mahalakshmimahalakshmiengineeringcollege.com/pdf/mec/iiisem... · 12 kg, calculate the heat added...

L.VIJAYAKUMAR/A.P-MECH

MAHALAKSHMI

ENGINEERING COLLEGE

TIRUCHIRAPALLI-621213.

Department: Mechanical Semester: III

Subject Code: ME2202 Subject Name: ENGG. THERMODYNAMICS

UNIT III

1) A rigid tank of 0.03 m3 capacity contains wet vapour at 80 kPa. If the wet vapour mass is

12 kg, calculate the heat added and the quality of the mixture when the pressure inside

the tank reaches 7 MPa. (AUC DEC’05)

Given:

Rigid tank

Volume (V1) = V2 = 0.03m3 [rigid tank]

Pressure (P1) = 80 Kpa = 80 / 100 KN/ m2 => 0.8 bar

Mass (m) = 12 Kg

Pressure (P2) = 7Mpa × 1000 = 7000/100 KN /m2

= 70 bar

To find:

a) Heat added (Q)

b) Quality of mixture ( X1 & X2) (or) dryness fraction

Solution:

Step: 1

Specific volume (V) = V1 / m =0.03/ 12 =0.0025 m3 /Kg

Step:2

From steam tables, corresponding to (P1 =0.8 bar)

Read Vf, Vg, hf , hf g [pressure table pg: no: 9]

Vf = 0.001039 m3/kg hf = 391.7 KJ /kg

Vg = 2.0869 m3 / kg hf g = 2274.21 KJ /kg


Step:3

Specific volume (V) = Vf + x1 Vhf g

0.0025 = 0.001039 + x1 [ 2.0869 – 0.001039]

0.0025 = 0.001039 + x1 (2.0858)

0.0025 - 0.001039 = 2.0858 x1

0.001461 = 2.0858 x1

0.001461/ 2.0858 = x1

x1 = 0.0007

Step :4

for rigid vessel (V1 =V2)

V2 = 0.0025 m3 / kg

From steam tables , corresponding to (P2 = 70 bar) [pg:no:13]

Vf = 0.001351 m3/kg hf = 1267.4 KJ /kg

Vg = 0.027368 m3 / kg hf g = 1506.0 KJ /kg

Vg > V2

0.027368 > 0.0025

The steam is in wet condition

(V2) = Vf 2 + (x2 . Vg2)

0.0025 = 0.001351 + (x2) . [0.027368 – 0.001351]

0.0025 - 0.001351 = (x2) . (0.026017)

0.001149 = x2 . (0.026017)

0.001149 / 0.026017 = x2

x2 = 0.044

(h2) = hf 2 + (x2 . hf g2)

(h2) = 1267 .4 + [0.044 × 1506.0]

(h2) = 1333.9101 KJ/ kg


Step:5

By first law of thermodynamics

Q = W + ( U)

Where

W =0 [for rigid vessel]

Q = U

Total heat transfer (Q) = m [h2 – h1] – V [P2 – P1]

(Q) = 12 [1333.9101 – 393.2918] – 0.0025 [7000 – 80]

(Q) = 11287.4196 – 17.3

Q = 11270.1196 KJ


2) Steam enters the turbine at 3 MPa and 400oC and is condensed at 10 kPa. Some

quantity of steam leaves the turbines at 0.6 MPa and enters open feed water heater.

Compute the fraction of the steam extracted per kg of steam and cycle thermal

efficiency. (AUC DEC’05)

Given:

Regenerative rankine cycle

Pressure (P1) = 3Mpa × 1000 KN / m2 = 30 bar

100

Temp (T1) = 400oC

Pressure (P2) = 0.6 Mpa × 1000 KN / m2 = 6 bar

100

Pressure (P3) = 10 Kpa = KN / m2 = 0.1 bar

100

To find:

Thermal efficiency (ή)

Solution :

Step:1

From superheated steam table

At P1 = 30 bar and T1 = 4000C

S1 = 6.925 KJ /Kg . K [pg.no :22]

H1 = 3232 . 5 KJ /Kg . K [pg.no :19]

S1 = S2 => 6.925 KJ /Kg . K

But at P2= bar => pressure table

Sg = 6.758 KJ /Kg . K [pg.no :11]

S1 > S2

6.925 > 6.758 so the steam is again in superheated state


From mollier diagram, corresponding to S2= 6.925 KJ /Kg . K entropy at 6 bar the enthalpy is

found

H2 = 2630 KJ / Kg

Step:2

S1 = S3 => 6.925 KJ/kg.K

At P3 = 0.1 bar [pressure table pg.no:07]

