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L.VIJAYAKUMAR /A.P-MECH

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI-621213.

Department: Mechanical Semester: III

Subject Code: ME2202 Subject Name: ENGG. THERMODYNAMICS

UNIT - 1

1. 1 kg of gas at 1.1 bar, 27oC is compressed to 6.6 bar as per the law pv1.3 = const.

Calculate w ork and heat transfer, if

(1) When the gas is ethane (C2H6) w ith molar mass of 30kg/k mol and cpof2.1 kJ/kg K.

(2) When the gas is argon (Ar) w ith molar mass of 40kg/k mol and cpof 0.52 kJ/kg K.

(AUC DEC ’05)

Given: Polytropic process

Mass (m) = 1kg

Pressure (P1) = 1.1bar

=1.1×100

=110KN/m2

Temperature (T1) = 27°C

=27+273

=300K

Pressure (P2) = 6.6bar

=6.6×100

=660KN/m2

Molar mass of ethane (m) =30Kg/K mol

Cp of ethane = 2.1KJ/K mol

Molar mass of argon = 40Kg/K mol

Cp of argon = 0.52KJ/Kg.K

PV1.3 = C


To Find:

1. Work Transfer (W)

2. Heat Transfer (Q)

Solution:

Step:1

Case:1 (Ethane)

Work transfer (W) =

Where,

P1V1 = mRT1

V1=

V1 =

V1 = 680.2363m3

Where,

P1V1n = P2V2

V2 =

=

V2 = 802.2438m3

Work transfer (W) =

=

W = -1515516.892KJ

Heat transfer (Q) = W ×

= -1515516.892

Q = -378879.223KJ


Case:2 (Argon)

Work transfer (W) =

Where,

P1V1 = mRT1

V1=

V1 =

V1 = 906.9818m3

Where,

P1V1n = P2V2

V2 =

=

V2 = 1166.0760m3

Work transfer (W) =

=

W = -2232807.207KJ

Heat transfer (Q) = W ×

= -2232807.207

Q = 558201.8018KJ


2. In an air compressor, air f low s steadily at the rate of 0.5 kg/sec. At entry to the compressor,

air has a pressure of 105kPa and specif ic volume of 0.86 m3/kg and at exit of the

compressor those corresponding values are 705 kPa and 0.16 m3/kg. Neglect Kinetic and

Potential energy change. The internal energy of air leaking the compressor is 95 kJ/kg

greater than that of air entering. The cooling w ater in the compressor absorbs 60kJ/sec. of

heat from the air. Find pow er required to derive the compressor. (AUC MAY ’06)

Given:

Mass (m) = 0.5 kg/s

Entering velocity (C1) = 7 m/s

Entering Pressure (P1) = 100 KPa = 100 KN/m2

Entering volume (V1) = 0.95 m3/Kg

Leaving velocity (C2) = 5 m/s

Leaving Pressure (P2) = 700 KPa = 700 KN/m2

Leaving volume (V2) = 0.19 m3/Kg

Change in internal energy (U2- U1) = 90 KJ/Kg

Heat absorbs (Q) = 58 KW

To Find:

1. Compute the rate of shaft w ork input to the air in KW (or)

Work input

2. Ratio of inlet pipe to outlet pipe ( )

Solution:

Step:1

m (P1V1 + +Z1.g) + Q = m ((U2- U1)+P2V2 + +Z2.g) + W

Assume Z1= Z2

0.5 ((100×0.95)+ ) + 58 = 0.5 (90+(700×0.19) + ) + W

47.51225+58 = 111.50625 + W


W = 105.51225-111.50625

W = 5.994KW

Note;

“-“ sign indicates that the w ork is done on the system

Step:2 To f ind ratio of inlet to outlet dia ( )

From continuity equation

=

= =

=3.57

= 3.57

= 3.57

= √3.57

The ratio of inlet to outlet dia of the pipe( ) = 1.8894


3. In an isentropic f low through nozzle, air f low s at the rate of 600 kg/hr. At inlet to

the nozzle, Pressure is 2 MPa and temperature is 127oC. The exit pressure is 0.5 MPa. Initial air

velocity is 300 m/s determines (i) Exit velocity of air (ii) Inlet and exit area of nozzle.

