applied thermodynamics and fluids

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EI-6403/Applied Thermodynamics and Fl uid Dynamics Department of Mechanical Engineering 2014-2015

St.Joseph’s College of Engineering 1 I SO 9001 - 2008

UNIT I

LAWS OF THERMODYNAMICS AND BASIC IC ENGINE CYCLES

Part A

1. What is meant by thermodynamic system? How do you classify it?

Thermodynamic system is defined as the any space or matter or group of matter where the energy transfer orenergy conversions are studied.It may be classified into three types.(a) Open system

(b) Closed system

(c) Isolated system2. 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.

3. Define a open system, Give an example.

When a system has both mass and energy transfer it is called as open system.

Example: Air Compressor.4. Differentiate closed and open system.

Closed System Open System

1. There is no mass transfer. Only heat and

work will transfer.

1. Mass transfer will take place, in addition to

the heat and work transfer.

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

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

3. Air compressor, boiler

5. Define an isolated systemIsolated system is not affected by surroundings. There is no heat, work and mass transfer take place. In thissystem total energy remains constant.Example: Entire Universe

6. What is meant by thermodynamic property?Thermodynamic property is any characteristic of a substance which is used to identify the state of the system

and can be measured, when the system remains in an equilibrium state.

7.

How do you classify the property?

Thermodynamic property can be classified into two types.

1. Intensive or Intrinsic and2. Extensive and Extrinsic property.

8. Define Intensive and Extensive properties.The properties which are independent on the mass of the system is called intensive properties.

e.g., Pressure, Temperature, Specific Volume etc.,The properties which are dependent on the mass of the system is called extensive properties.e.g., Total energy, Total volume, weight etc.

9. Differentiate Intensive and Extensive properties.

Intensive Properties Extensive Properties

1. Independent on the mass of the system Dependent on the mass of the system.

2. If we consider part of the system these

properties remain same.

e.g. pressure, Temperature specific volumeetc.,

If we consider part of the system it will have a

lesser value.

e.g., Total energy, Total volume, weight etc.,

3. Extensive property/mass is known asintensive property

--

10. What do you understand by equilibrium of a system?When a system remains in equilibrium state, it should not undergo any charges to its own accord.

11. What is meant by thermodynamic equilibrium?

When a system is in thermodynamic equilibrium, it should satisfy the following three conditions.

(a) Mechanical Equilibrium :- Pressure remains constant(b) Thermal equilibrium :- Temperature remains constant(c) Chemical equilibrium : There is no chemical reaction.

12. State the First law of thermodynamics (MU – Apr‟95) First of thermodynamics states that when system undergoes a cyclic process the net heat transfer is equal to

work transfer.

13.

What is meant by open and closed cycle.

In a closed cycle, the same working substance will re circulate again and again.In a open cycle, the same working substance will be exhausted to the surroundings after expansion.





14. What is meant by reversible and irreversible process.

A process is said to be reversible, it should trace the same path in the reverse direction when the process isreversed. It is possible only when the system passes through a continuous series of equilibrium state.If a system does not pass through continuous equilibrium state, then the process is said to be irreversible.

15. What is Quasi – Static process?

The process is said to be quasi – static, it should proceed infinitesimally slow and follows continuous series ofequilibrium states. Therefore, the quasi static, it should proceed infinitesimally slow and follows continuousseries of equilibrium states. Therefore, the quasi static process may be an reversible process.

16. Explain Zeroth Law of thermodynamics?

Zeroth law of thermodynamics states that when two systems are separately in thermal equilibrium with a thirdsystem, then they themselves are in thermal equilibrium with each other.

17. Define the term enthalpy?

The Combination of internal energy and flow energy is known as enthalpy of the system. It may also bedefined as the total heat of the substance.Mathematically, enthalpy (H) = U + pv KJ)

Where, U – internal energy p – pressure

v – volumeIn terms of C p & T → H = mC p (T2-T1)KJ

18. Define the term internal energyInternal 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.

19. 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 surroundings.

20. Define an isentropic process.

Isentropic process is also called as reversible adiabatic process. It is a process which follows the law of pVy = C

is known as isentropic process. During this process entropy remains constant and no heat enters or leaves thegas.

21. What is the difference between steady flow and non – flow process?During the steady flow process the rate of flow of mass and energy across the boundary remains constant. In

case of non – flow across the system and boundary.

22.

State the Kelvin – Plank statement of second law of thermodynamicsKelvin – Plank states that it is impossible to construct a heat engine working on cyclic process, whose only

purpose is to convert all the heat energy given to it into an equal amount of work.

23. State Clausius statement of second law of thermodynamics.

It states that heat can flow from hot body to cold without any external aid but heat cannot flow from cold bodyto hot body without any external aid.

24. State Carnot‟s theorem. No heat engine operating in a cyclic process between two fixed temperature, can be more efficient than areversible engine operating between the same temperature limits.

25. Write the expression for efficiency of the otto cycle?

1Efficiency n =1 - ---------

(r)r-1

26. What is meant by cutoff ratio?

Cutoff ratio is defined as the ratio of volume after the heat addition to before the heat addition. It is denoted bythe letter ‘p’

27. What are the assumptions made for air standard cycle.

1. Air is the working substance.2. Throughout the cycle, air behaves as a perfect gas and obeys all the gas laws.3. No chemical reaction takes place in the cylinder4. Both expansion and compression are strictly isentropic5. The values of specific heats of the air remain constant throughout the cycle

28. What is meant by air standard efficiency of the cycle?

It is defined as the ratio of work done by the cycle to the heat supplied to the cycle.

Work doneEfficiency n = --------------

Heat supplied





29. Define Mean effective pressure of an I.C. engine.

Mean effective pressure is defined as the constant pressure acting on the piston during the working stroke. It isalso defined as the ratio of work done to the stroke volume or piston displacement volume.

30. The Brayton cycle is mainly used in _______

Gas turbine power plant.

31. Give the expression for efficiency of the Brayton cycle.1

Efficiency n = 1 - ---------- where R p – pressure ratio.

(R p)y-1

32. What is a Gas turbine? How do you classify.

Gas turbine is an axial flow rotary turbine in which working medium is gas.Classification of gas turbine.1. According to the cycle of operationa) open cycle b) closed cycle and c) semi – closed cycle.

2. According to the processa) constant volume and b) constant pressure process.

33. What is meant by closed cycle gas turbine?

In closed cycle gas turbine, the same working fluid is recirculated again and again.

34. What is meant by open cycle gas turbine?In open cycle gas turbine, the exhaust gas form turbine is exhausted to the atmosphere and fresh air is taken incompressor for every cycle

35. What is the function of intercooler in gas turbines? Where it is placed?The intercooler is placed between L.P. and H.P. compressors. It is used to cool the gas coming form L.P.

compressor to its original temperature..

36. What is the function of regenerator in gas turbine?

The main function of heat regenerator is to exchange the heat from exhaust gas to the compressed air for preheating before combustion chamber. It increases fuel economy and increase thermal efficiency.

37. What is meant by single acting compressor?

In single acting compressor, the suction, compression and delivery of air take place on one side of the piston.

38. What is meant by double acting compressor?

In double acting reciprocating compressor, the suction compressin and delivery of air take place on bothside of the piston.

39. What is meant by single stage compressor?

In single stage compressor, the compression of air from the initial pressure to the final pressure is carriedout in one cylinder only.

40. Define clearance ratio

Clearance ratio is defined as the ratio of clearance volume to swept volume (or) stroke volume.

Vc Vc – clearance volumeC = -------- Vs – swept volume

Vs

41. What is compression ratio?

Compression ratio is defined as the ratio between total volume and clearance volume.

Total volumeCompression ratio = -------------------

Clearance Volume42. What are the factors that effect the volumetric efficiency of a reciprocating compressor?

1) Clearance volume 2) Compression ratio

43. What is meant by Point and Path function?

The quantities which is independent on the process or path followed by the system is known as point functions.Example: Pressure, volume, temperature, etc.,The quantities which are dependent on the process or path followed by the system is known as path functions.Example: Heat transfer, work transfer.

PART – B

1. When a system is taken from state l to state m, in Fig., along path lqm, 168 kJ of heat flows into the

system, and the system does 64 kJ of work :(i) How much will be the heat that flows into the system along path lnm if the work done is 21 kJ?





(ii) When the system is returned from m to l along the curved path, the work done on the system is 42 kJ.

Does the system absorb or liberate heat, and how much of the heat is absorbed or liberated?(iii) If Ul = 0 and Un = 84 kJ, find the heat absorbed in the processes ln and nm.

Ql – q – m = 168 kJ

Wl – q – m = 64 kJ

We have, Ql – q – m = (Um – Ul) + Wl – q – m

168 = (Um – U

l) + 64

Um – Ul= 104 kJ. (Ans.)

(i) Ql – n – m = (Um – Ul) + Wl – n – m

= 104 + 21

= 125 kJ. (Ans.)

(ii) Qm – l = (Ul – Um) + Wm – l

= – 104 + ( – 42)

= – 146 kJ. (Ans.)

The system liberates 146 kJ.

(iii) Wl – n – m = Wl – n + Wn – m

= Wl – m = 21 kJ [Wn – m = 0, since volume does not change.]

Ql – n = (Un – Ul) + Wl – n = (84 – 0) + 21

= 105 kJ. (Ans.)

Now Ql – m – n= 125 kJ = Ql – n + Qn – m

Qn – m = 125 – Ql – n = 125 – 105 = 20 kJ. (Ans.)

2. A fluid system, contained in a piston and cylinder machine, passes through a complete cycle of four

processes. The sum of all heat transferred during a cycle is – 340 kJ. The systems completes 200

cycles per min. Complete the following table showing the method for each item, and compute the

net rate of work output in kW.

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

1 — 2 0 4340 —

2 — 3 42000 0 —





3 — 4 – 4200 — – 73200

4 — 1 — — —

Sum of all heat transferred during the cycle = – 340 kJ. Number of cycles completed by the system = 200 cycles/min.Process 1 — 2 :

Q = Δ E + W0 = Δ E + 4340

∴ Δ E = – 4340 kJ/min.

