Tutorial Plant Solution

Jabatan Kejuruteraan Loji dan AutomotifFakulti Kejuruteraan Mekanikal dan PembuatanUniversiti Tun Hussein Onn Malaysia

SolutionTutorial 1/ ws

BDA 3043

Tutorial 1 – SolutionSteam Power Plant

Winardi Sani

¶ State 3 : p3 = 3 MPa. t3 = 350oC. Condenser pressure of p4 = 75 kPa.

a) Sketch this cycle in p−h diagram and read the value of each state variable:

qin

h4h1 h2 h3

boiler

condenser

turb

ine

pum

p

qout

wp

wout

3

1

2

4

h1 = 350 kJ/kg; h2 = 390 kJ/kg; (1)

h3 = 3150 kJ/kg; h4 = 2400 kJ/kg (2)

b) Determine the thermal efficiency of this cycle.

ηth =|wt| − |wp||qin|

(3)

The 1. law of T/D applied on the pump 1 to 2 :

h2 − h1 = wp (isentropic, no kin. and pot. energy) (4)

The 1. law of T/D applied on the turbine 3 to 4 :

h4 − h3 = −wt (isentropic, no kin. and pot. energy, work released)(5)

| − wt| = |wt| = |h4 − h3| = |h3 − h4| (6)

The 1. law of T/D applied on the boiler 2 to 3 :

h3 − h2 = qin (no work, no kin. and pot. energy) (7)

The 1. law of T/D applied on the condenser 4 to 1 :

h1 − h4 = −qout (no work, no kin. and pot. energy, heat is rejected) (8)

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BDA 3043


Substitute equations (4) – (7) into equation (3) yields:

ηth =|h3 − h4| − |h2 − h1|

|h3 − h2|(9)

=|3150− 2400| − |390− 350|

|3150− 390|

=750− 40

2760=

7102760

(10)

= 0.257

= 25.7% (11)

c) Compare your result with the thermal efficiency of the Carnot cycle.Carnot cycle operates between two isothermal processes, TL (condensertemperature) and TH (boiler temperature) with efficiency: .

ηth = 1− TLTH

(12)

= 1− 85 + 273.15350 + 273.15

= 1− 358.15623.15

= 0.425

= 42.5% (13)

ηth,Carnot > ηth,Rankine

You may compare the results using the data taken from the steam table (seethermodynamics book by Cengel). Hereby you must calculate the dryness fac-tor x to determine the state properties (h4 for example) at point 4 with formula:

x =s3 − s4fs4fg

(14)

p-h diagram is a convenient tool for energy analysis in Rankine cycle.

· State 3 : p3 = 8 MPa, saturated vapor, =⇒ t3 = 300oC (read p-h diagram). Con-denser pressure of p4 = 8 kPa, saturated liquid and turbine power delivered atWt = 100 MW.

a) The thermal efficiency


(15)

The same procedures in the previous solution are applied here to obtain:

ηth =|h4 − h3| − |h2 − h1|

|h3 − h2|(16)

=|1750− 2725| − |200− 180|

|2725− 200|

=975− 20

2525=

9552525

(17)

= 0.378

= 37.8% (18)

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BDA 3043

qin

qout

wpwout

100 MW

pum

p

turb

ine

condenser

boiler

h3h4h1 h2

4

3

1

2

b) The back work ratio(bwr, pump work divided by turbine work)

bwr =|wp||wt|

(19)

=|h2 − h1||h4 − h3|

(20)

=|200− 180||1750− 2725|

(21)

=20975

= 2% (22)

c) The mass flow rate of the steam in kg/hr

Wt = m ∗ wt (23)

= m(h3 − h4) (24)

=⇒ m =Wt

h3 − h4(25)

=100 MW

975 kJ/kg=

105 kJ/s975 kJ/kg

= 102.6 kg/s

= 102.6 kg/s ∗ 3600 s1 hr

=⇒ m = 3.7 ∗ 105 kg/hr (26)

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BDA 3043


d) The rate of heat transfer, Q into the working fluid as it passes through theboiler

Qin = m ∗ qin (27)

= m(h3 − h2) (28)

= 3.7 ∗ 105 kg/hr ∗ (2725− 200) kJ/kg

= 934, 250 kJ/hr = 934, 250kJhr∗ 1 hr

3600 s= 259, 514 kW = 259.5 MW (29)

e) The mass flow rate of the condenser cooling water in kg/hr if cooling waterenters the condenser at 25 oC and exists at 40 oC.

Qout = m ∗ qout (30)

= m(h4 − h1) (31)

= 3.7 ∗ 105 kg/hr ∗ (1750− 180) kJ/kg

= 580, 900 kJ/hr = 580, 900kJhr∗ 1 hr

3600 s= 161.4 MW (32)

.Q cw

oC40

cooling waterco

olin

g w

ater

in out

1 4

Q.

out

m

25oC

.