(AUC DEC ’06)

Given:

Mass of f low rate (m1) = 600Kg/hr

=600÷3600

=0.1666Kg/sec

Inlet pressure (P1) = 2MPa×10^6 N/m2

Inlet temp (T1) = 127°C + 273

=400K

Outlet pressure (P2) = 0.5MPa×10^6 N/ m2

Inlet air velocity (C1) = 300m/s

To Find:

1.Exit velocity of air (C2)

2. inlet and exit area (d1,d2)

Solution:

Step:1

= ( )^

=

= 1.4859

T2= 1.4859× T1

T2= 1.4859×400


T2= 594.3977 k

Step:2

C2 = √{2×m[Cp(T2-T1) + ]

C2 = √{2×0.1666[1.005(594.3977-400) + ]

C2 =282.4937m/s

Step:3

Inlet mass f low rate (m) =

Where,

P1V1 = mRT1

V1=

V1=

V1 = 0.0574 m3/Kg

M1 =

A1 =

A1 =

A1 =3.1876×10^-5

A1 = ×d1^2

3.1876×10^-5 = ×d1^2

d1 = √

d1 = 0.0063m×1000


d1 = 6.3706mm

Step:4

=

Where,

= ( )^

= ( )^

= 2.692

V2 = 2.692 V1

V2 = 2.692×0.0574

V2 = 0.155m3/Kg

=

A2 =

A2 =

A2 = 3.1410×10^-5

×d2^2 = 3.1410×10^-5

d2^2 =

D2 = 0.01078m ×1000

D2 = 10.7883mm


4. A centrifugal pump delivers 2750 kg of w ater per minute from initial pressure of 0.8 bar

absolute to a f inal pressure of 2.8 bar absolute. The suction is 2 m below and the delivery is5 m

above the centre of pump. If the suction and delivery pipes are of 15 cm and 10 cm diameter

respectively, make calculation for pow er required to run the pump. (AUC DEC ’06)

Given:

Mass (m) = 2750 Kg/min = 2750÷60

= 45.8333Kg/s

Initial pressure (P1) = 0.8bar ×100

= 80KN/m2

Final pressure (P2) = 2.8bar ×100

= 280 N/m2

Z1 = -2m (below the centre of pump)

Z2 = 5m (above the centre of pump)

Dia , d1 = 15cm÷100

=0.15m

d2 = 10cm÷100

=0.1m

To Find:

Pow er (P) or Work (W)

Solution:

Step:1

The steady f low energy equation is,

m [ + +P1V1] = m [ + + P2V2] + W


W = m [ + +(P1V1- P2V2)]

Where,

M = ×d12×C1×ρ

C1 =

=

C1 = 2.5936m/s

C2 =

=

C2 = 5.8356m/s

Step:2

W = 45.83 [ + +(80×1000 -280×1000)]

W =[-0.06867-0.01366-200×10^3]

W = 9166003.773KJ/Kg

W =

W = 91.6600KJ/Kg


6. A blow er handles 1 kg/sec of air at 293 K and consumes a pow er of 15kw .Theinletandoutletvelocitiesofairarel00 m/sec and 150 m/sec respectively. Find the exit air

temperature, assuming adiabatic conditions. Take Cp of air as 1.005 kJ/kg-K. (AUC DEC ’07)

Given:

Mass(m) = 1Kg/s

Temp (T1) = 293K

Pow er (P) or Work (W) = 15 KW

Inlet velocity (C1) = 100m/s

Outlet velocity (C2) = 150m/s

Cp = 1.005KJ/Kg.k

To Find:

Exit air temp (T2)

Solution:

Step:1

m (h1 + +Z1.g) + Q = m (h2 + +Z2.g) + W

Note;

