ME6301- ENGINEERING THERMODYNAMICS

UNIT-I

PART-B

1. During a flow process 5kw paddle wheel work is supplied while the internal energy of the

system increase in one minute is 200kJ.Find the heat transfer when there is no other

form of energy transfer.

Given data:

Work done=5kW (since work is supplied to the system)

Internal energy, ∆u=200kJ/min=200kJ/sec=3.33kJ/sec

To Find:

Heat transfer (Q)

Solution:

By first law of thermodynamics,

Q=w+∆u

=-5+3.33

Q=-1.67kw

2. In an internal combustion engine, during the compression stroke the heat rejected to the

cooling water is 50kj/kg and the work input is 100kJ/kg. Calculate the change in internal

energy of the working fluid starting whether it is a gain or loss.

Given data:

Q=-50kj/kg(heat is rejected)

Work input, W=-100kJ/kg

To Find:

∆E

Solution:

Q=∆E+W

-50=∆E-100

∆E=50kj/kg

Gain in internal energy=50kJ/kg.

3. A Fluid system, contained in a piston and cylinder machine, oases through a complete

cycle of four processes The sum of all heat transferred, during a cycle is -340kJ. The

system completes 200 cycles per minute. Complete the following table showing the

method for each item and complete the net rate of work output in KW.

PROCESS Q(KJ/min) W(KJ/min) ∆E(KJ/min)

1-2 0 4340 -

2-3 42000 0 -

3-4 -4200 - -73200

4-1 - - -

Solution:

Using First law of thermodynamics,

PROCESS 1-2

Q=∆E+w

O=∆E+4340.

∆E=-4340KJ/min

PROCESS 2-3

Q=∆E+w

42000=∆E+0

∆E=42000KJ/min

PROCESS 3-4

Q=∆E+w

-4200=-73200+w

-4200+7200=w

W=69000KJ/min

∑Q=-340KJ

The system completes 200 cycles/min

∑Qcycle×200=Q1-2+Q2-3+Q3-4+Q4-1

-340×200=0+42000-42000+ Q4-1

Q4-1=-105800KJ/min

Since cycle integral of any property is zero

∆E1-2+∆E2-3+∆E3-4+∆E4-1=0

-4340+420004(-732000)+ ∆E4-1=0

∆E4-1=35540KJ/min

Q4-1=∆E4-1+W4-1

-105800=35540+ W4-1

W4-1=-141340KJ/min

Since

∑Q cycle = ∑W cycle

∑Q=42000-4200-105800

∑Qcycle-68000KJ/min

Rate of work output=-68000KJ/min

=-68000/60KJ/sec

= 1133.3Kw

The complete table is given below,

PROCESS Q(KJ/min) W(KJ/min) ∆E(KJ/min)

1-2 0 4340 -4340

2-3 42000 0 42000

3-4 -4200 69000 -73200

4-1 -105800 -141340 35540

4. A Piston and cylinder machine contains a fluid system which process through a complete

cycle of four processes. During a cycle , the sum of all heat transfer is -170KJ. The

system completes 100 cycles/min .complete the following table showing the method for

item and completes the net rate of work output in KW

PROCESS Q (KJ/min) W(KJ/min) ∆E(KJ/min)

a-b 0 2170 -

b-c 21000 0 -

c-d -2100 - -36600

d-a - - -

[Similar as above problem]

5. Mass of 15 kg of air in a piston cylinder device is heated from 25°to90°by passing

current through a resistance heater inside the cylinder .the pressure inside the cylinder

is held constant at 300Kpa during the process and a heat loss of 60KJ occurs.

Determine the electrical energy supplied in kw-hr and changing in internal energy.

Given data:

M=15kg; T1=25°=25+273=298k

T2=90°=90+273=363k

P1=P2=300kpa=300KN/m²

Q=-60Kj

To find:

a) Electrical energy supplied

b) Changing in internal energy

Solution:

Work done, W=mR(T2-T1)

=15×0.287×(363-298)

=279.825KJ

Work done, W in KW-hr=workdone×3600

=279.825×3600

Electrical energy supplied =1007.37×10³Kw-hr

Change in internal energy=∆U=Q-W

=-60-279.825

∆U=-339.825 KJ

6. 5Kg of air at 40 °and 1 bar is heated in a reversible non-flow constant pressure until the

volume is doubled find (a) Change in volume(b)Work done(c) change in internal energy

and (d)Change in enthalpy.

Given data:

M=5Kg

T1=40 °c

P1=1bar=100kN/m²

V2=2V1

P=constant

To find:

a) V2 -V1=?

b) W=?

c) ∆U=?

d) ∆H=?

Solution

From ideal gas equation,

P1 V1=mR T1

V1=5×0.287×313

100

V1=4049m³

The final volume V2= V1

=2×4.49

V2 =8.98 m³

1) Change in volume

V2 -V1=8.98-4.49

=4.49m²

2) Work done transfer

W=P(V2 -V1)

=100(4.49)

=449KJ

3) Change in internal energy

∆U=mCv(T2-T1)

V1/ V2= T2/ T1

T2= T1( V1/ V2)

=313 ( 2 V1/ V2)

=626k

4) Change in enthalpy

∆H= mCp(T2-T1)

=5×1.005(626-313)

=1572.825KJ

7. An ideal gas of molecular weight 30 and specific heat ratio 1.4 is compressed according to the

law pv1.25=C from 1 bar absolute and 27°c to a pressure of 16 bar calculated the temperature

at the end of compression the heat received or rejected work done on the gas during the

process and changes in enthalpy, assume mass of the gas as 1kg

Given data:

Molecular weight (M) =30

Cp/Cv=ɤ=1.4

M=1kg

P1=1bar=100KN/m²

P2=16 bar=1600KN/m²

T1=27°c+273=300k

Pv1.25=C

To find;

T2, Q and w

Solution:

For polytrophic process the P, V and T relation.

