# Basic Thermodynamics - WordPress.com

Mechanics

Oct 27, 2013 (4 years and 6 months ago)

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1

Basic Thermodynamics

Syllabus:
-

Zeroth, first and second law of thermodynamics, thermodynamic system and
processes, car not cycle, inerversibility and availability, behavior of ideal gas and real
gases, properties of pure substance, calculation of wo
rk and heat in ideal processes,
analysis of thermodynamic, cycles related to energy conversion.

Gate
→ 2003

1.

A 2 kw, 40 liters water heater is switched on for 20 minutes. The heat capacity

for water is 4.2 kj/kg k. Assuming all the electrical energy has gone in to
hearting the water, increase of the water temperature in degree

(a) 2.7

(b) 4.0

(Mark = 1)

(c) 14.3

(d) 25.25

Solution

Heat supplied by heater in 20 minutes

As all the electric energy has gone into heating the water

So Heat taken by water = 2400 KJ

But heat ta
ken by water

Mass of water

Specific heat in kJ/kg K = 4.2.

2.

Considering the relationship Tds = Pdv between the entropy (s), internal energy
(u), pressure (P), temperature (T) and Volume (v) which of the following
statements is correct ?

(a)

It is applicable only for a reversible process.

(b)

For an irreversible process Tds

> du + Pdv.

(c)

It is valid only for an ideal gas.

2

(d)

It is equivalent to 1
st

Law, for a reversible process.

( Mark

2)

Solution. (c) Tds = du + Pdv

Is valid only for an ideal gas both for reversible and irreversible process undergone by a
closed

system, sin
ce it is a relation among properties which are independent of the path.

Common data question

Nitrogen g
a
s
( molecular weight 28 ) is enclosed in a cylinder by a piston, at the initial
condition
of 2 bar, 298 k and 1 m
3
. In a
particular process, the gas
slowly expands under
isothermal condition, until the volume becomes 2 m
3
. Heat exchange occurs with the
atmosphere at 298 k during this process.

3.

The work interaction for the nitrogen gas is

(a) 200 KJ

(b) 138
-
6 KJ

(c) 2 KJ

(d)
-
200KJ

(Mark

2)

4.

The entropy change for the universe during the process in KJ/K is

(a) 0.4652

(b) 0.0067

(c) 0

(d)
-
0.6711

(Mark
-
2 )

Solution

Work done

In isothermal process T = Constant

For an ideal gas Pv = RT

Where R is
constant

So Pv = Constant

Work done

3

Entropy change of the system

{ for reversible proc
ess}

For isothermal process = du = 0

So from first law of
thermodynamics

As T is constant

Entropy change of surrounding

=
-

0.4641

So entropy change of universe is zero.

Gate 2004

5.

A gas contained in a cylinder is compressed, the work required for compression
being 5000 KJ. During the process, the heat interaction of 2000 KJ causes the
surrounding to be heated
. The change in internal energy of the gas during the
process is

(a)

7000 KJ

(b)

3000 KJ

(c) + 3000 KJ

(d) + 7000 KJ

(Mark

1 )

Solution.

(c) Us
ing first law of thermodynamics.

4

{
work done is negative when it is done on the

S
ystem

}

{
Heat transferred is negative when it is taken from

the system or it is rejected}

6.

A Steel billet of 2000 kg mass is to be cooled

from 1250 k to 450 k . The heat
released during process is to be used as a source of energy. The ambient
temperature is 303 k and specific heat of steel is 0.5 kJ/kg K. The available
energy of this billet is.

(a) 490.44 MJ

(b) 30.95MJ

(c) 10.35 MJ

(
d) 0.1 MJ

Area under curve AB

Entropy change from B to A

Area under C
-
D
=

So Area ABCD = 800

309.56

5

= 490MJ.

Gate 2005

7.

The following four figures have been drawn to represent a fictitious
thermodynamic cycle, on the P

v and T

S planes.

According to first Law of thermodymics, equal areas are e
nclosed by

(a) Figure 1 and 2

(b) Figure 1 and 3

(c) Figure 1 and 4

(d) Figure 2 and 3

8.

A Reversible thermodynamic cycle containing only three processes and producing
work is to be constructed. The constraints are

(i)

There must be one isothermal proce
ss (ii) There must be one isentropic
process (iii) The maximum and minimum cycle pressures and clearance
volume are fined (iv) Polytrophic processes are not allowed. The number of
possible cycle are

(a) 1

(b) 2

(c) 3

(d) 4

(Mark

2)

Poss
ible processes

Constant pressure

Isothermal Compression

Isothermal Compression

P

6

V

Isothermal Expansion

Isothermal Compression

P

Isothermal Compression

Constant Pressure Process

2

V

Isothermal Expansion

P

3

1

V

Isothermal

Compression

Isothermal Compression

P

Constant Pressure

7

9.

Nitrogen at an initial state of 10 bar, 1 m
3

and 300 K is expanded isothermally to a
final volume of 2 m
3
, the P

v

T relation is

Where a > 0. The fi
nal pressure

(a) Will be slightly less than 5 bar.

(b) Will be slightly more less than 5 bar.

(c) Will be exactly 5 bar.

(d) Can not be ascertained in the absence of the value of a.

(Mark

2)

Solution. (b)

As proc
ess is isothermal

RT = constant

So

is slightly more than 5 bar as a is positive
.