Vf3 = 0.001010 m3 /kg hf3 = 191.8 KJ/kg

Vg3 = 14.675 m3 /kg hfg3 = 2392.9 KJ/kg

Sf3 = 0.649 KJ/kg.K Sfg3 = 7.502 KJ/kg.K

S1 = S3 => Sf3 + (x3 × Sfg3)

6.925 = 0.649 + (x3 × 7.502)

6.925 - 0.649 = 7.502 x3

6.925 - 0.649 /7.502 = x3

x3 = 0.8365

h3 = hf3 + (x3 × hfg3)

h3 = 191.8 + (0.8365 × 2392.9)

h3 = 2193.6448 KJ/kg

h4 = hf3 => 191.8 KJ/Kg

h5 - h4 = Vf3 [P2 - P3]

h5 - h4 = 0.001010 [600 – 10]

h5 - h4 = 0.5959 KJ/Kg

h5= 0.5959 + h4

h5= 0.5959 + 191.8

h5= 192.3959 KJ/Kg


Step:4

Amount of steam bleed

M = h6 – h5 / h2 – h5

h6 = hf = 670.4 KJ /Kg => [P2 =6 bar use pg.no:11]

M = 670.4 – 192.4 / 2630 – 192.4

M = 0.1960 KJ / Kg of steam

Step:5

At 6 bar

Vf = 0.001101 m3 / kg

Wp 6-7 = h7 – h6 => Vf 2 [P1 – P2]

= 0.001101 [3000 – 600]

h7 – h6 = 2.6424 KJ /Kg

h7 = 2.6424 + h6

h7 = 2.6424 + 670.4

h7 = 673.0424 KJ /Kg

Step:6

Regenerative rankine cycle efficiency

ήreg = (h1- h7) – (l – m) (h3- hf 3)

(h1- h7)

= (3232.5 – 673.0424) – (1 – 0.1960) (2193.6448 – 191.8)

3232.5 – 673.0424

= 2559.4576 – 1609.4832

2559.4576

= 0.3711 × 100

ήreg = 37.11%


3) 1 kg of steam initially dry saturated at 1.1 MPa expands in a cylinder following the law

PV1.13 = C. The pressure at the end of expansion is 0.1MPa. Determine

(i) The final volume

(ii) Final dryness fraction

(iii) Work done

(iv) The change in internal energy

(v) The heat transferred. (AUC DEC’06)

Given:

Poly tropic process

Mass (m) = 1Kg

Pressure (P1) =1.1 Mpa × 1000 => 1100 / 100 KN/m2 = 11bar

“ (P2) =0.1 Mpa × 1000 => 100 / 100 KN/m2 = 1 bar

PV1.13

= C

To find:

(i) The final volume (V2)

(ii) Final dryness fraction (x2)

(iii) Work done (W)

(iv) The change in internal energy ( U)

(v) The heat transferred (Q)

Solution:

Step:1

From steam table at (P1 = 11 bar)

V1 = Vg1 = 0.17739 m3 / kg

H1 = hg1 =2779.7 KJ / Kg = [Pg.no:11]

Step:2

Polytropic process

V2 = [P1 /P2] 1/n

× V1

V2 = [11/1] 1/1.13 x 0.17739

Final volume (V2) = 1.4808 m3


Step:3

From steam table at (P2 = 1 bar)

Hf 2 = 417.5 KJ/Kg Vf 2= 0.001043 m3 / kg

Hf g2 = 2257.9 KJ/Kg Vg2 = 1.6938 m3 / kg

Vfg2 = Vg2 - Vf 2 => 1.6938 – 0.001043

Vfg2 = 1.6927 m3 / kg

V2= Vf2 + x2 . Vfg2

1.4808 = 0.001043 + X2 . (1.6927)

1.4808 - 0.001043 = 1.6927 X2

1.4808 - 0.001043 / 1.6927 = X2

Final dryness fraction (X2) = 0.8742

Step:4

Work done (W) = P1 V1 – P2 V2

n-1

= (1100 x 0.17739) –(100 x 1.4808)

1.13 – 1

(W) = 361.9153 KJ

Step:5

Change in internal energy ( U) = u1 – u2

( U) = (h1 – h2) – (P1 V1 – P2 V2 )

Where

h2 = hf2 + x2 . hfg2

h2 = 417.5 + (0.8742 x 2257.9)

h2 = 2391.3561 KJ/Kg

( U) = (h1 – h2) – (P1 V1 – P2 V2 )

( U) = (2779.7 – 2391.3561) –(1100 x 0.17739 -100 x 1.4808)

( U) = 388.3438 – 47.049

( U) = 341.2948 KJ


Step:6


Heat transfer (Q) = W + U

(Q) = 361.9153 + 341.2948

(Q) = 703.2101 KJ


4) Steam at a pressure of 2.5 MPa and 500oC is expanded in a steam turbine to a

Condenser pressure of 0.05 MPa. Determine for Rankine cycle:

(i) The thermal efficiency of Rankine cycle

(ii) Specific steam consumption.

