 2010] (b) 50% of the value of the initial temperature difference (c) 63.2% of the value of the initial temperature difference (d) 98.8% of the value of the initial temperature difference

Work a Path Function IES-20. Assertion (A): Thermodynamic work is path-dependent except for an adiabatic

process. [IES-2005] Reason(R): It is always possible to take a system from a given initial state to

any final state by performing adiabatic work only. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-20a Work transfer between the system and the surroundings [IES-2011]

(a) Is a point function (b) Is always given by pdv (c) Is a function of pressure only (d) Depends on the path followed by the system


18

Free Expansion with Zero Work Transfer IES-21. Match items in List-I (Process) with those in List-II (Characteristic) and select

the correct answer using the codes given below the lists: List-I List-II [IES-2001] A. Throttling process 1. No work done B. Isentropic process 2. No change in entropy C. Free expansion 3. Constant internal energy D. Isothermal process 4. Constant enthalpy Codes: A B C D A B C D (a) 4 2 1 3 (b) 1 2 4 3 (c) 4 3 1 2 (d) 1 3 4 2 IES-22. The heat transfer, Q, the work done W and the change in internal energy U are

all zero in the case of [IES-1996] (a) A rigid vessel containing steam at 150C left in the atmosphere which is at 25C. (b) 1 kg of gas contained in an insulated cylinder expanding as the piston moves slowly

outwards. (c) A rigid vessel containing ammonia gas connected through a valve to an evacuated

rigid vessel, the vessel, the valve and the connecting pipes being well insulated and the valve being opened and after a time, conditions through the two vessels becoming uniform.

(d) 1 kg of air flowing adiabatically from the atmosphere into a previously evacuated bottle.

pdV-work or Displacement Work IES-23. One kg of ice at 0C is completely melted into water at 0C at 1 bar pressure.

The latent heat of fusion of water is 333 kJ/kg and the densities of water and ice at 0C are 999.0 kg/m3 and 916.0 kg/m3, respectively. What are the approximate values of the work done and energy transferred as heat for the process, respectively? [IES-2007]

(a) 9.4 J and 333.0 kJ (b) 9.4 J and 333.0 kJ (c) 333.0 kJ and 9.4 J (d) None of the above IES-24. Which one of the following is the

correct sequence of the three processes A, B and C in the increasing order of the amount of work done by a gas following ideal-gas expansions by these processes?


19

[IES-2006] (a) A B C (b) B A C (c) A C B (d) C A B IES-25. An ideal gas undergoes an

isothermal expansion from state R to state S in a turbine as shown in the diagram given below:

The area of shaded region is 1000 Nm. What is the amount is turbine work done during the process?

(a) 14,000 Nm (b) 12,000 Nm (c) 11,000 Nm (d) 10,000 Nm

[IES-2004]

IES-26. Assertion (A): The area 'under' curve on pv plane, pdv represents the work of reversible non-flow process. [IES-1992]

Reason (R): The area 'under' the curve Ts plane Tds represents heat of any reversible process.

(a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true IES-27. If pdv and vdp for a thermodynamic system of an Ideal gas on valuation

give same quantity (positive/negative) during a process, then the process undergone by the system is: [IES-2003]

(a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal IES-28. Which one of the following expresses the reversible work done by the system

(steady flow) between states 1 and 2? [IES-2008]

2 2 2 2

1 1 1 1(a) (b) (c) (d)pdv vdp pdv vdp

Heat Transfer-A Path Function IES-29. Assertion (A): The change in heat and work cannot be expressed as difference

between the end states. [IES-1999] Reason (R): Heat and work are both exact differentials. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true


20

Previous 20-Years IAS Questions

Thermodynamic System and Control Volume IAS-1. The following are examples of some intensive and extensive properties: 1. Pressure 2. Temperature [IAS-1995] 3. Volume 4. Velocity 5. Electric charge 6. Magnetisation 7. Viscosity 8. Potential energy Which one of the following sets gives the correct combination of intensive and

extensive properties? Intensive Extensive (a) 1, 2, 3, 4 5, 6, 7, 8 (b) 1, 3, 5, 7 2, 4, 6, 8 (c) 1, 2, 4, 7 3, 5, 6, 8 (d) 2, 3, 6, 8 1, 4, 5, 7

Zeroth Law of Thermodynamics IAS-2. Match List-I with List-II and select the correct answer using the codes given

below the lists: [IAS-2004] List-I List-II A. Reversible cycle 1. Measurement of temperature B. Mechanical work 2. Clapeyron equation C. Zeroth Law 3. Clausius Theorem D. Heat 4. High grade energy 5. 3rd law of thermodynamics 6. Inexact differential Codes: A B C D A B C D (a) 3 4 1 6 (b) 2 6 1 3 (c) 3 1 5 6 (d) 1 4 5 2 IAS-3. Match List-I with List-II and select the correct answer: [IAS-2000] List-I List-II A. The entropy of a pure crystalline 1. First law of thermodynamics substance is zero at absolute zero temperature B. Spontaneous processes occur 2. Second law of thermodynamics in a certain direction C. If two bodies are in thermal 3. Third law of thermodynamics equilibrium with a third body, then they are also in thermal equilibrium with each other D. The law of conservation of energy 4. Zeroth law of thermodynamics. Codes: A B C D A B C D (a) 2 3 4 1 (b) 3 2 1 4


21

(c) 3 2 4 1 (d) 2 3 1 4

International Temperature Scale IAS-4. A new temperature scale in degrees N is to be defined. The boiling and

freezing on this scale are 400N and 100N respectively. What will be the reading on new scale corresponding to 60C? [IAS-1995]

(a) 120N (b) 180N (c) 220N (d) 280N

Free Expansion with Zero Work Transfer IAS-5. In free expansion of a gas between two equilibrium states, the work transfer

involved [IAS-2001] (a) Can be calculated by joining the two states on p-v coordinates by any path and

estimating the area below (b) Can be calculated by joining the two states by a quasi-static path and then finding

the area below (c) Is zero (d) Is equal to heat generated by friction during expansion. IAS-6. Work done in a free expansion process is: [IAS-2002] (a) Positive (b) Negative (c) Zero (d) Maximum IAS-7. In the temperature-entropy diagram

of a vapour shown in the given figure, the thermodynamic process shown by the dotted line AB represents

(a) Hyperbolic expansion (b) Free expansion (c) Constant volume expansion (d) Polytropic expansion

[IAS-1995] IAS-8. If pdv and vdp for a thermodynamic system of an Ideal gas on valuation

give same quantity (positive/negative) during a process, then the process undergone by the system is: [IAS-1997, IES-2003]

