UNIT – I

PART – A

1. Define the term thermodynamics.

It is the branch of physics concerned with the conversion of different forms of energy

It is the science of the conversions between heat and other forms of energy

It is the study of energy 2. What is meant by thermodynamics system? How do you classify it? Thermodynamic system is defined as any space or matter or group of matter where the energy transfer or energy conversions are studies. It may be classified into three types. a) Open system b) Closed system c) Isolated system 3. What is meant by closed system? Give an example. When a system has only heat and work transfer, but there is no mass transfer, it is called as closed system. Example: Piston and cylinder arrangement. 4. Define an open system. Give an example. When a system has heat, work and mass transfer, it is called as open system. Example: Air compressor 5. Distinguish between Open and Closed systems.

S.No. Closed System Open System

1, There is no mass transfer. Only heat and work will transfer.

Mass transfer will take place, in addition to the heat and work transfer.

2. System boundary is fixed one. System boundary may or may not change.

3. Ex: Piston & cylinder arrangement, thermal power plant.

Air compressor, boiler.

6. Define an Isolated System. Isolated system is not affected by surroundings. There is no heat, work and mass transfer takes place. In this system total energy remains constant. 7. Define specific heat capacity at constant pressure. It is defined as the quantity of heat energy required for raising or lowering the temperature of a unit mass of the substance through one degree when the pressure is kept constant. It is denoted by Cp. Its unit is kJ/kg K 8. Define specific heat capacity at constant volume. It is defined as the quantity of heat energy required for raising or lowering the temperature of a unit mass of the substance through one degree when the volume is kept constant. It is denoted by Cv. Its unit is kJ/kg K

9. What is meant by surroundings? Any other matter out side of the system boundary is called as surroundings. 10. What is boundary? System and Surroundings are separated by an imaginary line called as the boundary. 11. What is meant by thermodynamic property? A quantity which is either an attribute of an entire system or is a function of position which is continuous and does not vary rapidly over microscopic distances, except possibly for abrupt changes at boundaries between phases of the system; examples are temperature, pressure, volume, concentration, surface tension, and viscosity. 12. How do you classify the property? Thermodynamic property can be classified into two types.

1. Intensive or Intrinsic property 2. Extensive and Extrinsic property

13. Define Intensive and Extensive properties. The properties which are independent of the mass of the system are called intensive properties. E.g: Pressure, Temperature, Specific Volume etc. The properties which are dependent on the mass of the system are called extensive properties. Eg: Total energy Total volume, weight etc. 14. Differentiate Intensive and Extensive properties.

Sl. No.

Intensive Properties Extensive Properties

1. Independent on the mass of the system.

Dependent on the mass system.

2. If we consider part of the these properties remain same e.g. Pressure, temperature, specific volume, etc.

If the consider part of the system it will have a lesser value e.g. Total energy, Total Volume, weight, etc.

15. What do you understand by equilibrium of a system? When a system remains in equilibrium state, it should not under go any changes to its own accord. 16. What are the types of equilibrium of a system? a. Mechanical Equilibrium b. Thermal Equilibrium c. Chemical equilibrium 17. When a system is said to be in “Thermodynamic Equilibrium”? When a system is in thermodynamic equilibrium, it should satisfy the following three conditions

b. Mechanical : Pressure remains constant b. Thermal Equilibrium : Temperature remain constant c. Chemical equilibrium : There is no chemical reaction 18. Define Zeroth law and first law of thermodynamics. Zeroth law of thermodynamics states that when two systems are separately in thermal equilibrium with a third systems, then they themselves are in thermal equilibrium with each other. First Law of thermodynamics states that when system undergoes a cyclic process net heat transfer is equal to work transfer gQ = gW 19. (a)(i) State First Law of thermodynamics and any two of its corollaries. First Law of thermodynamics states that when system undergoes a cyclic process net heat transfer is equal to work transfer. gQ = gW Corollaries of first law of thermodynamics Corollary I There exists a property of a closed system such than a change in its value is equal to the difference between the heat supplied and the work done any change of state. Corollary II The internal energy of a closed system remains unchanged if the system is isolated from its surroundings. Corollary III A perpetual motion machine of first kind (PPM-1) is impossible. 20. What is meant by “Perpetual Motion Machine of First Kind”? PMM I -“Perpetual Motion Machine of First Kind” delivers work continuously without any input. It violates first law of thermodynamics. It is impossible to construct an engine working with this principle. 21. What is meant by “Perpetual Motion Machine of Second Kind”? PMM II -“Perpetual Motion Machine of Second Kind” converts all the input heat energy into work without any loss of energy as entropy. It violates second law of thermodynamics. It is impossible to construct an engine working with this principle. 22. Prove that for an isolated system, there is no change in internal energy. For any isolated system. There is no heat, work and mass transfer. i.e., Q=W=0

According to the second law of thermodynamics, Q=W+U

U=0

23. Determine the molecular volume of any perfect gas at 600 N/m2 and 30C. Universal gas constant may be taken as 8314 j/kg mole – K. Given data: P = 600N/m2

T = 30C=30+273=303K

R=8314 J/kg mole – K To find: Molecular volume, V=? Solution: Ideal gas equation, PV=mRT

600 V = 1 8314 303 V=4198.57 m3/kg mole Result: Molecular volume, V=4198.57 m3/kg mole 24. An insulated rigid vessel is divided into two parts by a membrane. One part of the vessel contains air at 10 Mpa and other part is fully evacuated. The membrane ruptures and the air fills the entire vessel. Is there any heat and / or work transfer during this process? Justify your answer. For rigid vessel and unrestrained expansion Change in volume dV = 0

Work transfer. W = pdV = 0 For insulated vessel, heat transfer, Q = 0 According to the first law of thermodynamics the sum of work transfer is equal to the sum of heat transfer.

W = Q = 0 25. Define the term process. It is define as the change of states undergone by a system due to energy flow. 26. What are the classifications of process? i. Reversible process or Quasistatic process ii. Irreversible process 27. What are the types of process? i. Constant Volume Process ii. Constant Pressure Process iii. Constant Temperature Process iv. Adiabatic Process v. Polytropic Process 28. What is a Quasi- static Process? The process is said to be quasi – static, if it proceeds infinitesimally slow and follows continuous series of equilibrium states. Therefore the quasi-static process may be a reversible process. 29. What is meant by reversible and irreversible process? A process is said to the reversible. It should traces the same path in the reverse direction when the process is reversed. It is possible only when the system passes through a continuous series of equilibrium state. If a system does not passes through continuous

equilibrium state, then the process is said to be irreversible. 30. Define Constant Volume process. A constant volume process is also called as iso-choric process. During a constant volume process the volume of the system does not change (i.e., remains in equilibrium) , hence the name. V1 = V2 = V3 = …….= constant

31. Define Constant Pressure process. A constant pressure process is also called as iso-baric process. During a constant pressure process the pressure of the system does not change (i.e., remains in equilibrium) , hence the name. P1 = P2 = P3 = …….= constant

32. Define Constant Temperature process. A constant temperature process is also called as iso-thermal process. During a constant temperature process the temperature of the system does not change (i.e., remains in equilibrium) , hence the name. T1 = T2 = T3 = ……. = constant

33. Define adiabatic process. An adiabatic process is also called as isentropric process. During this process the entropy of the system remains the same hence its name. During an adiabatic process the net heat

transferred to the system is zero (i.e. Q = 0). Hence an adiabatic process does not have any heat transfer. S1 = S2 = S3 = ……. = constant

34. Define the term cycle. When a system undergoes a series of processes and return to its initial condition (state), it is known as a cycle. 35. What are the types of cycle? i. Open cycle ii. Closed cycle 36. What is meant by open and closed cycle? In a closed cycle, the same working substance will recirculate again and again. In an open cycle, the working substance will be exhausted to the surrounding after expansion. 37. What is meant by point and path function? The quantity which is independent on the process or path followed by the system is known as path functions. Example: Heat transfer, work transfer 38. Define the term enthalpy. The combination of internal energy and flow energy is known as enthalpy of the system. Mathematically, enthalpy (h) = U + pV Its unit is kJ Where U –Internal energy P – Pressure V – Volume

In terms of Cp&T H = mCp (T2 – T1)kJ 39. Define the term internal energy. Internal energy of a gas is the energy stored in a gas due to its molecular interactions. It is also defined as the energy possessed by a gas at a given temperature. 40. What is meant by thermodynamic work? It is the work done by the system when the energy transferred across the boundary of the system. It is mainly due to intensive property difference between the system and

surrounding. 41. Sketch isothermal expansion on p-V diagram and state the properties that remain constant.

The following properties remain constant. 1. Temperature 2. Internal energy 3. Enthalpy 42. Prove that the difference in specific heat capacities equal to Cp – Cv = R. Consider a gas heated at constant pressure So, heat supplied, Q = mCp (T2 – T1) Work done, W = p(V2 – V1)= m R (T2 – T1)

Change in internal energy, U = mCv (T2 – T1) According to the first law of thermodynamics,

Q = W + U So, mCp (T2 – T1) = mR (T2 – T1)+ mCv(T2 – T1)

Cp = r + Cv Cp – cv = R

PART – B

1. A mass of gas is compressed in a quasi-static process from 100 kPa 0.3m3 to 500 kPa, 0.05m3. Assuming that the pressure and volume are related by PV” = constant, find the workdone by the gas system. System : Closed system Process : PV” = C Known : Initial pressure = 100 kPa Initial Volume = 0.3m3

Final Pressure = 500 kPa Final Volume = 0.05m3 To find : Work done Diagrams : 0.3 m3 0.05 m3 100 kPa 500 kPa Initial State Final State P (kPa) 2

500 pVn = C 100 1

2

1

pdV

0.05 0.3 V(m3

P-V Diagram

In this equation all the values are known except the polytrophic index n.

Consider the process relation P1V1

n = P2V2n

1 2

2 1

n

V p

V p

Taking natural log on both sides

n In 1 2

2 1

V pIn

V p

1 2

2 1

( / )

( / )

In V Vn

In p P

Substituting the numerical values

500

100

0.3

0.05

IN

n

In

n = 0.898 Hence the work done

1W2 = 2

1

PdV

2 2 1 1

1

500 0.05 100 0.3

0.898 1

49.01

p V pV

n

KJ

Comment: Negative sign indicates that the work is done on the system.

2. A certain fluid expands from 10 bar, 0.05m3 to 2 bar, and 0.2m3 according to linear law. Find the work done in the process. System : Closed system Process : Linear Law Known : Initial pressure = 10 bar = 1000 kPa Initial volume = 0.05m3 Final pressure = 2 bar = 200 kPa Final volume = 0.2m3 To find : Work done Diagrams: 10 bar 2 bar 0.05 m3 0.2 m3

Initial State Final State

P (kPa) 1 1000 A1 200 2 A2 0.05 0.2 V (m3)

P-V Diagram

Analysis : 1W2 = 2

1

pdV

= Area A1 + Area A2

= 1

(0.2 0.05)(1000 200) (0.2 0.05)2002

= 90KJ 3. Following diagram shows a cycle executed by a closed system on a p-V diagram. Calculate the work done by the system for each of three processes and also the network done.

P(kPa)

1

1000

3 2

500

0.1 0.2

p-V Diagram

System : Closed System

Known : Initial and final pressure, Initial and final Volume of all the processes.

Process 1-2 Linear Law 2-3 Isobaric (p=C) 3-1 Isochoric (V=C)

Work done 2

1

pdV in KJ

Area 12ba 1 1

(0.2 0.1) (1000 500) (0.2 0.1) 5002

75

= 500(0.1– 0.2) -= -50

=0 = p2 (V3 –V2)

Net work:

Wnet = 1W2 + 2W3 + 3W1 = 75 – 50 + 0 = 25KJ 4. A spherical balloon of 0.5 m diameter contains air at a pressure of 500 kPa. The diameter increases to 0.55m in a reversible process during which pressure is proportional to diameter. Determine the work done by the air in the balloon during this process. Also calculate the final pressure.

System : Closed system Process : Expansion of air in a spherical balloon such that pressure is proportional to the diameter. Known : 1. Initial diameter = 0.50m 2. Initial Pressure = 500 kPa 3. Final diameter = 0.55m To find : 1. Work done 2. Final weight Diagram :

Air Dia = 0.5m P = 500 kPa

Air

Dia = 0.55m

P (kPa) 2 1 500

0.5 0.55 dia (m)

Analysis:

1. Work done

2W1 = 2

1

pdV

From the condition of the process P = CD Where p is the pressure (kPa) C is a constant D is the diameter (m) From the initial condition

500 = C 0.5 C = 500 / 0.5 = 1000 Therefore P = 1000 D Substituting p we get,

1W2 = 2

1

1000DdV

To substitute dV in terms of dD Consider the volume of the sphere

V = 3

6D

Upon different we get

dv = 236

D dD

Substituting dV in terms of dD we get

1W2 = 2

2

1

1000 36

D D dD

0.55 33

0.50

0.554

0.50

4 4

1000

6

5004

500[0.55 0.50 ]

4

11.39

D dD

D

KJ

2. Final pressure P2 = 100 D2

= 1000 0.55 = 550 kPa 5. A rubber balloon flexible is to be filled with hydrogen from a storage cylinder which contains hydrogen at 5 bar pressure until it has a volume of 1 m3. The atmospheric pressure may be taken as 101.32 kPa. Determine the work done by the system comprising the hydrogen initially in the bottle. System : Closed Process : Hydrogen expanding from a cylinder against the atmospheric pressure. Known : 1. Volume of expansion dv = 1 m3

2. Atmospheric pressure Patm = 101.32 kPa 3. Storage cylinder pressure = 500 kPa.

Diagram: Balloon after filling Balloon before filling Hydrogen Cylinder

To find : Work done by the hydrogen gas

Analysis : Work done = 2

1

pdV

2

1

0.10

22 1

1 0.15

52.0

5 2.0( ) 100

V

VIn V V

V

As the pressure is in bar, it has to be multiplied by 100, to get it in kpa Substituting the limits we get

1W2 = 5 In 0.10

2.0(0.1 0.15)0.15

= 2.027 + 0.1

= 1.027 KJ

6. A compressor receives 10kW of power at 600 rpm from a motor. What is the torque acting on the shaft connecting the motor and the compressor?