If the steam pressure is reduced to 1 MPa and the temperature is kept same 500oC.Determine the thermal efficiency and the specific steam consumption. Neglect feed pump work. (AUC DEC’06)

Given:

Rankine cycle

P1 = 2.5Mpa x 1000 = 2500/100 KN / m2

T1 = Tsup = 5000C

P2 = 0.05 Mpa x 1000 = 50 /100 KN/m2

To find:

Case:1



Case:2



Solution:

case:1

Step:1

From super heated steam table at [P1 = 2.5 bar , T1= 5000C] (pg.no:22)

H1 = 3462.9 + 3460.6 / 2

H1 = 3461.75 KJ/Kg

S1 = 7.344 + 7.305 / 2

S1 = 7.3245 KJ/Kg.K

From steam table at (P2 = 0.5 bar)

Vf2 = 0.001030 m3/ Kg hf2 = 340.6 KJ/Kg Sf2 = 1.091 KJ/Kg.k

Vg2 = 3.2401 m3/ Kg hfg2 = 2305.4 KJ/Kg Sfg2 = 6.504 KJ/Kg.k

S1 = S2 = Sf2 + X2 Sfg2


7.3245 = 1.091 + (6.504 x X2)

7.3245 – 1.091 / 6.504 = X2

X2 = 0.9584

H2 = hf2 + ( X2 hfg2)

H2 =340.6 + (0.9584 x 2305.4)

H2 = 2550.1188 KJ/Kg

Heat supplied (Qs) = h1 – h2

(Qs) = 3461.75 – 340.6

(Qs) = 3121.15 KJ/Kg

Work done (W) = h1 – h2

(W) =3461.75 – 2550.1188

(W) = 911.6312 KJ

Thermal efficiency rankine = W/Qs = 911.6312 / 3121.15

ήth = 0.2920 x 100

ήth = 29.20 %

Specific steam consumption

= 3600 / W =3600 / 911.6312

= 3.9489 Kg/Kw .hr

Case:2

Step:2

Condition:

The steam pressure is reduced to

P1 = 1Mpa x 1000 =1000/100 KN /m2 = 10 bar

T1 = 5000C

From super heated steam table at P1 = 10bar and T1 = 5000C

H1 = 3478.3 KJ/Kg [pg.no:19]

S1 = 7.763 KJ/Kg.K [pg.no:22]


From steam table P1 = 0.5bar

Vf2 = 0.001030 m3 /kg hf2 = 340.6 KJ/kg Sf2 = 1.091 KJ/Kg.K

Vg2 = 3.2401 m3 /kg hfg2 = 2305.4 KJ/kg Sfg2 = 6.504 KJ/Kg.K

Sg2 = 7.595 KJ/Kg.K

S1 = S2 => 7.763 KJ / Kg.K

S2 > Sg2

7.763 > 7.595 => steam is again super heated condition

Again from super heated table and [mollier diagram use]

Corresponding ( P2 = 0.5 bar and super heated steam 4000C meet)

H2 = 3279.0 KJ/Kg [pg.no: -18]

Heat supplied (Qs) = h1 – h2

(Qs) = 3478.3 – 340.6

(Qs) = 3137.7 KJ/Kg

Work done (W) = h1 – h2

(W) =3478.3 – 3279

(W) = 199.3 KJ

Thermal efficiency rankine = W/Qs = 199.3 / 3137.7

ήth = 0.06351 x 100

ήth = 6.35 %


= 3600 / W =3600 / 199.3

= 18.0623 Kg/Kw .hr


5) A cyclic steam power plant is to be designed for a steam temperature at turbine inlet

of633K and an exhaust pressure of 8 kPa. After isentropic expansion of steam in the

turbine, the moisture content at the turbine exhaust is not to exceed 157o. Determine the

greatest allowable steam pressure at the turbine inlet, and calculate the Rankine cycle

efficiency for these steam conditions. Estimate also the mean temperature of heat

addition.(AUC DEC’07)

Solution:

Rankine cycle

T1 = 633 K

After isentropic expansion ( X2) = 85/100 =>0.85

Exhaust pressure (P2) = 8Kpa / 100 =0.08 bar

To find:

a) Thermal efficiency (ήth )

b) Mean temp (Tmean)

Solution:

Step:1

From steam table ( P2 = 0.8 bar) [pg.no:07]

Sf2 = 0.593 KJ/Kg.K Vf2 = 0.001008 m3 /kg hf2 = 173.9 KJ/Kg

Sfg2 = 7.637 KJ/Kg.K Vg2 =18.105 m3 /kg hfg2 = 2403.2 KJ/Kg

Sg2 = 8.230 KJ/Kg.K hg2 = 2577.1 KJ/Kg

S2 = Sf2 + (x2 Sfg2)

S2 = 0.593 + (0.25 x 7.637)

S2 = 7.0844 KJ/Kg.K

h2 = hf2 + (x2 hfg2)

h2 =173.9 + (0.85 x 2403.2)

h2 = 2216.62 KJ/Kg

S1 = S2 = 7.0844 KJ/Kg.K


S1 = 7.0844 => compare steam table [Pg.no: -22]

6.960 – 7.130 KJ/Kg.K

Select 20 bar

H1 = 3138.6 + 3248.7 / 2 => at 20 bar [pg.no:20]

H1 = 3193.65 KJ/Kg

H3 = hf2 => 173.9 KJ/Kg

pump work (Wp) = Vf2 ( P1 – P2) (P1 = 20 bar x 100 = 2000 KN /m2)

(Wp) = 0.001008 (2000 – 8)

(Wp) = 2.0079 KJ/Kg

heat supplied (Qs) = h1 (hf2 + Wp)

(Qs) = 3193.65 – (173.9 + 2.0079)

(Qs) = 3017.7421 KJ/Kg

work done (W) = (h1 – h2) – Wp

(W) = (3193.65 – 2216.62)

(W) = 975 .0221 KJ/Kg

Thermal efficiency = W/Qs = 975 .0221 / 3017.7421

ήth = 0.3230 x 100

ήth = 32.30 %

Mean temp of heat addition (Tmean)

= h1 – h3 / s1 – s3 = 3193.65 – 173.9 / 7.0844 – 0.593

(Tmean) = 465 K


6) 3 kg of steam at 18 bar occupy a volume of 0.2550 m3. During a constant volume

process, the heat rejected is 1320 kJ. Determine final internal energy. Find dryness

fraction and pressure, change in entropy and work done. (AUC MAY’08)

Given:

Constant volume process

V1 = 0.2550 m3

P1 = 18 bar x 100 = 1800 KN / m2

M1 = 3 kg

Heat rejected (Q) =1320 KJ

To find:

a) Internal energy (u2)

b) Initial dryness fraction (X)

c) Work done (W)

Solution:

Step:1

Specific volume (V1 ) = V / m

= 0.2550 / 3

(V1 ) = 0.085 m3 / kg

Step:2

From steam table , corresponding to 18 bar

Vf1 = 0.00168 m3 / kg hf1 = 884.5 KJ/kg

Vg1 = 0.11033 m3 / kg hfg1 = 1910.3 KJ/Kg

Sf1 = 2.398 KJ/Kg.K Sfg1 = 3.977 KJ/Kg.k

V1 = X1 x Vg1

0.085 = X1 x 0.11033

0.085 / 0.11033 = X1

X1 = 0.7704

H1 = hf1 + (X1 hfg1)

H1 = 884.5 + (0.7704 x 1910.3)

H1 = 2356.2257 KJ/kg.K


S1 = Sf1 + (X1 Sfg1)

S1 = 2.398 + (0.7704 x 3.977)

S1 = 5.4618 KJ/Kg.k

Step:3

For closed vessel (V1 = V2)

For constant volume process (W) = 0


Q = W + u

Where

Q = u2 – u1

Q = m (u2 – (n1 – P1 V1))

1320 = 3 (u2 – (2356.2257 – (1800 x 0.085))

1320 / 3 = u2 – 2203.2257

440 = u2 – 2203.2257

440 + 2203.2257 = u2

u2 = 2643.2257 KJ/Kg


7) A steam power plant runs on a single regenerative heating process. The steam enters

the turbine at 30 bar and 400oC and steam fraction is withdrawn at 5 bar. The remaining

steam exhausts at 0.10 bar to the condenser. Calculate the efficiency, steam fraction

and steam rate of the power plant. Neglect pumps work. (AUC MAY’08)

Given:

Single regenerative heating process

P1 = 30 bar

T1 = 4000C

With draw pressure (P2 ) = 5 bar

Exhaust pressure to the condenser (P3 ) = 0.1 bar

To find:

a) Efficiency

b) Steam fraction

c) Steam rate of the power plant

Solution:

Step:1

At 30 bar and T1 =4000C => from super heated steam table

H1 =3232.5 KJ/Kg

S1 = 6.925 KJ/Kg.K

P2 = 5 bar => from pressure table [pg.no:10]

Tsat = 151.80C hf2 =640.1 KJ/kg

Vf2 = 0.001093 m3 / kg hfg2 = 2107.4 KJ/kg

Vg2 = 0.37466 m3 / kg hg2 = 2747.5 KJ/kg

Sf2 =1.860 KJ/kg.K Sfg2 = 4.959 KJ/kg.K

Sg2 = 6.819 KJ/kg.K

P3 = 0.1 bar => from pressure table [pg.no:07]

Tsat = 45.830C

Sf3 =0.649 KJ/kg.K Hf3 =191.8 KJ/kg

Sfg3 = 7.052 KJ/kg.K Hfg3 = 2392.9 KJ/kg

Sg3 = 8.151 KJ/kg.K Hg3 =2584.7 KJ/kg


Step:2

1 – 2 => isentropic expansion

S1 = S2 = 6.925 KJ/kg.K

S2 > Sg2 => at 5 bar

6.925 > 6.819

So, steam is at super heat condition

At 5 bar and S2 = 6.925 KJ /Kg.K

Use mollier diagram => interaction point

P2 = 5 bar => T= 1700 C

:. H2 = 2794 KJ/Kg

Step :3

3 – 2 Isentropic expansion

S2 = S3 = 6.925 KJ/Kg.K

S3 < Sg3

6.925 < 8.151 KJ/kg => at P3 = 0.1 bar

:. Steam is wet condition

:. S3 = Sf3 + (X3 . Sfg3)

6.925 = 0.649 + (X3 . 7.502)

6.925 – 0.649 / 7.502 = X3

X3 = 0.8365

H3 = hf3 + ( X3 . hfg3)

H3 = 191.8 + (0.8365 x 2392.9)

H3 = 2193.6448

H4 =hf3

H4 = 191.8 KJ/kg

Step:5

Mass of steam bled (m)

M= hf2 – hf4 / h2 – h4 => 640.1 – 191.8/ 2794 – 191.8

M= 0.1722 KJ/kg of steam

Work done by the turbine with generation (W reg)

Wreg = (h1 – h2) + ( l – m) (h2 – h3 )


Wreg = (3232.5 – 2794) + (1 – 0.1722) (2794 – 2193.64)

Wreg = 935.4740 KJ/Kg

Efficiency of cycle with regeneration (ήreg)

(ήreg) = W / h1 – h2 = (935.4740 / 3232.5 – 640.1) x 100

(ήreg) = 0.3608 x 100

(ήreg) = 36.08%

Steam rate with regeneration

SSCreg = 3600 / WREG = 3600 / 935.4740

= 3.8483 Kg/Kw - hr


8) Ten kg of water at 45oC is heated at a constant pressure of 10 bar until it becomes

Superheated vapour at 300oC. Find the changes in volume, enthalpy, internal energy

and entropy. (AUC DEC’09)

Given:

Constant pressure process

Mass (m) =10 kg

(T1) = 450C

(P1) = (P2) = 10 bar x 100 => 1000 Kpa

(T2) = 300oC

To find:

a) V => change in volume

b) H => change in enthalpy

c) U => change in internal energy

d) S => change in entropy

Solution:

Step:1

From steam table at T1 = 450C [use temp table] (pg.no:02)

Vf = V1 = 0.001010 m3 / kg Sf = S1 = 0.638 KJ/Kg.K

Hf = h1 = 188.4 KJ /Kg

Step:2

From steam table P2 = 10 bar and T2 = 3000C

Use super heated steam table

V2 = 0.2580 m3 / kg [pg.no:16]

H2 = 3052.1 KJ/Kg [pg.no:19]

S2 = 7.125 KJ/Kg.K [pg.no:22]


Step:3

Change in volume ( V) = m (V2 – V1)

= 10 (0.258 – 0.001008)

( V) = 2.57 m3

Change in enthalpy ( H) = m(h2 – h1)

= 10 (3052.1 – 188.4)

( H) = 28637 KJ

Change in internal energy ( u) = m [(h2 – h1) + P2 (V2 –V1)]

( u) = 10 [(3052.1 – 188.4) + 1000 (0.258 – 0.001008)]

( u) = 31206.92 KJ

Change in entropy ( s) = m(S2 – S1)