(a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal

IAS-9. For the expression pdv to represent the work, which of the following conditions should apply? [IAS-2002]

(a) The system is closed one and process takes place in non-flow system (b) The process is non-quasi static (c) The boundary of the system should not move in order that work may be transferred (d) If the system is open one, it should be non-reversible


22

IAS-10. Air is compressed adiabatically in a steady flow process with negligible change in potential and kinetic energy. The Work done in the process is given by:

[IAS-2000, GATE-1996] (a) pdv (b) +pdv (c) vdp (d) +vdp IAS-11. Match List-I with List-II and select the correct answer using the codes given

below the lists: [IAS-2004] List-I List-II A. Bottle filling of gas 1. Absolute Zero Temperature B. Nernst simon Statement 2. Variable flow C. Joule Thomson Effect 3. Quasi-Static Path D. pdv 4. Isentropic Process 5. Dissipative Effect 6. Low grade energy 7. Process and temperature during phase change. Codes: A B C D A B C D (a) 6 5 4 3 (b) 2 1 4 3 (c) 2 5 7 4 (d) 6 1 7 4

pdV-work or Displacement Work IAS-13. Thermodynamic work is the product of [IAS-1998] (a) Two intensive properties (b) Two extensive properties (c) An intensive property and change in an extensive property (d) An extensive property and change in an intensive property

Heat Transfer-A Path Function IAS-14. Match List-I (Parameter) with List-II (Property) and select the correct answer

using the codes given below the lists: List-I List-II [IAS-1999] A. Volume 1. Path function B. Density 2. Intensive property C. Pressure 3. Extensive property D. Work 4. Point function Codes: A B C D A B C D (a) 3 2 4 1 (b) 3 2 1 4 (c) 2 3 4 1 (d) 2 3 1 4


23

Answers with Explanation (Objective)

Previous 20-Years GATE Answers GATE-1. Ans. (a) GATE-2. Ans. (a, b, c) For an isolated system no mass and energy transfer through the system. 0, 0, 0 or ConstantdQ dW dE E= = = = GATE-2a. Ans. (d)

GATE-3. Ans. (a) Iso-thermal work done (W) = 211

ln VRTV

21 1

1ln

0.030800 0.015 ln0.015

8.32kJ/kg

VP VV

= =

=

GATE-4. Ans. (c) It is free expansion. Since vacuum does not offer any resistance, there is no work transfer involved in free expansion.

Here, 2

10 = and Q1-2=0 therefore Q1-2 = U + W1-2 so, U = 0

GATE-5. Ans. (c) For closed system W = pdv+ , for steady flow W = vdp GATE-6. (c) GATE-7. Ans. (b) W = Resistance pressure. V = 1 V = 100 0.1 kJ = 1kJ GATE-8. Ans. (b) W vdp=

Previous 20-Years IES Answers IES-1. Ans. (b) IES-2. Ans. (d) Intensive property: Whose value is independent of the size or extent i.e. mass of

the system. Specific property: It is a special case of an intensive property. It is the value of an

extensive property per unit mass of system (Lower case letters as symbols) e.g., specific volume, density (v, ).

IES-2a. Ans. (c) Kinetic energy 21 mv2

depends on mass, Entropy kJ/k depends on mass so

Entropy is extensive property but specific entropy kJ/kg K is an intensive property.


24

IES-3. Ans. (a) Extensive property is dependent on mass of system. Thus volume is extensive

property. IES-4. Ans. (a) Extensive property: Whose value depends on the size or extent i.e. mass of the

system (upper case letters as the symbols) e.g., Volume, Mass (V, M). If mass is increased, the value of extensive property also increases.

IES-4a Ans. (d) The properties like temperature, viscosity which are Independent of the MASS of the system are called Intensive property IES-5. Ans. (d)

But remember 100% heat cant be convertible to work but 100% work can be converted to heat. It depends on second law of thermodynamics.

A thermodynamic system is defined as a definite quantity of matter or a region in space upon which attention is focused in the analysis of a problem.

The system is a macroscopically identifiable collection of matter on which we focus our attention

IES-5a Ans. (d) IES-6. Ans. (b) In closed thermodynamic system, there is no mass transfer but energy transfer

exists. IES-7. Ans. (c) IES-8. Ans. (d) Isentropic means reversible adiabatic. Heat transfer in any finite temp difference is

irreversible. IES-9. Ans. (a) The energy transfer as heat and work during the forward process as always

identically equal to the energy transfer is heat and work during the reversal or the process is the correct reason for maximum efficiency because it is conservative system.

IES-9a. Ans. (b) IES-10. Ans. (b) Isolated System - in which there is no interaction between system and the

surroundings. It is of fixed mass and energy, and hence there is no mass and energy transfer across the system boundary.

IES-10a Ans. (a) IES-10b. Ans. (c) IES-10c. Ans. (d) IES-11. Ans. (a) All temperature measurements are based on Zeroth law of thermodynamics. IES-12. Ans. (a) Entropy - related to second law of thermodynamics. Internal Energy (u) = f (T) only (for an ideal gas) Van der Wall's equation related to => real gas. IES-13. Ans. (d) IES-14. Ans. (d)

IES-15.Ans. (d) 0 300 0 150 C100 300 100 0

C C = =


25

IES-16. Ans. (b) Both A and R are correct but R is not correct explanation for A. Temperature is independent of thermometric property of fluid.

IES-17. Ans. (d) IES-18. Ans. (d) But it will occur at absolute zero temperature. IES-19. Ans. (c) Time Constants: The time constant is the amount of time required for a

thermocouple to indicated 63.2% of step change in temperature of a surrounding media. Some of the factors influencing the measured time constant are sheath wall thickness, degree of insulation compaction, and distance of junction from the welded cap on an ungrounded thermocouple. In addition, the velocity of a gas past the thermocouple probe greatly influences the time constant measurement. In general, time constants for measurement of gas can be estimated to be ten times as long as those for measurement of liquid. The time constant also varies inversely proportional to the square root of the velocity of the media.

IES-20. Ans. (c) IES-20a Ans. (d) IES-21. Ans. (a) IES-22. Ans. (c) In example of (c), it is a case of free expansion heat transfer, work done, and

changes in internal energy are all zero.