T2/T2= (P1/ P2)n-1

T2= T1 (P1/ P2)n-1/n

T2=300(1600/100) 1.25-1/1.25

T2=522.33k

Work done

W=mR (T2-T1)

n-1

Gas constant R=Ru/M

=8.314/30

R=0.277KJ/kgk

W=1×0.277×(300-522.33)

1.25-1

W=-246.34KJ

Heat transfer

Q= W×(r-n/r-1)

=246.34×(1.4-1.25/1.4-1)

Q=-92.378KJ

Change in enthalpy

∆H= mCp(T2-T1)

=1×1.005(522.38-300)

∆H=233.49KJ

8. 50kg/min of air enters the control volume in a steady flow system at 2 bar and 100°c and

at an envelope elevation of 100m above the datum. The same mass leaves the control

volume at 150m elevation with a pressure of 10 bar and temperature of 300°c. The

entrance velocity as 2400m/min and the exit velocity is 1200m/min. During the process

50000kj/hr of heat is transferred to the control volume and the rice in enthalpy is 8kj/kg.

Calculate the power developed.

Given data:

1. M=50kj/kg=50/60=0.83kj/sec

2. P1=2bar=200kN/m2

3. T1=100°c=373k

4. Z1=100m

5. Z2=150m

6. P2=10bar=1000KN/m2

7. T2=300°c=573k

8. C1=2400m/min=2400/60=40m/sec

9. C2=1200m/min=1200/60=20m/sec

10. Q=50000KJ/hr=50000/3600=13.89KJ/sec

11. H2-h1=8KJ/kg

To find:

Power developed P=?

Solution:

SFEE is

Gz1+c12/2+u1+p1v1+Q=gz2+c2

2/2+u2+p2v52+W

W=g (z1-z2) +c12-c2

2/2+ (h1-h2) +Q

W=9.81(100-150) +402- 20

2/2+8×10

3+13.89×10

3

W=5999.5J/kg

W=6Kj/kg

Power developed

P=W×mass

=6.0×0.83

P=4.98KJ/sec

P=4.98KN

9. In a steady flow of air through a No 22 the enthalpy decreases by 50kj between two

sections. Assuming that there are no other energy changes than the kinetic energy

determine the increase in velocity is 90m/s.

Given data:

Enthalpy decrease (h1-h2) = 50kj

=50×103J

Velocity at section 1 C1=90m/sec

To find:

Increases in velocity (C1-C2) =?

Solution:

We know that

Steady flow energy equation for a nozzle is

H1-h2=C22-C1

2/2

Velocity at exit C2=√2(H1-h2)+ C12

=√1×50×103+90

2

C2=328.78m/s

Increase velocity

C1-C2= 328.78-90

=238.78m/s.

10. A room for four persons has two for each consuming 0.18KW power and three lamps.

Ventilation air at the rate of 80kg/hr enters with an enthalpy of 84kj/kg. In each person

put out heat of the rate of 630Kj/hr. Determine the rate at which heat is to be removed

by a room cooler so that a steady state is maintained in the room.

Given data:

Np=4(person); nf=2

Wf=0.18kw

Wl=100w

Mass of air m=80 kg/hr=80/3600=0.022kg/s

h1=84kj/kg

h2=59kj/kg

Qp=630kj/hr

To find:

Rate of heat is to be removed =?

Solution:

E=m (h1-c12/2+gz1+Q)-m (h2+C2

2/2+Z2g+w)

Assuming that,

C12-C2

2/2=0. (z1-z2)g=0.

Q=E-m (h1-h2)-w

E=-npQp

=-4×630/3600

W=-0.7Kw

m (h1-h2)=80/3600(84-59)

=0.55kw

W=electrical energy in put

W=nfwf+newe

=2×0.8+3×100/1000

W=0.66kw

Q= E- m (h1-h2)-w

=-0.7-0.5-0.66

Q=-1.916kw.

UNIT - II

PART-B

1. A heat engine is used to drive a heat pump. The heat transfer from the heat engine and

from the heat pump is used to heat the water circulating through the radiators of

building. The efficiency of the heat engine is 27% and COP of the heat pump is 4. (i)

Draw the heat diagram of the arrangement and (ii) evaluate the ratio of heat transfer to

the circulating water to the heat transfer to the heat engine.

[Oct – 95]

Given data:

H.E = 27%

COPH.E = 4

TO FIND:

= ?

SOLUTION:

H.E = =

0.27 = 1-

= 0.73

QR1 = 0.73 QS1 (1)

COP H.P =

=

4 = (2)

Substituting (1) (2)

4 =

=

QR2 = 1.08 QS1 (3)

Total heat supplied to the water, Q = QR1 + QR2

= 0.73 QS1 + 1.08 QS1

= 1.81 QS1

= 1.81

RESULT:

The ratio of heat transfer to the circulating water to the heat transfer to the heat engine,

= 1.81

2. A Carnot heat engine takes heat from an infinite reservoir at 550°c and rejects it to a

sink a 275°c. Half of the work delivered by the engine is used to run generator and the

other half is used to run heat pump which takes heat at 275°c and rejects it at 440°c.

Express the heat rejected at 440°c by the heat pump as % of heat supplied to the engine at

550°c. If the operation of the generator is 500KW, find the heat rejected per hour by the

heat pump at 440°c.

GIVEN DATA:

T1 = 550°c

T2 = 275°c

T4 = 440°c

Wg = 500 KW

TO FIND:

1. = ?

