CANKAYA UNIVERSITY

FACULTY OF ENGINEERING AND ARCHITECTURE

MECHANICAL ENGINEERING DEPARTMENT

ME 212 THERMODYNAMICS II

CHAPTER 8

EXAMPLES SOLUTIONS

Example 1:

Consider a steam power plant that operates on a simple ideal Rankine cycle. Steam enters the

turbine at 7 MPa and 450 oC and is condensed in the condenser at a pressure of 10 kPa by

running cooling water from a lake through the tubes of the condenser at a rate of 1600 kg/s.

Rankine cycle has a net power output of 40 MW. Show the cycle on T-s diagram with respect

to saturation lines, and determine (a) the thermal efficiency of this cycle, (b) the back work

ratio, (c) the mass flow rate of the steam, in kg/s, and (d) the temperature rise of the cooling

water, in oC.

Solution:

Assumptions : (1) Steady state conditions exits.

(2) Kinetic and Potential energy changes are negligible.

Analysis:

State 1

For P1=7 MPa and T1=450 °C , Table A-4 gives

kg

kJ57.3286h1 and

K.kg

kJ6359.6s1

State 2

K.kg

kJ6359.6ss 12

From Table A-3

K.kg

kJ1502.8s kPa10@g

K.kg

kJ6493.0s kPa10@f

7981.06493.01502.8

6493.06359.6

10@

10@2

2

kPafg

kPaf

s

ssx

So,

s

T

3 2

4 7MPa=70 bar

10kPa=0.1 bar

450°C

1

kPa10@fg2kPa10@f2 hxhh

From Table A-3, kg

kJ83.191h kPa10@f and

kg

kJ8.2392h kPa10@fg

Thus,

kg

kJ52.2101)8.2392(7981.083.191h 2

State 3: From table A-3

kg

kJ83.191hh kPa10@f3 and

kg

m100102.1

33

kPa10@f3

State 4:

)PP(hh

m

W34334

p

)PP(hh 34334

kg

kJ89.198

m

kN)107000)(

kg

m100102.1(

kg

kJ83.191h

2

33

4

Thus;

kg

kJ05.1185)52.2101()57.3286(hh

m

W21

t

kg

kJ06.7)83.191()89.198(hh

m

W34

p

kg

kJ68.3087)89.198()57.3286(hh

m

Q41

in

kg

kJ69.1909)83.191()52.2101(hh

m

Q32

out

kg

kJ99.117769.190968.3087

m

W

m

W

m

Q

m

Q

m

W ptoutinnet

(a) The thermal efficiency of the cycle is

)2.38(%382.068.3087

69.19091

m

Q

m

Q

1

in

out

(b) The back-work ratio is determined from

)60.0(%0060.005.1185

06.7

m

W

m

W

bwr

t

p

(c) The mass flow rate of the steam is

s

kg96.33

kg

kJ99.1177

s

kJ40000

m

W

Wm

net

net

(d) The rate of heat rejection to the cooling water and its temperature rise are

s

kJ07.853,64)

kg

kJ69.1909)(

s

kg96.33(

m

QmQ out

out

C70.9

)K.kg

kJ18.4)(

s

kg1600(

s

kJ07.853,64

)cm(

QT o

watercoolingp

out

watercooling

Example 2:

A Rankine steam power plant uses water as the working fluid. Steam enters the turbine at 7

MPa and 450 oC and is condensed in the condenser at a pressure of 10 kPa by running cooling

water from a lake through the tubes of the condenser at a rate of 1600 kg/s. Cycle has a net

power output of 40 MW. Isentropic efficiency of the turbine is 85 percent, and the isentropic

efficiency of the pump is 90 percent. Assuming no pressure losses in the condenser and boiler,

show the cycle on T-s diagram with respect to saturation lines, and determine (a) the thermal

efficiency of this cycle, (b) the back work ratio, (c) the mass flow rate of the steam, in kg/s,

and (d) the temperature rise of the cooling water, in oC.

Solution:

Assumptions: (1) Steady state conditions exist.

(2) Kinetic and potential energy changes are negligible.