IES-23. Ans. (a) Work done (W) = P V = 100 (V2 V1) = 1002 1

m m

= 100 kPa 1 1999 916

= 9.1 J

IES-24. Ans. (d) 4 (2 1) 4kJ= = =AW pdV

1 3 (7 4) 4.5kJ21 (12 9) 3kJ

= = == = =

B

C

W pdV

W pdV

IES-25. Ans. (c) Turbine work = area under curve RS

( )

( )3

5

1 bar 0.2 0.1 m 1000 Nm

10 0.2 0.1 Nm 1000Nm 11000Nm

pdv= = += + =

IES-26. Ans. (b) IES-27. Ans. (d) Isothermal work is minimum of any process.

0[ is onstant]pv mRTpdv vdp T c

pdv vdp

=+ =

=

IES-28. Ans. (b) For steady flow process, reversible work given by 2

1vdp .

IES-29. Ans. (c) A is true because change in heat and work are path functions and thus can't be expressed simply as difference between the end states. R is false because both work and heat are inexact differentials.


26

Previous 20-Years IAS Answers IAS-1. Ans. (c) Intensive properties, i.e. independent of mass are pressure, temperature, velocity

and viscosity. Extensive properties, i.e. dependent on mass of system are volume, electric charge, magnetisation, and potential energy. Thus correct choice is (c).

IAS-2. Ans. (a) IAS-3. Ans. (c) IAS-4. Ans. (d) The boiling and freezing points on new scale are 400 N and 100N i.e. range is

300N corresponding to 100C. Thus conversion equation is N = 100 + 3 C = 100+ 3 60 = 100 + 180 = 280 N

IAS-5. Ans. (c) IAS-6. Ans. (c) Since vacuum does not offer any resistance, there is no work transfer involved in

free expansion. IAS-7. Ans. (b) IAS-8. Ans. (d) Isothermal work is minimum of any process. IAS-9. Ans. (a) IAS-10. Ans. (c) For closed system W = pdv+ , for steady flow W = vdp IAS-12. Ans. (b) Start with D. PdV only valid for quasi-static path so choice (c) & (d) out.

Automatically C-4 then eye on A and B. Bottle filling of gas is variable flow so A-2. IAS-13. Ans. (c) W = pdv where pressure (p) is an intensive property and volume (v) is an

extensive property IAS-14. Ans. (a) Pressure is intensive property but such option is not there.

S

FSt

AlWhwoMa

In

In

K Mon

2. Theo

irst Lawtatemen

When asum of interact

Mathem

( The sum

lternate sthen a closed ork. athematical

other word

nternal

Firdals

First

ory at a

w of Tht:

a closed systeheat interacttions. matically

Q)cyclmmations be

tatement: system und

lly

ds for a two

AQ

Energy

rst Law

t Law

a Glan

hermody

em executes ations is equa

le = (eing over the

: ergoes a cycl

v Q process cy

+A1 2 Qy A Pr

W

w of T

28

w of T

ce (Fo

ynamic

a complete cyal to the sum

W)c entire cycle.

le the cyclic

=Q cle

=B2 1Qroperty

Which can be

Therm

herm

or GAT

cs

ycle the of work

ycle

integral of h

v +A1 2W

y of Sys

e written as

odyna

modyn

TE, IES

heat is equal

W

+ B 2 1Wstem

s

amics Cha

namic

S & PS

to the cyclic

1

apter 2

cs

SUs)

c integral of

S

We

K Mon

Since AB) the the samthemse

This impropert

This prall the e

e enunciate t

An isolTheref

Let us r Work d Work d Since t

area.

Firdals

A and B are adifference Q

me. The diffeelves depend mplies that thty of the systroperty is calenergies at a

the FIRST LA

lated systemfore, E remareconsider th

done during tdone during these two are

rst Law

arbitrarily chQ W remaerence depenon the path f

he difference em. led the energ

a given state.

AW for a proc

Q

m which doains constanhe cycle 1-2 ahe path A = Ahe path B = Aeas are not e

w of T

29

hosen, the conains a constads only on thfollowed. But between the

gy of the syst

cess as

W

oes not internt for such

along path A Area under 1Area under 1equal, the ne

Therm

nclusion is, aant as long ahe end pointt their differe

e heat and wo

tem. It is des

W =

ract with th a system. and 2-1 alon1-A-2-3-4 1-B-2-3-4 et work inter

odyna

as far as a pras the initial s of the procence does notork interactio

signated as E

dE

he surround

ng path B as s

raction is tha

amics Cha

rocess is conc and the finacess. Note tht. ons during a

E and is equa

dings Q = 0

shown in fig.

at shown by

apter 2

cerned (A or al states are at Q and W

process is a

al to some of

and W = 0.

the shaded

S

Thsubmo

Re

an

Q

[W BuMitraAllthe

K Mon The net Therefo First la

when tthe syst

In othout, yo

Thus, tsense.

In the e The cyc

Obviouenviron

For a p We can

of the s We can

the ene Energy

he internal enbstance doesovement. Tha

ecognize that

d similarly u

QW = = [

Where C = Vel

ut Remeicroscopic vieanslational l these energe gas.

Firdals t area is 1A2ore some woraw compels there is alsotem and the

her words, iou must givethe first law

example showcle shown hasly, the net h

nment. (as perocess we can

n extract worsystem. n add heat toergy of the syy of a system

nergy depend not include at so why in S

h = u + pu1 + p1v

[(h2 + (h2 - hlocity (m/s), h

ember: ew of a gas is energy, rotgies summed

rst Law

B1. rk is derived that this iso heat inter surroundingif you havee heat in. can be const

wn the area uas negative heat interacter the sign con have Q = 0

rk without su

the system ystem. m is an exte

ds only upon any energy tSFEE (Stead

pv from whicv1 = h1

C22/2 +1) + (Ch = Specific e

a collection tational ened over all the

w of T

30

by the cycle. s possible oaction betwe

gs. e to get wo

trued to be a

under curve Awork output

tion is also neonvention). 0 or W = 0 upplying hea

without doin

ensive prop

the initial anthat it may pdy flow energ

ch u2 + p2

+ gZ2) -22/2 - C

enthalpy (J/k

of particles iergy, vibrate particles of

Therm

only een

ork

statement o

A < that undt or it will regative. This

t (during a

ng work (in p

perty

nd final statepossess as a rgy equation) C

2 v2 = h2

- (h1+CC12/2) +kg), z = elevat

n random motional energf the gas, form

odyna

of conservatio

der B eceive work s implies tha

process) bu

process) wh

es of the systresult of its mC2/2 and gz is

C12/2 + + g(Z2 -

tion (m)

otion. Energygy and specim the specifi

amics Cha

on of energy

from the sut this cycle w

ut sacrificing

hich will go to

tem. Internalmacroscopics there.