2. Heat rejected, QR2 =?

SOLUTION:

For Carnot heat engine

=

= QR1

= 1.502 QR1 (1)

Work done by the heat engine, W = QS1 – QR1

= 1.502 QR1 – QR1

= 0.502 QR1

Generator input, Wg =

= 0.251 QR1

Work input to the heat pump,

WHP = 0.251 QR1

Heat rejected by the heat pump,

QR2 = QS2 + WHP

= QS2 + 0.251 QR1 (2)

For reverse heat pump,

QS2 = QR2

= QR2

QS2 = 0.77 QR2

Substituting QS2 (2)

QR2 = 0.77 QR2 + 0.251 QR1

= 0.77 QR2 + 0.251 × QS1

From characteristic gas equation, PV = mRT

V1 =

=

= 2.17 m3

From =

V2 =

=

= 0.274 m3

Similarly, =

T2 = 303

= 458.47 K

= 185.47°c

Change in entropy during compression,

S2 – S1 = mcV Ln

= mcP Ln

S2 – S1 = 5 × 0.718 Ln

= -1.483 KJ/K

Process 2 -3 is a constant volume process

Change in entropy ,S3–S1= mcVln

= 5×0.718ln

= -1.487 KJ/K

RESULT:

1. Final volume at the end of compression, V2 = 0.274 m3

2. The corresponding temperature, T2 = 185.47°c

3. Change of entropy during compression S2 – S1 = 1.483 KJ/K

4. Change of entropy during constant volume, S3 – S2 = 1.487 KJ/K

3. One Kg of Ice at -5°c is exposed to the atmosphere which is at 20°c. The ice melts and

comes into thermal equilibrium with the atmosphere (i) Determine the entropy increase

of the turbine (ii) what is the minimum amount of work necessary to convert the water

back to ice at -5°c ? Assume CP for Ice as 2.093 KJ/Kg K and the latent heat of fusion of

ice as 333.3 KJ/Kg.

Given data:

Ti = -5°c = 273-5 = 268 K

Ta = 20°c = 273 + 20 = 293 K

CPi = 2.093 KJ/Kg K

L = 333.3 KJ/Kg

To Find:

1. Entropy increase of universe (S)univ = ?

2. Minimum amount of work W/min = ?

Solution:

Heat absorbed by air from atmosphere ( ) = Heat absorbed in solid phase + Latent

heat + Head absorbed in liquid phase

= mcpi (To-Ti) + mL + mcpw (Tu-To)

Assuming m =1 Kg and cpw = 4.187 KJ/Kg K

Q = 1 × 2.093 (0-1-5) + 1 × 333.3 + 1 × 4.187 (20-0)

= 427.535 KJ

Entropy change of atm (S) atm = -

= = -1.46 KJ/K

Entropy change of system,

(S) system = (S)ice + (S)fusion + (S)liquid

= +

= 1 × 2.093 ln + +1×4.187 ln

= 1.556 KJ/K

Entropy of universe,

(S)univ = (S)sys + (S)atm

= 1.556-1.46

(S)univ = 0.096

KJ/K

If water is to be converted back to ice using a reversible refrigerator heat to be removed from

water.

Q = 427.535 KJ

Now, (S)sys = -1.556 KJ/K

But (S)atm =

(S)ref = 0 (as the refrigerant operates in a cycle)

(S)univ = (S)sys + (S)ref +(S)atm 0

-1.556 + 0 +

Q + W 445.908 KJ

W 445.908 – 427.535

, so,

Wmin = 28.373 KJ

4. Two reversible heat engines A and B are arranged in series. A rejecting heat directly to

B. engine receive 200 KJ at a temperature of 421°c from a hot source while engine B is in

communication with a cold sink at a temperature of 4.4°c if the work output of A is

twice that of B, find

(i) The intermediate temperature b/w (A) & (B)

(ii) The efficiency of each engine, and

(iii) The heat rejected to the cold sink

Given data:

1. TH = 421°c = 421 + 273 = 694 K

2. TL = 4.4°c = 4.4 +273 = 277.4K

3. QS1 = 200 KJ, WA = WB

To find:

1. The intermediate temperature b/w (A) & (B), T = ?

2. The efficiency of each engine A & B = ?

3. The heat rejected to the cold sink iQR2 = ?

Solution:

Work output from engine A,

WA = QS1 – QR1

= 200- QR1

For reversible heat engine,

Equating equations (4) and (5)

100 – 0.1447 = 0.2887 – QR2

QR2 = 0.4327 – 100 -------------------(5)

Similarly, for reversible engine B,

--------------(1)

o, =

QR1 = 0.2887 -----------(2)

o, WA = 200 – 0.2887

But WB = 100 – 0.1447 -------(3) ( ) ( )

and also WB = QS2 – QR2

= 0.2887 – QR2 -----(4)

so,

T = 416.42 K (or) T = 143.42°c

So, QR1 = 0.288 × 416.42 = 119.93 KJ

and QR2 = 0.432 × 416.42 - 100

QR2 = 79.89 KJ

Efficiency of engine A, A = 1-

=

= 40.04%

Efficiency of engine A, B = 1-

=

= 33.39%

Result:

1. The intermediate temperature between A and B, T = 143.42°c

2. The efficiency of each engine A = 40.04% and B = 33.39%

3. The heat rejected to the cold sink QR2 = 79.89 KJ.

UNIT-III

PART-B

1. A vessel of volume 0.04m3 contains a mixture of saturated water and steam at a

temperature of 250C. The mass of the liquid present is a 9 Kg find the pressure, mass,

specific volume, enthalpy, entropy and Internal energy.