System : Compressor receiving power from a motor through a shaft

Known : 1. Power = 10kW

2. Speed = 600 rpm

To find : 1. Torque

Diagrams:

Compressor

Motor

Fin

al Le

ngth

Analysis : 1. w = 2nt Rate of work transferred

= 10 103 KJ/s Speed = 600 rpm = 600 / 60 rps = 10 rps

3

/ 2

10 10 =

2 10

= 159.15 N m.

T W n

7. An elastic linear spring of spring constant 144 N/cm is compressed from an initial unconstrained length to a final of 6cm. If the work required on the spring is 648J, determine the initial length of the spring in centimeters:

System : Elastic spring Process : Stretching the elastic spring Known : 1. Spring constant = 144 N/cm 2. Final length = 6 cm 3. Work supplied = 648 KJ To find : Initial length in cm Diagrams: -

Fre

e L

en

gth

Analysis :

1

2

144648

2

W k

cm

Initial length = final length + deflection ()

= 6 + 3 = 9 cm 8. Calculate the work required to lift 25 kg from an evaluation of 208m above mean sea level to an elevation 80m higher in (a) min (b) 10 min.

System : A body of mass 25 kg

Process : Lifting the body against gravity

Known : 1. Mass = 25 kg

2. Initial height = 208m

3. Change in height = 80m

4. Time a) min b) 10 min

To find : Work done

Diagrams: - Final state

Initial state 80 m

208 m

Mean sea level

Analysis : Work done = -mgh

: The negative sign indicates that work is done on the system.

Work is independent on time and hence for both cases work done is – 19620J.

9. Determine the work required to accelerate a body of mass 100 kg from rest to a velocity of 100 m/s.

System : A body of mass 100 kg

Process : Acceleration of a body from rest

Known : 1. mass = 100 kg

2. Initial velocity

3. Final velocity = 100 m/s.

To find : Work

Analysis

2 2

2 1

2 2

1( )

2

1 100 (100 0 )

2

= 500

W m C C

KJ

Comment: The negative sign indicates that work done on the body to accelerate it.

10. A perfect gas for which the ratio of specific heats is 1.4, occupies a volume of 0.3m3 at

100 kpa and 27C. The gas undergoes compression to 0.06m3. Find the heat transfer by the gas in each of the following process.

a) Isobaric b) pV1.1 = Constant Assume R = 0.287 KJ/ kgK.

Case(a)

System : Closed system

Working fluid : A perfect gas

Process : Isobaric

Known : 1. Initial volume (V1) = 0.3m3

2. Initial pressure (p1) = 100 pa

3. Initial Temperature (T1) = 27C = 300 K

4. Ratio of sp. Heat () = 1.4

5. Final Volume (V2) = 0.06m3

6. Characteristic gas constant = 0.287KJ/kg K

To find : Heat transfer Q12

Diagram:-

0.3 m3 0.06 m3

100 kPa 100 kPa

300 K

Initial State Final State

P

(kPa)

2 1

2

1

pdV

0.06 0.3 V(m3)

p-V diagram

Analysis : Q12 – 1W2 = U

Where 1W2 = 2

1

pdV

= p(V2 – V1) for an isobaric process

= 100 (0.06 – 0.3)

= - 24KJ

U = 2

1

dU

= 2

1

. 3.23mC dT from eqn

= mC(T2 – T1)

To find T2

Since the working substance is an ideal gas

P1V1 = mRT1

P2V2 = mRT2

By rearranging the above equation

1 1 2 2

1 2

pV PV

T T

Applying the given process condition p1 = p2

1 2

1 2

22 1

1

2

0.06300

0.3

60

V V

T T

VT T

V

T k

To find C,

Cp – Cv = R

1

1

1

0.287

0.4

0.717

p

v

v

v

p

v

CC R

C

RC

C

C

R

v

KJ

kgK

To find m

m = 1 1

1

pV

RT [ The given substance is a perfect gas]

= 100 0.3

0.287 300

= 0.348 kg

Substituting the values of m and T2

U = 0.348 0.717 (60 – 300)

= 59.93KJ

Therefore Q12 = 1W2 + U

= - 24 – 59.93

= - 83.93KJ

Comment: The negative sign in work indicates work is Done on the system.

The negative sign in U shows that internal energy decrease.

The negative sign in heat transfer shows that heat is rejected by the system.

Case (b)

System : Closed system

Process : Polytropic of index 1.1 (pV1.1 = C)

Working fluid : A perfect gas

Known : 1. Initial volume (V1) = 0.3m3

2. Initial pressure (p1) = 100 kPa

3. Initial Temperature (T1)

4. Ratio of sp. heat () = 1.4

5. Final Volume (V2) = 0.06m3

6. Characteristic gas constant = 0.287 KJ/kg K.

Diagram:-

0.3 m3

100 kPa 0.06m3

300 K

Initial State Final State

P

(kPa) 2

pV1.1 = C

1

2

1

pdV 0.06 0.3 V(m3)

p-V diagram

Analysis : Q12 – W2 = U

1W2 = 2

1

pdV

= 2 2 1 1

1

p V pV

n

for a polytropic process

since the working fluid is a perfect gas

1W2 = 2 1

1

mRT mRT

n

= 2 1( )

1

mR T T

n

To find m

m = 1 1

1

[ ]pV

RTThe given substance is a perfect gas

100 0.3

0.287 300

0.348 kgb

To find T2

Since the working substance is an ideal gas

p1V1 = mRT1

p2V2 = mRT2

By rearranging we get

1 1 2 2

1 2

pV p V

T T

or

2 2 2

1 1 1

T p V

T pV

The process equation gives

p1V1n = p2V2

n

2 1

1 2

n

p V

p V

Combining the above equations

2 2 2

1 1 1

1

1 1

2 2

1

1

2

1

12 1

2

0.40.3

3000.06

571.1

n

n

n

T P V

T P V

V V

V V

V

V

VT T

V

K

Substituting the values of m and

1W2 = 0.348 0.287(571.1 300)

1.1 1

T2 = 270.76KJ

U = mCv (T2 – T1)

= 0.348 0.717 (571.1-300)

= 67.64 KJ

Q12 = 1W2 + U

= 270.76 + 67.64

= 203.12 KJ

Comment : The negative sign in work indicates that work is done on the gas.

The negative sign in heat transfer shows that heat is coming out of the system.

11. Calculate the change in internal energy, heat transferred and change in enthalpy for 0.5

kg of air expanding according to law pv1.2 = C from 1 Mpa and 300C to 100 kPa. What will be the workdone by air during the expansion?

System : Stationary system of fixed mass

Process : Polytropic (pV1.2 = C)

Work fluid : Air (ideal gas)

Known : 1. Mass (m) = 0.5 kg

2. Initial pressure (p1) = 1Mpa = 1000kPa

3. Initial Temperature (T1) = 300C = 573 K

4. Ratio of sp. heat () = 1.4

To find : 1. Change in internal energy U

2. Heat transferred Q12

3. Change in enthalpy H 4. Work done 1W2

Diagrams:

1 Mpa 100 kPa

300C

Final State Initial State

Analysis :

U = 2

1

dU

= mCv (T2 – T1)

To find T2

Since the working substance is an ideal gas

P1V1 = mRT1

P

(kPa)

1000

100

1

pV1.2 = C

2 2

1pdv

V(m3)

p-V Diagram

P2V2 = mRT2

By rearranging we get

2 2 2

1 1 1

T PV n

T pV

From the process equation

p1V1n = P2v2

n

1

2 1

1 2

1

2

1

n

n

V P

V P

P

P

combining the above equation

2 2 2

1 1 1

1

2 2

1 1

11

2

1

1

22 1

1

0.2

1.2

P =

100 573

1000

390.4

n

n

n

n

T P V

T P V

P

P P

P

P

PT T

P

K

12 1 2

12 1 2

0.5 0.7178 (390.4 573)

65.47

New U

KJ

Q W U

Q can be found only if W is known

1W2 = 2

1

pdV

2 2 1 1

2 1

1

( )

1

0.5 0.287(390.4 573)

1.2 1

PV PV

n

mR T T

n

for a polytropic process

since the working fluid is an ideal gas

= + 131.02 KJ

1Q12 = 1W2 + U

= 131.02 + (-65.47)

= 65.53 KJ

Comment :

The positive sign in work indicates that work is done by the system.

The positive sign in heat indicates that heat is given into the system.

12. A space vehicle has a mass of 500kg and is moving towards the moon. Calculate the kinetic and potential energies relative to the earth when it is 40 km from the launching place and travelling at 2400 km / hr and when the acceleration of the earths gravitational field is 792.5cm/sec2.

System : A moving system of fixed mass

Known : 1. Height h = 40 km

= 40 103 m

2. Velocity C = 2400 km/hr.

32400 10

3600

666.67 / .

m

s

m s

.

3. Acceleration due to gravity

= 792.5 cm /sec2

= 7.925 m/Sec2

To find : a) Kinetic energy

b) Potential energy

Analysis : a) Kinetic energy

2

2

6

1

2

1500 666.67

2

111.11 10

111.11

mC

J

MJ

b) Potential energy = mgh

= 500 7.925 40 103

= 158.5 106 J

= 158.5MJ

13. During a short period of time the following observation were made for a system of mass 50 kg.

a. Heat received = 107 J

b. Work produced = 4 106 J

c. Initial Velocity = 10 km/s.

d. Final velocity = 25 m/s

e) Initial elevation = 20 m

f) Final elevation = 12 m

Find the change in internal energy of the system.

Solution:

System : A moving system of fixed mass

Known : a. Heat received = 107 J

b. Work produced = 4 106 J

c. Initial velocity = 10 km / s

d. Final velocity = 25 m /s

e. Initial elevation = 20 m

f. Final elevation = 12 m

To find : U change in internal energy

Diagrams :

Analysis : 1Q12 – 1W2 = E

Initial State

Q = 1 107 J

10 103 m/s

50 kg

w = 4 106 J

20m

25 50 kgs.

during the process 12m

Datum

Note: Since a moving system of fixed mass is analyzed here E U

Q12 = 1W2 + U + KE + PE

or U = Q12 – 1W2 - KE - PE

2 2

12 1 2 2 1 2 1

2 27 6 3 3

7 6 10

10

1( ) ( )

2

11 10 4 10 50 25 10 10 10 50 9.81(12 20)

2

1 10 4 10 1.3125 10 3924

1.3 10

Q W m C C mg Z Z

U J

14. Air enters a compressor with a velocity of 60 m/s, pressure 100 kPa, temperature 40C

and leaves the compressor with a velocity of 90 m/s, 500 kPa and 120C. Consider the system as adiabatic. Find the power of the motor for a mass flow rate of 40 kg per minute. Write the assumption made.

System : Open System

Working Fluid : Air

Process : Steady flow, adiabatic

Known :

Property At inlet At exit

Velocity 60 m / s 90 m / s

Pressure 100 kPa 500 kPa

Temperature 40C 120C

and the mass flow rate = 40 kg / min.

Diagrams :-

Analysis : [ ]Q W m h ke pe

Q = 0 sincde the flow is adiabatic.

Unless the height from the datum of the inlet and out are mentioned pe = 0.

Therefore W = 2 2

1 21 2

2

C Ch h

2 2

1 21 2

2 2

2

60 9040 1005 40 120

2

C Cm Cp T T

6

6

3.306 10min

3.306 10

60

55.1 /

55.1

J

J

s

KJ s

kW

In

Q = 0

Control

Out

W

Volume

Comment : Since the working fluid is an ideal gas h1 - h2 is replaced by Cp (T1 – T2).

15. In a steam power station, steam flows steadily through a 0.32 m diameter pipeline from the boiler to the turbine. At the boiler end, the steam conditions are found to be P = 4 Mpa,

t = 400C, h = 3214 kJ / kg and v = 0.073 m3 / kg. At the turbine end p = 3.5 Mpa, t = 392C, h = 3202.6 kJ / kg and v = 0.084 m3 / kg. There is a heat loss of 8.5 KJ/kg from the pipeline. Calculate the steam flow rate.

System : Open system

Working fluid : Steam

Process : Steady flow process

Known :

Property At inlet At exit

Pressure 4 Mpa 3.5 Mpa

Temperature 400C 392C

Enthalpy 3214

KJ

kg 3020.6

KJ

kg

Sp. Volume 0.073 m3 / kg 0.84 m3 / kg

and heat loss of Q = 8.5 KJ/ kg.

To find : Steam flow rate

Diagram:-

Control volume

Exit Inlet

Boiler Turbine

Analysis : Q – W = m [h + ke + pe]

W = 0 Since it involves only the flow through pipes.

As the flow is steady mass flow rate is same all the inlet and exit.

in outm m

. .

. .

in exit

in exit

Volume flow rate Volume flow rate

Sp Volume Sp Volume

Area velocity Area velocity

Sp Volume Sp Volume

1 21 2

1 2

22 1

1

1

2 1

2 2

2 1

2 2

1 1

2

1

2

1

0.084

0.073

1.15

1

2

11.15

2

10.32

2

0.16

C CSince A A

V V

VC C

V

C

C C

ke C C

C C

C

C

where

11

1

2

0.073

0.2

2.32

mVC

A

m

m

ke = 0.16 (2.324 m)2

= 0.86 m2

pe = 0

Now Q – W = m [h + ke + pe] becomes

2

2 1

2

2 3

0 0.86 0

8.5 103 = (3202.6 - 3214) 103 + 0.86m

0.86 2.9 10

58.07 / .