gZ1)] - Z1)]

y of a particlific electronic internal en

apter 2

- in a broad

urroundings. will heat the

g the energy

o increasing

l energy of a c position or

or

e consists of nic energy. nergy, e , of

S

PThdes Thfora p Thcon

ETh

It i

Sp

It i Intwopresys

W

K Monerpetua

he first law ststroyed, but

here can be nrm of energy perpetual mo

he converse ntinuously co

nthalpyhe enthalpy o

is an extens

ecific Enthal

is an intensi

ternal energyork other thaessure procestem of unit m

At constan Therefo

Where H = USpecific enth

Where h = sp

Firdals al Motiotates the genonly gets tra

no machine disappearingotion machine

of the abovonsume work

A P

y f a substance

sive propert

lpy

ive propert

y change is an pdv work.ss involving mass of a pu d Q =

nt pressure pdv =

ore ( )dQP

= or ( )dQ

P =

or ( )dQP

=U + PV is thehalpy h = H/m

pecific entha

rst Law

on Macneral principlansformed fro

which wouldg simultaneoe of the first

e statementk without som

PMM1

e H is defined

Hty of a system

hty of a system

equal to the It is possibl no work othre substance

= du + pdv

d(pv) ( )du d pv= +

= d(u+pv)

= dh enthalpy, a

m, kJ/kg and

lpy, kJ/kg

w of T

31

chine ofle of the consom one form t

d continuousously (Fig. shkind, or in br

t is also trume other form

d as

H = Um and its uni

h = um and its unit

heat transfele to derive aher than pdve.

property of s also h = u +

Therm

f the Fiservation of eto another.

sly supply mhown in belowrief, PMM1.

ue, i.e. therem of energy a

The

U + Pit is kJ.

u + pt is kJ/kg.

erred in a coan expressionv work. In su

system. pdv

odyna

rst Kinenergy. Energ

mechanical wow). Such a ficA PMM1 is t

e can be noappearing sim

e Converse o

PV

pv onstant volumn for the heauch a proces

amics Cha

d-PMMgy is neither

ork without titious machthus imposs

machine wmultaneously

of PMM1

me process iat transfer inss in a closed

apter 2

M1 created nor

some other hine is called

ible.

which would y (Fig.).

involving no n a constant d stationary

First Law of Thermodynamics S K Mondals Chapter 2

32

u = specific internal energy, kJ/kg dv = change in specific volume, m3/kg. Specific heat at constant volume

The specific heat of a substance at constant volume Cv is defined as the rate of change of specific internal energy with respect to temperature when the volume is held constant, i.e.,

= v vuCT

For a constant volume process

( ) 21

.T

vvT

u C dT = The first law may be written for a closed stationary system composed of a unit mass of a pure

substance. Q = u + W or d Q = du + d W For a process in the absence of work other than pdv work d W = pdv Therefore d Q = du + pdv

Therefore, when the volume is held constant

( ) ( )( ) 2

1

v v

v

u

.T

vT

Q

Q C dT

=

=

Since u, T and v are properties, Cv is a property of the system. The product m

Cv is called the heat capacity at constant volume (J/K).

Specific heat at constant pressure

The specific heat at constant pressure pC is defined as the rate of change of specific enthalpy with respect to temperature when the pressure is held constant.

PC P

hT

= For a constant pressure process

( ) 21

.T

PPT

h C dT = The first law for a closed stationary system of unit mass

dQ = du + pdv Again, h = u + pv Therefore dh = du + pdv + vdp = d Q + vdp Therefore dQ = dh vdp Therefore ( dQ )P = dh


33

or ( ) ( h)ppQ = Form abow equations ( ) 2

1

P .T

PT

Q C dT= pC is a property of the system, just like Cv. The heat capacity at constant pressure is equal to m

pC (J/K).

Application of First Law to Steady Flow Process S.F.E.E S.F.E.E. per unit mass basis

(i) + + + = + + +

2 21 2

1 1 2 2C Cd Q d Wh g z h g z2 d m 2 d m

[h, W, Q should be in J/Kg and C in m/s and g in m/s2]

(ii) + + + = + + +2 21 1 2 2

1 2gZ Q gZ Wh h

2000 1000 dm 2000 1000 dmC d C d

[h, W, Q should be in KJ/Kg and C in m/s and g in m/s2] S.F.E.E. per unit time basis

+ + + = + + +

21

1 1 1

22

2 2 2

w z2

w z2

x

C dQh gd

dWCh gd

Where, w = mass flow rate (kg/s) Steady Flow Process Involving Two Fluid Streams at the Inlet and Exit of the Control Volume

S

Ma

Wh En

SoThengNoA ndrobel

K Mon

ass balance

here v = spec

nergy balan

ome examplhe following egineering sysozzle and Dinozzle is a dop, whereas low a nozzle

Firdals

+1 11

A Cv

cific volume (

nce

+= +

1 1

3 3

w h

w h

le of steadyexamples illustems. iffuser:

device which a diffuser inwhich is insu

11 2Ch +

rst Law

++

1

2 2

2

w wA C

v(m3/kg)

+ +

+ +

21

1

23

3

2

2

C Z g

C Z g

y flow proceustrate the a

increases thncreases the ulated. The s21

12QZ g

dm+ + d

F

w of T

34

==

2

3 3

3

w wA C

v

+ + +

2 2

4 4

w h

g w h

esses:- pplications o

he velocity opressure of a

steady flow en22

2 2Ch Z= + +

Fig.

Therm

++

3 4

4

4

wA C

v

+ +

+ +

22

2

24

4

2

2

C Z g

C Z g

of the steady

or K.E. of aa fluid at thenergy equati

2xWZ g

dm+ d

odyna

4

+ +

x

dQd

dWgd

flow energy

a fluid at thee expense of on of the con

amics Cha

equation in

e expense of its K.E. Figu

ntrol surface

apter 2

some of the

its pressure ure show in gives


35

Here 0; 0,xdWdQdm dm

= = and the change in potential energy is zero. The equation reduces to 2 21 2

1 22 2C Ch h+ = + (a)

The continuity equation gives 1 1 2 2

1 2

A Aw = = C Cv v

(b)

When the inlet velocity or the velocity of approach V1 is small compared to the exit velocity V2, Equation (a) becomes

22

1 2

2 1 2

22( ) /

Ch h

or C h h m s

= +=

where (h1 h2) is in J/kg. Equations (a) and (b) hold good for a diffuser as well. Throttling Device: When a fluid flows through a constricted passage, like a partially opened value, an orifice, or a porous plug, there is an appreciable drop in pressure, and the flow is said to be throttled. Figure shown in below, the process of throttling by a prettily opened value on a fluid flowing in an insulated pipe. In the steady-flow energy equation-