Process 2 — 3 : Q = Δ E + W42000 = Δ E + 0

Δ E = 42000 kJ/min.Process 3 — 4 : Q = Δ E + W

– 4200 = – 73200 + W

∴W = 69000 kJ/min.Process 4 — 1 : ΣQcycle= – 340 kJ The system completes 200 cycles/min

Q1 – 2 = Q2 – 3 + Q3 – 4 + Q4 – 1

= – 340 × 200= – 68000 kJ/min

0 + 42000 + ( – 4200) + Q4 – 1 = – 68000

Q4 – 1 = – 105800 kJ/min. Now, ∫ dE= 0, since cyclic integral of any property is zero.

Δ E1 – 2 + ΔE2 – 3 + Δ E3 – 4 + Δ E4 – 1 = 0 – 4340 + 42000 + ( – 73200) + Δ E4 – 1 = 0

∴ Δ E4 – 1 = 35540 kJ/min.

∴W4 – 1 = Q4 – 1 – Δ E4 – 1

= – 105800 – 35540= – 141340 kJ/min

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

1 — 2 0 4340 – 4340

2 — 3 42000 0 42000

3 — 4 – 4200 69000 – 73200 4 — 1 – 105800 – 141340 35540

Since ΣQcycle= ΣWcycle

Rate of work output = – 68000 kJ/min

= 1133.33 kW. (Ans.)3. A fluid system undergoes a non-flow frictionless process following the pressure-volume relation as

where p is in bar and V is in m3. During the process the volume changes from 0.15 m

3

to 0.05 m3 and the system rejects 45 kJ of heat. Determine :

(i) Change in internal energy ;

(ii) Change in enthalpy.

Initial volume, V1 = 0.15 m3

Final volume, V2 = 0.05 m3

Heat rejected by the system, Q = – 45 kJ

Work done is given by,





= – 5.64 × 105 N-m = – 5.64 × 105 J [1 Nm = 1 J]

= – 564 kJ

(i) Applying the first law energy equation,

Q = Δ U + W

– 45 = Δ U + ( – 564)

∴ΔU = 519 kJ. (Ans.)

This shows that the internal energy is increased.

(ii) Change in enthalpy,

Δ H = Δ U + Δ ( pV)

= 519 × 103 + (p2V2 – p1V1)

= 34.83 × 105 N/m2

= 101.5 bar = 101.5 × 105 N/m2

∴ Δ H = 519 × 103 + (101.5 × 10

5 × 0.05 – 34.83 × 10

5 × 0.15)

= 519 × 103 + 103(507.5 – 522.45)

= 103(519 + 507.5 – 522.45) = 504 kJ

∴ Change in enthalpy = 504 kJ. (Ans.)

4. The following equation gives the internal energy of a certain substance u = 3.64 pv + 90 where u is

kJ/kg, p is in kPa and v is in m3/kg.A system composed of 3.5 kg of this substance expands from an

initial pressure of 500 kPa and a volume of 0.25 m3 to a final pressure 100 kPa in a process in which

pressure and volume are related by pv1.25 = constant. (i) If the expansion is quasi-static, find Q, ΔUand W for the process. (ii) In another process, the same system expands according to the same

pressure-volume relationship as in part (i), and from the same initial state to the same final state as

in part (i), but the heat transfer in this case is 32 kJ. Find the work transfer for this process.

(iii) Explain the difference in work transfer in parts (i) and (ii).





Internal energy equation : u = 3.64 pv+ 90


Initial pressure, p1 = 500 kPaFinal pressure, p2 = 100 kPaProcess: pv

1.25= constant.

(i) Now, u = 3.64 pv+ 90

Δ u = u2 – u1 = 3.64 (p2v2 – p1v1) ...per kg

∴ Δ U = 3.64 (p2V2 – p1V1) ...for 3.5 kg

Now, p1V11.25

= p2V21.25

= 0.906 m3

ΔU= 3.64 (100 × 103 × 0.906 – 500 × 103 × 0.25) J [1 Pa = 1 N/m2]

= 3.64 × 105 (0.906 – 5 × 0.25) J

= – 3.64 × 105 × 0.344 J = – 125.2 kJ

i.e., ΔU = – 125.2 kJ. (Ans.)

For a quasi-static process

= 137.6 kJ

Q = ΔU+ W

= – 125.2 + 137.6

= 12.4 kJ

i.e., Q = 12.4 kJ. (Ans.)

(ii) Here Q = 32 kJ

Since the end states are the same, ΔU would remain the same as in (i)

∴W = Q – ΔU= 32 – ( – 125.2) = 157.2 kJ. (Ans.)

(iii) The work in (ii) is not equal to ∫ p dV since the process is not quasi-static.

5. 0.2 m3 of air at 4 bar and 130°C is contained in a system. A reversible adiabatic expansion takesplace till the pressure falls to 1.02 bar. The gas is then heated at constant pressure till enthalpy

increases by 72.5 kJ. Calculate :

(i) The work done ;





(ii) The index of expansion, if the above processes are replaced by a single reversible polytropic

process giving the same work between the same initial and final states.

Take cp = 1 kJ/kg K, cv = 0.714 kJ/kg K.


Initial pressure, p1 = 4 bar = 4 × 105 N/m

2

Initial temperature, T1 = 130 + 273 = 403 K

Final pressure after adiabatic expansion,

p2 = 1.02 bar = 1.02 × 105 N/m2

Increase in enthalpy during constant pressure process = 72.5 kJ.

(i) Work done :

Process 1-2 : Reversible adiabatic process :

Also

= 0.53 m3

= 272.7 K

Mass of the gas,

where, R = (c p – cv) = (1 – 0.714) kJ/kg K

= 0.286 kJ/kg K

= 286 J/kg K or 286 Nm/kg K





Process 2-3. Constant pressure :

Q2 – 3 = m c p (T3 – T2)

72.5 = 0.694 × 1 × (T3 – 272.7)

T3 = 377 K

Also,

V3= 0.732 m3

Work done by the path 1-2-3 is given by

W1 – 2 – 3 = W1 – 2 + W2 – 3

Hence, total work done = 85454 Nm or J.

(ii) Index of expansion, n :

If the work done by the polytropic process is the same,

n = 1.062

Hence, value of index = 1.062. (Ans.)

6. A cylinder contains 0.45 m3 of a gas at 1 × 10

5 N/m

2 and 80°C. The gas is compressed to a volume of

0.13 m3, the final pressure being 5 × 10

5 N/m

2. Determine :

(i) The mass of gas ;

(ii) The value of index „n‟ for compression ;

(iii) The increase in internal energy of the gas ;

(iv) The heat received or rejected by the gas during compression.

Take γ = 1.4, R = 294.2 J/kg°C.

Initial volume of gas, V1 = 0.45 m

Initial pressure of gas, p1 = 1 × 105 N/m

2

Initial temperature, T1 = 80 + 273 = 353 K

Final volume after compression, V2 = 0.13 m3

The final pressure, p2 = 5 × 105 N/m2.

(i) To find mass ‘m’ using the relation

(ii) To find index ‘n’ using the relation





(3.46)n = 5

Taking log on both sides, we get

n loge 3.46 = loge 5

n = loge 5/loge 3.46 = 1.296. (Ans.)

(iii) In a polytropic process,

∴ T2 = 353 × 1.444 = 509.7 K

Now, increase in internal energy,

Δ U = mcv (T2 – T1)

= 49.9 kJ. (Ans.)

(iv) Q = Δ U + W

= – 67438 N-m or – 67438 J = – 67.44 kJ

∴ Q = 49.9 + ( – 67.44) = – 17.54 kJ

7. 0.1 m of an ideal gas at 300 K and 1 bar is compressed adiabatically to 8 bar. It is then cooled atconstant volume and further expanded isothermally so as to reach the condition from where it

started. Calculate :

(i) Pressure at the end of constant volume cooling.

(ii) Change in internal energy during constant volume process.

(iii) Net work done and heat transferred during the cycle. Assume

cp = 14.3 kJ/kg K and cv = 10.2 kJ/kg K.

Given: V1 = 0.1 m3 ; T1 = 300 K ; p1 = 1 bar ; c p = 14.3 kJ/kg K ; cv = 10.2 kJ/kg K.

(i) Pressure at the end of constant volume cooling, p3:

Characteristic gas constant,





R = c p – cv = 14.3 – 10.2 = 4.1 kJ/kg K

Considering process 1-2, we have :

= 544.5 K

Considering process 3 – 1, we have

p3V3 = p1V1

(ii) Change in internal energy during constant volume process, (U3 – U2) :

Mass of gas,

Change in internal energy during constant volume process 2 – 3,

U3 – U2 = mcv(T3 – T2)

= 0.00813 × 10.2 (300 – 544.5) (Since T3 = T1)

= – 20.27 kJ (Ans.)( – ve sign means decrease in internal energy)

● During constant volume cooling process, temperature and hence internal energy is reduced. This

decrease in internal energy equals to heat flow to surroundings since work done is zero.

(iii) Net work done and heat transferred during the cycle :

W2 – 3 = 0 ... since volume remains constant





= 14816 Nm (or J) or 14.82 kJ

∴ Net work done = W1 – 2 + W2 – 3 + W3 – 1

= ( – 20.27) + 0 + 14.82

= – 5.45 kJ

– ve sign indicates that work has been done on the system. (Ans.)

For a cyclic process :

∴ Heat transferred during the complete cycle = – 5.45 kJ

– ve sign means heat has been rejected i.e., lost from the system. (Ans.)

Flow of fluid = 10 kg/min

Properties of fluid at the inlet :

Pressure, p1 = 1.5 bar = 1.5 × 105 N/m

2

Density, ρ1 = 26 kg/m3

Velocity, C1 = 110 m/s

Internal energy, u1 = 910 kJ/kg

Properties of the fluid at the exit :

Pressure, p2 = 5.5 bar = 5.5 × 105 N/m2

Density, ρ2 = 5.5 kg/m3


Internal energy, u2 = 710 kJ/kg

Heat rejected by the fluid,

Q = 55 kJ/s

Rise is elevation of fluid = 55 m.