= 10 (7.125 – 0.638)

( s) = 64.87 KJ/Kg.K


9) A steam power plant running on rankine cycle has steam entering H P turbine at 20

Mpa, 5000C and leaving L P turbine at 90% dryness. Considering condenser pressure of

0.005Mpa and reheating occurring upto the temp of 5000C

Determine

a) The pressure at which steam leaves H P turbine

b) The thermal efficiency

c) Work done (AUC MAY 2011)

Given:

Rankine cycle

H.P turbine

P1 = 20 Mpa => 200 bar

T1 = 5000C

L.P turbine

Dryness (X4) = 0.9

Condenser pressure (P3) = 0.005 Mpa

(P3) = 0.05 bar

To find:

a) The pressure at which steam leaves H P turbine (P2)

b) The thermal efficiency

c) Work done (W)

Solution:

Step:1

P1 = 200 bar & T1 = 5000C => from super heated steam table

H1 = 3241.1 KJ/Kg [pg.no:20]

S1 = 6.146 KJ / Kg.K [pg.no:23]

P4 = 0.05 bar => from pressure table

Vf4 = 0.001005 m3 /kg hf4 = 137.8 KJ/Kg Sf4 = 0.476 KJ/Kg.K

Vg4 = 28.194 m3 /kg hfg4 = 2423.8 KJ/Kg Sfg4 =7.920KJ/Kg.K


H4 = hf4 + (X4 x hfg4)

H4 = 137.8 + (0.9 x 2423.8)

H4 = 2319.22 KJ/Kg

S4 = Sf4 + (X4 x Sfg4)

S4 = 0.476 + (0.9 x 7.920)

S4 = 7.604 KJ/Kg.K

S1 =S2 = 6.146 KJ/Kg.K

S4 =S3 = 7.604 KJ/Kg.K

S3 =7.604 KJ/Kg.K & T = 5000C => use super heated steam table

P2 = 14 bar [pg.no:22]

H3 = 3473.9 KJ/Kg

H5 = hf4

H5 = 137.8 KJ/Kg

P2 = 14 bar => use pressure steam table

Hf2 = 830.1 KJ/Kg Sf2 = 2.284 KJ/Kg.K

Hfg2 = 1957.7 KJ/Kg Sfg2 = 4.181 KJ/Kg.K

Hg2 = 2787.8 KJ/Kg Sg2 = 6.465 KJ/Kg.K

S2 = Sf2 + (X2 . Sfg2)

6.146 = 2.284 + ( X2 . 4.181)

6.146 – 2.284/4.181 = X2

X2 = 0.92

H2 = hf2 + (X2 x hfg2)

H2 = 830.1 + (0.92 x 1957)

H2 = 2631.184 KJ/Kg

H6 = hf4 + Vf4 (P1 – P2)

H6 = 137.8 + 0.001005 (20x1000 – 0.005 x1000)

H6 = 157.8949 KJ/Kg


(ήrankine) = (h1 – h2) + (h3 – h4) – (h6 – h5) / (h1 – h6) + (h3 – h2)

= (3241.1 – 2631.184) + (3473.9 – 2319.22) – (107.8949 -137.8)

(3241.1 – 2631.184) + (3473.9 – 2613.184)

= 609.916 + 1154.68 – 20.0949

3083.2051 + 842.716

= 1744.5011

3925.9211

= 0.4443 x 100

(ήrankine) = 44.43%

Work done by the pump (Wp) = Vf4 (P1 –P2)

= 0.001005 (20000 – 1400)

(Wp) = 18.693 KJ/Kg


10) A rankine cycle works between 40 bar and 0.2 bar with saturated steam at turbine inlet

determine cycle efficiency and the ratio of pump work and turbine work

[AUC DEC 11]

Given:

Rankine cycle

P1 = 40 bar x 100 = 4000 KN/m2

P2 = 0.2 bar x 100 = 20 KN/m2

To find:

a) Cycle efficiency

b) The ratio of pump work and turbine work

Solution:

Step:1

From steam tables, at 40 bar

H1 = hg1 = 2800.3 KJ/Kg

Sg1 = S1 = 6.069 KJ/Kg.K

Step:2

From steam tables at 0.2 bar

Vf2 = 0.0010107 m3 / kg hf 2 = 251.5 KJ/kg sf2 = 0.832 KJ/Kg.K

Vg2 = 7.6498 m3 / kg hfg

2 = 2358.4 KJ/kg sfg2 = 7.077 KJ/Kg.K

hg 2 = 2609.9 KJ/kg sg2 = 7.909 KJ/Kg.K

Step:3

Pump work (Wp) = Vf2 (P1 –P2)

(Wp) = 0.001017 (4000 – 20)

(Wp) = 4.0476 KJ/Kg


Step:4

Turbine work (WT ) = h1 – h2

Where

H2 = ?