Neglect datum head (Z1) = (Z2) = 0

Q = 0 (Adiapatic process

m (h1 + ) = m (h2 + ) + W

1 (h1 + ) = 1 (h2 + ) + 15

(h1 – h2) =[ ] + 15

(h1 – h2) =21.25

Cp(T1 – T2) = 21.25


1.005(293 – T2) = 21.25

293 – T2 =

293 – T2= 21.1442

293 –21.1442 =T2

T2 = 271.8557K

7. A room for four persons has tw o fans, each consuming 0.18 kW pow er and three 100W

lamps. Ventilation ant at the rate of 0.0222 kg/sec enters w ith an enthalpy of 84 kJ/kg and

leaves w ith an enthalpy of 59 kJ/kg. If each person puts out heat at the rate of0.175 kJ/sec,

determine the rate at w hich heat is to be removed by a room cooler, so that a steady state is

maintained in the room. (AUC DEC ’07)

Given:

No of person(np) = 4

No of fan (nf) = 2

(Wf) = 0.18KW (each)

(Wl) = 0.1KW (each)

Mass of air (m) = 80Kg/hr

=80÷3600

=0.022Kg/s

Enthalpy of air entering (h1) = 84KJ/Kg

Enthalpy of air leaving (h2) = 59KJ/Kg

Heat (Qp) = 630KJ/hr

=630÷3600

=0.175KJ/s

To Find:

Rate of heat is to be removed

Solution:

Step:1

Rate of energy increase = Rate of energy in f low - Rate of energy out f low


Q1 = -Ƞp Qp

Q1 = -(4×0.175)

Q1 = - 0.7KJ/s or KW

Step:2

=M(h1- h2)

=0.022(84- 59)

=0.55KJ/s

Step:3

W = electrical energy input

W = Ƞf Wf + Ƞl Wl

W = (2×0.18)+( 3×0.1)

W = 0.66KW

Step:4

Rate of heat is to be removed (Q)

Q = Q1 – 0.55 – W

Q = - 0.7 – 0.55 – 0.66

Q = - 1.916KW


8. One liter of hydrogen at 273 K is adiabatically compressed to one half of its initial volume. in

the change in temperature of the gas, if the ratio of tw o specif ic heats for hydrogen is 1.4.

(AUC DEC ’07)

Given: adiabatic process

Initial volume (V1) = 1lit = 1÷1000

= 0.001m3

Temp (T1) = 273K

Initial volume (V2) = one half of it’s Initial volume

(V2) = 0.001÷2 = 0.0005 m3

To Find:

Change in temp of gas (T2 – T1)

Solution:

Step:1

= ( )^( )

= ( )^( )

= 1.3195

T2 = 1.3195×T1

T2 = 1.3195×273

T2 = 360.2256K

Step:2 To f ind change in temp (T2 – T1)


= T2 – T1

= 360.2256 – 273

T2 – T1= 87.2256K

9. The velocity and enthalpy of f luid at the inlet of a certain nozzle are50 m/sec and2800

kJ/kg respectively. The enthalpy at the exit of Nozzle is 2600 kJ/kg. The nozzle is

horizontal and insulated so that no heat transfer takes place from it' Find

(1) Velocity of the f luid at exit of the nozzle

(2) Mass f low rate, if the area at inlet of nozzle is 0.09 m2

(3) Exit area of the nozzle, if the specif ic volume at the exit of the Nozzle is 0.495 m3/kg.

(AUC DEC ’07)

Given:

Inlet velocity of nozzle (C1) = 50m/s

Inlet enthalpy of nozzle (h1) = 2800KJ/Kg

h1 = 2800×10^3J/Kg

no of heat transfer take place (Q) = 0

exit enthalpy (h2) = 2600KJ/Kg

h2 = 2600×10^3J/Kg

To Find:

1. C2

2. m

3. A2

Solution:

Step:1 To f ind velocity of the f luid at exit of the nozzle (C2)

C2 =√[2 (h1 - h2) + C1^2]

C2 =√[2 (2800 - 2600)×1000 + 50^2]