Given data

V=0.04m3

T=250oC

M1=9kg

To find:

P, m, v, h, s and U

Solution:

From steam tables corresponding + 250oC read

Vf = V1 = 0.001251 m3/Kg

Vg = Vs = 0.050037 m3/Kg

P = 39.776 bar

Total volume occupied by the liquid

V1 = m1v1 = 9 x 0.001251

V1 = 0.0113m3

Total volum of the vessel

V = Volume of liquid + volume of steam

V = V1 + Vs

0.04 = 0.0113 + Vs

Vs = 0.0287 m3

Mass of steam, Ms = =0.574 kg

Mass of mixture of liquid and steam, m = m1 + ms

M = 9+0.574=9.574 kg

Total Specific volume of mixture

V=

We know that

V = Vf + xVfg

0.00418 = 0.001251 + x (0.050037 – 0.001251)

X = 0.06

From steam tables corresponding to 250oC,

hf = 1085.8 KJ/Kg hfg=1714.6 KJ/kg

sf = 2.794 KJ/KgK sfg = 3.277 KJ/Kg K.

Enthalpy of mixture,

h = hf + xhfg

= 1085.8 + 0.06 x 1714.6

H = 1188.67 KJ/kg

Entropy of mixture,

S = sf + xsfg

= 2.794 + 0.06 x 3.277

S = 2.99 KJ/Kg K

Internal energy u = h-Pv

= 1188.67 – 39.776 x 102 x 0.00418)

u = 1172 KJ/kg

2. In steam generator compressed water at 10 MPa, 30oC enters a 30mm diameter tube at the

rate of 3 litres1sec. Steam at 9 MPa, 400oC exist the tube. Find the rate of heat transfer.

(November -2003)

Given data

P1 = 10 bar

TW = 30oC

D = 30mm = 0.03m

N1 = 3 litres / sec = 0.003 m3/sec

P2 = 9 bar

T2 = 400oC

To find : Q

Solution:

From steam tables corresponding to 30oC

Vf1 = 0.001004 m3/kg

Hf1 = h1 = 125.7 KJ/Kg

V1 = mvf1

Mass flow rate of steam m =

M = 2.988 Kg/s

Area of the tube A =

A = 7.068x10-4

m2

Inlet Velocity C1

C1

From steam tables corresponding to 9MPa and 400oC

V2 = 0.02993 m3/kg

H2= 3121.2 KJ/kg

Final velocity, C2=

C2 = 126.53 m/s.

From SFEE (steady How energy equation)

Q = 9874.45 KW

Result:

The rate of heat Transfer to the water Q = 8974.45KW

3. Steam at 0.8MPa, 250oC and flowing the rate of 1 Kg / s passes into a pipe carrying wet

steam at 0.8MPa, 0.95 dry, after adiabatic mixing the flow rate is 2.3 kgs /s. Determine the

properties of the steam after mixing.

Given data

P1 = 0.8 MPa = 8 bar

T1 = 250oC

M1 = 1 Kg/s

P2 = 0.8 MPa = 8 bar

X2 = 0.95

M3 = 2.3 kg/s

To find

Properties

Solution

The sum of mass of the steam before mixing = The sum of mass of

the steam after mixing.

M1 + m2 = m3

M2 = m3 – m1 = 2.3 – 1 = 1.3 Kg/sec

The energy balance equation for adiabatic mixing.

M1h1 + m2h2 = m3h3 1

Corresponding to 8 and 250oC

h1 = 2950.4 KJ/Kg

Corresponding to8 bar, read hf and hfg

hf2 = 720.9 KJ/kg. hfg2=2046.5KJ/kg

h2 = hf2+ x2hfq2

= 720.1 + 0.95 x 2046.5

h2 = 2664.27 KJ/Kg

Sub all the values to the equation 1

1x2950.4 +1.3 x 2664.27 = 2.3 x h3

H3 = 2788.67 KJ/kg

Corresponding to 8 bar, read,

hg = 2767.4 KJ/Kg

Since h37hg the steam is in super heated condition from the molier chart, corresponding

to 8 bar and h3 = 2788.67 KJ/kg

Super heated temperature T3 = 180oC

Entropy, S3 = 6.645 KJ/Kg K

Specific volume, V3 = 2.5m3 / Kg

The conditions of steam, after mixing is 0.8 MPa and180oC

Result:

The conditions of steam after mixing is 0.8 MPa and 180oC

Enthalpy, h3 = 2788.67 KJ/kg

Entropy, S3 = 6.645KJ/Kg K

Specific volume V3 = 2.5 m3/kg

4. In a steam power plant operating on an ideal reheat Rankine cycle, the steam enters the

High-pressure turbine at 3 MPa and 400oC after expansion to 0.6MPa. The steam is

reheated to 400oC and then expanded the logo – pressure Turbine to the condenser

pressure of 10kPa. Determine the thermal efficiency of the cycle and the quality of the

steam at the outlet of the pressure turbine

Given data:

P1 = 3 MPa and T1 = 400oC

P2 = 0.6 MPa

P3 = 10 kPa

To find

Reheat x4.

Solution:

From Super heated steam table.

A+ P1 = 30bar and 400oC

H1 = 3232.5 KJ/kg

S1 = 6.925 KJ/Kg K

From saturated steam table.

Sq2 = 6.758 KJ/KgK; Sf2 = 1.931 KJ/Kgk

Sfg2 = 4.827 KJ/Kg K: hfg2 = 2085.1 KJ / Kg

Hf2 = 670.4 KJ/Kg

Since S1>Sg2 the steam is in super heated conditions,

So equating the superheated entropy at 6 bar you will get superheated temperature.