Qh h m

m

m

m kg s

Comment: Though the flow is steady through the pipe the increase in velocity is due to in increase in specific volume.

16. The stream of air and gasoline vapour in the ratio of 14:1 by mass enters a gasoline

engine at a temperature of 30C and leaves as combustion products at a temperature of

790. The engine has a specific fuel consumption of 0.3 kg/k Whr. The net heat transfer rate from the fuel – air stream to the jacket cooling water and to the surroundings in 35 KW. The shaft power delivered by the engine is 26KW. Compute the increase in the specific enthalpy 0 fuel air stream.

System : Open system

Process : Steady flow, process

Known : 1. Air fuel ratio = 14:1

2. Inlet temperature t1 = 30C

3. Exit temperature t2 = 790C

4. Specific Fuel Consumption = 0.3kg/kwhr.

5. Heat rejection rate 35Q KW

6. Shaft work output W = 26 KW

To find : Increase in specific enthalpy h

Diagram :-

Analysis : [ ]Q W m h ke PE

ke and pe terms can be neglected unless the relevant data are given.

Fuel air Stream

26kW

ENGINE Shaft work W =

Exhaust

35Q kW

To find m

Specific fuel consumption = 0.3 kg/kwhr One whr is equivalent to 3600KJ. Hence, fuel consumption is 0.3 kg for an output of 3600 KJ. Since the output is 26kW, the rate of fuel consumption.

3

0.326

3600

2.17 10 /

r

kg kJm

kJ s

kg s

The rate of air consumption = 14 rate of fuel consumption

3

3

14 2.17 10

0.03033 /

0.03033 2.17 10

0.0325 /

r

a f

m

kg s

m m m

kg s

Substituting the numerical values in the SFEE we get

- 35 – 26 = 0.0325 h

h = -1877 KJ / kg

Comment: Each kg of the mixture contain 1

15kg of fuel and

14

15kg of air. This mixture

undergoes combustion in the engine. Out of the energy liberated during combustion only 1877 KJ is extracted within the engine and the remaining is taken away by the exhaust gases. This is the reason for higher exhaust temperature.

17. A steam turbine operates under steady flow conditions receiving steam at the following state: Pressure 15 bar. Specific volume 0.17m3/kg internal energy 2700 KJ/kg velocity 100 m/s. The exhaust of steam from the turbine is at 0.1bar with internal energy 2175 KJ/kg. specific 15 m3/kg and velocity 300 m/s. The intake is 3 m above the exhaust. The turbine develops 35 kW and heat loss over the surface of turbine is 20 KJ /kg. Determine the steam flow rate through the turbine.

System : Open System

Process : Steady flow process

Known :

Properties At inlet At exit

Pressure 15 bar = 15 103 Pa bar = 0.1 103 Pa

Internal energy 2700 103J/Kg 2175 103 J/Kg

Sp. Volume 0.17 m3 / kg 15 m3 / kg

Height from the datum 3m 0m

Velocity 100 m/s 300 m/s

Work output W = 35 103 W

Heat rejected work 320 10 /Q

q J kgm

Diagram:

Analysis : [ ]Q W m h ke pe

(or)

In

q = 20 103 J/Kg

Out

Control Volume

W = 35 103W

2 2 2 2

2 2 2 1 1 1 2 1 2 1

3 5

3 5 3

2 2

1( )

2

2175 10 0.10 10 15

(2700 10 15 10 0.17) 20 10

1300 100 9.81(3)

2

wq u p v u p v C C g Z Z

m

w

m

3

3

3

3

2325000 2955000 4000 29.43 20 10

569.970 10

35 10

569.970 10

0.0614 / .

wm

m

w

m

m

m kg s

18. Steam enters a nozzle operating at steady state pressure of 2.5 Mpa and a temperature

of 300C (h1 = 3008.8 KJ/kg) and leaves at a pressure of 1.7 Mpa with a velocity of 470 m/s. The rate of flow of steam through the nozzle is 1360 kg/hr. Neglecting the inlet velocity of the steam and considering the flow in the nozzle to be adiabatic, find:

a) the enthalpy hexit

b) the nozzle exit area, if Vexit = 0.132 m3/kg.

System : Open system

Process : Steady flow, adiabatic

Known :

Properties At inlet At exit

Pressure 2.5 mPa -

Temperature 300C -

Enthalpy 3008 KJ/kg -

Velocity Negligible 470 m/s

Sp. Volume - 0.132 m3/kg

The mass flow rate in = 1360 Kg/hr

= 1360

/3600

kg s

= 0.378 kg/s

To find : Exit enthalpy – h2

Exit Area – A2

Diagram :

Analysis : Q W m h ke kpe

Flow is given as adiabatic hence 0Q

Nozzle will not involve W

Change in potential energy can be neglected in nozzles and diffusers

h = ke

2 2

1 2

2 2

1 2

2

2

C C

C Ch

As the inlet velocity is negligible

In Out

h2 – h1 = 2

2

2

C

= 2470

2

= - 110450 J/kg

h2 = h1 2

2

2

C

= 3008.8 103 – 110450

= 2898.4 103 J/kg

= 2898.4 KJ/kg

b)

2 2

2

22

2

4 2 2

0.378 0.132

470

1.062 10 1.062

A Cm

V

mVA

C

m or cm

19. Air enters the diffuser of a jet at 10C, and 80 kPa, with a velocity of 200 m/s. The inlet area of the diffuser is 0.4m2. Neglecting the exit velocity of air, determine the mass flow rate of air and the temperature of air at the exit.

System : Open system

Process : steady flow

Known : At the inlet

Temperature T1 = 283 K

Pressure P1 = 80 kPa

Velocity C1 = 200 m/s

Area A1 0.4 m2

To find : 1. Mass flow rate m

2. Exit temperature T2

Diagram :

Analysis : 1. .

Area Velocitym

Sp Volume

1 1 1 1

1 1 1

/

AC ACor

V RT P

= 80 0.4 200

0.287 283

= 78.8 kg/s.

2. Q W m h ke pe

Heat transfer across the control surface and change in potential energy can be neglected in a diffuser.

A diffuser will not involve w

h = -ke

(h2 – h1) = - 2 2

2 1

2

C C

Since air is an ideal gas

2 2

2 12 1

2

12 1 1

( )2

( ) 2

p

p

C CC T T

CC T T Since C is negligible

2

12 1

2

2

200 283

2 1005

303

p

CT T

C

K

20. Hellium at 300 kPa, 60C, enters a nozzle with negligible velocity and expands steadily without heat transfer in a quasi equilibrium manner to 120 kPa. The process is such that Pv1.67 = constant. Calculate the exit velocity.

System : Open system

Process : Steady flow, adiabatic, pV1.67 = C

Known : 1. Inlet pressure (p1) = 300 kPa

2. Inlet Temperature (t1) = 60C

3. Exit pressure (p2) = 120 kPa

Diagram:

Analysis : Q – W = m [h + ke + kpe]

We can’t proceed with this equation since the value of enthalpy drop is not known. It can be substituted by Cp (T2 – T1), but still Cp is an unknown. In problems of this kind, Equation can be used.

2 2 2

2 12 1

1

( )2

flow

C CW vdp Z Z g

Since these is no work interaction in a nozzle and (Z2 – Z1) term can be dropped.

22 2

2 1

12

C Cvdp

Where C1 is negligible

22

2

12

Cvdp

In Out

Control Volume

From the process equation pVn = constant

1

1

2 1 12

2

1

21

11

2

1

2

11

n

n n

nn

Cv

p

CC p dp

pC

n

1 11

2 1

1 1 1

1 1 2 1

1 1 11

1 1 1 2 1

1 1

2 11 1 1 1

1 1

1

1

1( )

1since

n n

n nn

n n

n n n

n n

n n n

n n

n n

n n

n n

p pC

n

n

np v p p

n

np p v p p

n

p pnp v

np p

1

21 1

1

1

22 1 1 1 1

1

0.67

1.67

11

21 [ )

1

2 1.67 8314 120 333 1

1.67 1 4 300

1030 / .

for an ideal gas

n

n

n

n

pnp v

n p

pnC RT p v RT

n p

n

m s

21. Imagine two curves intersecting at some point on a p – v diagram. The curves are representing reversible processes undergone by a perfect gas; one is an adiabatic process and the other an isothermal process. Show that the ratio of slope of the adiabatic curve to that of the isothermal curve at the point of intersection is equal to:-

System : Closed / Open

Process : case (i) Isothermal

case (ii) Adiabatic

Diagram :

To prove : Adiabatic

Isothermal

dp

dv

dp

dv

Proof : For case (i) Isothermal Process

pv = RT = constant

By differentiating we get,

Pdv + vdp = 0

vdp = dv

(1)Isothermal

dp p

dv v

For case (ii) Adiabatic Process

pv = constant

case (i)

case (ii)

P

V

Taking natural log we get

In (p) + In (v) = In (C)

By differentiating we get

(2)Adiabatic

dp dv

p v

dp p

dv v

From (1) & (2) we get Adiabatic

Isothermal

dp

dv

dp

dv

Hence proved.

22. A piston cylinder arrangement as shown in the diagram contains 5g of air at 250 kPa,

300C. The piston is of mass 75 kg and diameter 10 cm. Initially the piston is kept pushed

against the stops. The atmospheric pressure is 100 kPa and temperature is 20C. The

cylinder now code to 20C. Calculate the heat transfer.

Process : Constant volume / constant pressure

When heat is removed from the system temperature decreases and volume remains constant until the upward pressure exerted on the bottom side of the piston just balances the atmospheric pressure and the weight of the piston. Further cooling results in decreases of volume at constant pressure.

Stop

Piston

Air Cylinder

Known : T1 = 300C

P1 = 250 kPa

DP = 0.1 m

Patm = 100 kPa

Tatm = 293 K

T2 = 293 K

Diagram :

To find : Heat transfer Q12

Analysis : To find the pressure which just balance the atmospheric pressure and the weight of the piston.

Pair APiston = mPiston g + Patm APiston

3

2

3

75 9.81 100 10

0.1

193.68 10

193.68

Piston g

air atm

Piston

mP P

A

Pa

kPa

Therefore cooling from 250 kPa to 193.68 kPa will be taking place at constant volume. Temperature at the stage can be computed as follows:-

P

2

1

21

V

2 2 1 1

2 2

2 1

22 2

1

193.68 573

250

443.9

PV PV

T T

V V

PT T

P

K

since

Therefore further cooling from 443.9 K to 293K will be a constant pressure process. Thus,

Q12 = Q12’ + Q2’-2

Q12 = 1W2 + 2W2 + (U2 – U2’) + (U2 – U1)

Q12 = 1W2 + 2W2 + (U2 – U1)

Since the process 1-2’ is a constant volume process 1W2 = 0 and as the process 2’-2 is isobaric.

2’W2 = P2’ (V2’ – V2)

where V2 = 3.2 10-3 m3

Substituting we get

2’W2 = 193.68 (2.171 10-3 – 3.2 10-3)

= - 0.199 KJ.

U2 – U1 = mCv (T2 – T1)

= 5 10-3 0.717 (293-573)

= 1.0038 KJ.

Therefore Q12 = -0.199 – 1.0038

= - 1.2028 KJ.

23. A closed system undergoes a cycle consisting of two processes. During the first process,

40 KJ of heat transferred to the system while the system does 60 KJ of work in the second process, 45 KJ of work is done on the system.

a) Determine the heat transfer during the second process.

b) Calculate the network done and net heat transfer for the cycle.

Solution:-

System : Closed System

Process : Cycle consisting of two processes

Known :

To find : Q21, Q, W

Analysis : 1

Q21 = 2W1 + U2-1

Where 2W1 = -45 KJ.

To find (U2-1)

For cyclic process the algebraic sum of any property of the system must be zero.

(U12) + (U21) = 0

(U12) = (U)12

= - [Q12 – 1W2]

= -[40 – 60]

= +20 KJ

U21 = -45 +20

Q21 = -25KJ

2 Network done = (1W2 + 2W1)

= 40 – 25

= 15 KJ

Net heat transfer = (Q12 + Q21)

= 60 – 45

= 15 KJ.

Comment :

The negative sign indicates Q21 that 25 KJ of heat is supplied by the system to the surroundings during the process 2 to 1.

24. A system is capable of executing a cyclic process as indicated in the pV sketch; it may be executed either clockwise abca or counter clockwise adca.

a. When going clockwise to state C, 80 KJ of heat flow to the system and 35 KJ of work are done by it. When returning to state a for c, 60 KJ of heat flow from the system. Find work along the path ca.

b. When going counter clockwise to state C, 70KJ of heat flow to the system. Find the work during the process adc.

Solution :

System : Closed (as the total volume changes)

Process : Cyclic with two possible cycles.

Process Q(KJ) W(KJ)

1-2

2-1

40

unknown

60

-45

P

b

a d

c

V

Known :

Cycle abca Cycle adca

Process Q(KJ) W(KJ) Process Q(KJ) W(KJ)

a – c

c – a

80

-60

35

Unknown

a – c

c – a

70

Unknown

Unknown

Unknown

To find : 1. Work done in the process c – a in path abca

2. Work done in the process c – a in path adca

Analysis : 1. dQ dw for the cycle abca

80 – 60 = 35 +cWa

cWa = 20 – 35

= 15 KJ

2. Qac = Wac + Uc – Ua

Wac – Qac – (Uc - Ua)

To find : Uc – Ua.

Irrespective of the path followed U between also are some.