0, 0xWdQdm dm

= =d And the changes in P. E. are very small and ignored. Thus, the S.F.E.E. reduces to

2 21 2

1 22 2C Ch h+ = +

(Fig.- Flow Through a Valve)

Often the pipe velocities in throttling are so low that the K. E. terms are also negligible. So

1 2h h= or the enthalpy of the fluid before throttling is equal to the enthalpy of the fluid after throttling. Turbine and Compressor: Turbines and engines give positive power output, whereas compressors and pumps require power input. For a turbine (Fig. below) which is well insulated, the flow velocities are often small, and the K.E. terms can be neglected. The S.F.E.E. then becomes


36

(Fig.-. Flow through a Turbine)

1 2

1 2

x

x

dWh hdm

Wor h hm

= +

=

The enthalpy of the fluid increase by the amount of work input. Heat Exchanger: A heat exchanger is a device in which heat is transferred from one fluid to another, Figure shown in below a steam condenser where steam condense outside the tubes and cooling water flows through the tubes. The S.F.E.E for the C.S. gives

c 1 s 2 c 3 s 4s 2 4 c 3 1

w w w w, w ( ) w ( )h h h h

or h h h h+ = +

= Here the K.E. and P.E. terms are considered small, there is no external work done, and energy exchange in the form of heat is confined only between the two fluids, i.e. there is no external heat interaction or heat loss.

Fig. -

Figure (shows in below) a steam desuperheater where the temperature of the superheated steam is reduced by spraying water. If w1, w2, and w3 are the mass flow rates of the injected water, of the steam entering, and of the steam leaving, respectively, and h1, h2, and h3 are the corresponding enthalpies, and if K.E. and P.E. terms are neglected as before, the S.F.E.E. becomes 1 1 2 2 3 3w h w h w h + = and the mass balance gives w1 + w2 = w3

S

Thpraun (1)

(2)

(b)

(c)

K Mon

he above lawactical situat

nity.

) Work deve(a) Water In this cas

1 1 1v +z g p

(b) Steam In this cas

( 1hW =

) Work abso

(a) Centrifug The system

In this sys

1 1 1v +z g p

Centrifuga

21

1 h2C + +

Blowers

Firdals

w is also calltions as work

eloping syst turbines se Q = 0 and

21

2 z g 2C+ = +

or gas turbinse generally

) 211 2 h C+ orbing syste

gal water pumm is shown in

stem Q = 0 a

2W z g + = +

al compress22

2CW Q =

In this case

rst Law

led as steadk developing

tems

U = 0 and e2 2 p v W+ +

nes Z can be as

22 Q

2C +

ems

mp the Figure b

Fig. nd U = 0; th

22

2 2 p v 2C+ +

sor In this s22

2 h + we have z =

w of T

37

dy flow ene system and

equation bec

sumed to be

below

he energy equ

system z =

= 0, 1 1v p =

Therm

ergy equati work absorp

omes

zero and the

uation now b

0 and the equ

2 2 p v and Q

odyna

ion. This canption system

e equation be

becomes,

uation becom

= 0; now the

amics Cha

n be applied. Let the ma

comes

mes,

e energy simp

apter 2

d to various ass flow rate

plifies to

S

(d)

(e)

(3) (a)

(b)

(c)

(viMacanunwitsur

K Mon 1u +W

) Fans In fand hence th

22

2CW =

Reciprocatequation app

or

) Non-work

Steam boiequation for

Steam conare very sma(h1 h2) andheat lost by

Steam nozz

In this systepossible heat The ene

ii) Unsteadyany flow procn be analyzed

nder non-steathin the contrface, as give

Firdals

22

2 u as2C= +

fans the temhe energy equ22

ting compreplied to a rec

(1

2

h hh

QW Q

== +

developing

iler In this a boiler beco

ndenser Inall. Under sted this heat is steam will be

zle:

em we can at loss also zeergy equation

1

2

h

or C

+=

y Flow Analcesses, such ad by the cont

ady state condtrol volume isen below:

rst Law

2 1s C C mperature risuation for fan

essor In a iprocating co

)2

1

h h

W

g and absorb

s system we omes Q = (h2

n this systemeady conditio also equal te equal to he

ssume Z anro. n for this cas

2 21 2

2

21 1

2 22(

C Ch

C h

+ = += +

lysis:- as filling up atrol volume teditions (Figus accumulate

w of T

38

e is very smns becomes,

reciprocatingompressor is

bing system

neglect Z, 2 h1)

m the work doons the chango the changeat gained by

nd W to be z

se becomes. 22

2 )h

and evacuatiechnique. Co

ure-shown in ed is equal to

Therm

all and heat

g compressor

ms

KE and W

one is zero age in enthalpe in enthalpy the cooling w

zero and hea

ing gas cylindonsider a dev below). The ro the net rate

odyna

loss is negle

r KE and P

(i.e.) Z = K

and we can apy is equal toy of cooling wwater.

at transfer w

ders, are not vice through wrate at which

e of mass flow

amics Cha

ected (i.e.) h

PE are neglig

KE = W = 0;

also assume o heat lost bywater circulat

which is noth

steady, Suchwhich a fluidh the mass ofw across the c

apter 2

h = 0, q = 0

gibly energy

; the energy

Z and KE y steam. Q = ted (i.e.) the

ing but any

h processes d is flowing f fluid control

S

Wh ThacrRa

Wh

Fo

an

Or

K Mon

here vm is th Over an vm =

he rate of accross the contate of energy

vdE =d

U= vE

here m is the

21

1

dEd

h +2

=

v

C

llowing Figu

vE = Q

Equatio

d the equatiovdE

d =r dEv = d

Firdals

1vdm w w

d =

he mass of fluny finite peri

1 2 m m cumulation ofrol surface. I increase = R

21

1 1

2

h + +Z2

mCU + + m2

Cw

e mass of flui

21 1

1

d mCU + d 2

dm+Z gd

= +

ure shows all

1h + + Q W

on (A) is genevdE = 0

d

on reduces FdQ dW d d

dQ dW or

rst Law

F1

2dm dwd d=

uid within thiod of time f energy withIf Ev is the en

Rate of energy

1

v

dQZ g +d

mgZ

w

id in the cont2

v

2

C + mgZ2

dQ h +d 2

=+

C

these energy21

1+ +Z g dm2

C

eral energy e

For a closed s

dQ = dE + d

w of T

39

Fig. 2dm

d

he control vol

hin the contrnergy of fluidy inflow Ra

22

2 2 2h + +Z2

Cw

trol volume a

22 2

2dm+Z g

2 d

=

C

y flux quantit22

1 2m h + 2

C

equation. For

system w1 = 0

dW

Therm

ume at any i

rol volume isd within the cate of energy

2dWgd

at any instant

dWd

ties. For any

2 2+Z g dm

Fig. r steady flow

0, w2 = 0, the

odyna

instant.

s equal to thecontrol volum outflow.

equati

t

(equa

y time interva

,

en from equat

amics Cha

e net rate of me at any ins

...ion A

)........ation B

al, equation (

tion (A),

apter 2

energy flow tant,

(B) becomes

S

Flo ExVatec theansup

pp

Sywh

Wh

bee

vol

Us

K Monow Processes

xample of a ariable flow pchnique, as il

Considee beginning td gas flows inpply to the pi

P P P,T , v , h ,

ystem Technhich would ev

Energy

1E m=here ( 2m m

2E m

E E

= =

The P.E. ten omitted.