(i) The change in enthalpy,

Δh = Δu + Δ(pv)





= 1 × 105 – 0.0577 × 10

5

= 105 × 0.9423 Nm or J

= 94.23 kJ

Δu = u2 – u1

= (710 – 910)

= – 200 kJ/kg

Substituting the value in eqn. (i), we get

Δh = – 200 + 94.23

= – 105.77 kJ/kg. (Ans.)

(ii) The steady flow equation for unit mass flow can be written as

Q = Δ KE + Δ PE + Δ h + W

where Q is the heat transfer per kg of fluid

= 55 × 6 = 330 kJ/kg

= 12000 J or 12 kJ/kg

ΔPE = (Z2 – Z1) g = (55 – 0) × 9.81 Nm or J

= 539.5 J or ≈ 0.54 kJ/kg

Substituting the value in steady flow equation,

– 330 = 12 + 0.54 – 105.77 + W or W

= – 236.77 kJ/kg.

= – 39.46 kJ/s = – 39.46 kW. (Ans.)

9. At the inlet to a certain nozzle the enthalpy of fluid passing is 2800 kJ/kg, and the velocity is 50 m/s.

At the discharge end the enthalpy is 2600 kJ/kg. The nozzle is horizontal and there is negligible

heat loss from it.(i) Find the velocity at exit of the nozzle.(ii) If the inlet area is 900 cm2 and thespecific volume at inlet is 0.187 m

3/kg, find the mass flow rate.(iii) If the specific volume at the

nozzle exit is 0.498 m3/kg, find the exit area of nozzle.





Conditions of fluid at inlet (1) :

Enthalpy, h1 = 2800 kJ/kg


Area, A1 = 900 cm2 = 900 × 10 – 4 m2

Specific volume, v1 = 0.187 m3/kg

Conditions of fluid at exit (2) :

Enthalpy, h2 = 2600 kJ/kg

Specific volume, v2 = 0.498 m3/kJ

Area, A2 =?

Mass flow rate, =?

(i) Velocity at exit of the nozzle, C2 :

Applying energy equation at ‘1’ and ‘2’, we get

were Q = 0, W = 0, Z1 = Z2

= 201250 N-m

∴ C22 = 402500

∴ C2 = 634.4 m/s. (Ans.)

(ii) Mass flow rate :

By continuity equation,

∴ Mass flow rate = 24.06 kg/s. (Ans.)

iii) Area at the exit, A2 :





A2 = 0.018887 m2 = 188.87 cm2

Hence, area at the exit = 188.87 cm2. (Ans.)

10. Air at a temperature of 20°C passes through a heat exchanger at a velocity of 40 m/s where its

temperature is raised to 820°C. It then enters a turbine with same velocity of 40 m/s and expands

till the temperature falls to 620°C. On leaving the turbine, the air is taken at a velocity of 55 m/s to

a nozzle where it expands until the temperature has fallen to 510°C. If the air flow rate is 2.5 kg/s,calculate :

(i) Rate of heat transfer to the air in the heat exchanger ;

(ii) The power output from the turbine assuming no heat loss ;

(iii) The velocity at exit from the nozzle, assuming no heat loss.

Take the enthalpy of air as h = c pt, where cp is the specific heat equal to 1.005 kJ/kg°C and t the

temperature.

Temperature of air, t1 = 20°C

Velocity of air, C1 = 40 m/s.

Temperature of air after passing the heat exchanger, t2 = 820°C

Velocity of air at entry to the turbine, C2 = 40 m/s

Temperature of air after leaving the turbine, t3 = 620°C

Velocity of air at entry to nozzle, C3 = 55 m/s

Temperature of air after expansion through the nozzle, t 4 = 510°C

Air flow rate, = 2.5 kg/s.

(i) Heat exchanger :

Rate of heat transfer :

Energy equation is given as,

Here, Z1 = Z2, C1, C2 = 0, W1 – 2 = 0

∴ mh1 + Q1 – 2 = mh2

or Q1 – 2 = m(h2 – h1)

= mc p (t2 – t1)

= 2.5 × 1.005 (820 – 20) = 2010 kJ/s.





Hence, rate of heat transfer = 2010 kJ/s. (Ans.)

(ii) Turbine :

Power output of turbine :

Energy equation for turbine gives

[Since Q2 – 3 = 0, Z1 = Z2]

= 2.5 [201 + 0.7125] = 504.3 kJ/s or 504.3 kW

Hence, power output of turbine = 504.3 kW. (Ans.)

(iii) Nozzle:

Velocity at exit from the nozzle :

Energy equation for nozzle gives,

[Since W3 – 4 = 0, Q3 – 4 = 0, Z1 = Z2]

C4 = 473.4 m/s.

Hence, velocity at exit from the nozzle = 473.4 m/s. (Ans.)

12. A single stage single acting reciprocating air compressor has air entering at 1 bar, 20°C and

compression occurs following polytrophic process with index 1.2 upto the delivery pressure of 12

bar. The compressor runs at the speed of 240 rpm and has L/D ratio of 1.8. The compressor has

mechanical efficiency of 0.88. Determine the isothermal efficiency and cylinder dimensions. Also

find out the rating of drive required to run the compressor which admits 1 m3 of air per minute.





13. An engine working on Otto cycle has a volume of 0.45 m3, pressure 1bar and temperature 30°C

at the beginning of compression stroke. At the end of compression stroke, the pressure is 11 bar. 210

kJ of heat is added at constant volume. Determine : (i) Pressures, temperatures and volumes at

salient points in the cycle, (ii) Percentage clearance, (iii) Efficiency, (iv) Net work per cycle, (v)

Mean effective pressure and (vi) Ideal power developed by the engine if the number of working

cycles per minute is 210. Assume the cycle is reversible.

V2 = 0.081 m3

Temperature and pressure at point 3 can be determined using above equations.Similarly, calculate temperature and pressure at point.





14 .An engine with 200 mm cylinder diameter and 300 mm stroke works on theoretical Diesel cycle. The

initial pressure and temperature of air used are 1 bar and 27°C. The cut-off is 8% of the stroke. Determine:

(i) Pressures and temperatures at all salient points, (ii) Theoretical air standard efficiency, (iii) Mean

effective pressure and (iv) Power of the engine if the working cycles per minute are 380. Assume that

compression ratio is 15 and working fluid is air. Consider all conditions to be ideal.

1. Determine stroke volume

2. Use gas equation to find mass of air in the cylinder

3. Process 1-2 (Adiabatic) and Process 3-4 (Isentropic) – Use relations between pressure, volume andtemperature



EI -6403/Applied Thermodynamics and Fl uid Dynamics Department of Mechanical Engineeri ng 2014-2015

St.Joseph’s College of Engineering 19 ISO 9001 - 2008

UNIT II

THERMODYNAMICS OF REFRIGERATORS AND PUMPS

PART- A

1. Define Quality of steam. What is the various quality of steam?

The Quality of steam is the ratio of mass of dry steam actually present to the mass of wet steam whichcontains it. Various qualities of steam are (i) Wet steam (ii) Dry steam (iii) Superheated steam.

2. Define dryness fraction.

It is a measure of quality of steam. It is the ratio of mass of vapour to the total mass of the steam. It isdenoted by X.

3. Define heat engine, heat pump, and refrigerator.Heat engine: A system which by operating in a cyclic manner produces net work from a supply ofheat. (Or) A heat engine is any continuously operating thermodynamic system across the boundaries ofwhich flow only heat and work.Heat Pump: A device that transfers heat from a low temperature medium to a high temperature

medium is called a heat pump.Refrigerators: A device that transfers heat from low temperatures to high temperature region is knownas refrigerator.

4. What is the difference between boiler mountings and accessories?

5. What is the function of pressure relief valve?

The pressure is relieved by allowing the pressurised fluid to flow from an auxiliary passage out of thesystem. The relief valve is designed or set to open at a predetermined set pressure to protect pressure

vessels and other equipment from being subjected to pressures that exceed their design limits.6. What is the function of safety valve?

It is a mechanical device used to safe guard the boiler, in case the pressure inside the boiler rises aboveits normal working pressure.

7. Define C.O.P of refrigeration. What is the unit of refrigeration?

It is defined as the ratio of net refrigeration effect to the work required to produce the effect.

Capacity of refrigeration unit is generally defined in ton of refrigeration. A ton of refrigeration isdefined as the quantity of heat to be removed in order to form one ton (1000 kg) of ice at 0

0C in 24 hrs,

from liquid water at 00

C. This is equivalent to 3.5 kJ/s (3.5 kW) or 210 kJ/min.

8. Define refrigeration effect.

The amount of heat extracted in a given time is known as refrigeration effect.9. Differentiate vapour compression system and vapour absorption system.

S. No. Vapour Compression System Vapour Absorption System

1 Due to compressor and fan more wear andtear.

Only moving part is liquid pump, less wearand tear.

2Electrical power is essential to operate thesystem

Electrical power is not essential to operatethe system (heat energy is used).

3Compressor is used to compress theRefrigerant.

Compressor is used to compress theRefrigerant.

10. What are the various modes of heat transfer?

Modes of Heat Transfer – (i) Conduction (ii) Convection (iii) Radiation

11. Define Conduction

Energy transfers across a system boundary due to a temperature difference by the mechanism of intermolecular interactions. Conduction needs mater and does not require any bulk motion of matter.

12. Define Convection.

An energy transfer across a system boundary due to a temperature difference by the combinedmechanism of intermolecular interactions and bulk transport. Convection needs fluid mater.

13. Define black body.

A black body is one which absorbs heat radiation of all wave length falling on it.

Boiler Mountings Boiler Accessories

The necessary devices installed or

mounted for the safety of boiler and itscontrol are called boiler mountings.

The devices which are installed in the boiler

for their efficient operation and smoothworking are called boiler accessories.

1. Water level indicator.2. Safety valves.3. Fusible plug.4.Pressure gauge

1. Water heating devices.2. Water feeding devices.3. Super heater4. Economizer

http://en.wikipedia.org/wiki/Pressure_vessel









14. Define grey body.If a body absorbs definite percentage of incident radiations irrespective of their wavelength, then the body is known as grey body.