(U)ac in cycle abc = (U) ac in cycle adc

(Uc – Ua) in adc = (Qac – aWc) in path abc

= 80 – 35

= 45 KJ

cWa = Qac – (Uc – Ua)

= 70 – 45

= 25 KJ.

25. For a certain system executing a cyclic process 250J of heat is absorbed by the system and 100J of heat are rejected. The system also receives 30W-s of electrical power while it moves a 3 kg mass vertically by means of a pulley arrangement. How far does the mass move? Take local g=9.65 m/s2.

System : Closed / Open

Process : Cyclic

Known : Qin = 250 J

Qout = 100 J

EEin = 30 w-s

To find : Height through which is a mass to be lifted.

Analysis : Energy in = Energy out

250 + 30 = 100 + W

W = 180J

This must be equivalent to the increase in potential energy of the mass that is mgh = 180.

180

3 9.64

6.22 .

h

h m

26. A three process cycle operating with nitrogen as the working substance has constant temperature compression 1-2 (T=40C, p1 = 110 kPa); constant volume heating 2-3; and polytropic expansion 3-1 (n=1.35), the isothermal compression requires – 67 KJ/kg of work.

Determine.

a. P, T and v around the cycle:

b. The heats in and out;

c. The net work. Take Cv = 0.7431 KJ/kg.

System : Closed System

Process : Cycle consisting of there processes

Known : P1 = 110 kPa

T1 = 313K

Process Type QKJ/Kg WKJ/Kg

1 – 2

2 – 3

Isothermal

Isochoric

-

-

-67

-

3 – 1 PV1.35 = C - -

To find : 1. p.T.v. at state points 1, 2 and 3

2. Q, w in all the processes

3. net work

State 1:

Analysis : a) Finding p, T, v of the state points

Consider state 1 and process 1-2 (Isothermal)

P1 = 110 kPa

T1 = 40 + 273

= 313 K

11

1

3

1 2

8.314313

28

110

0.845 /

67 /

RTv

p

m kg

W KJ kg

2since N , is an ideal gas

That is

21 1

1

2

1 1 1

67

110 0.8452 1

3

67

67

0.411 /

vp v In

v

vIn

v p v

v v e

m kg

Also p1v1 = p2v2

1 11

2

12 1 2 12

226.17

[

67 /

p vp

v

kPa

and Q W U

KJ Kg

12

since the process is isothermal]

Q

Consider state 3 and the process 2-3 (Isochoric)

2W3 = 0

V3 = V2

0.411 m3 /Kg

T2 = 313 K

From process 3-1 )pV1.35=C)

p1v11.35 = p3v3 1.35

1.35

13 1

3

1.350.845

1100.411

291.0

vp p

v

kPa

Knowing p3, T3 can be found

T3 = T2 3

2

p

p

23 3 2

31 1 2

291.0313

226.17

402.72

( )

0.7431(402.72 313)

66.67 /

( )

0.7431(313 402.72)

66.67 /

v

v

K

U C T T

KJ kg

U C T T

KJ kg

1 1 3 33 1

1

p v p vW

n

31 3 1 31

110 0.845 291 0.411

1.35 1

76.2

76.2 66.67

9.5 / .

KJ

Kg

Q W U

KJ Kg

Tabulating the property values:

State P(kPa) T(K) V(M3/kg)

1. 110 313 0.845

2. 226.17 313 0.41

3. 291.0 402.72 0.411

b) Heats in and out.

Process QKJ/Kg W(KJ/kg) U(KJ/Kg)

1 – 2 - 67 - 67 0

2 – 3 66.67 0 66.67

3 – 1 9.5 76.2 - 66.67

c) Wnet = - 67 + 76.2

= 9.2 KJ/Kg

Check : Q = - 67 + 66.7 – 9.5

= 9.8 KJ/Kg

Qnet = Wnet from first law

U = 0 + 66.7 – 66.67

Also

U = 0

27. In a steady flow process air is passing through a series of components as shown in the following diagram. Assuming the heat addition process is reversible isobaric and all others are reversible adiabatic processes obtaining the exit velocity. The power developed by the turbine is just enough to run the compressor.

System : Open

Process: Steady flow through a compressor, a heat exchanger, a turbine and a nozzle in sequence.

Known : P1 = 100 kPa

T1 = 300K

M = 0.2 Kg/s

P3 = 600KPa

P5 = 100 Kpa

o o

r cW W

To find : Exit velocity from the nozzle

Diagram :-

100 kPa

Heat exchanger

2 3

1

Com

pre

ssor

Turb

ine

4

5

Nozzle

Q = 100 kW

Common Shaft

600 kPa

600 kPa 100 K 0.2 kg/s

Analysis :

Process : 12

2 1

1 1

322 1 1

1 1

0.4

1.4

( )

600300

100

500.6

o o

c pW mC T T

PPWhere T T T

P P

K

(ii) Process 23

23

3 2

3

3

100

( ) 100

100500.6

0.2 1.005

993

o

o

p

Q kw

mC T T

T

T K

(iii) Process 34

4 3 2 1( )

998 (500.6 300)

7975

T T T T

K

o o

T cW W

3 4 2 1( ) ( )o o

p pmC T T mC T T

3 4 2 1

4 3 2 1( )

998 (500.6 300)

7975

T T T T

T T T T

K

Also 1

4 4

3 3

P T

P T

P4 =

1.4

0.47975600

998

= 273.6 Kpa

(iv) Process 4-5

1

5 5

4 4

0.4

1.4

5

1007975

273.6

598.2

T P

T P

T

K

For the flow through the nozzle

2 2

5 44 5

2

C Ch h

Neglecting C4 compared C5 we get

5 4 52 ( )pC C T T

2 1005(7975 598.2

632.9 / .m s

The exit velocity of air from the nozzle is 632.9 m/s.

28. An air compressor is used to supply air to a rigid tank that h as a volume of 4m3.

Originally, the air pressure and temperature in the tank are 101 kPa and 35C. The supply pipe is 5 cm diameter and the velocity of air in the inlet pipe remains at 15 m/s. The

pressure and temperature of the air in the inlet pipe are constant at 750 kPa and 35C. Calculate the following:

a) the time rate of change of mass inside the tank.

b) the mass of air added to the tank if the pressure of the tank reaches 400 kPa and 50C.

c) time required to reach the tank pressure of 400 Kpa at 50C.

System : Open

Process : Unsteady flow

Known : 1. Volume of the tank V1 = 4m3

2. Initial pressure P1 = 101 Kpa

3. Initial temperature T1 = 273 + 35 = 308 K

4. Supply pipe diameter D1 = 0.05 m

5. Velocity of inlet air Ci = 15 m/s.

6. Pressure at the inlet Pi = 750 Kpa

7. Temperature at the inlet Ti = 273 + 35 = 308 K

Diagram :-

To find : 1. Rate of increase in mass of the control volume.

2. Mass of the air in the tank when the pressure and temperature reach 400 kPa and 50C respectively.

3. Time required for the pressure of the tank to reach 400 kPa at 50C.

Analysis :

Air from the compressor

Tank of volume 4m3 Insulation

1. Rate of increase in mass of the control volume = Rate of mass inflow into the control volume.

That is 1 1

dmAC

dt

Where i

i

p

RT

= 750

0.287 308

= 8.48 kg/m3

2

2

3 2

0.05

1.963 10

i iA D

m

Therefore 38.48 1.963 10 15cv

dm

dt

= 0.25 kg/s.

2. Mass of air added = Final mass – Initial mass

= (m2 – m1)cv

Where m1 = 1 1

1

pV

RT

2 22

2

101 4

0.287 308

457Kg

p Vm

RT

Final volume V2, is the same as V1 since the vessel is rigid.

400 4

0.287 323

17.26 kg

Substituting m1, and m2 we get

(m2 – m1) = 17.26 – 4.57

= 12.69 kg

3. Time required to reach 400 kPa and 50C

2 1

( )

12.69

0.25

50.76

cv

cv

m m

dm

dt

Seconds

Increase in mass

Rate of air supply

29. A rigid vessel of volume 250 litre containing at air 1 bar and 300 k is being filled adiabatically by connecting it to a pipe supplying air 5 bar and at unknown temperature T1. Once the vessel completely filled, it is disconnected from the main temperature and pressure of the air inside the vessel when it is full is T2 and 5 bar, respectively. Determine the temperature T2. Take T1 = T2.

System : Open

Process : Unsteady flow

Known : 1. Volume of the vessel V1 = 250 lit = 0.25 m3

2. Initial pressure p1 = 1 bar = 100 Kpa.

3. Initial temperature T1 = 300 K

4. Final pressure p2 = 5 bar = 500 KPa

5. Pressure of the supply air p1 = 5 bar = 500 KPa

6. Final Temperature T1 = Supply air temperature.

Diagram :-

Vessel or Volume 250 litre

Supply air

To find : Final temperature

Analysis : Balancing mass over the filling process

(m2 – m1)cv = mihi

Balancing energy over the filling process

(E)cv = mihi

m2u2 – m1u1 = mihi

m2CvT2 – m1CvTi = m1CpTi

m2T2 – m1T1 = miTi

Since T2 – Ti and mi = m2 – m1

m2 T2 ( - 1) = m1T2 - m1T1

m2T2 ( - 1) = m1T1 2

1

T

T

Consider this equation

m2T2 = 2 2p V

Rand m1T1 = 1 1pV

R

substituting in the above equation we get

2 2 1 1 2

1

( 1) ( 1)p V pV T

R R T

As we vessel in rigid V2 = V therefore

22 1

1

12 2 1

1

22 1 1

1

2 12 1

1

1 2

1

( 1) ( 1)

( 1) ( )

( 1)

1( 1)

1 ( 1)

Tp p

T

pp T T

T

pT T T

p

PTT T

P

T P

P

substituting the numerical values we get

2

300 5001 (1.4 1)

1.4 100

642.9

T

K

30. What are the Limitations of First Law thermodynamics?

(1) First Law of thermodynamics does not specify the direction of flow of heat and work. i.e., whether the heat flows from hot body to cold or from cold body to hot body.

(2) The heat and work are mutually convertible the work can be converted fully into heat energy but heat cannot be converted fully into mechanical work. This violates the foresaid statements. A Machine which violates the First law of thermodynamics is known as perpetual motion machine (PMM - !) of the first kind which is impossible.

PMM – 1 is a machine which delivers work continuously without any input. Thus, the machine violates first law of the thermodynamics. 31. During a flow process a 5 kW paddle wheel work is supplied while the internet energy of the system increases in one minute is 200 kJ. Find the heat transfer when there is no other form of energy transfer. Given data: Work done, W=5kW(since work is supplied to the system)

Internal energy, U= 200

3.33 /60

kj see ]

From first law of Thermodynamics.

Q=W+U Q=5+3.33 Q=-kW Heat transfer, Q=-1.67kw (Note:-ve sign indicates that the heat is transferred from the system.) 32. A liquid of mass 18 kg is heated from 250C to 850C. How much heat transfer is required? Assume Cp for water is 4.2kJ/kgK. M=18kg T1=250C=25+273=298K T2=850C=85+273=358k Heat transfer, Q=mCp(T2-T1) Q=18x4.2x(358-298) Q=4536kJ 33. A closed system receives an input of 450kJ and increases the internal energy of the

system for 325kJ. Determine the work done by the system. Given Data Heat received, Q=450kJ

U=325kJ. By fist law of thermodynamics

Q=W+U 450=W+325 W=450-325=125kJ 34. During the compression stroke of reciprocated compressor, the work done to the air in the cylinder is 95kJ/kg and 43lJ/kg of heat is rejected to the surroundings. Determine the change in internal energy. Given data: W=-95kg(-ve sign indicates that the work is done on the system) Q=-43kJ(since the heat is rejected form the system) Solution: By first law of Thermodynamics

Q=W+U

-43=-95+U

U=-43-(-95)

U=52kJ/kg The internal energy increases by 52kj/kg. 35. Calculate the distance moved by a locomotive from consuming 2 tone of coal if 10% of the heat generated by the coal is converted into to coal gas then into mechanical work. The tractive effort required is 30N/tonne of dead mass of the locomotive where the dead mass of the locomotive is 2500 tones. Assume 1 kg of coal liberates 35000kJ of heat on burning. Given data: Mass of coal, Q=2 tonne =2000kg Available mechanical work=10%Q Tractive effort =30N/tonne Mass of the locomotive =25000 tonnes. For 1kg of coal, 35000kJ of heat is generated.

Calorific value of coal. CV=35000kJ/kg Heat generated, Q=mcxCV=2000x35000=70x6kJ

6

5 8

1070 10

100

70 10 70 10

W

kJ J

we know that, work done = Force x distance

Where work done is nothing but available mechanical work 70x105x103=75000xx x=93333.3m =93.3km. Distance moved by a locomotive, x=93.3km 36. The following data refer to a closed system, which under goes a thermodynamics cycle consisting of four processes.

Process Heat transfer kJ/min Work transfer kJ/min

a-b b-c c-d d-a

50,000 -5,000

-16,000

- 34,200 -2,200 -3,000

Show that the data is consistent with the first law thermodynamics and calculate:

a) Net rate of work outputs is MW b) Efficiency of the cycle

Given data: Qa-b =50,000kJ/min Qb-c =50,000kJ/min Qc-d =-16,000kJ/min Qd-a=0 Wd-a=0 Wb-c=34,200kJ/min Wc-d=-2,200kJ/min Wd-a=3,000kJ/min To Find

1. ?W and

2. =? Solution: Cyclic heat transfer of the cycle,

Q=50000-5000-16000+0 =29000kJ/min similarly, cyclic work transfer of the cycle,

W=0+34200-2200-3000 =29000kJ/min

From first law of the thermodynamics Q=W

The gives data is consistent with the first law of thermodynamics.