By interpolation superheated temp is192oC from super heated table at 6 bar and 192

oC.

h2 = 2831.41 KJ/kg

From super heated steam table at 6 bar and 400oC

h3 = 3270.6 KJ/kg: S3 = 7.709 KJ/KgK

From Saturated steam table at 0.15 bar.

Hf4 = 191.8 KJ/Kg hfg4=2392.9 KJ/kg

Sf4 = 0.649 KJ/Kgk Sfg4 = 7.502 KJ/KgK

Vf4 = 0.001010m3/Kg

We know that

S3 = S4 = Sf4+ X4 x Sfg4

7.709 = 0.649 + X4 x 7.502

X4=0.941

Dryness fraction of stream at the end of the Turbine.

X4 = 0.941

h4 = hf4 + X4 x hfg4

= 191.8 + 0.941 x 2392.9

h4 = 2443.52 KJ/kg

Enthalpy of steam the end of the Turbine.

h4 = 2443.52 KJ/kg

h5 = hf4 = 191.8 KJ/kg

Pump work

Wp = Vf4 (P1-P4)

= 0.001010(3000-10)

WP = 3.0199 KJ/kg

Efficiency of reheat Rankine Cycle

Reheat =

=

Reheat - 0.47154 = 47.154%

Result

Efficiency of reheat Rankine cycle -47.154% Quality of steam at outlet of L.P. Turbine

X4 = 0.941 = 94.1% dry)

5. A reheat cycle operating between 30 and 0.04 bar has a superheat and reheat temperature

of 450oC. The first expansion takes place till the steam is dry saturated and then reheat is

given. Neglecting feed pump work, determine the ideal cycle efficiency.

Given data:

P1 = 30 bar, P4 = 0.04 bar, T1 = 450oC, T3 = 450

oC, X2 = 1

To find:

Efficiency of the Cycle

Solution

From steam tables at 30 bar and 450oC

H1 = 3344.35 KJ/Kg

S1 = 7.08 KJ/Kgk

A = 0.04 bar

Hf4 = 121.4 KJ/Kg

sf4 = 0.423 KJ/Kgk

hfg4 = 2433.1 KJ/Kg

Sfg4 = 8.053 KJ/Kgk

1-2 – isentropic process

S1 = S2 = 7.087 KJ / KgK

S2 = Sg = dry saturated steam

P2 = Psat at sg

(From steam table) P2 = 2.3 bar

At 2.3 bar

h2 = 2712.6 KJ/Kg

At 2.3 bar and450oC

h3 = 3381.46 KJ/Kg, S3 = 8.3061 KJ/Kgk

3-4 isentropic process,

S3 = S4 = 8.3061 KJ/Kg k

S4 = Sf4 + x4 x sfg4

=0.423 + X4 x 8.053

S4 = 0.423 + X4 x 8.0523

X4 = 0.98

h4 = hf4 + x4 x hfg4

= 121.4 + 0.98 x 2433.1

h4 = 2505.84 KJ/Kg

The cycle efficiency

M =

M =

M = 0.3873 = 38.73%

Result

The cycle efficiency = 38.73%

6. In a generative cycle, the steam pressure at Turbine inlet is 30 bar and the exhaust is at

0.04 bar. The steam is initially saturated. Enough steam is bled off at the optimum pressure

of 3 bar to heat the feed water. Determine the cycle efficiency, neglect pump work.

Given data

P1 = 30 bar

P2 = 3 bar

P3 = 0.04 bar

To find:

cycle

Solution:

From Steam tables

A + 30 bar and dry

h1 = 2802.3 KJ/Kg

S1= 6.184 KJ/Kg K

At 0.04 bar

hf3 = 121.4 KJ/kg

sf3 = 0.423 KJ/Kg K

hfg3 = 2433.1 KJ/kg

Sfg3 = 8.053 KJ/ KgK Vf3 = 0.001004 m3/kg

A+3 bar

Hf2 = 561.4 KJ/Kg ; hfg2 = 2163.3 KJ/kg

Sf2 = 1.672 KJ/Kgk ; Sfg2 = 5.319 KJ/Kg k

Sg2 = 6.991 KJ/Kgk ; Ffb = 0.00107 m3 / kg

1-2 = isentropic process

S1 = S2 = 6.184 KJ/KgK

Since Sg2>S1 : the condition of steam is wet

S1 = Sf2 + X2 x Sfg2

6.184 = 1.672 + X2 x 5.319

X2 = 0.85

h2 = hf2 + x2 x hfg2

= 561.4 + 0.85 x 2163.3

h2 = 2400.205 KJ/Kg K

Similarly process 1-3 insentropic process

S1 = S3 = 6.184 KJ/Kg K

S3 = Sf3 + X3 x Sfg3

6.184 = 0.423 + X3 x 8.053

X3 = 0.72

h3 = hf3 + X3 x hfg3

h3 = 121.4 + 0.72 x 2433.1

h3 = 1873.23 KJ/Kg

h4 = hf3 = 121.4 KJ/Kg

Pump work during 4-5

Wp = (1-m)(h5-h4) = (1-m) Vf3 (P5-P4)

h5-h4 = 0.001004 x (300-4) = 0.297184

h5 = 0.297184 + 121.4 (h4 – hf3)

h5 = 121.7 KJ/Kg

Similarly for pump work during 6-7

h7-h0 = V6(P7-P6) = Vf2 (P1-P2)

= 0.001074 (3000-300)

h7-h6 = 2.8998 (h6 = hf2 = 561.4 KJ/Kg)

h7 = 561.4 + 2.8998

h7 = 564.29 KJ/kg

The amount of steam bled (m)

M =

M = 0.193

Thermal efficiency of the Regenerative cycle

=

=

regenerative = 0.3683 = 36.83%

07. A Steam Power plant uses steam at boiler pressure of 150 bar and temperature 550o

with reheat at 40 bar and 500oC at condenser pressure of 0.1 bar. Find the quality of

steam at Turbine exhaust, cycle efficiency and steam rate. (Apr – 2004)

Given data:

P1 = 150 ba

P2 = 40 bar

P3 = 0.1 bar

T1 = 550oC

T3 = 550oC

To find:

X4 =? : SSC =?