15. What is the difference between forced and free convection? If the fluid motion is produced due to change in density resulting from temperature gradients, the modeof heat transfer is said to be free or natural convection.If the fluid is artificially created by means of an external force like a blower or fan, that type of heat

transfer is known as forced convection.16. Define Emissivity.

Emissivity is defined as the ratio between energy radiated by the surfaces at temperature and the energyradiated by black surface at same temperature.

17. What is meant by radiation shape factor?

The shape factor is defined as the fraction of the radiative energy that is diffused from one surfaceelement and strikes the other surface directly with no intervening reflections.

18. Write the difference between water tube and fire tube boiler.

S.No Fire tube boiler Water tube Boiler

1. Hot Gases flows through the tubes. Water flows through the tube

2. Compact and easy design. Complicated design

3. Benson boiler Babcock and Wilcox boiler

19. How the boiler efficiency is calculated?

For a steam generating unit, efficiency is defined as the ratio of heat absorbed by the boiler fluid to thefuel fired.

Boiler = Heat absorbed by boiler fluid X100 Fuel fired (or) fuel supplied

20. What are fins? Why fins are used?

A fin is a surface that extends from an object to increase the rate of heat transfer to or from theenvironment by increasing convection. Examples of fins: 1.Thin rods on the condenser in back of

refrigerator.2. Honeycomb surface of a car radiator.3. Corrugated surface of a motor cycle engine. 4. Coolers of PC boards.

21. Sketch the vapour compression refrigeration cycle process.

22. What are the processes that constitute a Rankine cycle?Process 1 – 2: Isentropic expansion of the working fluid through the turbine from saturated vapour atstate 1 to the condenser pressure.

Process 2 – 3: Heat transfer from the working fluid as it flows at constant pressure through thecondenser with saturated liquid at state 3.Process 3 – 4: Isentropic compression in the pump to state 4 in the compressed liquid region.Process 4 – 1: Heat transfer to the working fluid as it flows at constant pressure through the boiler tocomplete the cycle.

23. How do you determine the state of steam?If V > vg then super-heated steam, V= vg then dry steam and V < vg then wet steam.

24. Define enthalpy of steam.It is the sum of heat added to water from freezing point to saturation temperature and the heat absorbedduring evaporation.

http://en.wikipedia.org/wiki/Convection

http://en.wikipedia.org/wiki/Convection





25. Explain the term critical point, critical temperature and critical pressure.In the T-S diagram the region left of the waterline, the water exists as liquid. In right of the dry steamline, the water exists as a super heated steam. In between water and dry steam line the water exists as a

wet steam. At a particular point, the water is directly converted into dry steam without formation of wetsteam. The point is called critical point. The critical temperature is the temperature above which asubstance cannot exist as a liquid; the critical temperature of water is 374.15

oC. The corresponding

pressure is called critical pressure.

26. Find the mass of 0.7 m3of wet steam at 150

0C and 90% dry.

The specific volume of dry steam at 1500C, vg = 0.3928 m3/kg;

The mass of 0.7 m3of wet steam at 1500C and 90% dry is: 0.7 / (0.3928×0.9) = 1.98 kg.27. Determine the enthalpy and sp. volume of steam at a pressure of 6 bar having a quality of 0.85.

h = hf + x hfg = 1213.35+.85 X 1571.0 = 2548.7 kJ/kg ; v = x·vg = 0.85 X 0.03244=0.027574 m3/kg

28. What are the major components in a steam power plant?

29. Show Rankine cycle on T-s diagram.

30. What are the ways by which Rankine cycle efficiency may be improved?

(i) Operating the boiler at high pressure with reheat and (ii) regeneration

31. Write down the expression for efficiency of Rankine cycle without considering pump work.

32. What is reversed carnot heat engine? What are the limitations of carnot cycle?

1. No friction is considered for moving parts of the engine.2. There should not be any heat loss

33. What is the difference between a heat pump and a refrigerator?

Heat pump is a device which operating in cyclic process, maintains the temperature of a hot body at atemperature higher than the temperature of surroundings.

A refrigerator is a device which operating in a cyclic process, maintains the temperature of a cold bodyat a temperature lower than the temperature of the surroundings.

34. Why a heat engine cannot have 100% efficiency?

For all the heat engines there will be a heat loss between system and surroundings. Therefore we can’tconvert all the heat input into useful work.

35. What are the processes involved in Carnot cycle.Carnot cycle consist of

i) Reversible isothermal compression

ii) Isentropic compressioniii) Reversible isothermal expansion

iv) Isentropic expansion





35. Write the expression for COP of a heat pump and a refrigerator?COP of heat pump

Heat Supplied T2

COP HP = ------------------- = --------Work input T2-T1

COP of RefrigeratorHeat extracted T1

COP Ref = --------------- = --------Work input T2-T1

36. Why Carnot cycle cannot be realized in practical?(i) In a Carnot cycle all the four processes are reversible but in actual practice there is no process isreversible.(ii) There are two processes to be carried out during compression and expansion. For isothermal process the piston moves very slowly and for adiabatic process the piston moves as fast as possible.

This speed variation during the same stroke of the piston is not possible.(iii) It is not possible to avoid friction moving parts completely.

Unit – II Part – B

1. A reversible heat engine operates between two reservoirs at temperatures 700°C and

50°C. The engine drives a reversible refrigerator which operates between reservoirs at

temperatures of 50°C and – 25°C. The heat transfer to the engine is 2500 kJ and the net

work output of the combined engine refrigerator plant is 400 kJ. (i) Determine the heat

transfer to the refrigerant and the net heat transfer to the reservoir at 50°C. (ii) Reconsider

given that the efficiency of the heat engine and the C.O.P. of the refrigerator are each 45

per cent of their maximum possible values.

Ans)

Temperature, T1 = 700 + 273 = 973 K

Temperature, T2 = 50 + 273 = 323 KTemperature, T3 = – 25 + 273 = 248 K

The heat transfer to the heat engine, Q1 = 2500 kJThe network output of the combined engine refrigerator plant,

W = W1 – W2 = 400 kJ.(i) Maximum efficiency of the heat engine cycle is given by

W1 = 0.668 × 2500 = 1670 kJ





Since, W1 – W2 = W = 400 kJW2 = W1 – W= 1670 – 400= 1270 kJ

∴ Q4 = 3.306 × 1270= 4198.6 kJQ3 = Q4 + W2 = 4198.6 + 1270= 5468.6 kJQ2 = Q1 – W1 = 2500 – 1670= 830 kJ.

Heat rejection to the 50°C reservoir= Q2 + Q3 = 830 + 5468.6

= 6298.6 kJ. (Ans.)

(ii) Efficiency of actual heat engine cycle,

η = 0.45 ηmax

= 0.45 × 0.668

= 0.3

∴ W1 = η × Q1

= 0.3 × 2500

= 750 kJ

∴ W2 = 750 – 400

= 350 kJ

C.O.P. of the actual refrigerator cycle,

= 0.45 × 3.306 = 1.48

∴ Q4 = 350 × 1.48

= 518 kJ. (Ans.)

Q3 = 518 + 350

= 868 kJ

Q2 = 2500 – 750

= 1750 kJ

Heat rejected to 50°C reservoir

= Q2 + Q3

= 1750 + 868

= 2618 kJ. (Ans.)

2. (i) A reversible heat pump is used to maintain a temperature of 0°C in a refrigerator

when it rejects the heat to the surroundings at 25°C. If the heat removal rate from the

refrigerator is 1440 kJ/min, determine the C.O.P. of the machine and work input

required. (ii) If the required input to run the pump is developed by a reversible engine

which receives heat at 380°C and rejects heat to atmosphere, then determine the overall

C.O.P. of the system.





Ans) (i) Temperature, T1 = 25 + 273 = 298 K

Temperature, T2 = 0 + 273 = 273 K

Heat removal rate from the refrigerator,

Q1 = 1440 kJ/min = 24 kJ/s

Now, co-efficient of performance, for reversible heat pump,

W = 2.2 kW

i.e., Work input required = 2.2 kW. (Ans.)

Q2 = Q1 + W = 24 + 2.2 = 26.2 kJ/s

(ii) Refer Fig.

The overall C.O.P. is given by,

For the reversible engine, we can write





298(Q4 + 2.2) = 653 Q4

Q4(653 – 298) = 298 × 2.2

Q3 = Q4 + W

= 1.847 + 2.2

= 4.047 kJ/s

Substituting this value in eqn. (i), we get

If the purpose of the system is to supply the heat to the sink at 25°C, then

3. A Rankine cycle operates between pressures of 80 bar and 0.1 bar. The maximum cycletemperature is 600°C. If the steam turbine and condensate pump efficiencies are 0.9

and 0.8 respectively, calculate the specific work and thermal efficiency. Relevant steam

table extract is given below.





Ans)

At 80 bar, 600ºC :

h1 = 3642 kJ / kg ;

s1 = 7.0206 kJ / kg K.

Since s1 = s2,

∴ 7.0206 = sf2 + x2 sfg2

= 0.6488 + x2 × 7.5006

= 0.85

Now, h2 = hf2 + x2 hfg2

= 191.9 + 0.85 × 2392.3

= 2225.36 kJ/kg

Actual turbine work

= ηturbine × (h1 − h2 )

= 0.9 (3642 – 2225.36)= 1275 kJ/kg

Pump work = vf ( p2 )( p1 − p2 )

= 10.09 kJ/kg

Specific work (Wnet ) = 1275 – 10.09

= 1264.91 kJ / kg. (Ans.)

where, Q1 = h1 – hf4

But hf4 = hf3 + pump work

= 191.9 + 10.09





= 202 kJ/kg

∴ Thermal efficiency, ηth =

= 0.368 or 36.8 %. (Ans.)

4. (i) Find the enthalpy and internal energy of unit mass of steam at temperature of 200ºC

when (a) Dry saturated (b) Steam delivered at 0.7MPa and (c) quality is 0.85.

(ii) Steam enters a steam turbine at a pressure of 10 bar and 300˚C with a velocity of 50m/s.The steam leaves the turbine at 1.5 bar and with a velocity of 200m/s. Determine the work

done per kg of steam flow through the turbine. Assume that the process to be reversible and

neglect the change in potential energy.