Analysis:

State 1:

With P1 = 7 MPa and T1 = 450 0C, Table A-4 gives

h1 = 3286.57 kJ/kg and s1 = 6.6359 kJ/kg

State 2:

s2s = s1 = 6.6359 kJ/kg

From Table A-3

sg@10kPa = 8.1502 kJ/kg.K

sf@10kPa = 0.6493 kJ/kg.K

X2s =

So,

h2s = hf@10kPa + X2s.hfg@10kPa

From Table A-3

hf@10kPa = 191.83 kJ/kg

hfg@10kPa = 2392.8 kJ/kg

Thus,

h2s = 191.83 + 0.7981(2392.8) = 2101.52 kJ/kg

Isentropic turbine efficiency is

Ƞt = =

h2 = h1 - Ƞt (h1-h2s)

= 3286.57-0.85(3286.57-2101.52)

= 2279.28 kJ/kg

State 3:

h3 = hf@10kPa = 191.83 kJ/kg

v3 = vf@10kPa = 1.0102 x 10-3

m3/kg

State 4:

s = h4s – h3 = v3(P4s - P3)

h4s = h3 + v3(P4s – P3)

= 191.83 kJ/kg + (1.0102 x 10-3

m3/kg) (7000 - 10) kN/m

2

= 198.89 kJ/kg

Isentropic pump efficiency is

Ƞp = =

h4 = + h3 = + 191.83

h4 = 199.67 kJ/kg

Hence,

= h1 – h2 = 3286.57 – 2279.28 = 1007.29 kJ/kg

= h4 – h3 = 199.67 – 191.83 = 7.84 kJ/kg

= h1 – h4 = 3286.57 – 199.67 = 3086.9 kJ/kg

= h2 – h3 = 2279.28 – 191.83 = 2087.45 kJ/kg

3086.7 - 2087.45 = 999.25 kJ/kg

(a) The thermal efficiency of the cycle is

Ƞ = 1 - = 1 – = 0.324 or % 32.4

(b) The back-work ratio is determined from

bwr = = = 0.0078

(c) The mass flow rate of the steam is

= 40.03 kg/s

(d) The rate of heat rejection to the cooling water and its temperature rise are

= = (40.03 kg/s) (2087.45 kJ/kg) = 83,560.62 kJ/s

= = = 12.49 0C

Example 3:

In a Rankine cycle, saturated liquid water at 10 kPa is compressed in a pump isentropically to

8 MPa. It is then heated, first in a boiler and then by superheating at a constant pressure of 8

MPa, to a temperature of 600 oC. After an adiabatic reversible expansion to 3 MPa, the steam

is reheated to 600 oC, and a second adiabatic reversible expansion to 10 kPa occurs.This is

essentially a reheat cycle. (a) What is the total work (kJ/kg) generated. (b) What is the

efficiency of the cycle (%)? (c) Sketch the cycle on a T-s diagram.

Solution:

(c)

State 1;

P1=7MPa

T1=700°C Table A-4 gives

kg

kjh 0.36421 and

Kkg

kjs

.0206.71

State 2:

P2=3MPa

Kkg

kjss

.0206.712

Kkg

kjs MPag

.1869.63@ , From Table A-3,

MPagss 3@2

600°C 1

2

3

4

6

5

8MPa

3MPa

10kPa= 0.10 bar

s

T

P2=3MPa

Kkg

kjs

.0206.72 ,

kg

kjh 8.32992 , From Table A-4

State 3;

P1=3MPa

T1=600°C Table A-4 gives

kg

kjh 3.36823 and

Kkg

kjs

.5085.71

State 4;

Kkg

kjs kPag

.1502.810@ From Table A-3

Kkg

kjs kPaf

.6493.010@

Kkg

kjss

.50853.734

kPagkPaf sss 10@410@

9145.06493.01502.8

6493.05085.7

10@10@

10@4

4

kPafkPag

kPaf

ss

ssx

kPafgkPaf hxhh 10@410@4

kg

kjh 1.2380)8.2392)(9145.0(83.1914 (Table A-3)

State 5;

kg

kjhh kPaf 83.19110@5 ,

kg

mkPaf

33

10@5 100102.1

State 6;