29000Net work output, W= / sec

60

=483.3kW

=0.48MW

kJ

Heat supplied, Qs = 50,000kJ/min( Taking the positive heat only)

Efficiency of the cycle,=

supplied

29000 =

50,000

=58%

s

WWork done

Heat Q

Result:

1. Total work output of the cycle, W=0.48MW

2.Efficiency of the cycle, =58%

37. A paddle wheel fixed to the shaft of an engine revolves in a closed hollow vessel containing water. This closed vessel is connected freely on the shaft and restraint to its turning moment is proved by mass attached to its side. Find the temperature rise for the following observations. Engine rpm=650 Load applied = 60kg at a leverage of 1.2m Quantity of water =200kg. Duration of test = 20minutes Given data: Speed, N=650kg

Load, W=60kg

Leverage,1-1.2m

Mass of water, mw=200kg.

Time, t=20 minutes. Solution Torque=WxI

=60x10x12 [1kg=10N] =720N-m

Break power or work done = 2x650x720x20 =58810614.48J

Heat gained by the water in 20min Q=mwCpw tw

58810614.48=200x4.18x103xtw [Cpw=4.18kJ/kgK=4.18x103J/kgK]

tw=70.350C. Result:

Rise in temperature, tw=70.350C 38. During a non flow process, the temperature of the system changes form 150C. The work done by the system and heat transfer/0C rise in temperature, at each temperature is given by

0 06 0.065 / 1.005 / .w Q

T kJ C and kJ CdT dT

Determine change in internal energy of the system during the process. Given data : T1=150C=15+273=288k T2=500C=50+273=323k

0

0

6 0.065 /

1.005 / .

wT kJ C

dT

QkJ C

dT

Solution:

.............(1.2)

1.005 ..............(1.3)

6 0.065 ...........(1.4)

Q W

U Q dT

W T dT

Substituting (1.3) & (1.4) in (1.2)

2 2

1 1

2 2

1 1

2 2

2 1 2 1 2 1.

2 2

1.005 6 0.065

0.0651.005[ ] 6

2

0.0651.005[323 288] 6 323 288 323 288

2

520.188

T T

T T

T T

T T

U Q W

dT T dT

T T T T T T

kJ

Result :

The change in internal energy, U=520.188kJ. 39. A 1kW string motor is applied to tank or water. The tank contains 15kg of water and the string action is applied for ½ hr. if the tank is perfectly insulated calculate the change in internal energy of water assuming that the process occurs at constant pressure and that Cv for water may be taken as 4 kJ/kg K. Also calculate rise in temperature. Given data: Power P=1kW Time, t=1/2 hr. Mass of water, mw =15kg. For perfect insulation, Q=0. Cv=4. 18kJ/kg K Solution: By First law of Thermodynamics,

Q=W+U Q=0 for perfect insulation

W=-U Work done, W=1 x ½ x3600[ 1 hr= 3600 sec] =1800kJ

U=1800kJ -ve sign indicates that the internal energy is decreased. For calculates, we can consider only magnitude.

U=mwCwwt

t=28.7080C Result:

1. The change in internal energy U=-1800kJ

2. The rise in temperature, t=28.7080C 40. During summer season a room measuring 10x13x6m3 is cooled electrically from initial temperature 280C to 20C. the air pressure inside the room is same as that of surroundings and is equal to 72cm of Hg. The pressure remains constant during the cooling process. The cooling capacity of furniture and wall is 35kJ/k. The specific heat of air is 1.005 kJ/kg k. calculate the amount of electric energy needed cooling the room. How much air comes out through gaps and windows during cooling period? Given data: Volume of air =10x13x6m3 Temperature range =280C to 20C at 72cm of Hg Cooling capacity of a/c and wall = 35kJ/K Cpof air, Cpa = 1.005kJ/kg k. Solution: We know that, 76cm of Hg = 1.013 bar

For 72cm of Hg = 1.013

7276

=0.9596 bar Volume of air, Va = 10x13x6 =780m3

11

1

PVm

RT

Mass of air before cooling, 1

0.9596 100 780

0.28 301

=888.097kg

m

11

2

PVm

RT

Mass of air after cooling, 1

0.9596 100 780

0.28 273 2

=972.06kg.kg

m

The amount of air coming into the room during cooling. M=m2-m1 M=972.06-888.097 =83.963kg Cooling effect required from 280C to 20C Q=mCp(T1-T2)

1 2

1

m 888.097 972.06 mass, m=

2 2

=930.08kg

Q 930.08 1.005(28 2)

=24302.99kJ

mMean

Cooling effect required to cool the walls and furniture Q2=35x(28-2) =910kJ

The total cooling effect Q=Q1+Q2 Q=24302.99+910 =25212.99kJ

Q needed, E=

3600

25212.99 =

3600

=7kW-hr

Electricity

Result: 1. Amount of electricity needed, E=7kW-hr. 2. Amount of air comes through gaps & windows 83.963kg.

41. 25 people attended a farewell party in a small room of size 10x8m and have a 5m ceiling. Each person gives up about 350 kJ of heat per hour. Assuming that the room is completely sealed off and insulated, calculate the air temperature rise occurring within 10 minutes. Assume Cv of air 0.718kJ / kg k and R=0.287kJ/kg k and each person occupies a volume of 0.05m3. Given data: No. of persons =25 Room size = 10x8m Ceiling height =5m Heat /Person=350kJ/hr. Time, t==10min Cv0.718kJ/kg K. R=0.287kJ/kg K. Each person volume, Vp=0.05m3 Solution: Volume of the room, Vr=10 x 8 x 5 =400m3

The volume of air, Va=Vr -Vp x n =400-0.05 x 25 =398.75M3

0

1.013 100 398.75 of air,=

0.287 293

=472.29kg

[ Assuming, p=1.013 bar & T=20 ]

apVm

RT

Mass

C

By first law of thermodynamics,

Q=W+U Assume heat addition at constant volume process

W=0

Q=U=Heat/ person x no. of persons =350 x 25 =8750 kJ/hr

Heat loss for 10 minutes, 8750

10 1458.3360

Q kJ

Heat gained by air, Q=mCv=T

1458.33=472.29x0.718xT

T=4.220C. Result:

The rise in temperature, T=4.220C. 42. Mass of 15kg of air in a piston cylinder device is heated from 250C to 900C by passing current through a resistance heater inside the cylinder. The pressure inside the cylinder is held constant at 300 kPa during the process and heat loss of 60kJ occurs. Determine the electrical energy supplied in kW – hr and changed in internal energy. Given data: M=15kg T1=250C90+273=298K T2=900C=90+273-363K P1=P2=300Pa=300kN/m2 Q=-60kJ To find:

(i) Electrical energy supplied (ii) Change in internal energy

Solution: Work done, W=mR(T2-T1)

=15x0.287(363-298) =279.825kJ Work done in kW-hr= Work done x 3600 =279.825x3600 =1007.37x103kW-hr Change in internal energy

U=Q-W =-60..279.825=-339.825kJ. Result:

(i) Electrical energy =1007.37x103kW-hr

(ii) Change in internal energy,U=-339.825kJ 43. In a vessel 10kg of O2 is heated in a reversible non-flow constant volume process, so that the pressure of O2 is increased two times that of initial value. The initial temperature is 200C. Calculate the final temperature, change in internal energy, change in enthalpy, and heat transfer. Take R=0.259kJ/kg K and Cv=0.625 kJ/kg K for oxygen. Given data: M=10kg Process: constant volume P2=2p1 T1=200C=20+273=293K R=Cp=Cv=0.259kJ/kgK So1 Cp=R+Cv=0.259+0.652=0.911kJ/kg K To find:

T2U,H and Q Solution:

(i) temperature at the end of the process (T2): for constant volume process

1 1

2 2

1

1 2

2

293

2

586

P T

P T

P

P T

T K

(b) Change in internal energy

U=mxCvx(T2-T1) =10x0.625x(586-293) =2669.23kJ By first law of thermodynamics

Q=W+U For constant volume process, w=0

Q=U Q=1910.36kJ Result: Final temperature T2 =586K

Change in internal energy, U=1910.36kJ

Change in enthalpy, H=26669.23kJ Work done, w=0 Heat transfer, Q=1910.36kJ. 44. A certain gas of volume 0.4m3, pressure of 4.5 bar and temperature of 1300C is heated in a cylinder to a 9 bar when the volume remains constant. Calculate (i) Temperature at the end of the process, (ii) the heat transfer, (iii) change in internal energy (iv) work done by the gas, (v) change in enthalpy and (vi) mass of the gas, assume Cp = 1.005kJ/kg K. and Cv = 0.7kJ/kg K. Given data: V1=0.4m3 P1=4.5bar=450kN/m2 T1=1300C=130+273=403K P2=9bar=900kN/m2 Cp=1.005kJ/kg K Cv=0.71 kJ/kg K and It is given that the process ion constant volume process To find:

T2,Q,U, W,H and m. Solution:

(i) Temperature at the end of the process (T2): Relation between p.v.T for constant volume process is

1 1

2 2

22 1

1

900403 806

450

P T

P T

PT T k

P

Heat transfer (Q): Q=mCv(T2-T1)kJ First find out the mass of the gas by using characteristic gas equation. P1V1=mRT1

Where R=Cp-Cv=1.005-0.71=0.295kJ/kg K

1 1

1

450 0.41.51

0.295 403

PVm kg

RT

Then, Q=1.51 x 0.71 x(806-403) =433.2kJ

(c) Change in internal Energy, (U):\

From First law, Q=W+U In this process, W=0

Therefore, Q=U

U=433.2kJ. (d) Work done by gas (w): For the constant volume process work done by the gas zero. W=0 (e) Change in Enthalpy(H)

H=mCv(T2-T1) =1.51 x 0.71 (806-403)=432.06kJ

H=432.06kJ (f) Mass of the gas (m): Already, we find that mass, m=1.51kg. Result:

(i) Temperature at the end of the process (T2)=806K (ii) The heat transfer, (Q)=433.2kJ

(iii) Change in Internal energy, (U)=433.2kJ (iv) Work done by the gas, (W)=0

(v) Change in Enthalpy, (H)=432.06kJ (vi) Mass of the gas, (m)=1.51kg.

45. 2 5kg of air at 400 C and 1 bar is heated in reversible non-flow constant pressure until the volume is doubled. Find (a) change in volume (b) work done (c) change in internal energy and (d) change in enthalpy. Given data: M=5kg T1=400C P1=1bar =100kN/m2 V2=2V1 P=constant To find:

1. V2-V1

2. W=?

3. U=?

4. H=?

Solution: From ideal gas equation

1 1 1

1

3

5 0.287 313

100

=4.49m

pV mRT

V

The final volume V2=2v1 =2x4.49 =8.98m3

(i) Change in volume: V2-V1=8.98-4.49 =4.49m3

(ii) Work transfer:

2

1

2 1

=100(8.98-4.49)

=449kJ

v

v

W pdv p v v

(iii) change in internal energy:

U=mCv(T2-T1) For constant Pressure process

2 2

1 1

22 1

1

1

1

2V =313

V

=626K

V T

V T

VT T

V

U=5x0.714(626-313) =1117.41kJ

(iv) Change in enthalpy

H=mCp(T2-T1) -5x1.005(626-313) =1572.825kj.

Result: 1. Change in volume, V2-V1=4.49m3 2. Work done, W=449kJ

3. Change in internal energy, U=1117.41kJ

4. Change in enthalpy, H=1572.83kJ. 46. One kg of gas is expands at constant pressure from 0.085 m3 to 0.13 m3. If the initial temperature of the gas is 2250C, find the final temperature, net heat transfer, change in internal energy and the pressure of gas. Given data: M=kg V1=0.085m3 V2=0.13m3 T1=2250C=25+273=498K Assume Cp=1.005kJ/kg k; Cv=0.71kJ/kg K. To find:

T2,Q,U&P Solution:

(i) Final temperature (T2): By using p, V&T relation

1 1

2 2

22 1

1

2

0.13 498

0.085

761.6

V T

V T

VT T

V

T K

(ii) Heat transfer(Q): Q=mCp(T2-T1) =1x0.7(761.6-498) Q=264.9kJ

(iii) Change in Internal energy (U):

U=mCv(T2-T1) =1x0.7(761.6-498)

U=187.16kJ. (iv) Pressure(p):

p1V1=mRT1

11

1

2

1 2

1 0.295 498 =1.005-0.71

0.085 R=0.295kJ/kg K

1728.3 /

p v

p

p p

R C CmRT

V

kN m

Result: (i) Final Temperature (T2=761.6K (ii) Heat transfer (Q) =264.9kJ

(iii) Change in internal energy (U)=187.16kJ (iv) Pressure, p1=p2=1728.3kN/m2

47. 0.25kg of air at a pressure of 1 bar occupies as volume of 0.3m3. if this air expands isothermally to a volume of 0.9m3. Find (i) the initial temperature, (ii) the final temperature (iii) external work done, (iv) Heat absorbed by the air, (v) change in internal energy. Assume R=0.29kJ/kg K. Given data: M=0.25kg P1=1 bar = 100kN/m2 V1=0.3m3 V2=0.9m3 To find:

1. T1=? 2. T2=? 3. W=? 4. Q=?

5. U=? Solution: From ideal gas equation.

P1V1 = mRT1

1

100 0.3T

0.25 0.287

For isothermal process T1 = T2 = 418.12K

Work done, W, mp1V, in 1

2

Vp1 orp1v1in

p2 V

0.9100 0.3 in

0.3

=32.96kJ

Heat absorbed, Q = W = 32.96kJ

Change in internal energy, U = 0. Result:

1. Work done, W = 32.96kJ 2. Heat absorbed, Q = 32.96kj

3. Change in internal energy, U = 0. 48. A mass of 1.5kg of air is compresses in a quasi static process from 0.1MPa to 0.7 MPa for which pV = constant. The initial density of air is 1.16kg/m3. Find the work done by the piston to compress the air. Given data: M = 1.5kg Process = Quasi – static, pV = C (isothermal) P1 = 0.1MPa = 100kN/m2 P2 = 0.7MPa = 700kN/m2

Initial density (1) = 1.16kg/m3 To find: Work done, W Solution:

1

3

mass mDensity =

Volume V

mass m 1.5Volume 1.293m

Density 1 1.16

Volume (V1) = 1.293m3

21

1

VForisothermalprocess,W p1Vin

V

0.1847100 1.293 in 251.6kJ

1.293

(-ve sign indicates that the work is done on the system) Result: Work done, W = -251.6kJ 49. 2 kg of gas at a pressure of 1.5 bar occupies volume of 2.5m3. If this gas compress isothermally to 1/3 times the initial volume. Find (i) initial temperature (ii) final temperature (iii) work done and (iv) heat transfer. Assume R = 0.287 kJ/kg K.