5. Explain with neat sketch about steam power plant.Ans) Steam is an important medium of producing mechanical energy. Steam has the advantage that, it can be raised from water which is available in abundance it does not react much with the materials of theequipment of power plant and is stable at the temperature required in the plant. Steam is used to drivesteam engines, steam turbines etc. Steam power station is most suitable where coal is available in





abundance. Thermal electrical power generation is one of the major methods. Out of total power developed in India

about 60% is thermal. For a thermal power plant the range of pressure may vary from 10 kg/cm2 to super

critical pressures and the range of temperature may be from 250°C to 650°C.Essentials of Steam Power Plant Equipment

A steam power plant must have following equipment:(a) A furnace to burn the fuel.(b) Steam generator or boiler containing water. Heat generated in the furnace is utilized to

convert water into steam.(c) Main power unit such as an engine or turbine to use the heat energy of steam and perform

work.(d) Piping system to convey steam and water.

In addition to the above equipment the plant requires various auxiliaries and accessories depending uponthe availability of water, fuel and the service for which the plant is intended.

The flow sheet of a thermal power plant consists of the following four main circuits:

4. Feed water and steam flow circuit.

5. Coal and ash circuit.

6. Air and gas circuit.

7. Cooling water circuit.

Steam Power Plant Layout

Air taken from the atmosphere is first passed through the air pre-heater, where it is heated by fluegases. The hot air then passes through the furnace. The flue gases after passing over boiler and superheater tubes, flow through the dust collector and then through economizer, air pre-heater and finally theyare exhausted to the atmosphere through the chimney.

6. What is meant by fire tube boiler? Explain with neat sketch.Ans) A fire-tube boiler is a type of boiler in which hot gases from a fire pass through one or more tubesrunning through a sealed container of water. The heat of the gases is transferred through the walls of thetubes by thermal conduction, heating the water and ultimately creating steam.Types of fire-tube boiler Cornish boiler

Lancashire boiler

Scotch marine boiler

Locomotive boiler

Vertical fire-tube boiler

http://en.wikipedia.org/wiki/Boiler

http://en.wikipedia.org/wiki/Heat

http://en.wikipedia.org/wiki/Thermal_conduction

http://en.wikipedia.org/wiki/Fire-tube_boiler#Types_of_fire-tube_boiler


http://en.wikipedia.org/wiki/Fire-tube_boiler#Lancashire_boiler

http://en.wikipedia.org/wiki/Fire-tube_boiler#Scotch_marine_boiler

http://en.wikipedia.org/wiki/Fire-tube_boiler#Locomotive_boiler

http://en.wikipedia.org/wiki/Fire-tube_boiler#Vertical_fire-tube_boiler



http://en.wikipedia.org/wiki/Fire-tube_boiler#Locomotive_boiler

http://en.wikipedia.org/wiki/Fire-tube_boiler#Scotch_marine_boiler

http://en.wikipedia.org/wiki/Fire-tube_boiler#Lancashire_boiler


http://en.wikipedia.org/wiki/Thermal_conduction

http://en.wikipedia.org/wiki/Heat

http://en.wikipedia.org/wiki/Boiler





Advantages of fire tube boilers• Low cost • Fluctuations of steam demand can be met easily

• It is compact in size. Cochran Boiler Salient features

• The dome shape of the furnace causes the hot gases to deflect back and pass through the flue. The

un‐ burnt fuel if any will also be deflected back.• Spherical shape of the top of the shell and the fire box gives higher area by volume ratio.• It occupies comparatively less floor area and is very compact.

• It is well suited for small capacity requirements.• Very compact and requires minimum floor area• Any type of fuel can be used with this boiler • Well suited for small capacity requirements gives about 70% thermal efficiency with coal firing andabout 75% with oil firing.

7. What are the various boiler mountings? Explain any two with neat sketch.

1. Various boiler mountingsMainly seven (7) mountings are required and essential to a Boiler.

1. Water level indicator.2. Main Steam stop valve.

3. Pressure gauge.

4. Feed check valve.

5. Fusible plug.

6. Blow down valve.7. Safety valve.

Water level indicator

• To know the water level in the boiler the handles of the steam cock and water cock are kept invertical

positions.

• Water rushes through the bottom casting and steam rushes through the upper casting to the gauge

glass

tube.

• The level of water corresponds to the water level in the boiler





Fusible plug

Under normal working conditions, the fusible plug is completely covered with water.

Hence the temperature of the plug is not increased appreciably during combustion process.

When the water level falls below the safe limit the fusible plug is uncovered from water andexposed to steam.

The furnace heat over heats the plug and it melts the fusible metal and copper plug falls

down.

Due to this water steam mixture rushes into the furnace and the fire is extinguished

8. What are the various boiler accessories? Explain any two with neat sketch.

Ans) Various boiler accessories• Economiser • Air Preheater • Super Heater • Steam Separator • Steam Trap • Feed Pump

i) Economiser

• The feed water is pumped to the bottom header and this water is carried to the top header number of

vertical tubes.

• Hot flue gases are allowed to pass over the external surface of the tubes.• The feed water which flows upward in the tubes is heated by the flue gases.• This preheated water is supplied to the water.

• Scrappers are moved slowly moved up and down to clean the surface of the tubes.

Economiser





(ii) Air Preheater

• Hot flue gases pass through the tubes of air preheater after leaving the boiler or economiser.• Air and flue gases flow in opposite directions.

• Baffles are provided in the air preheater and the air passes number of times over the tubes.• Heat is absorbed by the air from the flue gases.• This preheated is supplied to the furnace to aid combustion.

Air Preheater





9. Steam at 480oC, 90 bar is supplied to a Rankine cycle. It is reheated to 12 bar and 480

oC.

The minimum pressure is 0.07 bar. Find the work output and cycle efficiency using steam

tables with and without considering pump work.









10. What are the various types of boilers? Explain any one type of high pressure boiler

with neat sketch.Ans) A boiler is basically a closed vessel in to which water is heated until water is converted in top

steam at required pressure.1. Classification of Boilers.

A. According to relative position of water and hot gases.B. According to geometric orientation of boiler.

C. According to location of furnace.

D. According to method of water circulation.E. According to working pressure.F. According to mobility of boiler.

Benson Boiler

(a) Natural circulation boilers require expansion joints but these are not required for Benson asthe pipes are welded. The erection of Benson boiler is easier and quicker as all the parts arewelded at site and workshop job of tube expansion is altogether avoided.

(b) The transport of Benson boiler parts is easy as no drums are required and majority of the parts are carried to the site without pre-assembly.

(c) The Benson boiler can be erected in a comparatively smaller floor area. The space problemdoes not control the size of Benson boiler used.

(d) The furnace walls of the boiler can be more efficiently protected by using small diameterand close pitched tubes.

(e) The super heater in the Benson boiler is an integral part of forced circulation system,therefore no special starting arrangement for super heater is required.

(f) The Benson boiler can be started very quickly because of welded joints.

(g) The Benson boiler can be operated most economically by varying the temperature and pressure at partial loads and overloads. The desired temperature can also bemaintained constant at any pressure.

(h) Sudden fall of demand creates circulation problems due to bubble formation in the naturalcirculation boiler which never occurs in Benson boiler. This feature of insensitiveness to loadfluctuations makes it more suitable for grid power station as it has better adaptive capacity tomeet sudden load fluctuations.

(i) The blow-down losses of Benson boiler are hardly 4% of natural circulation boilers of samecapacity.

(j) Explosion hazards are not at all severe as it consists of only tubes of small diameter and hasvery little storage capacity compared to drum type boiler.





11. A simple Rankine cycle works between pressures 28 bar and 0.06 bar, the initial

condition of steam being dry saturated. Calculate the cycle efficiency, work ratio and

specific steam consumption.

From steam tables,

At 28 bar : h1 = 2802 kJ/kg,

s1 = 6.2104 kJ/kg K

At 0.06 bar : hf2 = hf3 = 151.5 kJ/kg,

hfg2 = 2415.9 kJ/kg,

sf2 = 0.521 kJ/kg K,

sfg2 = 7.809 kJ/kg K

vf = 0.001 m3/kg

Considering turbine process 1-2, we have :

s1 = s2

6.2104 = sf2 + x2 sfg2

= 0.521 + x2 × 7.809

x2 = 0.728

h2 = hf2 + x2 hfg2

= 151.5 + 0.728 × 2415.9

= 1910.27 kJ/kg

∴ Turbine work, Wturbine = h1 – h2

= 2802 – 1910.27

= 891.73 kJ/kg

Pump work, W pump = hf4 – hf3

= vf (p1 – p2)





= 2.79 kJ/kg

[Since, hf4 = hf3 + 2.79 = 151.5 + 2.79 = 154.29 kJ/kg]

∴ Net work, Wnet = Wturbine – W pump

= 891.73 – 2.79

= 888.94 kJ/kg

Cycle efficiency

= 0.3357 or 33.57%. (Ans.)

= 0.997. (Ans.)

Specific steam consumption =

= 4.049 kg/kWh.

12. What is meant by water tube boiler? Explain with neat sketch.Ans) The boilers can be classified according to the following criteria. According to of water and hot gases:

(a) Water tube

(b) Fire tube.

In water tube boilers, water circulates through the tubes and hot products of combustion flow overthese tubes. In fire tube boiler the hot products of combustion pass through the tubes, which aresurrounded, by water. Fire tube boilers have low initial cost, and are more compacts. But they are morelikely to explosion, water volume is large and due to poor circulation they cannot meet quickly the change





in steam demand. For the same output the outer shell of fire tube boilers is much larger than the shell ofwater-tube boiler. Water tube boilers require less weight of metal for a given size, are less liable toexplosion, produce higher pressure, are accessible and can respond quickly to change in steam demand.

Tubes and drums of water-tube boilers are smaller than that of fire-tube boilers and due to smaller size ofdrum higher pressure can be used easily. Water-tube boilers require lesser floor space. The efficiency ofwater-tube boilers is more.