S1 = S2 => 6.069 KJ/Kg.K

S1 = S2 + (X2 . Sfg2)

6.069 = 0.832 + (X2 . 7.077)

6.069 – 0.832 / 7.077 = X2

0.7399 = X2

H2 = hf2 + (X2 . hfg2) note: P2 = bar use mollier diagram

h2 = 251.5 + (0.7399 x 2358.4) h2 = 2010 KJ/Kg

h2 = 1996.48016 KJ/Kg

Turbine work (WT) = h1 – h2

(WT) = (2800.3 – 2010)

(WT) = 790.3 Kw

The ratio of pump work and turbine work = Wp / WT = 4.0476 / 790.3= 0.005

Cycle efficiency (ήrank) = Wnet / Q1

Wnet = WT – Wp

(WT) = 790.3 – 4.0476

(WT) = 786.2524

Q1 = h1 – h4

Where

Hf2 = h3 => h3 = 251.5 KJ/kg

Wp = h4 – h3

:. H4 = 4.0476 + 251.5

H4 = 255.476 KJ/kg

Q1 = h1 – h4 => 2800.3 – 255.476

Q1 = 2544.7524 KJ

ήcy cle = Wnet / Q1 = 786.2524 / 2544.7524

ήcy cle = 0.3089 x 100=30.89%


11) One kg of steam at 10 bar exists at the following condition (i) wet and 0.8 dry (ii) dry and

saturated (iii) at a temp of 199.90 C . determine the enthalpy, specific volume , density ,

internal energy and entropy in each case take Cps = 2.25 KJ/kg (AUC JUNE 2012)

Given:

P1 = 10 bar x 100 = 1000 KN/m2

X = 0.8

T= 199.90 C

Cps = 2.25 KJ/kg

To find:

Enthalpy (h)

specific volume (V)

density (P)

internal energy

entropy

Solution:

Step:1

From steam table (P1) = 10 bar

Ts = 179.90C hf = 762.6 KJ/Kg sf=2.138 KJ/Kg.K

Vf = 0.001127 m3/kg hfg = 2013.6 KJ/Kg sfg=4.445 KJ/Kg.K

Vg = 0.19430 m3/kg hg = 2776.2 KJ/Kg sfg=6.583 KJ/Kg.K

Condition:

When steam is 0.8 bar

H1 = hf1 + (x . hfg)

H1 = 762.6 + (0.8 x 2013.6)

H1 = 2373.48 KJ/Kg


V1 = X . Vg1

V1 = 0.8 x 0.19430

V1 = 0.15544 m3 / kg

ρ = 1 / V1 = 1/ 0.15544

ρ = 6.4333 Kg/m3

U = h1 – P x V1

U = 2373.48 – (1000 x 0.15544)

U = 2218.08 KJ/Kg

S=Sf1 + (X1 x Sfg1)

S= 2.138 + (0.8 x 4.445)

S=5.694 KJ/Kg.K

Condition:2

When steam is dry and saturated

Hg1 = h

H = 2776.2 KJ/kg

Vg1 = V1

Vg1 = 0.19430 m3 / kg

ρ = 1 / V1 = 1/0.19430

ρ = 5.1466 Kg/m3

U = h – P1V1

U = 2776.2 – (1000 x 0.19430)

U = 2581.9 KJ/Kg

S1 = Sg1

S1 = 6.583 KJ/Kg.K


Condition:3

When steam is superheated

H= hg1 + Cp (T – Ts)

= 2776.2 + 2.25 (199.9 – 179.9)

H= 2821.2 KJ/Kg

V1 = Vg1 (T+ 273 / Ts + 273)

=0.19430 (199.9 + 273 / 179.9 + 273)

V1 = 0.2028 m3 / Kg

U = h –P1V1

U= 2821.2 – (1000 x 0.2028)

U= 2618.3197 KJ/Kg

S = Sg1 + CPs ln (T+ 273 / Ts + 273)

S = 6.583 + 2.25 ln (427.9 / 452.9)

S = 6.6802 KJ/Kg.K


12) In a steam generator compressed liquid water at 10 Mpa , 300C enters a 30mm dia tube

at the rate of 3 lit/sec, steam at 9 Mpa, 4000C exists the tube. Find the rate of heat

transfer to the water [AUC JUNE 2012]