Now, therlume. Then t

sing the firstQ = E = 2m

Firdals

s

variable floprocesses mayllustrated beer a process ithe bottle connto the bottleipeline is ver

P P u and v .

nique: Assumventually enty of the gas be

(1 1 2m u m +)1m is the ma

2 2

2 1 2

m u

E E m=terms are ne

re is a changthe work don

W = = =

t for the proc + W

(2 2 1 1u m u

rst Law

ow problemy be analyzedlow. in which a gantains gas of e till the masry large so th

me an enveloter the bottleefore filling.

) 21m 2P PC u

+ass of gas in t

(2 2 1 1u m uglected. The

ge in the voe

( 2 1 p V= p V( p 0 mp =

= ( 2 m mess

( )2 1m m C

w of T

40

m: d either by th

as bottle is fi mass m1 at sss of gas in thhat the state

ope (which is e, as shown in

the pipeline a

) 21 1m u 2PC

gas in the bo

lume of gas

)1 )2 1 Pm m v

)1 Pm p v p

(2P P+ u m2C

Therm

he system tec

lled from a pstate p1, T1, vhe bottle is mof gas in the

extensible) on Figure abov

and tube whi

Pu+

ottle is not in

because of t

P

)2 1m m P Pp v

odyna

chnique or th

pipeline (Figuv1, h1 and u1. m2 at state p2 pipeline is c

of gas in the pve.

ich would en

n motion, and

the collapse

amics Cha

he control vol

ure shown in The valve is, T2, v2, h2 anonstant at

pipeline and

ter the bottle

d so the K.E.

of the envel

apter 2

lume

below). In s opened nd u2. The

the tube

e.

terms have

lope to zero

S

wh CoFigwr

Sin

No

Q

DLeapp

As

Ag

or

or

whqu

K Mon = 2m

hich gives the

ontrol Volumgure above, Aritten on a tim

nce hP and CP

ow v 2

2 2

E U U

Q m u m

==

Dischargt us considerplying first la

suming K.E.

gain

hich shows tasi-static.

For cha

Firdals

(2 2 1 1u m u e energy bala

me TechniqApplying the me rate basis

vdE dQd d

= +P are constan

vE Q h = +

1 2 2

1 1 P

U m u

m u h +

=

ging anr a tank discaw to the con

and P.E. of td(m

mdu+ udm

( )

== =

+=

= + == 0

dm dum pv

V vm covdm mdvdm dvm v

du dvpv vd u pvdQ

that the pro

arging the tan

rst Law

( ) 2P2 1m m 2C

ance for the p

que: Assume energy equas -

2P

Pdmh +

2 dC

nt, the equati

(2PP 2h + m2

C

(1 12P

2

m u

+ m m2

C

nd Charcharging a fluntrol volume,

= VdU dQ

the fluid to bu) = hdm

= udm+ pv d

=.

0

0

onst

cess is adiab

nk

w of T

41

2P

P+ h2

process.

a control volation in this

m

ion is integra

)1m

)1m

rging a uid into a su,

+ + Q h

be small and

dm

batic and

Therm

lume boundecase, the foll

ated to give f

Tank upply line (Fi

+ 2

o

gz2

C

dQ = 0

Chargi

odyna

d by a controlowing energy

for the Total

igure). Since

outut

dm

ng and Disc

amics Cha

ol surface as y balance ma

process

e xdW = 0 an

charging a

apter 2

shown in ay be

nd dmin = 0,

Tank


42

( ) = = = 2 2 1 1in

2 2 1 1

V

p p

hdm U m u m u

m h m u m u

where the subscript p refers to the constant state of the fluid in the pipeline. If the tank is initially empty, m1 = 0. = 2 2p pm h m u Since

==

2

2

p

p

m mh u

If the fluid is an ideal gas, the temperature of the gas in the tank after it is charged is given by

==

2

2

p p v

p

c T c Tor T T

PROBLEMS & SOLUTIONS Example 1 The work and heat transfer per degree of temperature change for a closed system is given by

1 1/ ; /30 10dW dQkJ C kJ CdT dT

= = Calculate the change in internal energy as its temperature increases from 125C to 245C. Solution:

( ) ( )

( )

2

1

2

1

2 1

301 1 245 125

30 30 30

101 245 125 12

10 10

T

T

T

T

dTdW

dTW T T

dTdQ

dTQ kJ

=

= = =

=

= = =

Applying First Law of Thermodynamics Q = W + U U = Q W = 12 4 = 8kJ. Example 2 Air expands from 3 bar to 1 bar in a nozzle. The initial velocity is 90 m/s. the initial temperature is 150C. Calculate the velocity of air at the exit of the nozzle. Solution: The system in question is an open one. First Law of Thermodynamics for an open system gives

2 21 2

1 1 1 2 2 22 2C Cw h Z g Q w h Z g W

+ + + = + + +

Since the flow is assumed to be steady. 1 2w w=

Flow in a nozzle is adiabatic flow.


43

Hence Q = 0 Also W = 0 The datum can be selected to pass through axis; then Z1 = Z2. Hence

( ) ( )

2 21 2

1 2

2 22 1

1 2

1

22 1

1

10.4/1.4

2

2 2

2 2

for air = 1.4T 150 273 423

1T 423 3093

C Ch h

C Cor h h

and

pT Tp

K

+ = +

= +

=

= + = = =

For air Cp = 1.005 kJ/kgC Cv = 0.718 kJ/kgC. R = 287 J/kg K = 0.287 kJ/kg K S.F.E.E. : - We have (h1 h2) = Cp(T1 - T2)

( )2 2 232 11 22

901.005 10 (423 309)2 2 2

, 487 / .