Given data: M = 2kg P1=1.5 bar = 150k/m2 V1=2.5m3

3

2 1

1 1V V 2.5 0.83m R 0.287kJ/kgK

3 3

To find: T1, T2, W and Q Solution: (i) Initial and final temperature (T1 & T2) P1V1 = mRT1

11

p1V 150 2.5T

mR 2 0.287

T1=653.3K

This is a isothermal T1 = T2 = 653.3K (ii) Work done (W):

21

1

VW mRT in

V

0.832 0.287 653.3 in

2.5

W = -413.5kJ Note: Here, negative sign indicates that the work is done on the system. (iii) Heat transfer (Q):

From First law, Q = W + U

For isothermal process, U = 0 Q = W = -413.5kJ Note: Here negative sign indicates that the heat is rejected by the system. Result:

i. Initial and Final temperature, T1 = T2 = 653.3K ii. Work done (W) = -413.5kJ iii. Heat transfer (Q) = -413.5kJ

50. 10kg of gas at 10 bar and 4000C expands reversibly and adiabatically to 1 bar. Find the work done and change in internal energy.

Given data: M = 10kg P1 = 10bar = 1000kN/m2 T1 = 4000C = 400 + 273 = 673K P2 = 1 bar = 100 kN/m2 To find

W and U Solution By adiabatic relation,

1

2 2

1 1

14 1

14

2

T p

T p

100T 673 348.58K

1000

Work done,

1 2mR T TW

1

10 0.287 673 348.582327.74kJ

1.4 1

Change in internal energy

U = m x Cv x (T2 – T1) = 10 x 0.718 x (348.58 – 673) = -2327.74kJ

Alternately, from first law, Q = W + U In adiabatic process, Q = 0

W = -U = -2327.74kJ Result:

1. Work done, W = 2327.74 kJ

2. Change in internal energy, U = -23 27.74kJ 51. 1.5kg of certain gas at a pressure of 8 bar and 200C occupies the volume of 0.5 m3. It expands adiabatically to a pressure of 0.9 bar and volume 0.73 m3. Determine the work done during the process, gas constant, ratio of the specific heats, values of two specific heats, change in internal energy and change in enthalpy.

Given data: M = 1.5kg P1 = 8 bar = 800kN/m3 V=1 = 0.15m3 T1=200C = 20+273 = 293K P2=0.9bar = 90kN/m2 V2 = 0.73m3

To find: W,R, Cp / Cv, Cp, Cv, U and H Solution: From characteristics gas equation P1V1 = mRT1

1

1

p1VR

mT

800 0.15R

1.5 293

R = 0.273kJ / kgK Ratio or two specific heat.

p

v

10

110

2

C

C

p2log

p1

Vlog

V

10

10

90log

800

0.15log

0.73

p

v

p v

p v

v v

v

v

p

1.38.

C1.38

C

C 1.38C

R C C

1.38C C

0.273 C 1.38 1

0.273C 0.718kJkgK

0.38

C 1.38 0.78 0.99kJ/kgK

By using Relation between T & V for isentropic process

1 1

2 1 12 1

1 2 2

1.38 1

2

T V VT T

T V V

0.15T 293 160.59K

0.73

Change in internal energy (U)

U=mCv(T2 – T1) =1.5 x 0.718 x (160.59 – 293)

H = -196.6kJ Work done (W):

1 2p1V p2V 800 0.15 90 0.73W

1 1.38 1

W .89kJ

Alternately, from first law of thermodynamics, Q = W + U For isentropic process, Q = 0

W = -U = 142.5kJ Results:

i. Work done (W) = 142.89kJ

ii. Gas constant = 0.273kJ/kg K.

iii. Ratio of two specific heat p

v

C1.38

C

iv. Values of two specific heat, Cp=0.99kJ/kg Cv = 0.718kJ/kg K

v. Change in internal energy (U) = -142.5kJ

vi. Change in enthalpy (H) = -196.6kJ 52. A gas of mass 0.35g pressure 1535 kN/m3 as temperature of 3350C is expanded adiabatically to pressure of 126 kN/m2. The gas is than heated constant volume until it reaches 3350C, when its pressure is found to be 275kN/m2. Finally the gas is compressed isothermally until the original pressure 1535 kN/m2 obtained. Draw the p-V diagram and find out the following (i) the value of adabatic index (ii) change in internal energy during adabatic process and (iii) heat transfer during constant volume process Take Cp = 1.005kJ/kg k. Given data: M = 0.35kg P1 = 1535 kN/m2 Process 1-2: Adiabatic T1 = 3350C = 608K P2 = 126kN/m2 Process 2 – 3: Constant volume T2 = T1 = 608K

P3=275kN/m2 Process 3-1: Isothermal To find:

0, U and Q Solution: Consider process 2-3: Constant volume process

2 3

2 3

2

2 33

p p

T T

p 126T T 608

p 275

T2 = 278.57K Consider process 1-2: Adiabatic process

13

1

2

2

T p

T p

Taking log on both sides

110 10

2

10 10

T 1 p1log log

T p2

608 1 1535log log

278.57 126

10.339 1.0857

pV=C

pV=C

p1

2

V1 V2

Volume (V)

Pre

ss

ure

(p

)

0.339 10.3122

1.0857

0.3122 1

0.6877 1

11.454

0.6877

1.454

Change in internal energy (U) during adiabatic process

U=mCv(T2 – T1)

p

v

v

C

C

C kJ/kgK

U = 0.35 x 0.691 x (278.57 – 608)

U = -79.69kJ Heat transfer during Constant volume process

U=mCv(T3 – T2) = 0.35 x 0.691 x (608 – 278.57)

U=79.69KJ

From first law of thermodynamics, Q = W + U

Q = U [W=0] =79.69kJ Result:

i. Value of adiabatic index, = 1.454

ii. Change in internal energy during adiabatic process (U) =-79.69kJ iii. Heat transfer during constant volume process (Q) = 79.69kJ

53. A cylinder contains 1 m3 of gas at 100 kPa and 1000C, the gas is polytropically compressed to a volume of 0.25 m3, the final pressure is 600kPa. Determine (a) mass of the gas (b) the value of index ‘n’ for compression (c) change in internal energy of the gas (d)

heat transferred by the gas during compression. Assume R=0.287 kJ/kg and 1.4. Given data: V1 = 1m3 P1 = 100kPa = 100KN/m2 T1 = 1000C = 100+273=373K V2 = 0.25m3 P2 = 600kPa = 600kN/m2 R = 0.287kJ/kgK

= 1.4 To find:

M,n U and Q Solution: By general gas equation

P1V1 = mRT1 ====>

1

1

p1Vm

RT

100 1m 0.93kg

0.287 373

The value of index ‘n’ is given by

10 10

11010

2

p1 600log log

p2 1001.29

1V loglog0.25V

From polytrophic equation,

n 1

n2

1

1

1.29

2

T p2

T p2

600T 373 558K

100

Change in internal energy.

U = m x Cv x (T2 – T1) = 0.93 x 0.718 x (558 – 373) =123.53kJ

1 2mR T TW

n 1

0.93 0.287 373-558Work done, =

1.29 1

170.19 1

nQ W

1

1.4 - 1.9Heat transfer, = -170.1 46.802kJ

1.4 - 1

Result: Mass of gas, m = 0.93kg Value of index, n = 1.29

Change in internal energy, U = 123.53kJ Heat transfer, Q = -46.802kJ

54. An ideal gas of molecular weight 30 and specific heat ratio 1.4 is compressed according to the law pV1.25 = C from 1 bar absolute and 270C to a pressure of 16 bar (abs). Calculate the temperature at the end of compression, the heat received or rejected, work done on the gas during the process and change in enthalpy. Assume mass of the gas as 1 kg. Given data: Molecular weight, M = 30

p

v

C1.4

C

m 1kg

p1 1bar 100kN/m2

p2 16bar 1600kN/m2

0

1

1.25

T 27 C 27 273 300K

PV C

To find: T2, Q, and W Solution: For polytropic process, the p,V and T relation

n 1 n 1

n n2

2 1

1

1.25 1

1.25

2

T p2 p2T T

T p1 p1

600T 300 522.33K

100

Work done:

1 2

u

mR T TW

n 1

R Universal as constantGas constant, R=

M molecular weight

0.277kJ/kgK

1 0.277 300 522.33W

1.25 1

246.34kJ

(-ve sign indicates that the work is done on the system) Heat transfer,

nQ W

1

1.4 1.25246.34 92.3782kJ

1.4 1

(-ve sign indicates that the heat is rejected from the system)

Change in enthalpy:

p 2 1H m C T T

1 1.005 522.38 300

233.49kJ

Result: Temperature at the end of the compression, T2=522.38K Work done on the system, W = -246.34kJ Heat transferred from the system, Q = -92.378kJ

Change in enthalpy, H = 223.49kJ 55. A perfect gas for which ratio of specific heats is 1.4 occupies a volume of 0,3 m3 at 100 kPa and 270C. The gas undergoes compression of 0.06m3. Find the heat transfer during the compression for the following methods (a) pV = constant (b) Isentropic and (c) pV1.1 = C. Given data: P1=1000kPa=100kN/m2 V1=0.3m3 T1=270C+273=300K

= 1.4 V2=0.06m3 Process (i) Constant pressure

(i) pV = C (ii) pV1=C and (iii) pV1.1 = C

To find: Heat transfer (Q) Solution:

a. pV = C Heat transfer,

1 2pIV p2V nQ

n 1 1

100 0.3 587.3 0.06 1.4 1.1

1.1 1 1.4 1

39.285kJ

Result: Heat transfer for a. pV = C process, Q=-48.28kJ b. pV1 = C process, Q=0 c. pV1.1 = C process, Q=-39.285kJ

56. In steady flow process, 125 kJ of work is done by each kg of working fluid. The specific volume, velocity and pressure of the working fluid at inlet are 1.41 m3/kg,15.5 m/s and 6 bar respectively. The inlet is 31m above the ground, and the exhaust pipe is at ground level. The discharge conditions of the working fluid are 0.64 m3/kg, 1 bar and 263 m/s. The total heat loss between inlet and discharge is 8.7 kJ/kg of fluid. In flowing through this apparatus, does the specific internal energy increases or decreases and by how much? Given data: W = 125kJ/kg V1=0.41m3/kg C1=15.5m/s P1=6bar=600kN/m2 Z1=31m Z2 = 0 V2 = 0.64m3/kg C2 = 263 m/s P2 = 1 bar = 1000kN/m2 Q = 8.7 kJ/kg To find: Whether the internal energy increases or decrease and how much? Solution: The steady flow energy equation is

2 2

1 2

1 1 1 1 2 2 2 2

2 2

1 2

1 2 2 1 2 2 1 1

2 2

3 3

1 2

1 2

1 2

2 1

2 2

2

263 15.59.81 0 31 10 10 100 0.64 600 0.4

2

125 8.7

0.304 34.46 182 125 8.7

14.14 /

14.14 /

C Cgz u Pv Q gz u Pv W

C Cu u g z z Pv Pv W Q

u u x

U U

U U kJ kg

U U kJ kg

Result: Internal energy increases by 14.14kJ/kg

57. 50 kg/min of air enters the control volume in a steady flow statement at 2 bar and 100 and at an elevation of 100m above the datum. The same mass leaves the control volume at

150m Elevation with a pressure of 10 bar temperature of 300C. the entrance velocity as 2400 m/min. during the process 50,000 kJ/hr. calculate the power developed.

Given data:

m = 50 kg/ min = 50

60=0.83kg/sec

p1 = 2 bar = 200kN/m2

T1=100C = 373K Z1= 100m Z2 = 150m P2 = 10 bar = 1000 kN/m2

r2 = 300C = 573k

C1m = 2400m / min = 2400

60= 40m/sec

C2m = 1200m / min = 1200

60s=20m/sec

Q = 50,000kJ / hr = 50,000

3600= 13.89kJ / sec

To Find: Power developed, P = ? Solution: SFEE is

2 2

1 21 1 1 1 2 2 2 2

2 2

1 22 1 1 2

2 23 3

2 2

2

40 209.81 100 150 8 10 13.89 10

2

C Cgz u p v Q gz u p v W

C CW g z z h h Q

W

W=5999.5J/kg = 6kJ/kg K Power developed, P = W x mass = 6.0 x 0.83 P = 4.98kJ / sc-4.98kW Result: Power developed P = 4.98kW

58. A boiler produces steam from water at 35C. enthalpy of steam is 2675kJ/kg. Calculate the head transferred per kg. Neglect the potential and kinetic energies. Give data:

Tw=35C = 35+273 = 308k H2 = 2675 kJ/kg Cpw = 4.19kJ/kg K.