)( 56556 PPhh

m

Wp

)( 56556 PPhh

kg

kj

m

kN

kg

m

kg

kjh 9.199)108000)(100102.1(83.191

2

33

6

(a) The total work generated

m

W

m

W

m

W ttgeneratedtotal

21

kg

kjhh

m

W t2.342)8.3299()0.3642(21

1

kg

kjhh

m

W t2.1302)1.2380()3.3682(43

2

kg

kj

m

W generatedtotal

4.16442.13022.342

(b)

kg

kjhhhh

m

Qin 6.3824)8.32993.3682()9.1990.3642()()( 2361

kg

kjhh

m

W p07.8)83.191()9.199(56

428.06.3824

07.82.13022.342

21

m

Q

m

W

m

W

m

W

m

Q

m

W

in

ptt

in

net

or (%42.8)

Example 4:

Consider the steam power plant with an open feedwater heater operating under the conditions

shown in the following diagram. Saturated liquid exits the open feedwater heater at 1 MPa,

and saturated liquid exits the condenser. Each pump operates isentropically. Net power output

of the cycle is 100 MW.

Using the conditions shown on the diagram above and values obtained from the steam tables,

determine

a. the enthalpy values at all eight stations,

b. the mass fraction, y, of steam bled from the turbine set at station (2),

c. the thermal efficiency, , of this power plant, and

d. the mass flow rate of steam entering the first turbine stage, in kg/h.

e. Show the cycle on T-s diagram with respect to saturation lines.

Solution:

Assumptions: (1) Steady-state

(2) Each pump operates isentropically.

(3) The turbines, pumps, and feedwater heater operate adiabatically.

(4) Kinetic and potential energy changes are negligible.

(5) Saturated liquid exits the open feedwater heater, and saturated liquid exits the condenser.

Analysis:

a) State 1: T1 = 520 0Cand P1 = 10 MPa, Table A-4 gives h1 = 3425.1 kJ/kg

State 2: With P2 = 1 MPa and T2 = 200 0C, Table A-4 gives h2 = 2827.9 kJ/kg

State 3: With 1 MPa and 520 0C, Table A-4 gives h3 = 3522.1 kJ/kg

State 4: With 6 kPa and 50 0C, Table A-4 gives h4 = 2593.5 kJ/kg

State 5: With 6 kPa, Table A-3 gives h5 = hf@6kPa = 151.53 kJ/kg

v5 = vf@6kPa = 1.0064 x 10-3

m3/kg

State 6: h6 = h5 + v5 (P6 – P5)

= 151.53 kJ/kg + (1.0064 x 10-3

m3/kg) (1000 – 6) kN/m

2

= 152.53 kJ/kg

State 7: With 1 MPa, Table A-3 gives

h7 = hf@10bar = 762.81 kJ/kg

v7 = vf@10bar = 1.1273 x 10-3

m3/kg

State 8: h8 = h7 + v7 (P8 – P7) = 762.81 kJ/kg + (1.1273 x 10-3

m3/kg) (10000 – 1000) kN/m

2

= 772.96 kJ/kg

(b) Energy rate balance for open feedwater heater gives

yh2 + (1-y)h6 = h7

(c) On the basis of a unit of mass passing through the high pressure turbine, the total turbine

work output is

(h1 – h2) + (1-y) (h3 – h4)

= (3425.1 – 2827.9) + (1 – 0.2281) (3522.1 – 2593.5)

= 1314.0 kJ/kg

The total pump work per unit mass passing through the high pressure turbine is

(h8 – h7) + (1 – y) (h6 – h5)

= (772.96 – 762.81) + (1 – 0.2281) (152.53 – 151.53)

= 10.92 kJ/kg

The heat added in the steam generator per unit of mass passing through the high pressure

turbine is

(h1 – h8) + (1 – y) (h3 – h2)

= (3425.1 – 772.96) + (1 – 0.2281) (3522 – 2827.9)

= 3187.9 kJ/kg

The thermal efficiency is then

ƞ = or % 40.9

(d) The mass flow rate of the steam entering the high pressure turbine, ṁ1, can determined

using the given value for the net output, 100 MW. Since Ẇcycle = Ẇt - Ẇp

and

1314.0 kJ/kg and 10.32 kJ/kg

it follows that

ṁ1 = = = 76.74 kg/s

= 2.76 x 105 kg/h

(e)

Example 5:

Consider a water-ammonia binary vapor cycle consisting. In the steam cycle,

superheated vapor enters the turbine at 7 MPa, 450 oC, and saturated liquid exits the

condenser at 55 oC. The heat rejected from the steam cycle is provided to the ammonia

cycle, producing saturated vapor at 45 oC, which enters the ammonia turbine. Saturated

liquid leaves the ammonia condenser at 1 MPa. For a net power output of 24 MW from

the binary cycle, determine (a) the mass flow rates for the steam and ammonia cycles,

respectively, in kg/s, (b) the power output of the steam and ammonia turbines,

respectively, in MW. (c) the rate of heat input to the ammonia cycle, in MW, (d) the

rate of heat addition to the binary cycle, in MW, and (e) the thermal efficiency of the

binary vapor cycle. (f) Show the cycle on T-s diagram with respect to saturation lines.

Solution:

Assumptions: (1) Each component is analyzed as a control volume at steady state

(2) All process of the working fluids is internally reversible, except in the interconnecting heat

exchanges.

(3) There are no stray heat transfers from the turbines or the heat exchanger.

(4) Kinetic and potential energy effects can be neglected.

Analysis: First, fix each of principle states. For steam, use tables A-2 and A-4.

State 1:

P1 = 70 bar, T1 = 450 0C → h1 = 3286.57 kJ/kg

s1 = 6.6359 kJ/kg.K

State 2:

T2 = T3 = 55 0C, s2 = s1 = 6.6359 kJ/kg

sg@55C = 7.9913 kJ/kg.K

sf@55C = 0.7679 kJ/kg.K

X2 =

hf@55C = 230.23 kJ/kg

hfg@55C = 2370.7 kJ/kg

h2 = hf@55C + X2.hfg@55C

h2 = 2309.23 + 0.8124 (2370.7) = 2156.19 kJ/kg

State 3:

T3 = 55 0C, saturated liquid.

h3 = 230.23 kJ/kg

v3 = 1.0146 x 10-3

m3/kg

P3 = 0.1576 bar = 15.76 kPa

State 4:

= h4 – h3 = v3 (P4 – P3)

h4 = h3 + v3 (P4 – P3)

= 230.23 + (1.0146 x 10-3

m3/kg) (7000 – 15.76) kN/m

2

= 237.32 kJ/kg

Now, for Ammonia, use Tables A-13 and A-14

State a:

Ta = 45 0C, saturated vapor

ha = 1470.96 kJ/kg

sa = 4.8125 kJ/kg.K

Pa = 17.819 bar = 1781.9 kPa

State b:

Pb = Pc = 10 bar

sb = sa = 4.8125 kJ/kg.K

sf@10bar = 1.1191 kJ/kg.K

sg@10bar = 5.0294 kJ/kg.K

Xb =

hb = hf@10bar + Xb.hfg@10bar

hf@10bar = 297.76 kJ/kg

hfg@10bar = 1165.42 kJ/kg

hb = 297.76 + 0.9445 (1165.22) = 1398.50 kJ/kg

State c:

Pc = 10 bar, saturated liquid.

hc = 297.76 kJ/kg

vc = 1.6584 x 10-3

m3/kg

State d:

hd = hc + vc (Pd – Pc)

= 297.76 + (1.6584 x 10-3

m3/kg) (1781.9 - 1000) kN/m

2

= 299.06 kJ/kg

(a) The mass flow rates of the steam and ammonia can be obtained using mass and energy

rate balances for the inter-connecting heat exchanges.

0 =

or

For the steam cycle

And, for the ammonia cycle

Noting that and combining

Solving

=

(b) The power output of the steam and ammonia turbines are

(c) The rate of heat input to the ammonia cycle is

= 37.27 MW

(d) The rate of heat addition to the binary cycle is

(e) The thermal efficiency of the binary cycle is

ƞ = or % 40.7

(f)

6) Water is the working fluid in a Rankine cycle. Superheated vapor enters the

turbine at 10 MPa, 480 0C, and the condenser pressure is 6 kPa. The turbine and

pump have isentropic efficiencies of 80 and 70%, respectively. Determine the

rate of exergy input, in kJ per kg of steam flowing, to the working fluid passing

through the steam generator. Perform calculations to account for all outputs,

losses, and destructions of this exergy. Let 0

0 0T 15 C and p 0.1MPa T0 5

158C, p0 5 0.1 MPa