Solution:

Properties of steam from steam table at 150 bar and 550oC

h1 = 3445.2 KJ/kg, S1 = 6.5125 KJ/ Kg K

A + 40 bar and 550oC

h3 = 3558.9 KJ/Kg, S3 = 7.2295 KJ/Kg K

A + 40 bar

Tsat = 250.3oC = 523. K

hf = 1087.4 KJ/Kg : hfg = 1712.9 KJ/Kg K

:

A+0.1bar

:

sf = 0.649 KJ / Kg : Sfg = 7.502 KJ / Kg K

1-2 – isentropic

S1 = S =6.5125 KJ / Kg K

S2 > Sg at 40 bar

Exist of Hp turbine is super heat

Trup – 322oC

h2 = 3047.18 KJ/kg

S3 < Sg at 0.1 bar

Steam is at wet conditions

S4 = S3 = 7.2295 KJ / Kg K

S4 = Sf4 + X4 x Sfg4

X4 =

X4 = 0.877

h4 = hf + x4 x hfg4

h4 = 191.8 + 0.877 x 2392.9

h4 = 2290.37 KJ / Kg K

Cycle efficiency

=

=

= 0.4426 x 100

= 44.26%

Steam state

=

=

= 2.16 Kg / Kw – hr

= 2.16 kg / kw - hr

UNIT – IV

PART-B

1. Derive MAXWELL’S equations:-

The Maxwell’s equation relates entropy to the Three directly measurable properties P, V and T

for pure simple compressible substances.

From first Law of thermodynamics

Q= W+ U

Rearranging the Parameters

Q - U + W

Tds = du + pdv

[ ds = Q/T by second Law of thermodynamics

W = Pdv by first Law of thermodynamics]

du=Tds – Pdv -1

We know that h = u + pv

dh = du + d (pv)

= du +Vdp + pdv -2

Substituting the value du in equation(2),

dh = Tds-Pdv +Vdp+Pdv

dh = Tds + Vdp -3

By Helmotz’s function

a = U-Ts

da = du – d (Ts)

= du – Tds – SdT -4

(By differentiation rule, d(uv) – udv+vd)

Substituting the Value fo du in equation (4),

Da = Tds – Pdv – Tds – sdT

= - Pdv – SdT -5

By Gibbs functions

G = h – Ts

dg = dh – d (Ts)

dg = dh – Tds – SdT -6

Substituting the value of dh in equation (6),

So, dg becomes

dg = Tds _ Vdp – Tds - SdT (dh = Tds + Vd)

dg = Vdp – SdT -7

By inverse exact differential we can write equation (1) as

du = Tds – pdv

-8

Similarly, equation (3) can be written as

dh = Tds + Vdp

-9

Similarly, equation (5) can be written as

-10

Similarly,equation (7) can be written as

dg = Vdp – sdT

-11

These equations 8, 9, 10 and 11 are maxwell’s equation

2. Derive the ENTROPY relations (Tds Equation)

The entropy (s) of a pure substance canbe Expressed as a function of temperature (T) and

pressure (P)

S = f(T, P)

We know that

and

From Maxwell equation, we know that

Substituting in ds equation,

_ 1

Multiplying by T on both Sides of the equation,

_ 2

This is known as the first form of entropy equation (or) the first Tds equation

By considering The entropy of a pure substance as a function of temperature and specific

volume.

i.e. s = f(T,V)

We know that

From Maxwell Equations, We know that

Substituting in Ds Equation,

ds =

Multiplying by T,

Tds = _ 3

This is known as the second form of Entropy equation (or) the Second Tds equation

3. Derive CLAPEYRON equation

Clapeyron equation which involves the relationship between the saturation pressure,

Saturation, Temperature, the enthalpy of evaporation and the Specific volume of the two

phases involved. This equation provides a basis for calculation of properties in a two phase

region. It gives the slope of a curve Separating the two phases in the P-T diagram.

Let, Entropy (s) is a function of Temperature (T) and volume (V)

i.e. S = f (T, V)

ds = _ 5

When the phase is changing from saturated liquid to saturated Vapour temperature

remains constant. So, ds equation reduces to

ds = _ 6

[ Temperature remains constant dT = 0]

From Maxwell equations, We know that,

Substituting in Equation (6)

S = 7

The term is the slope of the saturation curve. Integrating the above equation between

saturated liquid (f) and saturated vapour (g),

_ 8

(Sfg = Sg – Sf, Vfg – Vg – Vf)

From Second Law of Tharmodynomics, We knowthat

ds =

For constant pressure process

dQ = dh

d3 =

Sfg =

Substituting in Equation _ 8

_ 9

This equation is known as clapeyron Euation.

4. Derive JOULE – THOMSON co-efficient

Joule – Thomson Coefficient is defined as the change in temperature with change in pressure

keeping the enthalpy remains constant. It is denoted by

------ 1

Throttling process:-

Throttling process is defined as the fluid expansion through a minute Orifice or slightly

opened Value. During the throttling process, Pressure and velocity are reduced. But there is

no heat transfer and no work done by the system. In this process, enthalpy remains constant.