To find: Heat transferred, Q Solution: Enthalpy of feed water

1

1

4.19 308

1290.52 /

pu wh C T

h kJ kg

SFEE for boiler heat transferred (Q) Q = h2-h1=2675 – 1290.52 Q = 1384.48 kJ/kg Result: Heat transferred per kg, Q = 1384.48 kJ/kg 59. In a steady flow of air through a nozzle, the enthalpy decreases by 50 kJ between two sections. Assuming that there are no other energy changes than the kinetic energy determine the increases in velocity at section2, if the initial velocity is 90m/s. Given data: Enthalpy decrease, (h2-h2) = 50kJ = 50x103J Velocity at section (1), C1 = 90m/s To find: Increase in velocity, (C2 – C1) = ? Solution: Steady flow energy equation for a nozzle is

2 2

2 11 2

2

C Ch h

2

2 1 2 1

3 2

2

at exit, C 2

= 2 50 10 90

C =328.78m/s

velocity h h C

Increase in velocity C2 – C1 = 238.78m/s Result: Heat transferred per kg. Q=1384.48 kJ/kg. 60. At the inlet of the nozzle, the enthalpy and velocity of the fluid are 3000 kJ/kg and 50m/s respectively. There is negligible heat loss from the nozzle. At the outlet of the nozzle enthalpy is 2450 kJ/kg. if the nozzle is horizontal, find the velocity of the fluid at exit.

Given data: h1=3000 kJ/kg C1= 50m/s Q = 0 z1= z2 h2 = 2450kJ/kg C2-C1 = 328.78-90 = 238.78m/s To find Velocity of the fluid at exit C2 = ? Solution: The SFEE for nozzle

2 2

1 21 1

2

2 1 2 1

2 2

2

2 2

C 2

= 2 3000 2450 10 50

C 1050 /

C Ch h

h h C

m s

Result: Velocity of fluid at exit, C2=1050 m/s. 61. In a thermal power station, the steam flows, steadily through a 0.3 m diameter pipeline from boiler to turbine. A boiler exhausts the steam at a pressure of 4.2 bar the temperature

of 420C and the enthalpy of 3216.3 kJ/kg. The specific volume of the steam boiler outlet is 0.076m3 /kg. After flowing through the turbine, the steam conditions are measured as the

pressure of 3.1 bar, temperature of 379C, the enthalpy of 3201.7 kJ/kg and the specific volume of 0.086 m3/kg. There is a heat loss of 8.3 kJ/kg from the pipeline between boiler and turbine. Calculate the rate of flow of steam. Give data: d1= 0.3m p1= 4.2 bar = 420kN/m2

T1=420C h1= 3216.3kJ/kg v1 = 0.076m3/kg p2 = 3.1

T2 = 379C h2 = 3201 v2 = 0.086m3/kg Q =-8.3kJ/kg (heat loss) To find:

Rate of flow of steam, m=? Solution: By mass flow equations

1 1 2 2

1 2

AC A C

v v

Since, pipe diameter 0.3m is constant (A1=A2)

1 2

1 2

1 12 2

1

2 1

0.0860.076

C 1.13

C C

v v

C CC v

v

C

The steady flow energy equation for the above system is

2 2

1 22 2 1 2

2 2

2 11 2

2 2

2 1

2 2 3

1 1

2 3

1

1

: 02 2

2

= 3216.3-3201.7 8.3

=6.3kJ/kg

C 2 6.3 12.6 /

1.13 12.6 10 /

45.5 10

213 /

C Ch Q h z z w

C Ch h Q

C kJ kg

C C j kg

C

C m s

Mass rate of flow statement

2

1 1

1

0.3 2134 198

0.076

ACm kg

v

Result: Mass rate of flow of steam, m=198kg/sec 62. A turbine operates under steady flow conditions receiving steam at the following state:

pressure = 1.2 mpa, temperature = 188C. The steam leaves the turbine at the following state. Pressure = 20 kpa, Enthalpy = 2512 kJ/s. if the rate of steam flow through the turbine

is 0.42 kg/s, what is the power out put of the turbine in kW? Given data: p1=1.2MPa

T1= 188C h1= 2785kJ/kg. C1= 33.3m/s z1=3 p2= 20 kpa h2 = 2512kJ/kg C2=100m/s z2 = 0m Q=-0.29kJ/s. m=0.42kg/s. To find: Power output W=? Solution: SFEE

2 2

1 21 1 2 2

2 2

2 2

33.3 9.81 3 1000.42 2785 0.29 0.42 2512 0

2000 1000 2000

C Cm h z g Q m h z g W

W

Result: The power out put of the turbine, W = 112.52W. 63. A steam turbine operates under steady flow conditions. It receives steam 7200 kg/hr from the boiler. The steam enters the turbine at enthalpy of 2800 kJ/kg, a velocity of 400 m/min and an elevation of 4m. The steam leaves the turbine at enthalpy of 2000 kJ, a velocity of 8000 m/min and an elevation of 1m. Due to radiation the amount of heat losses from the power output of the turbine.

Given data:

2

1

1 2

2

2

7200/ / sec

3600

2800 /

400400 / min 6.67 /

60

4 & 1

2000 /

8000 / min 133.3 /

1580 / 0.438 / sec

m kg hr kg

h kJ kg

C m m s

z m z m

h kJ kg

C m m s

Q kJ hr kJ

To find: Power output of the turbine, p=? Solution: The SFEE for the above system is

2 2

1 21 1 1 1 2 2 2 2

2 2

1 21 2 1 2

2 2

3 3

2 2

2

6.67 133.3# 9.81 4 1 2800 2000 10 0.438 10

2

C Cg u p v Q gz u p v W

C CW g z z h h Q

W = 791605.23 J/kg = 791.6kJ/kg Power output from the turbine, P=W x mass = 791.6 x 2 P=1583.2 kW Result: Power output from the turbine, P=1583.2 kW

64. Air flows steadily at the rate of 0.5 kg/s through an air compressor. Entering at 7 m/s velocity, 100 kPa pressure and 0.95 m3/kg, volume and leaving at 5m/s, 700kPa and 0.19 m3/kg. The internal energy of the air leaving is 90kJ/kg greater than that of the air entering. Cooling water in the compressor jackets absorbs heat from the air at the rate of 58kW. (a) Compute the rate of shaft work input is the air in kW. (b) Find the ration of the inlet pipe diameter to the outlet pipe diameter. Given data:

1

1

3

1

2

2

3

2

2 1

m 0.5kg/ s.

C 7m/ s.

P = 100kpa

v = 0.95 m /kg

C = 5m /s

P = 700kPa

v 0.19m /kg

U U 90kJ/kg

Q 58kN.

To find:

1. Work input, W =?

2. 1

1

D?

D

Solution: SFEE,

2 2

1 21 1 1 1 2 2 2 2

2 2

1 21 1 1 1 2 2 2 2

1 2

2 2

C Cu p v z g Q m u p v z g w

2 2

C Cp v z g Q m (u u )p v z g w

2 2

Assumez z

7 50.5 (100 0.95) 58 0.5 90+(700 0.19)+ W

2000 2000

workinput, W=-5.995K W

( -ve sign indicates that the work is done on the system) From continuity equation,

1 1 2 2

1 2

1 2 1

2 1 2

2

4 1

2

4 2

2

1

2

1

2

A C A C

v v

A C v 53.57

A C v 7 0.19

D3.57

D

D3.57

D

D1.89

D

Result: 1. Work input, W = -5.994k W

2. The ratio of the inlet to outlet pipe diameter, 1

2

D1.89

D

65. Air is compressed from 100 kpa and 220C to a pressure of 1 Mpa while being cooled at the rate of 16 KJ/kg by circulating water through the compressor casing. The volume flow rate of air inlet condition is 150m3/min and power input to compressor is 500 kW Determine (a) mass flow rate (b) temperature of air exit. Neglect datum head . Given data:

1

o

1

2

3

1

P 100kPa

T 22 C 22 273 295K

p IMPa

Q 16kJ/kg.

C 5m /min

W =500kw.

To find:

1. Mass flow rate, m=? 2. Temperature of air exit, T2=?

Solution:

2 2

1 21 1 2 2

C Cm h z g Q m h z g W.............(1)

2 2

From ideal gas equation, P1V1= mRT1

1 2

2 2

1 2

1 2

100 150m

0.287 295

177.17kg/min

2.953kg/ sec

Neglect datum head, Z Z 0

C CAssume change in velocity head, 0

2

The equation (1) reduces to

m(h ) Q m(h ) W

1 2

2 2

1 2

1 2

p

p 1 p 2

100 150

0.287 295

177.17kg/min

2.953kg/ sec

Neglect datum head, z z 0

C CAssume change in velocity head, 0

2

Theequation (1) reduces to

m(h ) Q m(h ) W

AssumeC 1.005kJ/kg K for air.

M(c T Q) mC T W

2

2

2

.953 (1.005 295-16)=(2.953 1.005T ) 500

T 110.6K

Result:

1. Mass flow rate, m=2.953kg/sec. 2. Temperature of air exit, T2=110.6K.

66. During the working stroke of an engine the heat transferred out of the system was 150kj/kg of the working substance, determine the work done, when the internal energy is decreased by 400 kj/kg. Also state whether the work done on or by the engine. Given data: Q=-150 kJ/kg (Heat is transferred out)

U=-400kj/kg. To find: Work done, W=? Solution: By first law of thermodynamics

Q=W+U

W=Q-U =-150-(-400) =250kJ/kg Result: Work output, W=250kJ/kg (Positive work indicates that the work done is by the engine) 67. A fluid is confined in a cylinder by a spring-loaded frictionless piston so that the pressure in the fluid is a linear function of the volume (p=a+bv). The internal energy of the fluid is

given by the following equation u = 34+3. 15pV, where us is kJ, p is in kPa and V in m3. If the

fluid change from an initial state of 170kPa. 0.03m3 to a final state of 400 kPa, 0.06m3, with no other work than that done on the piston, find the direction and magnitude of the work and heat transfer. Given data: P. a+bV U = 34+3.15pV P1=170kPa V1=0.03m3 P2=400kPa V2=0.06m3

To find: Work transfer, W =? Heat transfer =? Solution: Change in internal energy,

2 1 2 2 1 1

2 2 1 1

U U (34 3.15p V ) (34 3.15p V )

= 3.15(p V p V )

= 3.15(400 0.06-170 0.03)

= 59.535kJ

From pressure equation. P=a+bV P1= a+b V1.........(1.24) P2=a+bV2……….(1.25) At initial state, 170=a+b(0.03) =a+0.03b………….(1.26) At final state, 400 =a+0.06b………….(1.27) Equation (1.27)-(1.26) results, s30=0.03b B=7666.67 Substituting the value of b in (1.26) 170=a+0.03(7666.67) a=60.0 The pressure equation, p=-60+7666.67V

2 2

1 1

v v

v V

0.062

0.03

2 2

W p.dV ( 60 7666.67V)dV

V = -60+7666.67

2

7666.67W 60(0.06 0.03) (0.06 0.03 )

2

= 8.55kJ

By first law of Thermodynamics.

Q = W +U Q=8.55+59.535 =68.085kJ Result: 1. Work transfer, W=8.55kJ (+ve sign indicates that the work is done by the system.) 2. Heat transfer, Q = 68.085 J (+ve heat transfer indicates that the heat is transferred into the system.) 68. A Piston and cylinder machine contains a fluid system which passes through a complete cycle of four processes. During the cycle, the sum of all heat transfers is – 170kJ. The system completes 100cycles per minute. Complete the following table showing the method for each item, and compute the net rare of work out put in kW.

Process Q(kj/min) W(kj/min) E(kJ/min) a-b 0 2,170 - b-c 21,000 0 - c-d -2,100 - - d-a - - - Solution : Process a-b: By first law of thermodynamics

Q=W+UE

0=2170+E

E=-2170kJ/min Process b-c

Q=W+E

21,000=0+E

E=21,000kJ/min Process C-d -2,100=W-36,600 W=34, 500kJ/min Process d-a:

Q = -170k(given) No. of cycles/min = 100 =-17,000kJ/min

cycle Q=Qa-b+Qb-c+Qc-d+Qd-a

-17,000=0+21000-2100+Qd-a

Qd-a=-35900kJ/min. We know that, in a cyclic process

Q=W -17,000=Wa-b+Wb-c+Wc-d+Wd-a

=2170+0+34500+Wd-a

Wd-a = -53670kJ/min. Change in internal energy,

E=Q-W =-35900 –(-53670) =17770kJ/min.

Rate of work output,

W=-1700kJ/min 17000

60

=-283.3kW. Result:

Process Q(kJ/min) W(kJ/min) E(kJ/min)

a-b 0 2,170 -2170

b-c 21,000 0 21,000

c-d -2,100 34,500 -36,600

d-a -35,900 -53670 17770

69. Five kg of air is compressed poly tropically (n-1.3) from 1 bar and 270C to 3 bar. Find (i) work transfer (ii) heat transfer (iii) change in internal energy. Given data: P1=1bar=1100kN/m2

T1=270C=27+273=300K P2 = 3 bar = 300kN/m2 N=1.3 To find:

1. Work transfer, W=? 2. Heat transfer, Q=? 3. Change in internal energy, U=?

Solution: During poly tropic process,

n 1

n2

2 1

1

1.3 1

1.3

PT T

P

300 =300

100

=386.57K

1 2mR T TWork transfer, W=

n 1

1 .287 300-386.57 =

1.3 1

=-82.82kJ/kg

r-nheat transfer, Q= W

r-1

1.4-1.3 = 82.82

1.4-1

Q=20.705kJ/kg.

By first law of thermo dynamics, U=Q-W =-20.705-(82.82) =62.1155KJ/kg.

