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

IAS-20. A gas chamber is divided into two parts by means of a partition wall. On one side, nitrogen gas at 2 bar pressure and 20°C is present. On the other side, nitrogen gas at 3.5 bar pressure and 35°C is present. The chamber is rigid and thermally insulated from the surroundings. Now, if the partition is removed, (a) High pressure nitrogen will get throttled [IAS-1997]

(b) Mechanical work, will be done at the expense of internal energy (c) Work will be done on low pressure nitrogen

(d) Internal energy of nitrogen will be conserved

First Law of Thermodynamics

S K Mondal’s Chapter 2

55

Answers with Explanation (Objective) Previous 20-Years GATE Answers

GATE-1. Ans. (a) ₁

+

¹²

+ gZ

¹

+ dQ =

₂

+

¹²

+ gZ

²

+ dW

h h

2000 1000 dm 2000 1000 dm

C C

× ×

+ + = + + +

+ + = +

2 2

160 9.81 10 100 9.81 6 dW

3200 2600

2000 1000 2000 1000 dm

600 7.8 0.04 dW

dm GATE-2. Ans. (c)

( ) ( )

ν

= ₂− ₁ = 1 − ×

W P P 3000 70 kJ/kg = 2.93

1000

GATE-3. Ans. (a) Fig-1 & 2 both are power cycle, so equal areas but fig-3 & 4 are reverse power cycle, so area is not meant something.

GATE-4. Ans. (c)

2 1 2 1 2 1

dQ du dw

Q u u W or 2000 u u 5000 or u u 3000kJ

= +

= − + − = − − − =

GATE-4a. Ans. (a) Q = 0, W = –2.3, ΔU= +2.3

Tank is well insulated so Q = 0 Work is given to the system in the form of electric current.

So, W = = –2300 W = –2.3 kW By 1^st Law of Thermodynamics

=

0 =

= 2.3 kW GATE-5. Ans (a) The final Temp. (T2)=

γ T

₁

Previous 20-Years IES Answers

IES-1. Ans. (a) If we adiabatically mix two liquid then perfect gas law is not necessary. But entropy change in the universe must be calculated by Second law of thermodynamics. Final entropy of then system is also a property. That so why we need second law.

IES-2. Ans. (b) Using conservation of energy law we may find final temperature.

IES-3. Ans. (d) From First law of thermodynamics, for a closed system the net energy transferred as heat Q and as work W is equal to the change in internal energy, U, i.e. Q – W = U

I R I

2 2

I R 10 23

− = − ×

Q1 2₋ U₂ −U₁ +W_{1 2−}

2 1

U −U −2.3 U

Δ

S

tal heat rejec

w from equat

1 2

ange of inter ) Q = ΔE+Δ

addition is co tion is consta tion of state

nal energy = ΔW

decrease in in (work done b + 20 kJ n is isocho rk on the sys In an irrev

, the system rrounding at

.

w of T

56 onstant volum ant pressure

2 2 y the system 74 – 180 kJ/m m doesn't per

the expense

Therm

me heat addi heat rejectio

1

operty and it

−₂+W_{2 1}−

ts cyclic integ

60 kJ/s = 4.9

apter 2

gral must be

kW

S

(a) The inter ernal energy d specific el s, form the sp

rnal energy d y of a substa acroscopic

gy include po er:

w of a gas s of transla lectronic en ance does no position or m

osition or mo is a collectio ational ene nergy. All th al energy, e , V is cons tan t

Therm

y upon the in ot include an movement. T

vement then on of particl rgy, rotatio hese energies , of the gas.

nitial and fin ny energy th That so why

n why this v²/ les in rando onal energy

summed ove

( )

^dU_v

process 1-2 dQ = process 2-3

dQ = process 3-1

dQ = clic process

∑dQ om motion. E

y, vibration

the system.

possess as a /2 and gz is

ms is there.

Energy of a nal energy rticles of the

) 0 kJ,

0

= -20kJ

First Law of Thermodynamics

S K Mondal’s Chapter 2

58 IES-18. Ans. (c)

( )

2 150 2 2

100

dQ du dw

2.dt du 2 0.1T dT

0.1 0.1

or du 0.1TdT T 150 100 625kJ

2 2

= +

= + −

⎡ ⎤ ⎡ ⎤

= = ×⎣ ⎦ = ⎣ − ⎦=

∫ ∫

IES-19. Ans. (c) Change of internal energy from A to B along path ACB = 180 - 130 = 50 kJ. It will be same even along path ADB. ∴ Heat flow along ADB = 40 + 50 = 90 kJ.

IES-20. Ans. (d) dQ=dE+ dW dQ dE dW or dt = dt + dt Given: E 25 0.25t kJ and 0.75 /

then 0.25 /

Therefore 0.25 0.75 / 1.00 /

dW kJ k

dt dE kJ K

dt

dQ dE dW

kJ K kJ K dt dt dt

= + =

=

= + = + =

IES-21. Ans. (a)

IES-22. Ans. (a) A closed system does exchange work or energy with its surroundings. option ‘3’ is wrong. 4. “The law of conservation of entropy” is imaginary so option ‘4’ is also wrong.