= + = +=

C Ch h

or C m s

Example 3 An evacuated cylinder fitted with a valve through which air from atmosphere at 760 mm Hg and 25C is allow to fill it slowly. If no heat interaction is involved, what will be the temperature of air in the bottle when the pressure reaches 760 mm Hg? Use the following: (1) Internal energy of air u = 0u + 0.718T kJ/kg where T is temperature in C. (2) R = 0.287 kJ/kg K. Solution: Applying first law, ignoring potential and kinetic energy terms, to the vessel as control volume.


44

( )( )

( )

i i e e 2 2 1 1

e

1 2

2

0

0

0 2

2 0

2 0

Q m h = m h m u m u WHere Q 0, W 0, m 0 no mass leaving from control vol.

0 evacuated

0.718 0.287 2730.718 25 0.287 298103.48 /

103.48 /0.71

i

i

i i i i

m m mh u

h u pv u T Tuu kJ kg u

or u u kJ kgu u

+ + += = =

= = =

= + = + + += + + = + =

== + 2

2 02

8103.48 144.2

0.718 0.718

Tu uT C= = =

Example 4 A system whose mass is 4.5 kg undergoes a process and the temperature changes from 50 C to 100C. Assume that the specific heat of the system is a function of temperature only. Calculate the heat transfer during the process for the following relation ship.

801.25 /160nc kJ kg C

t= + + [t is in oC]

Solution:

[ ] ( )[ ] ( ) ( )( )

100 100

1 250 50100 100

50 50

10010050

50

804.5 1.25160

4.5 1.250.0125 2.0

14.5 1.25 ln 0.0125 2.00.0125

14.5 1.25 50 1.25 2.0 0.625 20.0125

4

= = + + = + + = + + = + + +

=

nQ mc dt dtt

dtdtt

t t

ln ln

1 3.25.5 62.5 3580.0125 2.625

+ = ln kJ

S

A

A CoThstearno

GA

GA

GA

K MonASKED

P

Applicati

ommon Dhe inlet andeam for ane as indic

otations are

ATE-1. If mthe(a)

ATE-2. Asswaene

(a)

ATE-3. Thethe

Acc(a)

Firdals

D OBJEC

Previou

ion of F

ata for Qud the outle

n adiabaticcated in t as usually

mass flow re turbine (in12.157

sume the abter at the ergy effects

0.293

e followingermodynam

cording to tFigures 1and

rst Law

CTIVE Q

us 20-Y

First Law

uestions Qet conditioc steam turthe figure. followed.

rate of stean MW) is:

bove turbininlet to the

s, the specifi

(

g four figmic cycle, on

the first lawd 2 (b) Fig

w of T

45

QUEST

Years

w to Ste

Q1 and Q2ns of rbine The

m through

(b) 12.941

ne to be pare pump is fic work (in

b) 0.35 1

ures have n the p-v and

w of thermogures 1and 3

Therm

IONS (G

GATE

eady Flo

2:

the turbin (c

rt of a simp1000 kg/m3. kJ/kg) supp

(c)

been drad T-s planes

dynamics, e(c) Figures

odyna

GATE

E Ques

ow Proc

e is 20 kg/s

c) 168.001

ple Rankine. Ignoring kplied to the

) 2.930

awn to res.

equal areass 1and 4

amics Cha

, IES,

stions

cess S.F

[GA

s the power [GA (d)

e cycle. The kinetic and pump is:

[G (d) 3

epresent a [G

are enclose(d) Figures

apter 2 IAS)

F.E.E

ATE-2009]

r output of ATE-2009]

168.785

density of d potential

GATE-2009] 3.510

fictitious GATE-2005]

ed by 2 and 3


46

Internal Energy A Property of System GATE-4. A gas contained in a cylinder is compressed, the work required for

compression being 5000 kJ. During the process, heat interaction of 2000 kJ causes the surroundings to the heated. The change in internal energy of the gas during the process is: [GATE-2004]

(a) 7000 kJ (b) 3000 kJ (c) + 3000 kJ (d) + 7000 kJ GATE-4a. The contents of a well-insulated tank are heated by a resistor of in

which 10 A current is flowing. Consider the tank along with its contents as a thermodynamic system. The work done by the system and the heat transfer to the system are positive. The rates of heat (Q), work (W) and change in internal energy during the process in kW are [GATE-2011]

(a) Q = 0, W = 2.3, = +2.3 (b) Q = +2.3, W = 0, = +2.3 (c) Q = 2.3, W = 0, = 2.3 (d) Q = 0, W = +2.3, = 2.3

Discharging and Charging a Tank GATE-5. A rigid, insulated tank is initially

evacuated. The tank is connected with a supply line through which air (assumed to be ideal gas with constant specific heats) passes at I MPa, 350C. A valve connected with the supply line is opened and the tank is charged with air until the final pressure inside the tank reaches I MPa. The final temperature inside the tank

(A) Is greater than 350C (B) Is less than 350C (C) Is equal to 350C (D) May be greater than, less than, or equal to

350C, depending on the volume of the tank

Previous 20-Years IES Questions

First Law of Thermodynamics IES-1. Which one of the following sets of thermodynamic laws/relations is directly

involved in determining the final properties during an adiabatic mixing process? [IES-2000]

(a) The first and second laws of thermodynamics (b) The second law of thermodynamics and steady flow relations (c) Perfect gas relationship and steady flow relations (d) The first law of thermodynamics and perfect gas relationship

23

( U) U U U U


47

IES-2. Two blocks which are at different states are brought into contact with each

other and allowed to reach a final state of thermal equilibrium. The final temperature attained is specified by the [IES-1998]

(a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Third law of thermodynamics IES-3. For a closed system, the difference between the heat added to the system and

the work done by the system is equal to the change in [IES-1992] (a) Enthalpy (b) Entropy (c) Temperature (d) Internal energy IES-4. An ideal cycle is shown in the figure. Its

thermal efficiency is given by

3 3

1 1

2 2

1 1

1 11(a)1 (b) 1

1 1

v vv vp pp p

( )( )

( )( )3 1 3 11 12 1 1 2 1 1

1(c)1 (b) 1

v v v vp pp p v p p v

[IES-1998] IES-5. Which one of the following is correct? [IES-2007] The cyclic integral of )( WQ for a process is: (a) Positive (b) Negative (c) Zero (d) Unpredictable IES-6. A closed system undergoes a process 1-2 for which the values of Q1-2 and W1-2 are

+20 kJ and +50 kJ, respectively. If the system is returned to state, 1, and Q2-1 is -10 kJ, what is the value of the work W2-1? [IES-2005]