Given:

P1 = 10 Mpa x 10 = 100 bar

Temp of water (Tw) = 300C

Dia (D) = 30mm / 1000 => 0.03

(V1) =3 / 1000 = 0.003 m3/sec

P2 = 9Mpa x 10 => 90 bar

Ttube (T2) = 4000C

To find:

Rate of heat transfer to the water (Q)

Solution:

Step:1

From steam table corresponding to 300C

V1 = Vf1 = 0.001004 m3 / kg

Hf1 = h1 = 125.7 KJ/Kg

Step:2

Mass flow rate of steam (m)

M = V / Vf1 = 0.003 / 0.001004

M= 2.9880 Kg/s

Step:3

Area of tube (A)

(A) = /4 x d2 => /4x (0.03)2

(A) = 7.0685 x 10-4 m2

Step:4

Inlet velocity (C1)


C1 = m x V1 / A = 2.9880 x 0.001004 / 7.0685 x 10-4 m2

C1 = 4.2440 m/s

Step:5

From steam tables corresponding to P2 = 9 bar and T2 = 4000C

V2 = 0.02993 m3/Kg

H2 = 3121.2 KJ/Kg

Step:6

Final velocity

C2 = m x V2 /A = 2.9880 x 0.02993 / 7.0685 x 10-4

C2 = 126.5202 m/s

Step:7

M h1 + C12 / 2000 + Z1 g + Q = m h2 + C2

2 / 2000 + Z2 g + W

M h1 + C12 / 2000 +Q = m h2 + C2

2 / 2000

2.9880 125.7 + (4.2440)2 / 2000 +Q = 2.9880 3121.2 + 126.5202 / 2000

375.6185 + Q = 9350.0606

Q = 9350.0606 – 375.6185

Q = 8974.4418 KJ/sec


13) A steam turbine receive steam at pressure 20 bar superheated at 3000C the exhaust

pressure 0.07 bar and expansion taken place isentropic calculated (i) heat rejected

(ii) heat supplied (iii) work done (iv) thermal efficiency [AUC JUNE 2013]

Given:

Isentropic process – rankine cycle

(P1) = 20 bar x 100 = 2000 KN/m2

(T1) = 3000C + 273 = 573K

(P2) = 0.07 bar x 100 = 7 KN/m2

To find:

a) heat supplied (Q)

b) heat rejected (Qr)

c) work done (W)

d) thermal efficiency (ήther)

Solution:

Step:1

From super heated steam table at P1 = 20 bar and T1 = 3000C

H1 = 3025.0 KJ/Kg => [pg.no:19]

S1 = 6.770 KJ/Kg.K => [pg.no:22]

Step:2

From steam table P2 = 0.07 bar => [pg.no:07]

Vf2 = 0.001007 m3 /kg hf2 = 163.4 KJ/Kg Sf2 = 0.559 KJ/Kg.K

Vg2 = 20.531 m3 /kg hfg2 = 2409.2 KJ/Kg Sfg2 =7.718 KJ/Kg.K

hg2 = 2572.6 KJ/Kg Sg2 =8.277 KJ/Kg.K

Step:3

S1 = S2

S2 = Sf2 + ( X2 x Sfg2)

6.770 = 0.559 + (X2 x 7.718)

6.770 – 0.559 / 7.718 = X2


X2 = 0.8047

H2 = hf2 + (X2 . hfg2)

H2 = 163.4 + (0.8047 x 2409.2)

H2 = 2102.1848 KJ/Kg

Step;4

Work done (Wp) = Vf2 (P1 – P2)

Wp = 0.001007 (2000- 7)

Wp = 2.0069 KJ/Kg

Step:5

Heat supplied (Qs) = h1 - (hf2 + Wp)

(Qs) = 3025.0 – (163.4 + 2.0069)

(Qs) = 2859.5931 Kg/KJ

Step:6

Heat rejected (QR) = h2 –hf2 (or) h2 – h3

(QR) =2102.1848 – 163.4

(QR) = 1938.7848 Kg/KJ

Step:7

Work done (W) = Qs - QR

(W) = 2859.5931 – 1938.7840

(W) = 920.8084 KJ/Kg

Step:8

Thermal efficiency (ήthermal) = W/Qs

= 920.8084 / 2859.5931

= 0.3220 x 100

(ήthermal) = 32.20%

Step:9


= 3600 / W => 3600 / 920.8084

= 3.9096 Kg/Kw - h

mahalakshmimahalakshmiengineeringcollege.com/pdf/mec/iiisem... · 12 kg, calculate the heat added...

Documents