C2 = 634.4288m/s

Step:2 Tofind mass flow rate (m)

m =

Where,

V1 = Specif ic volume is not given so it is assumed to be

V1 = 1m3/Kg

m = =

m = 4.5Kg/s

Step:3 To f ind exit area of nozzle (A2)

=

A2 =

=

A2 = 0.0035m2


10. A w ork done by substance in a reversible non-flow manner is in accordance w ith V = (15/p)

m3, w here p is in bar. Evaluate the w ork done on or by the system as pressure increases from

10 to 100 bar. Indicate w hether it is a compression process or expansion process. If the change

in internal energy is 500kJ, calculate the direction and magnitude of heat transfer. (AUC MAY

’08)

Given: reversible non-flow manner

V = (15/p) m3

Pressure (P1) = 10bar

Pressure (P2) = 100bar

Change in internal energy (Δu) = 500KJ

To Find:

The direction and magnitude of heat transfer (Q)

Solution:

Step:1

Heat transfer (Q) = W+ Δu

Where,


Work done = ʃ V dp

= ʃ 15/p dp

=15 ʃ 1/p dp

= 15 [log p ]

= 15 [log 100- log 10]

= 15 [2-1]

=15 [1]

W = 15KJ

Direction:

W = 15

Positive sign indicates the expansion process

Step:2

By first law of thermodynemics,

Q = W+ Δu

Q = 15+ 500

Q = 515KJ

Q = 515 KJ

Positive sign indicates the expansion process


11. In a Gas turbine installation, the gases enter the turbine at the rate of 5 kg/sec w ith a

velocity of 500 m/sec and enthalpy of 900 kJ/kg and leave the turbine w ith 150 m/sec, and

enthalpy of 400 kJ/kg. The loss of heat from the gases to the surroundings is 25kJ/kg. Assume

R = 0.285 kJ/kg.K, Cp = 1.004 kJ/kg.K and inlet conditions to be at 100 Kpa and

27oC.Determine the diameter of the inlet pipe. (AUC MAY ’08)

Given:

Mass flow rate (m) = 5Kg/s

Inlet velocity (C1) = 50m/s

Inlet enthalpy (h1) = 900KJ/Kg

Outlet enthalpy (h1) =400KJ/Kg

Outlet velocity (C1) = 150m/s

Heat loss (Q) = -25KJ/Kg

R = 0.285KJ/Kg

Cp = 1.004KJ/Kg

Inlet pressure (P1) = 100KPa ×1000 = 100000KN/m2

Inlet temperature (T1) = 27°C + 273 = 300K


To Find:

Work done (W)

Dia of the inlet pipe (d1)

Solution:

Step: 1

m (h1 + +Z1.g) + Q = m (h2 + +Z2.g) + W

Where,

Z2 = Z1

m (h1 + ) + Q = m (h2 + ) + W

5 (900 + ) - 25 = 5 (400 + ) + W

4481.25 = 2056.25 + W

4481.25 – 2056.25 = W

W = 2425KW

Step: 2 To find dia of the inlet pipe (d1)

m =

Where,

V1 =?

P1V1 = mRT1

V1 =

V1 =

V1 = 4.2875 m2/s


m =

A1 =

A1 =

A1 = 0.4275 m2

A1 = ×d12

d12

=

d1 = √

d1 = √

0.7377m = d1

d1 = 0.7377m ×1000

d1 = 737.7736mm


12. A frictionless piston-cylinder device contains 2 kg of nitrogen at 100 Kpa and 300 K.

Nitrogen is now compressed slow ly according to the relation PV 1.4 = C until it reaches a f inal

temperature of 360 K. Calculate the w ork input during this process. (AUC DEC ’09)

Given: Polytropic process

Molecular weight of nitrogen (m) = 28

Ɣ = 1.4

Inlet pressure (P1) = 100KN/m2

Inlet temperature (T1) = 300K

Outlet temperature (T2) = 360 K

PV1.4 = C

To Find:

Work input (W)

Solution:

Step: 1


W =

Where,

Gas constant (R) =

= = 0.297KJ/ Kg.K

W =

W =

W = -89.1KJ

Direction:

W = -89.1KJ

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

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