UNIT III BASIC CONCEPTS OF FLUID MECHANICS AND FLOW OF FLUIDS

Part A

1. Define fluids.

Fluid may be defined as a substance which is capable of flowing. It has no definite shape of its own, but

confirms to the shape of the containing vessel.2. What are the properties of ideal fluid?

Ideal fluids have following properties

It is incompressible; It has zero viscosity; Shear force is zero

3. What are the properties of real fluid?

Real fluids have following propertiesi)It is compressible; ii) They are viscous in nature; iii) Shear force exists always in such fluids.

4. Define density and specific weight.

Density is defined as mass per unit volume (kg/m3)

Specific weight is defined as weight possessed per unit volume (N/m3)

5. Define Specific volume and Specific Gravity.Specific volume is defined as volume of fluid occupied by unit mass (m3/kg)Specific gravity is defined as the ratio of specific weight of fluid to the specific weight of standard fluid.

6. Define Surface tension and Capillarity.

Surface tension is due to the force of cohesion between the liquid particles at the free surface.Capillary is a phenomenon of rise or fall of liquid surface relative to the adjacent general level of liquid.7. Define Viscosity.

It is defined as the property of a liquid due to which it offers resistance to the movement of one layer of

liquid over another adjacent layer.8. Define kinematic viscosity.

It is defined as the ratio of dynamic viscosity to mass density. (m²/sec)

9. Define Relative or Specific viscosity.

It is the ratio of dynamic viscosity of fluid to dynamic viscosity of water at 20°C.10. Define Compressibility.

It is the property by virtue of which fluids undergoes a change in volume under the action of external

pressure.11. Define Newton‟s law of Viscosity. According to Newton’s law of viscosity the shear force F acting between two layers of fluid is





proportional to the difference in their velocities du and area A of the plate and inversely proportional tothe distance between them.

12. What is cohesion and adhesion in fluids?

Cohesion is due to the force of attraction between the molecules of the same liquid.Adhesion is due to the force of attraction between the molecules of two different liquids or between themolecules of the liquid and molecules of the solid boundary surface.

13. State momentum of momentum equation?

It states that the resulting torque acting on a rotating fluid is equal to the rate of change of moment ofmomentum

14. What is momentum equation?It is based on the law of conservation of momentum or on the momentum principle It states that, the netforce acting on a fluid mass is equal to the change in momentum of flow per unit time in that direction.

15. Why is it necessary in winter to use lighter oil for automobiles than in summer? To what

property does the term lighter refer?

The term lighter refers to the property called viscosity. In winter, if heavy oil is used for automobiles, theoil becomes more viscous, and doesn’t serve lubrication purpose. So lighter oil is used. 16. If the pressure on the fluid is increased from 75 bar to 140 bar, the volume of liquid decreases by

0.15%. Find the bulk modulus of elasticity of the liquid.

4.33 x 109 N/m

2

17. At a certain point in flowing caster oil, the shear stress is 2 N/m2 and velocity gradient is 0.25/sec.

The mass density of the oil is 800kg/m3. Find the kinematic viscosity of oil in stokes.

18. State Pascal‟slaw.

This lawindicates that the pressure intensity at any point in a static liquid is equal in all directions.

19. Does viscosity vary with pressure and temperature?The value of μ of liquid or gas is practically independent of pressure for the range generally countered

in practice but it varies widely with temperature. The temperature has predominant effect on the viscosity

of the liquids. With the increase in temperature, the viscosity decreases rapidly.20. Discuss about Newtonian and Non Newtonian Fluids

Newtonian fluids. These fluids follow Newton’s viscosity equation. For such fluids does not changewith rate of deformation. Non Newtonian fluids: There fluids which do not follow the linear relationship between the shear stress and the rate of

deformation21. State Bernoulli‟s Theorem as applicable to fluid flow.. Bernoulli’s Theorem states that an ideal incompressible fluid when flow is steady and continuous, the sum of pressureenergy, kinetic energy and potential energy is constant along the stream line.

22. What are the three major assumptions made in the Bernoulli‟s Theorem.

The liquid is ideal and incompressible

The flow is steady and continuous

All the frictional losses are negligible

23. Mention the applications of Bernoulli‟s Equation Venturimeter

Orifice meter

Pitot Tube

24. What is Venturimeter?Venturimeter is a device used to measure the rate of flow through pipes.

25. Write the expression for discharge through a venturimeter.

Q= ghaa

aaC d 2**

2

2

2

1

21

PART B

1. The velocity distribution over a plate is given by u = (3/4) * y - y2, where u is velocity in m/s and at a

depth y in m above the plate. Fin d the shear stress at a distance of 0.3 m from the top of plate. Take dynamic viscosity of the fluid is taken as 0.95 Ns/m2.

(du/dy)=(3/4)-2y

=0.1425 N/mm2

2. Inside a 60 mm diameter cylinder a piston of 59 mm diameter rotates concentrically. Both thecylinder and piston are 80 mm long. If the space between the cylinder and piston is filled with oil of

V

dV

dp K

dy

du sm /

100

1 2

2

/825.0

2m Ns





viscosity of 0.3 N.s/m2 and a torque of 1.5 Nm is applied, find: the r.p.m. of the piston, and the powerrequired.

U=(DN/60) A=πdl =/(du/dy)

Shear force F = x A,

T=Fxd/2

Power required = 2 π NT/60

3. A cylinder 100mm diameter rotates in an annular sleeve 102mm internal diameter at 100 rpm. The

cylinder is 300mm long. If the dynamic viscosity of the lubricant between the two cylinders is

0.1poise, (assuming velocity profile linear between the cylinders) find the torque needed to drive the

cylinder against visco resistance.U=(DN/60)

=/(du/dy)


T=Fxd/2

4. A shaft of diameter 30cm rotates concentrically inside a sleeve having diameter 31cm and length30cm. Find the viscosity of the fluid that fills the gap between the cylinders, if a torque of 9.81Nm isrequired to maintain the speed at 60 rpm.

U=(DN/60) A=πdl =/(du/dy)


T=Fxd/2

5.





6. A plate 0.0254 mm distant from a fixed plate, moving at 61 cm/s requires a force of 0.2 kg(f)/m 2 to

maintain this speed. Find the dynamic viscosity of fluid between the plates.du=u-0

=/(du/dy)7. A shaft 300 mm in diameter revolves in a guide bearing 600 mm long at 500 rpm. If the oil film between the shaft and the bearing is 0.13mm and viscosity of oil is 0.32373 poise, what is the power

absorbed?The dynamic viscosity of oil is 0.03 Pas. (Pas = (N/m2) × s).

Shear stress on the shaft surface = τ = µ (du/dy) = µ (u/y)

u = π DN/60 = π × 0.3 × 400/60 = 6.28 m/s

τ = 0.03 {(6.28 – 0)/ 0.001} = 188.4 N/m2

Surface area of the two bearings, A = 2 π DL

Force on shaft surface = τ × A = 188.4 × (2 × π × 0.3 × 0.3) = 106. 6 N

Torque = 106.6 × 0.15 = 15.995 Nm

Power required = 2 π NT/60 = 2 × π × 400 × 15.995/60 = 670 W.

8.

A flat plate 0.75 m2 in area moves through the oil between large fixed parallel plates 150 mm apart isfilled with oil of kinematic viscosity of 0.8x10

-4m

2/s , specific gravity 0.75 and the plate moving with

the velocity of 1m/s, calculate the drag force when (i) The plate is from one of the planes and (ii) The

plate is midway between the planes.

Specfic gravity S=water

oil

Kinematic Viscosity

Total Shear Stress=21

ie...21

t

u

t

u





Now F=

21 t

u

t

u A

Find21

& F F

9. A cubical block of 20cm edge weighing 20 kg(f) is allowed to slide down a plane inclined at 20 tothe horizontal on which there is a thin film of viscosity 0.22 x 10-3 kg(f) s/m2. What terminal velocity the block will obtain, if the film thickness is 0.025 mm?

F=wcos ,u y

A F , U=(DN/60),

Kinematic Viscosity

10. Calculate the capillary effect in millimeters in a glass tube of 4 mm diameter, when immersed in (1)water and (2) mercury the temperature of the liquid is 20°C and the values of surface tension of water andmercury at 20°C in contact with air are 0.0735 N/m and 0.51 N/m respectively. The contact angle forwater θ = 0 and for mercury θ = 130°. Take specific weight of water at 20°C as equal to 9790 N/m3 and

specific gravity of mercury is 13.6.

, (i) h= 7.51 mm, (ii) h=-2.46 mm.

, d= 100 mm,

11.

12. Derive the continuity equation of differential form. Discuss whether the equation is valid for a steadyor unsteady flow, viscous or inviscid flow, compressible or incompressible flow.





For 3D incompressible fluid, ,

13. A pipe line carrying oil of specific gravity 0.87 changes in diameter from 200 mm diameter at position‘1’ to 500 mm diameter at position ‘2’ which is 4 meters at a higher level. If the pressures at 1 and 2 are100 kN/m2 and 60 kN/m2 respectively and the discharge is 0.2 m3/s, determine: (a) Loss of head, (b)

Direction of flow.

At point A

Total Energy 1

1

2

1

1 2 z w

p

g

v

E

At point B

Total Energy2

2

2

2

2

2 z

w

p

g

v E

Loss of head21

E E h f

If21

E E direction A to B,

If21

E E direction B to A,

14. A horizontal venturimeter with inlet and throat diameter 300 mm and 100 mm respectively is used tomeasure the flow of water. The pressure intensity at inlet is 130 kN/m

2 while the vacuum pressure head at

throat is 350 mm of mercury. Assuming that 3% head lost between the inlet and throat. Find the value ofcoefficient of discharge for the venturimeter and also determine the rate of flow.

Differential head,

; Q=

15. An orifice meter with orifice diameter 15 cm is inserted in a pipe of 30 cm diameter. The pressuredifference measured by a mercury oil differential manometer on the two sides of the orifice meter gives a

reading of 50 cm of mercury. Find the rate of flow of oil of sp.gr 0.9 and C d = 0.6.

= 137414.25 cm3/s

UNIT IV DIMENSIONAL AND MODEL ANALYSIS

Part A

1-What are the uses of dimensional homogeneity?