Joule – Thomson Experiment:-

In this experiment, a stream of gas at a pressure P1 and Temperature T1 is allowed to flow

continuously through a porous plug. The gas comes out from the other side of the porous plug

at a pressure and Temperature.

The whole apparatus is completely insulates. Therefore, no heat transfer takes place

i.e. Q = 0

The system does not exchange work with the surrounding.

So,

W = OF from steady flow energy equation, we know that,

---------- 1

Since There is no considerable change velocity

V1 = V2 and Z1 = Z2 Q = 0 W= 0 V1 = V2 and

Z1 = Z2 are applied in steady flow energy equation so the equation (1) becomes

h1 = h2

Enthalpy at inlet, h1 = Enthalpy at outlet hz

It indicates that the Enthalpy in constant for throttling process

It is assumed that a series of Experiment performed on a real gas keeping the initial

pressure P1 and Temperature T1 constant with various reduced down steam pressures. It is

found that the down steam temperature also changes. The results from these experiments can

be plotted as constant enthalpy curve on T-p plane.

The slope of a constant Enthalpy is known as Joule – Thomson Co-efficient-it it denoted

by

Case (i)

There is always pressure drop in throttling process - & 0P and temperature change are

negative. Therefore is positive.

This throttling process produces cooling effect

Case (ii)

There is always pressure drop in throttling process So p in negative.

When the temperature change in positive u is negative

This throttling process produces heating effect

Case (iii)

When is zero the temperature of the gas remains constant with throttling. The

temperature at which = 0 is called inversion temperature for a given pressure

Inversion Curve:

The maximum point on each curve is called inversion point and the lows of the inversion

point is called inversion curve.

A Generalized equation of the Joule Thomson Co-efficient can be derived by using

change of enthalpy equation.

WKT

dh =

=

=

Dividing by Cp on both sides

_ 3

Differentiating this equation with respect to pressure at Constant enthalpy.

_ 4

From eqn 4, we can determine the Joule – Thomson Co-efficient

Joule Thomson Coefficient for ideal gas:

The Joule Thomson. Coefficient is defined as the change in temperature with change in

pressure, keeping the enthalpy remains constant. It is denoted by

W.K.T

PV = RT

Differentiating the above equation of state with respect to T by keeping pressure p

constant

=0 It implies that the Joule – Thomson Coefficient is Zero.

Constant temperature coefficient, Lets Enthalpy is a function of pressure and temperature i.e.,

h = f (p, T)

---------- 5

For throttling process, enthalpy remains constant.

h = C

dh = 0

Sub-dk value in eqn – 5

---------- 6

Diving by dT.

The property Cp= is known as constant temperature to efficient.

5. A mixture of 2 kg Oxygen and 2 kg Argon is in an insulated piston cylinder arrangement

at 100Kpo. 300k. the piston now compresses the mixture to half its initial volume.

Molecular weight of oxygen is 32 and for organ is 40. Ratio of specific heats for oxygen

is1.39 and for argon is i-667.

Given data:

Moz = 2 kg

MAR = 2 kg

P1 = 100 kpa

T1 = 300k

V 2 = ½ V1

MO2 = 32

Mar = 40

roz= 1.39 +Par = 1.667

Solution

Mars fraction XOz =

PO2 = XO2 x p = 0.56 x100 = 556kpa

Similarly, XAr =

Pa = 0.444 x 100 = 44.4 kpa

From equation of state

Voz = =

= 2.804m3

= 2.81 M

3

Volume after compression

Ratio of Specified heat of mixture

V = XO2 VO2 + XAR VAR

= 0.55 x 1.39 + 0.444 x 1.667 = 1.513

Jnoulot process refers reersibel adiabatic process

P1V1 r = P2 V2r

P2 =

= 285 Kpa

Similarly

Tz =2040.19k

Piston Work W =

= - 1000.23KJ

= -IMJ

(-Ve sign indicates the work input to the piston)

6. 0.45 kg of Co and 1 Kg of air is contained in a vessel of volume 0.4m3 at 15

oC. Air has

23.3% of O2 and 76.7% of N2 by mass. Calculate the partial press are of each constitutent and

total pressure in the vessel Motor masses of Co. Co2 and N2 are 28.32 and 28kg/km

Given Date

Mco = 0.49kg

Mar = 1 kg

V = 0.4m3

T=150C

XO2 = 0.233

Xn2 = 0.767

Mco = 28kg / kmol

MCO2 = 32 kg kmol

MN2 = 28kg / kmol

Solution

Ra =

Xco = 0.6826

R = XcoRco + XaRa

= 0.3174 x

= 0.189 KJ/KgK

PCO = 0.3174 x 197.32 = 62.63 kpa

Pa = 0.6226 x 197.32 = 134.69kpa

WKT,

Air contains 23.3% O2 and 76.7% N2

partial pressure of 02 = 0.233 x 134.69

= 31.39 kpa

Partial pressure of N2 = 0.767 x 134.69

= 130.3kpa

UNIT-V

PART B

1. Air at 20oC, 40% RH is mixed adiabatically with air at 40

oC, 40% RH in the ratio of 1

Kg of the forces with 2kg of the Latter (on dry basic) find the final condition of air

(Nov’03)

Given data:-

Dry bulb temperature td1 = 20oC

Relative humidity Q1 = 40%

Dry bulb temperature td2 = 40oC

Relative humidity Q2 = 40%

Solution:-

KlKT By mass balance

M1 +m2 = m3

M1w1 + m2w2 = m3w3

By energy balance

M1h1 + m2h2 = m3h3

Sub the value of m3 ch eq

m1w1+m2h2 = (m1+m2)w3

m1w1 – m1w3 = m2w3 – m2w2

m1(w1-w3) – m2 (w3-w2)