IES-23. Ans. (a)

IES-24. Ans. (b) h₁ ^v21 gz₁ ^dQ h₂ ^v22 gz₂ ^dw 0

2 dm 2 dm

+ + + = + + + =

For boiler v1, v2 is negligible and z1 = z2 and dw 0 dm =

(

2 1

)

or dQ h h

dm = − IES-25. Ans. (d)

( )

mC T = 4 10 60P

20 4 T = 2400 T = 30C

Δ × ×

⇒ × × Δ

⇒ Δ °

IES-26. Ans. (c) Enthalpy of additional gas will be converted to internal energy.

Uf= miui+(mf-mi)hp = 0.25x200+(1-0.25)x400 = 350 kJ As total mass = 1kg, uf=350 kJ/kg

Note: You cannot simply use adiabatic mixing law here because it is not closed system.

This is a problem of variable flow process. If you calculate in following way it will be wrong.

Final internal energy of gas(mixture) is

u = ^{1 1 2 2}

1 2

m u m u

m m

+ +

u =

kJ kJ

(0.25kg) 200 (0.75kg) 300

kg kg

(0.25 0.75)kG

⎛ ⎞+ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

+ u = 275 ^kJ

kg

First Law of Thermodynamics

S K Mondal’s Chapter 2

59 It is valid for closed system only.

Previous 20-Years IAS Answers

IAS-1. Ans. (b)

∑

dQ=

∑

dW or 220 -25 -180 +50 = 15 -10 +60 +W4-1

IAS-2. Ans. (c) Area under p-v diagram is represent work.

Areas ΔPTS=1

2 Area (WVUR) ∴Work PTS=1

2×48=24 Nm IAS-3. Ans. (a) work output work out put 54 0.45

Heat input work output heat rejection 54 66

η = = = =

+ +

IAS-4. Ans. (b)

3 10 watts3

Thermal efficiency 0.3 30%

10.000 J/s W

Q

= = × = =

IAS-5. Ans. (c)

( )

( ) ( ) ( )

( )

( ) ( )

1 2 2 1 1 2

2 1 2 1

2 1 1 2 2 1

2 1 2 1 1 2 2 1 2 1

Q U U W

or 0 U U 5000 or U U 5000J

Q U U W

or W Q U U Q U U 1000 5000 6000J

− −

− −

− − −

= − +

= − + − − =

= − +

= − − = + − = + =

IAS-6. Ans. (c) Net work output = 3 + 10 – 8 = 5 unit and Heat added = 30 + 5 = 35 unit Therefore efficiency, 5

100% 14.33%

η=35× = IAS-7. Ans. (a)

IAS-8. Ans. (d) Q123 = U13 + W123 or, 100 = U13 + 60 or, U13 = 40 kJ And Q143 = U13 + W143 = 40+20 = 60 kJ

IAS-9.Ans. (b) Total work = 5 × 3 + 5 1 17.5J 2

1× × = or

δ

W =du+

δ

W =2.5+17.5=20J

IAS-10. Ans. (c) Efficiency Area under 500 and 1500 Area under 0 and 1500

=

{ }

{ }

1 (5 1) (4 2) (1500 500) 3000

2 0.6

1 (5 1) (4 2) (1500 500) (5 1) 500 5000 2

× − + − × −

= = =

× − + − × − + − ×

IAS-11. Ans. (a)

IAS-12. Ans. (b) Internal energy is a property of a system so du 0

∫

=

IAS-13. Ans. (a) dQ du dW if du= + = +30kJ then dQ= −50kJ and dW= −80kJ IAS-14. Ans. (b)

δ

W =du+

δ

W =du+ pdV

or 84 × 10³J = du + 1 × 10⁶ × (0.06 – 0.03) = du +30 kJ or du = 83 – 30 = 54 kJ

First Law of Thermodynamics

S K Mondal’s Chapter 2

60 IAS-15. Ans. (a) Q1-2 = U2 –U1 +W1-2

Or 0 = U2 –U1 - 6000 or U2 –U1 = +6000 Q2-1 = U1-U2+W2-1

or W2-1 = Q2-1 - (U1-U2)

=1000+6000=7000J

IAS-16. Ans. (c)

IAS-17. Ans. (a) Energy balance gives as

( ) ( )

= Δ + Δ +

= − −

=

air water

dW dQ

h h

dm dm

or dQ 90 30 40 dm

20kJ / kg of air compressed.

IAS-18. Ans. (a)

( ) ( )

( ) ( )

1 2

1 2

dQ dw

m h m h

dt dt

dw dQ

or m h h 2 100 200 50 2 300kW

dt dt

i.e. 300kW work have to given to the system.

+ = +

= − + = × − − × = −

IAS-19. Ans. (b) IAS-20. Ans. (a)

In document Thermodynamics Theory + Questions (Page 53-60)