To determine the dimension of a physical variable To convert units from one system to another.2. Name the different forces present in fluid flow

Inertia force, viscous force, Surface tension force, Gravity force

3. When in a fluid considered steady?

In steady flow, various characteristics of following fluids such as velocity, pressure, density,temperature etc at a point do not change with time. So it is called steady flow.

4. What is kinematic similarity?

It is the similarity of motion corresponds to the points in the model and the prototype5. Mention the types of models.

1. Undistorted models2. Distorted models6. Define Mach number.

It is the square root of Inertia force to the elastic force





2

22

kL

V L

7. Define Reynolds Number.

It is the ratio of Inertia force to the viscous force

VL Re

8. State the methods of dimensional analysis.

1. Rayleigh’s method; 2. Buckingham’s Π theorem

9. State Buckingham‟s Π theorem

It states that if there are ‘n’ variables in a dimensionally homogeneous equation and if these variablescontain ‘m’ fundamental dimensions (M, L, T), then they are grouped into (n-m), dimensionlessindependent Π-terms.

10. State the limitations of dimensional analysis.

1. Dimensional analysis does not give any due regarding the selection of variables.2. The complete information is not provided by dimensional analysis.3. The values of coefficient and the nature of function can be obtained only by experiments or from

mathematical analysis.11. Define Similitude

Similitude is defined as the complete similarity between the model and prototype.

12. State Froude‟s model law Only Gravitational force is more predominant force. The law states ‘The Froude’s number is same for both model and prototype’. 13. Define Weber Number.

It is the ratio of inertia force to the surface tension force

2 LV W e

14. What are the factors to the determined when viscous fluid flows through the circular pipe ?

Velocity distribution across the section; Ratio of maximum velocity to the average velocity; Shearstress distribution; Drop of pressure for a given length .

Using Buckingham’s π theorem, show that the drag DF of a supersonic aircraft is given by:

)M(Re,VLF 22D . Where,

VLRe = Reynolds number, cVM = Mach number, = fluid density,

c = sonic velocity = ,K V = velocity of aircraft, K = bulk modulus of fluid,

L = chord length,2L = wing area = chord x span, ρ = a functional notation.

15. What is meant by undistorted models?

The model which is geometrically similar to its prototype is known as undistorted models. In suchmodels, the conditions of similitude are fully satisfied.

16. Describe briefly the selection of repeating variables in Buckingham Π theorem. There is no separate rule for selecting repeating variables. But the number of repeating variables is

equal to the fundamental dimensions of the problem. Generally ρ, V , ℓ , or ρ, V , D are chosen as repeating

variables. In addition to this, the following points should be kept in mind while selecting repeatingvariables. (i) No more variable should be dimensionless (ii) The selected two repeating variables shouldnot have the same dimensions. (iii) The selected repeating variables should be independent as far as possible.

17. Define the term dimensional homogeneity.Dimensional homogeneity means, the dimensions of the terms on left hand side should be same as the

dimensions of the terms on right hand side.

18. Define Euler number It is defined as the square root of the ratio of inertia force to the pressure of a flowing fluid.

2

22

pL

V L

u E

16.

What is kinematic similarity? It is the similarity of forces. The flows in the model and prototype are of dynamic similar.

20. Define the term scale effect.





It is impossible to predict the exact behavior of the prototype by model testing alone. The two modelsof same prototype behavior will be different. Scale ratios will not be same. So discrepancy between modeland prototype will always occur. It is known as scale effect.

21. Distinguish between Rayleigh‟s method and Buckingham‟s Π theorem

S.No Rayleigh‟s method Buckingham‟s Π theorem

1 The expression is determined for a variable

depending upon maximum three or four

variables only.

It is easy for the number of independent

variables being more than three or four

2 This method is difficult for more than or

four variables.

In this method, ‘n’ variables can be usedwithout any difficulties.

22. What are the points to be remembered while deriving expressions using dimensional analysis?

(i) First, the variables controlling the phenomenon should be identified and expressed in terms of primary dimensions (ii) Any mathematical equation should be dimensionally homogeneous.(iii) Intypical cases a suitable mathematical model is constructed to simplify the problem with suitableassumptions

23. What are the similarities between model and prototype?(i) Geometric Similarity (ii) Kinematic Similarity (iii) Dynamic Similarity

24. State the Fourier law of dimensional homogeneity.The law of Fourier principle of dimensional homogeneity states ― an equation which expresses a physical phenomenon of fluid flow should be algebraically correct and dimensionally homogeneous.

25. State the Euler model law and give its significance.

Only pressure is more predominant force in addition to the inertia force. According to this law, theEuler number is same for both prototype and model. The Euler number itself is significant criterion inthe following phenomena: (i) Where the gravity and surface tension forces are fully absent and theturbulence is fully developed with negligible viscous force. (ii) It is applied in cavitations’ phenomena.

PART B1. The pressure difference P in a pipe of diameter D and length l due to turbulent flow depends on the

velocity V , viscosity μ, density ρ and roughness K . By using dimensional analysis, obtain an expressionfor the pressure difference P .

Solution:

Total number of variables n = 7

Number of fundamental dimensions m = 3

Number of dimensionless number of π-terms= n-m= 4

Dimension of variable involved in given problem

Variables Dimensions

D L

V LT-1

ρ ML-3

P ML-1

T-2

L L

µ ML-1

T-1

K L





2. The resisting force (R) of a supersonic flight can be considered as dependent upon the length of the aircraft 'ℓ', velocity 'v', air viscosity 'μ', air density 'ρ' and bulk modulus of air is 'k'. Express the functional

relationship between these variables and the resisting force.

Solution:






R MLT-2

L L

V LT-1

µ ML-1

T-1

Ρ ML-3

K ML-1

T-2

3. Consider force F acting on the propeller of an aircraft, which depends upon the variable U, , , D

and N. Derive the non – dimensional functional form )).U/DN(),/UD((f DUF 22

Solution:










Ρ ML-3

U LT-1

D L

T ML2T

-2

N T-1

µ ML-1

T-1

4. The discharge Q of a centrifugal pump depends upon the mass density of fluid ( ρ), the speed of the pump ( N ), the diameter of the impeller ( D), the manometric head ( H m) and the viscosity of fluid (μ). Show

that

222

3

ND,

DN

gHNDQ

.

Solution:



D L

N T-1

Ρ ML-3

Q L

3T

-1

G LT-2

H L

µ ML-1

T-1








5. An object of diameter 900 mm is to move in air at 60 m/s. Its drag is to be estimated from tests on ahalf scale model in water. The drag on the model is 1140 N. Estimate the speed of the model and drag on

the full scale object. Given, air = 1.2 kg/m3, air = 1.86 10

– 5 Ns/m

2,water = 1.0110

– 3 Ns/m

2, water

= 1000 kg/m3

Given:

Model half size tested in water:

Solution:Speed of the model:

Drag of the full scale object,(FD)P:





6. It is desired to obtain the dynamic similarity between a 30 cm diameter pipe carrying linseed oil at 0.5

m3/s and a 5 m diameter pipe carrying water. What should be the rate of flow of water in lps? If the

pressure loss in the model is 196 N/m2, what is the pressure loss in the prototype pipe? Kinematic

viscosities of linseed oil and water are 0.457 and 0.0113 stokes respectively. Specific gravity of linseed oil

= 0.82.

Given :

Dm = 30 cm

Q m = 0.5 m3/s

Dp = 5 m

Model fluid linseed oil

Prototype fluid –water

Pressure loss in model Pm =196 N/m2

Kinematic viscosity of oil,νm =0.457 strokes =0.457 x10

-4m

2/s

Kinematic viscosity of oil,νp = 0.0113 strokes =0.0113 x10-4

m2/s

Specific gravity for model,Sm =0.82

Specific gravity for prototype, Sp = 1.12

Solution:

Discharge of model Q m =Am x Vm

0.5 =

By dynamic similarity,

Discharge of prototype, Q p = Ap x Vp





Pressure drop scale ratio,

Pressure drop on the prototype,

7.

A model of a hydro electric power station tail race is proposed to built by selecting vertical scale 1 in50 and horizontal scale 1 in 100. If the design pipe has flow rate of 600 m 3/s and the allowable discharge

of 800 m3/s. Calculate the corresponding flow rates for the model testing.

Given:

(Lr) v =50

(Lr) H =100

Q P =600 m3/s

Q max =800 m3/s

Solution:

Discharge scale ratio, Qr = (Lr) H x [(Lr) v] 3/2

=100 x 503/2

= 35355.34

We know that, discharge scale ratio,

The allowable discharge of model





8. Model of an air duct operating with water produces a pressure drop of 10 kN/m2 over 10 m length. If

the scale ratio is 1/50. Density of water is 1000 kg/m3 and density of air is 1.2 kg/m

2. Viscosity of water is

0.001 Ns/m2 and viscosity of air is 0.00002 Ns/m2. Estimate corresponding drop in a 20 m long air duct.

Given:

Solution:

9.

Define and Explain Reynolds Number, Froude’s Number, Euler’s Number and Mach number.

Reynolds Number:

It is the ratio of Inertia force to the viscous force

VL R

e

Froude‟s Number:Only Gravitational force is more predominant force. The law states ‘The Froude’s number is same for both model and prototype’. Euler number

It is defined as the square root of the ratio of inertia force to the pressure of a flowing fluid.

2

22

pL

V L

u E

Mach number:

It is the square root of Inertia force to the elastic force





2

22

kL

V L

UNIT V PUMPS AND TURBINES

PART A

1. What is meant by pump?

It is defined as the hydraulic machine which converts the mechanical energy into hydraulic energy, which

is mainly in the form of pressure energy.2. Mention the components of centrifugal pump?

Impeller, suction pipe, strainer & foot valve, delivery pipe and delivery valve.3. Mention the classification of reciprocating pump.

Reciprocating pumps are classified asi. According to the water in contact with one side or both sides of the piston

a. Single Acting b. Double Acting

ii. According to the number of cylindersa. Single cylinder pump

b. Double cylinder pumpc. Triple cylinder pump

4. List the main parts of reciprocating pump.

A cylinder with a piston, piston rod, crack, connecting rod; Suction pipe, delivery pipe, suction valve and

delivery valve.5. What is meant by slip in reciprocating pump?