From Psychrometriy chart w1 = 0.0058 kg/kg of dry

w2 = 0.0187 kg/kg of dry

W3 = 0.0144 kg/kg of air

Similarly

m1h1 + m2h2 = (m1+m2) h3

m1h1 – m1h3 = m2h3 – m2h2

m1(h1-h3) = m2 (h3-h2)

From psychrometry

h1 = 35 KJ/kg

H2 = 90 KJ/Kg

½ =

h3 = 71.67 KJ/kg

2. An air water vapour mixure at 0.1 mpa, 30oC 80% RH has a volume of 50m

3, calculate

the specific humidity, dew point, wet bulb temp, mass of dry air andmass of water

vapour

Given data

P1 = 0.1 mpa

Td =30oC

Q =80%

Va = 50 m3

Solution:-

1. Relative humidity Q = Pv/PS

From steam table. K/e find that fro 30oC dry bulb temperature corresponding pressure is

0.04246 bar.

Ps = 0.0426 bar

Q = PV/PS

0.8 = PV / 0.04246

Vapour Pressure Pv = 0.03397 bar

Specific humidity w = 0.622 PV / Pb-Pv

= 0.622 x

W = 0.02187 Kj/Kg of dry air

From ST – corresponding to vapour Pr. Pv = 0.03397 bar temp is 31.06oC

So the (DBT) Dew point temp tdp = 31.06oC

From gas Law

Pava = maRaT

Ma =

When

Volume Va = 50m3

Gas constant Ra = 0.287 KJ/kg-K

Temperature T = 30+273=303K

Pb = Pa + Pv

Pa = Pb – Pv

= 1-0.03397

Pa = 0.966 bar

Pa = 0.966 x 100 kpa

= 96.6 kpa (1 bar = 100 kpa)

Ma =

Specific humidity

W=

0.02187 =

Mv = 1.215 kg

From psychometry chart, corresponding to dry bulb temperature 30oC and relative humidity

80% the wet bulb temp is 270C

Tw = 27oC

3. Air at 16oC and 25% relative humidity passes through a heater and then through a

humidify to reach final dry bulb temperature of 30oC and 50% relative humidity.

Calculate the heat and moisture added to the air what is the sensible heat factor?

Given:

Td1 = 16oC

Q1 = 25%

Td2 = 30oC

Q2 = 50%

Solution

Step 1

The dry of ai at 16oC- dry bulb temp and25% (RH) is marked on the psychometric chart

at point 1.

The dry of air 30oC dry bulb temp and 50% (RH) is marked on the psychometric chart at

point 2.

Step 2

Draw ahorizontal line from point 1 and draw a vertical line. From paint 2. There named

as point 3.

Step 3

Draw a inclined line from pant 1 to 2 read enthalpies and specific humidty values at point

1, 2 and 3 from psychonotric chart

At point 1 enthalpy h1 = 24.5 KJ/kg

Specific humidity w1 = 0.0025 kg/kg of dry air

At point 2, enthalpy h2 = 67.5 KJ/kg

Specific humidity w2 = 0.014 kg/kg of dry air

At point3, enthalpyh3 = 38KJ/kg

Heat added Q = h2-h1 = 67.5 – 24.5

= 43 KJ/kg

Moisture added w = w2 – w1 = 0.014-0.0025

= 0.0115 kg / kg of dry air

Sensible heat SH =- h3-h1 = 38-24.5

= 13.5 KJ/Kg

Latnet heat LH = h2-h3 =67.5 – 38 = 29.5 KJ/Kg

SHF =

=

= 0.314

4. Saturated air at 20oC at s rate of 1.16 m3/sec is mixed adiabatically with the outside at

ari 35oC and 50% relative humidity at a rate of 0.5 m

3/sec. Assuming a adiabatic mixing

condition at 1ctm, determine specific humidity, relative humidity, dry bulb temperature

and volume flow rate of the mixture.

Given:-

First Stream of air

DBT td1 = 20oC

Flow ratev1 = 1.167 m3/s

Second stream of ai:

DBT + d2 = 35oC

RH Q2 = 50%

Flow rate V2 = 0.5 m3/S

Solution:-

Step 1 :

The first steam of air i.e. 20oC dry bulb temp upto the saturation curve I marked on the

psychometric short at point 1.

Step 2

The second stream of air i.e. 35oC, dBT & 50o of (RH) is marked in the psychometric

chart at point 2.

Step 3

Joint the points 1 and 2 from the psychrometric chart WKT

Specific humidity of the first stream of air

W1 = 0.015 kg / kg of dry air.

Speciify humidity of the second stream of air

W2 = 0.01875 kg/kg of dry air

Step 4:

WKT

First stream flow rate v1 = 0.3 m

3/s (given)

Mass m1 =

From psychrometric chart WKT (Specific volun) (vs) passing through point 1is 0.863.

V1 = 0.863 m3 / kg

M1 =

Second stream flwo rate v2 = 0.5m3/3

From psychrometric chart

MKT (S.V) (V)

Passing through point 2 is 0.916

V2 = 0.916 m3/kg

M2 =

Sub m1 & m2 Value

0.0203 – 0.35 W3 = 0.546 w3 – 0.01024

1.896 w3 = 0.03054

W3 = 0.0161 kg/kg of dry air

Specific humidity after mixing w3 = 0.0161 kg/kg of dry air

Step 5

From psychrometric chart at point 3 is 0.878 m3/kg

V3 = 0.878 m

3/kg

Q = 82%

Td3 = 23.5oC