Result: 1. Work transfer, W=-82.82KJ/Kg 2. Heat transfer, Q=20.705kJ/Kg 3. Change internal energy, U=62.115kJ/kg

70. 5kg of air at 40C and 1 bar is heated in a reversible non-flow constant pressure until the volume is doubled. Find (a) change in volume (b) work done (c) change in internal energy and (d) change in enthalpy. Given data: M=5kg

T1=40C P1=1bar=100kN/m2 V2=2V1 P=Constant To find:

1. V2-V1=? 2. W=?

3. U=?

4. H=? Solution: From ideal gas equation. P1V1=mRT1

1

5 0.287 313

100V

=4.49m

The final volume V2=2V1 =2 x4.49 =8.98m3 Change in volume V2-V1=8.98-4.49 =4.49m3

2

1

r

2 1

r

transfer, W= ( )

100(8.98 4.49)

449 .

Work pdV p V V

kJ

Change in internal energy

U=mCv(T2-T1) For constant Pressure process

2 2

1 1

22 1

1

1

1

2313

626

5 0.714(626 313)

1117.41

V T

V T

VT T

V

V

V

K

U

kJ

Change in enthalpy

H=mCp (T2-T1) =5 x 1.005(626-313) =1572.825kJ. Result:

1. Change in Volume, V2-V1=4.49m3 2. Work done, W=449kJ

3. Change in internal energy,U=1117.41kJ

4. Change in enthalpy,H=1572.83KJ

71. A gas whose original pressure volume the temperature were 140kNm2,0.13 and 25 C respective. It is compressed such that new pressure is 700kN/m2 and its new temperature is

60C. Determine the new volume of the gas. Given data: P1=140kN/m2 V1=0.1m3

T1=25C=25+273=298K P2=700kN/m2

T2=60C =60+273=333K To find: V2=? Solution: From ideal gas equation,

1 1 2 2

1 2

2

3

2

700140 0.1

298 333

0.0223

pV p V

RT RT

V

V m

Result: New volume, 3

2 0.0223V m

72. 0.25kg of air at a pressure of 1 bar occupies a volume o 0.3m3. If this air expands isothermally to a volume o 0.9m3. Find (i) the initial temperature , (ii) The final temperature (iii) External work done, (iv) Heat absorbed by the air, (v) change in internal energy Assume R=0.29kJ/K. Given data: M=0.25kg P1=1 bar =100kN/m2 V1=0.3m3 V2=0.9m3 To find:

1. T1=? 2. T2=? 3. W=? 4. Q=?

5. U=? Solution: From ideal gas equation, P1V1=mRT1

1

100 0.3

0.25 0.287

=418.12K.

T

For isothermal process T1=T2=418.12.12K

11 1

2

V1, 1 1 In

2 V

0.9100 0.3

0.3

32.96 .

pWorkdon W mp VIn orp v

p

In

k J

Heat absorbed, Q=W =32.96kJ

Change in internal energy, U=0. Result:

1. Work done, w=32.96kJ 2. Heat absorbed, Q=32.96kJ

3. Change in internal energy, U=0. 73. In a steady flow system a working substance at a rate of 4kg/s enter a pressure of 620 kN/m2 at a velocity of 300m/s. The internal energy is 2100 kJ/kg and specific volume 0.37m3/kg. It leaves the system at pressure of 130kN/m2, a velocity of 150m/s, Inter energy of 1500 kJ/kg and specific volume of 1.2m3/kg. During its passage in the system, the substance has a heat transfer of loss of 30kJ/kg to its surroundings. Determine the power of the system. State that it is from (or) to the system. Given data: M=4kg/s P1=620kN/m2 C1=300m/s U1=2100kJ/kg

V1=0.37m3/kg P2=130kn/m2 C2=150m/s U2=1500kJ/Kg V2=1.2m3/kg Q=-30kJ/kg To find: Power of the system, W=? Solution; SFEE

2 2

1 21 1 1 1 2 2 2 2

1 2

2 2

1 21 1 1 1 1 2 2 2

2 2

2 2

2 2

300 1504 2100 (620 0.37) 30 4 1500 (130 1.2)

2000 2000

2708.6

C Cm u p v Z g Q m u p v Z g W

AssumeZ Z

C Cm u p v Z g Q m u p v Z g W

W

W KW

(+Ve sign indicates that the work is one by the system.) Result: Work output from the system, W=2708.6kW

74. A mass of air is initially at 260C and 700kPa and occupies 0.028m3. the air is expanded at constant pressure to 0.084m3. A polytrophic process with n=1 is then carried out, followed by a constant temperature process. All the process are reversible.

1. Sketch the cycle in the p-V and T-s planes 2. Find the heat received and heat rejected in the cycle. 3. Find the efficiency of the cycle

Given data;

T1=260C=273+260=533K P1=700kPa=p2 V1=0.028m2 V1=0.084m2 Process 2-3 is polytrophic Process 3-4 is constant temperature. To find:

1. Sketch p-V and t-s diagram 2. Heat received and heat rejected, Q=?

3. Efficiency of the cycle, =?

Solution; Process 1-2: Constant pressure process.

1 1

1 2

22 1

1

0.084533

0.028

1599

v T

T T

VT T

V

Mass of air,

1-2 2 1

1-2 2 1

700 0.028 =

0.287 533

=0.128kg

Work done, W ( )

=700(0.084-0.028)

=39.2kJ

Heat transfer, Q ( )

p

pVm

RT

p V V

mC T T

p

=0.128 1.005 (1599-533)

C air =1.005kJ/kgK

=137.13kJ

of

Process 2-3: Polytropic process

1

3 3

2 2

1.5 1

1.53

3

33

3

3

1 1 2 22 3

533

1599 700

25.93

0.128 0.287 533

25.93

0.755

,1

700 0.084-25.93 0.755 =

1.5 1

n

nT p

T p

P

P kPa

FrompV mRT

mRTV

P

m

pV p VPolytropicwork W

n

23 2 3

=78.446kJ

Heat transfer. Q1

=19.612kJ

nW

Process 3-1

Constant temperature process, T1=T3=260C=533K

Work transfer, W3-1=P3V3 In 1

3 1

P

P

25.9325.93 0.755

700

64.52 ( )

64.52 [ U=0]

In

kJ Workinput

Q kJ

Heat received in this cycle, Qs=137.13kJ (consider only +ve heat) Heat reject in this cycle QR=19.612+64.52(consider only –ve heat)=84.132kJ

1-2 2 3 3 1

work done of the cycle, =

sup

W =

39.2+78.446-64.52 =

137.13

s

EfficiencyHeat plied

W W

Q

R

s

=38.74%

Or

QEfficiency of the cycle, =

Q

84.132 =1.

137.13

=38.74%

Result: 1.p-V and T-s planes and drawn 2. Heat received by the cycle, Qs=137.13kJ 3.Heat rejected by the cycle, QR=84.132kJ

4. Efficiency of the cycle,=38.74%

75. Air at a temperature of 15C passes through a heat exchanger at a velocity of 30m/s and

expands until the temperature falls to 650C. On leaving the turbine, the air is taken at a

velocity of 60m/s to a nozzle where it expands until the temperature has fallen to 500C. If the air flow rate is 2kg/s, calculate: (i) The rate of heat transfer to the air in the heat exchanger, (ii) The power output from the turbine assuming no heat loss, and (iii) The velocity at exit from the nozzle, assuming no heat loss. Take the enthalpy of air as h=Cpt, where Cp is the specific heat equals to 1.005kJ/kg K and ‘t’ the temperature. Given data:

T1=15C=273+15=288K C1=30m/s

T2=800C=273+800=1073K C2=30m/s

T3=650C=273+650=923K C1=30m/s

T4=500C=273+500=773K m=2kg/s To find:

(i) The rate of heat transfer in the heat exchanger, Q1-2=? (ii) The velocity at exit from the nozzle C4=? (iii) Power out put from the turbine, W2-3=?

Solution: SFEE between 1-2,

2 2

1 21 1 2 2 1 2.............(1 28)

2 2

C Cm h z g Q m h z g W

Assuming that, Work transfer, W1-2=0 and Z1=Z2

So, the above equation reduces to

2 2

1 21 1 2 2

1 2

1 2 2 1

2 1

22

322 2 2 3 2 3 2 3

2 2

,

( )

( )

2 1.005 (1073 288)

1577.85

.....(1.29)2 2

p

C Cm h Q m h

Here C C

Q m h h

mC T T

W

CCm h z g Q m h z g W

Assuming that Q-3=0 and Z2=z3 Equation (1.29) reduces to

22

322 3 2 3

2 2

2 3

2-3

2 2

30 602 1.005 1073 2 1.005 923

2 1000 2 1000

W 298.8

CCm h m h W

x W

kW

Similarly, applying SFEE between 3-4

2 2

3 43 3 3 4 4 4 3 4.....(1.30)

2 2

C Cm h z g Q m h z g W

Assuming that Z3=Z4, Q3-4=0 (adiabatic nozzle) and W3-4=0

The Equation (1.30) reduces to

2 2

3 42 3

224

4

2 2

602 1.005 923 2 1.005 773

2 1000 2 1000

C 552.36 /

C Cm h m h

C

m s

Result:

1. The rate o heat transfer in the heat exchanger, Q1-2=1577.85kW 2. The power output from the turbine , W2-3=298.8kW 3. The velocity of nozzle exit, C4=552.36 m/s

76. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p=a+bv, when a and b are constants. The initial and final pressure are 1000kPa and

200kPa respectively and the corresponding volumes are 0.2m3 and 1.2m3. the specific internal energy of gas is given by the relation u=1.5pV-85kJ/kg. where p is in kPa and V is in m. Calculate the net heat transfer and the maximum internal energy of the gas attained during expansion

Given data: M=1.5 kg P=+bV P1=10000kPa P2=200kPa V1=0.2m3 V2=1.2m3 U=1.5 pV-85 To find:

1. Heat transfer, Q=?

2. Maximum internal energy,U=? Solution: P1=a+bV1 and p2=a+bV2 1000=a+0.2b………(1.31) 200=a+1.2b………(1.32) (1.32)-(1.31),b=800 From (1.31),a-1160

the relationship becomes P=1160-800V

2 1.2

1 0.2

1.22

0.2

2 2

transfer, W= (1160 800 )

800V = 1160V

2

=1160(1-2-0.2)-(1.2 0.2 )

=600kJ

Work pdV V dv

change in internal energy, U=U2-U1

=(1.5p2V2-85)-(1.5p1V1=85) =(1.5p2V2-1.5p1V1) =1.5(200 x 1.2-1000 x 0.2) =60KJ According to the first law of thermodynamics

Q=W+U =600+60 =660kJ Maximum internal energy, U2=mU2

1.5 200 1.21.5 85

1.5

232.5KJ

Result: Heat transfer, Q=660KJ

Maximum internal energy, U=232.5kJ 77. A room for four people has two fans, consuming 0.18kW power and three 100W

lamps Ventilation air at the rate of 80kg/hr enters with the enthalpy of 84kJ/kg and leaves with an enthalpy a 59KJ/Kg. If each person puts out heat at the rate 630kJ/hr. Determine the rate at which eat is to be removed by a room cooler so that steady state maintained in the room

Given data: Np=(persons)s, nf=2 Wf=0.81kW (each) W1=100 W(each)

Mass of air, m=80kg/hr=80

0.22 / sec3600

kg

Enthalpy of air entering, h1=84kJ/kg Enthalpy of air leaving, h2=59kJ/kg Qp=630kJ/hr (each person) To find: Rate of heat is to be removed =? Solution; Rate of energy increase = Rate of energy inflow-Rate of energy outflow

2 2

1 21 1 2 2 .....(1.33)

2 2

C CE m h z g Q m h z g W

Assuming that,

2 2

1 2

1 2

1 2

1 2

02

( ) 0

, equation (1.33) reduces to Q=Em (h )

6304

3600

0.7 / 0.7

80( ) (84 59)

3600

p p

C C

z z g

Now the h W

n Q

kJ s kW

m h h

0.55kJ/s=0.55kW W= electrical energy input =nfWf+n1W

1002 0.18 3

1000

0.66

0.70.556 0.66

1916

kW

Q

kW

Result; Rate of heat to be removed , Q=-1.916kW 78. Air flows steadily at the rate of 0.5kg/s through air compressor entering at 7m/s

velocity, 100kpa Pressure, and 0.953/kg specific volume, and leaving at 5m/s. 700kPa, and 0.19m3/kg. The internal energy of air leaving is 90kJ/Kg greater than that of the air entering. Cooling water in the compressor jackets absorb heat at the rate of 58kW. Calculate the rate of shaft work input to the compressor

Given data: M=0.5kg/s. C1=7m/s P1=100kPa V1=0.95m3/kg C2=5m/s P2=700kPa V2=0.19m3/kg U2-U1=90KJ/kg

Q=58kN. To find:

1. Work input, W=?

2. 1

2

?D

D

Solution: SFEE,

2 2

1 21 1 1 1 2 2 2 2

2 2

1 21 1 1 1 2 2 2 2

1 2

2 2

2 2

2 2

Z

7 5100 0.95 58 0.5 90+ 700 0.19)+

2000 2000

input W =-5.995kW

C Cu pv z g Q m u p v z g W

C Cpv z g Q m u u p v z g W

Assume Z

S W

work

(-ve sign indicates that the work is done on the system from continuity equation)

1 1 2 2

1 2

1 2 1

2 1 2

53.57

7 0.19

AC A C

V V

A C v

A CV

2

4 1

2

4 2

2

1

2

1

2

3.57

3.57

1.89

D

D

D

D

D

D

Result; Work input , w=-5.994 kw. 1. Work input, W =5.994kW

2. The ratio of the inict to outlet pipe diameter 1

2

D16

D