(a) + 20 kJ (b) 40 kJ (c) 80 kJ (d) +40 kJ IES-7. A gas is compressed in a cylinder by a movable piston to a volume one-half of

its original volume. During the process, 300 kJ heat left the gas and the internal energy remained same. What is the work done on the gas? [IES-2005]

(a) 100kNm (b) 150 kNm (c) 200 kNm (d) 300 kNm IES-8. In a steady-flow adiabatic turbine, the changes in the internal energy,

enthalpy, kinetic energy and potential energy of the working fluid, from inlet to exit, are -100 kJ/kg, -140 kJ/kg, -10 kJ/kg and 0 kJ/kg respectively. Which one of the following gives the amount of work developed by the turbine? [IES-2004]

(a) 100 kJ/kg (b) 110 kJ/kg (c) 140 kJ/kg (d) 150 kJ/kg


48

IES-9. Gas contained in a closed system consisting of piston cylinder arrangement is expanded. Work done by the gas during expansion is 50 kJ. Decrease in internal energy of the gas during expansion is 30 kJ. Heat transfer during the process is equal to: [IES-2003]

(a) 20 kJ (b) +20 kJ (c) 80 kJ (d) +80 kJ

IES-10. A system while undergoing a cycle [IES-2001] A B C D A has the values of heat and work transfers as given in the Table: Process Q kJ/min W kJ/min

AB BC CD DA

+687 -269 -199 +75

+474 0

-180 0

The power developed in kW is, nearly, (a) 4.9 (b) 24.5 (c) 49 (d) 98 IES-11. The values of heat transfer and work transfer for four processes of a

thermodynamic cycle are given below: [IES-1994] Process Heat Transfer (kJ) Work Transfer (kJ)

1 2 3 4

300 Zero -100 Zero

300 250 -100 -250

The thermal efficiency and work ratio for the cycle will be respectively. (a) 33% and 0.66 (b) 66% and 0.36. (c) 36% and 0.66 (d) 33% and 0.36. IES-12. A tank containing air is stirred by a paddle wheel. The work input to the

paddle wheel is 9000 kJ and the heat transferred to the surroundings from the tank is 3000 kJ. The external work done by the system is: [IES-1999]

(a) Zero (b) 3000 kJ (c) 6000 kJ (d) 9000 kJ

Internal Energy A Property of System IES-13. For a simple closed system of constant composition, the difference between the

net heat and work interactions is identifiable as the change in [IES-2003] (a) Enthalpy (b) Entropy (c) Flow energy (d) Internal energy IES-14. Assertion (A): The internal energy depends on the internal state of a body, as

determined by its temperature, pressure and composition. [IES-2006] Reason (R): Internal energy of a substance does not include any energy that it

may possess as a result of its macroscopic position or movement. (a) Both A and R are individually true and R is the correct explanation of A (b) Both A and R are individually true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false but R is true


49

IES-15. Change in internal energy in a reversible process occurring in a closed system is equal to the heat transferred if the process occurs at constant: [IES-2005]

(a) Pressure (b) Volume (c) Temperature (d) Enthalpy IES-16. 170 kJ of heat is supplied to a system at constant volume. Then the system

rejects 180 kJ of heat at constant pressure and 40 kJ of work is done on it. The system is finally brought to its original state by adiabatic process. If the initial value of internal energy is 100 kJ, then which one of the following statements is correct? [IES-2004]

(a) The highest value of internal energy occurs at the end of the constant volume process (b) The highest value of internal energy occurs at the end of constant pressure process. (c) The highest value of internal energy occurs after adiabatic expansion (d) Internal energy is equal at all points IES-17. 85 kJ of heat is supplied to a closed system at constant volume. During the next

process, the system rejects 90 kJ of heat at constant pressure while 20 kJ of work is done on it. The system is brought to the original state by an adiabatic process. The initial internal energy is 100 kJ. Then what is the quantity of work transfer during the process? [IES-2009]

(a) 30 kJ (b) 25 kJ (c) 20 kJ (d) 15 kJ IES-17a A closed system receives 60 kJ heat but its internal energy decreases by 30 kJ. Then the

work done by the system is [IES-2010] (a) 90 kJ (b) 30 kJ (c) 30 kJ (d) 90 kJ

IES-18. A system undergoes a process during which the heat transfer to the system per degree increase in temperature is given by the equation: [IES-2004]

dQ/dT = 2 kJ/C The work done by the system per degree increase in temperature is given by the equation dW/dT = 2 0.1 T, where T is in C. If during the process, the temperature of water varies from 100C to 150C, what will be the change in internal energy?

(a) 125 kJ (b) 250 kJ (c) 625 kJ (d) 1250 kJ IES-19. When a system is taken from state A to

state B along the path A-C-B, 180 kJ of heat flows into the system and it does 130 kJ of work (see figure given):

How much heat will flow into the system along the path A-D-B if the work done by it along the path is 40 kJ?

(a) 40 kJ (b) 60 kJ (c) 90 kJ (d) 135 kJ

[IES-1997]

IES-20. The internal energy of a certain system is a function of temperature alone and

is given by the formula E = 25 + 0.25t kJ. If this system executes a process for which the work done by it per degree temperature increase is 0.75 kJ/K, then the heat interaction per degree temperature increase, in kJ, is: [IES-1995]

(a) 1.00 (b) 0.50 (c) 0.50 (d ) 1.00


50

IES-21. When a gas is heated at constant pressure, the percentage of the energy supplied, which goes as the internal energy of the gas is: [IES-1992]

(a) More for a diatomic gas than for triatomic gas (b) Same for monatomic, diatomic and triatomic gases but less than 100% (c) 100% for all gases (d) Less for triatomic gas than for a diatomic gas

Perpetual Motion Machine of the First Kind-PMM1 IES-22. Consider the following statements: [IES-2000] 1. The first law of thermodynamics is a law of conservation of energy. 2. Perpetual motion machine of the first kind converts energy into equivalent

work. 3. A closed system does not exchange work or energy with its surroundings. 4. The second law of thermodynamics stipulates the law of conservation of

energy and entropy. Which of the statements are correct? (a) 1 and 2 (b) 2 and 4 (c) 2, 3 and 4 (d) 1, 2 and 3

Enthalpy IES-23. Assertion (A): If the enthalpy of a closed system decreases by 25 kJ while the

system receives 30 kJ of energy by heat transfer, the work done by the system is 55 kJ. [IES-2001]

Reason (R): The first law energy

Date post:	19-Oct-2015
Category:	Documents
Upload:	chandankrdumka
View:	427 times
Download:	11 times

IES previous year Thermodynamics solution

Documents