Slip in reciprocating pump can be defined as the difference of theoretical discharge and actual discharge of

the pump.Slip = (Q) theoretical-(Q) actual

6. Reason out for the negative slip occurs in reciprocating pump.

If the actual discharge is more than the theoretical discharge the slip of the will be negative. It happens if

the momentum of the liquid in the suction pipe is large enough to open the delivery valve before the beginning of the delivery stroke. It is possible in the case of the pumps having a short delivery pipe, long

suction pipe and especially when these are running at at high speeds.

7. Define the coefficient of discharge (Cd) of reciprocating pump.

It is the ratio of actual discharge to the theoretical discharge of the reciprocating pump.Cd= Q actual/Q theoretical

8. Delineate the term “Priming” in the centrifugal pump. The operation of filling the suction pipe, casing of the pump and a portion of delivery pipe completelyfrom the outsource with the liquid to be raised, before starting the pump, to remove any air, gas or vapour

from these parts of the pump is called Priming of a centrifugal pump.

9.

What is meant by indicator diagram in reciprocating pumps?

The indicator diagram of a reciprocating pump is the diagram which shows the pressure head of the liquidin the cylinder corresponding to any position during the suction and delivery strokes. It is the graph

between pressure head and stroke length of the piston for one complete revolution (pressure head is takenas ordinate and stroke length as abscissa).

10. List any two applications of reciprocating pump.

Generally, the reciprocating pumps are best suited for relatively small capacities and high headsapplications. Such asi. Light oil pumping

ii. Feeding small boilers condensate return, and

iii. Pneumatic pressure systems11. List the advantages of centrifugal pump over displacement (reciprocating) pump.

The advantages of centrifugal pump over the reciprocating have been listed below.

i. The cost of centrifugal is lessii. The installation and maintenance are easier and cheaper.

iii. Discharging capacity is much higher than the reciprocating pump.

iv. It can be operate at high speeds without any danger of separation and cavitation.

12. How can express the head of a centrifugal pump?

The head of a centrifugal pump can be expressed in terms of Static head; Manometric head; and Total,gross or effective head.

13. According to the head, how the turbines are classified? Write down an example for each one.According to the head the turbines are classified as follows.

a. Low head turbines Ex: Kaplan Turbine b. Medium head turbines Ex: Francis Turbine





c. High head turbines Ex: Pelton Wheel

14. Define the terms “Gross and Net head”.Gross head: The gross (total) head is the difference between the water level at the reservoir and the water

level at the tail race.Net head: The head available at the inlet of the turbine is known as the net or effective head.

15. List down the advantages of Kaplan turbine over the Francis turbine.

i. For the same power developed Kaplan turbine is more compact in construction

and smaller in size.ii. Part-load efficiency is considerably high.

iii. Low frictional losses (because of smaller number of blades used.)16. Mention the drawbacks of Francis turbine over the tangential turbine.

i. Water hammer effect is more troublesome with the Francis turbine.ii. Cavitation is an ever-present danger.

iii. The overhaul and inspection is much more difficulty comparatively.

17. Delineate the term “Specific Speed” in turbines.The specific speed of a turbine is defined as the speed of a turbine which is identical in shape, geometricaldimensions, blade angles, gate opening, etc. which would develop unit power when working under a unithead.

18.

Write down the mathematical expression for the specific speed of turbine.

Ns = ; Where, Ns = Specific Speed (in r.p.m); P = Shaft Power (Watts); H = Net head on the

turbine (in metre); N = Operating speed of turbine (in r.p.m)

19. A turbine is to operate under a head of 25m at 200 r.p.m with a discharge of 9 m3/s. If the 1986.5

kW power is generated, determine the specific speed of the turbine and identify the type of turbine.

Ns = = = 159.4 r.p.m

The calculated specific speed is lies between the 80 and 400, the turbine is a Francis turbine.

20. List down and define the various efficiencies of turbines.

Hydraulic Efficiency =

Mechanical efficiency =

Overall efficiency =

21. What is the function of spear and nozzle?

The nozzle is used to convert whole hydraulic energy into kinetic energy. Thus, nozzle delivers the jetwith high speed. A spear which is fitted inside the nozzle controls the water flow through the nozzle and at

same time, provides the smooth flow of the water with negligible energy loss.

22. Distinguish among Pelton wheel, Francis and Kaplan turbines with respect to at least three aspects.

Aspect Francis turbine Kaplan turbine Pelton Wheel

Type of tu rbine Radially inward or mixed

ward.

Partially axial

flow

Tangential flow (impulse

turbine)Head Medium ( 60 m to 250 m) Low (up to 60 m) High (above 250 m)

Specific speed 50-250 250-280 12 to 70

F low rate Medium high Low

Part A

1. Explain the working principle and construction of centrifugal pump with suitable sketches.





The main components of a centrifugal pump are:i) Impeller : Impeller is the rotating component of the pump. It is made up of a series of curved vanes. The

impeller is mounted on the shaft connecting an electric motor.ii) Casing : Casing is an air tight chamber surrounding the impeller. The shape of the casing is designed insuch a way that the kinetic energy of the impeller is gradually changed to potential energy. This is chieved by gradually increasing the area of cross section in the direction of flow.

iii) Suction pipe : Suction pipe It is the pipe connecting the pump to the sump, from where the liquid hasto be lifted up.iv) Foot valve with strainer: Foot valve with strainer the foot valve is a non-return valve which permits

the flow of the liquid from the sump towards the pump. In other words the foot valve opens only in theupward direction. The strainer is a mesh surrounding the valve, it prevents the entry of debris and silt intothe pump.v) Delivery pipe: Delivery pipe is a pipe connected to the pump to the overhead tank.

vi)Delivery valve: Delivery valve is a valve which can regulate the flow of liquid from the pump.Working of a centrifugal pump:A centrifugal pump works on the principal that when a certain mass of fluid is rotated by

an external source, it is thrown away from the central axis of rotation and a centrifugalhead is impressed which enables it to rise to a higher level.

2. List the main components of reciprocating pump and explain them sufficiently.





3. A centrifugal pump is to discharge 0.118m3/s at a speed of 1450 rpm against a head of 25m. The

impeller diameter is 250mm. Its width at outlet is 50mm and the manometric efficiency is 75%.

Find the vane angle at outer periphery of the impeller.

Tangential Velocity

Discharge222 f V B DQ

Manometric Efficiency22

uV

gH

w

mmano

Vane Angle

2

2

2

1tan

w

f

V u

V

Ans. Vane angle =59.74o

4. The diameter and width of a centrifugal pump impeller are 300 mm and 60 mm respectively.

The pump is delivering 144 litres of liquid per second with a manometric efficiency of 85

percent. The effective outlet vane angle is 30 . If the speed of rotation is 950 rpm. Calculate the

specific speed of the pump.

Find

2

2

2

1tan

w

f

V u

V

,

60

2

2

N Du

Here 2

1

2

2

22

120

D D

DV N

wmano

5. A double acting reciprocating pump running at 60 rpm is discharging 1.5 m3 of water per

minute. The pump has a stroke length of 400 mm. The diameter of the piston is 250 mm. The

delivery and suction heads are 20m and 5m respectively. Find the power required to drive the

pump and the slip of the pump.

Theoretical Discharge60

2 ALN Qth

Power Required P= 60000

2 d s hh ALN g

P

6. A single acting reciprocating pump has a bore of 200 mm and a stroke of 350 mm and runs at 45

rpm. The suction head is 8 m and the delivery head is 20 m. Determine the theoretical discharge

of water and power required. If slip is 10%, what is the actual flow rate?





Theoretical Discharge60

ALN Qtheo

th

act d

Q

QC

Slip= act theo QQ

%slip= 100

theo

act theo

Q

QQ

7. Explain the working principle of tangential turbine with neat sketch.

8. Explain the working principle of a Axial flow reaction turbines, how they are classified and

differentiate them.Axial flow reaction turbine: This is a reaction turbine in which the water flows parallel to the axisof rotation. The shaft of the turbine may be either vertical or horizontal. The lower end of the shaft

is made larger to form the boss or the hub. A number of vanes are fixed to the boss. When the

vanes are composite with the boss the turbine is called propeller turbine. When the vanes areadjustable the turbine is called a Kaplan turbine.

9. Explain in detail about the working principle of Francis turbine with suitable sketch.





10. Distinguish between the inward and outward flow reaction turbines with respect to at least five

aspects.

11. A Pelton wheel, working under a head of 500 m develops 13 MW when running at a speed of 430

rpm. If the efficiency of the wheel is 85%, determine the rate of flow through the turbine, thediameter of the wheel and the diameter of the nozzle. Take speed ratio as 0.46 and coefficient of

velocity for the nozzle as 0.98.

12. A Pelton wheel works under a gross head of 510 m. One third of gross head is lost in friction in

the penstock. The rate of flow through the nozzle is 2.2 m3/sec. The angel of deflection of jet is

165°. Find the (i) power given by water to the runner (ii) hydraulic efficiency of Pelton wheel.Take CV = 1.0 and speed ratio = 0.45

13. A reaction turbine works at 450 r.p.m. under a head of 120 m. Its diameter at inlet is 1.2 m and

the flow area is 0.4 m2. The angles made by absolute and relative velocities at inlet are 20 and

60 respectively with the tangential velocity. Determine: (i)the volume rate of flow, (ii) the power

developed, and (iii) the hydraulic efficiency.




15. A hub diameter of a Kaplan turbine, working under a head of 12m, is 0.35 times the

diameter of the runner. The turbine is running at 100rpm. If the vane angle of the runner at

outlet is 15deg. And flow ratio 0.6, find (i) diameter of the runner, (ii) diameter of the boss, and

(iii) Discharge through the runner. Take the velocity of whirl at outlet as zero.

16. A Kaplan turbine develops 24647.6kW power at an average head of 39m. Assuming the

speed ratio of 2, flow ratio of 0.6, diameter of the boss equal to 0.35 times the diameter of the

runner and an overall efficiency of 90%, calculate the diameter, speed and specific speed of the

turbine.

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applied thermodynamics and fluids