The following table of properties was printed out for saturated liquid

and saturated
v
apor of ammonia. The title for only the first two columns are available. All that we
know that the other columns (columns 3 to 8 ) contain data on specific properties,
namely, internal energy (KJ/kg ),enthalpy (KJ/kg ) and entropy (KJ/kg)

t
0

P(KPa)

-
20

190.2

88.76

0.3657

89.05

5.6155

1299.5

1418

0

429.6

179.69

0.7114

180.36

5.3309

1318

1442.2

8

20

587.5

272.89

1.0408

274.3

5.0860

1332.2

1460.2

40

1554.9

368.74

1.3574

371.43

4.8662

1341.0

1470.2

20. The specific enthalpy data are in colu
mns

(a) 3 and 7

(b) 3 and 8

(c) 5 and 7

(d) 5 and 8

10.

when saturated liquid at 40
0
C is throttled to

20
0
C the quality atenit will be

(a) 0.189

(b) 0.212

(c) 0.231

(d) 0.788

10. (d) We know that h = u + PV

h = enthalpy u = internal energy.

S
o all the value in column 5 are greater than
corresponding value in column 3 so
column 5 represent
specific enthalpy of saturate liquid and similarly all the values of
column 8 are greater than corresponding values of column 7.50 the column 8
represent

specific enthalpy of saturated steam.

11.

Enthalpy of saturated liquid at 40
0
C = 371.43 KJ/kg.

Enthalpy of vapor at

20
0
C = ( let x be dry ness fraction)

= 89.05+x (1418

89
-
05).

During throttling enthalpy remains constant

371.43 = 89.05 + x (1418

89.
05)

282.38 = x (1418

89.05)

x = 0.212.

Gate → 2006

12.

Given below is an extract from steam tables.

Temperature

In
o
C

P
Sat

(bar)

Specific volume

Enthalpy (KJ/kg)

Saturated
Liquid

Saturated

Vapour

Saturated

Liquid

Saturated

Vapour

45

0.09593

0.001010

15.26

188.45

2394.8

342.24

150

0.001
658

0.010337

1610.5

2610.5

9

Specific enthalpy of water in KJ/kg at 150 bar and 45
0
C is

(a) 203.6

(b) 200.53

(c) 196.38

(d) 188.45

Solution. (b) Specific enthalpy at 150 and 45
0
C

13.

Match items from groups I, II, III
, IV a
nd V.

Group I

Group II

Group III

Group IV

Group V

the system is

Differential

Function

Phenomenon

E

eeat

G P潳itive

I

Enact

K⁐ath

M⁔ransie湴

F⁷潲k

e P潳itive

J⁩n enact

L P潩nt

N⁂潵n摡ry

(搩 eeat when a摤d搠d漠ohe system

is 灯pitive an搠is in enact 摩ffere湴ial 扥cause it is

negative ⁡n搠is in enact 摩fferential 扥cause it is 灡th 摥灥n摥nt an搠is transient

Statement f潲 lin步搠 ans
wer 煵esti潮 ㄴ⁡n搠ㄵ⸠

A f潯o扡ll was inflate搠 t漠a⁧auge 灲essure 潦 ㄠ扡r⁷hen the am扩ent tem灥rature was

0
C. When the game started next day, the air temperature at the stadium was 5
0
C.
Assume that the volume of foot ball remains constant 2500 cm
3
.

14.

The amount of heat lost by the air in the football and the gauge pressure of air in
the football at the stadium respectively equal.

(a) 30.6J, 1.94 bar

(b) 21.8J, 0.93 bar

(c) 61.1J, 1.94 bar

d) 43.7J, 0.93 bar

15.

Gauge pressure of air to which the

ball must have been originally inflated so that
it would be equal 1 bar gauge at the stadium is

10

(a) 2.23 bar

(b) 1.94 bar

(c) 1.07 bar

(d) 1 bar

Solution. 14 (d) Gau
ge pressure with in foot ball

= 1.01 + 1

= 2.01

bat

Temperature when total pressure is 2.01 bar = 15
0
C or 288
0
k.

Volume of air with in foot ball = 2500 cm
3
.

Since the volume remains same

Next day temperature

or 278
0
k.

1.94 bar

= 0.93 bar.

Heat lost =

Heat lost

15 (c)

{ final pressure}

{ final pressure}

11

Gate 2007

16.

(d) Which of the following relationships is valid only for reversible processes
under gone by a

closed system of simple compressible substance ( neglect
changes in kinetic and potential energy ) ?

(a)

(b)

(c)

(d)

(d)

is true for reversible process and closed system.

37. Water has a critical specific volume of 0.003155 m
3
/kg. A closed and rigid steel
tank of volume .025m
3

contains a mixture of water and steam at 0.1 MPa. The mass
of the mixture is 10 kg. The

tank is now slowly heated. The liquid level inside the
tank.

(a)

Will rise.

(b)

Will fall.

(c)

Will remain constant.

(d)

May rise or fall depending on the amount of heat transferred.

Solution. Critical volume = 0.003155m
3

/ kg.

Specific volume of mixture

Which is less than critical volume

As it is a constant volume process

P

V

Critical Volume

v

Saturated Volume

v

Saturated Liquid

v

12

40. Which combination of the following statements is correct ?

P : A gas cools upon expansion only.