Energy balance (1. law of T/D) applied on the condenser system:[∆E∆t

]system

= mcw(hcw,in − hcw,out) + m(h4 − h1) (33)

0 = mcw(hcw,in − hcw,out) + m(h4 − h1) (34)

mcw =m(h4 − h1)

hcw,out − hcw,in(35)

=Qout

hcw,out − hcw,in(36)

The enthalpy values of saturated liquid at 25 oC and 40 oC from the satu-rated water table: ht=25oC = 104.88 kJ/kg, and ht=40oC = 167.57 kJ/kg.

mcw =161.4 MW

(167.57− 104.88)kJ/kg= 2, 575 kg/s (37)

=⇒ mcw = 9.3 ∗ 106 kg/hr (38)

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BDA 3043

¸ With ηt = ηp = 85%, determine for the modified cycle:

qin

qout

wp

wp,swp

pum

p

condenser

boiler

h4s h4

isent

ropi

c

h3

isen

trop

ic

wt,s

h1

h2s h2

wt100 MW

wt

3

2

1 4s

2s

not scaled

4

a) The thermal efficiency:


(39)

Turbine

ηt =wtwt,s

(40)

=⇒ wt = ηt ∗ wt,s (41)

Pump

ηp =wp,swp

(42)

=⇒ wp =wp,sηp

(43)

=1ηp∗ |h2s − h1| (44)

=1

0.85∗ |200− 180| kJ/kg (45)

= 23.5 kJ/kg; greater than wp,s = 20 kJ/kg (46)

5


BDA 3043


Substitute the equations (43) and (41) into the equation (39) yields:

ηth =|ηt ∗ wt,s| − |

wp,sηp|

|qin|(47)

=ηt ∗ |h4s − h3| − 1

ηp∗ |h2s − h1|

qin(48)

From the previous question we have calculated:

|h4s − h3| = |1750− 2725| kJ/kg = 975 kJ/kg and

|h2s − h1| = |200− 180| kJ/kg = 20 kJ/kg

qin = h3 − h2 (49)

The value of h3 remains constant, but h2 changes!. h2 must be calcu-lated using the the value of wp.

h2 = wp + h1 (50)

= (23.5 + 180) kJ/kg

= 203.5 kJ/kg (51)

=⇒ qin = (2725− 203.5) kJ/kg

=⇒ qin = 2, 521.5 kJ/kg (52)

All the data necessary has been prepared to solve the new efficiencyin equation (48):

ηth =0.85 ∗ 975− 1

0.85 ∗ 202, 521.5

= 0.319

= 31.9% (53)

The new efficiency must be lower compared to the ideal, isentropiccase (37.8%).

b) The back work ratio (pump work divided by turbine work)

bwr =|wp||wt|

=wp,sηp∗ 1ηt ∗ wt,s

=(wp,swt,s

)∗ 1ηt ∗ ηp

= bwrs ∗1

ηt ∗ ηp

=20975∗ 1

0.85 ∗ 0.85= 2.8%

It means, more investment compared to the isentropic process.

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BDA 3043

c) The mass flow rate of the steam in kg/hr. Energy balance applied on thesteam power plant.(∆E=

∑energy in−

∑energy out) per unit time.

0 = qin − qout + wp − wtwt − wp = qin − qout

m(wt − wp) = m(qin − qout)

Wcycle = m

[(h3 − h2)− (h4 − h1)

]=⇒ m =

Wcycle

(h3 − h2)− (h4 − h1)(54)

State property value [kJ/kg]h1 180h2 203.5h3 2725h4 ?

To obtain the value of h4, we can apply the 1. law of T/D on the turbine,from state 3 to 4 :

h4 = h3 − wt (55)

= h3 − ηt ∗ wt,s= h3 − ηt ∗ (h4,s − h3)

= (2725− 0.85 ∗ 975) kJ/kg

h4 = 1896.5 kJ/kg (56)

Because of irreversibilities inside the turbine, h4 > h4s = 1750 kJ/kg. WithWcycle = 100 MW, the mass flow rate is then:

m =100

(2725− 203.5)− (1896.5− 180)MW

kJ/kg

=100, 000

2521.5− 1716.5kW

kJ/kg

=100, 000

805J/s

J/kg

= 124.223kgs

=⇒ m = 4.47 ∗ 105 kghr

(57)

d) The rate of heat transfer, Q into the working fluid as it passes through theboiler

Qin = m ∗ qin (58)

= m(h3 − h2) (59)

= 124.223 kg/s ∗ (2725− 203.5) kJ/kg

=⇒ Qin = 313.2 MW (60)

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BDA 3043


More heat addition must be provided to overcome losses compared toQin,s in the previous isentropic case.

e) The mass flow rate of the condenser cooling water in kg/hr if cooling waterenters the condenser at 25 oC and exists at 40 oC.

Qout = m ∗ qout

= m(h4 − h1)

= 124.223kgs∗ (1896.5− 180) kJ/kg

=⇒ Qout = 213.2 MW (61)

f) Discuss the effects on the steam power plant of the irreversibilities within theturbine and pump.The effects of irreversibilities within the turbine and the pump can be gaugedby comparing the present values with their counterparts. The turbine workis less and the pump work greater than in the previous case. The thermalefficiency of the power plant is also less than in the ideal case. The mag-nitude of the heat transfer to the cooling water is greater than in the idealcase, consequently, so greater mass flow rate of cooling water would berequired. This conclusion is still valid eventhough the net power output inthe last case is not 100 MW. It is true because the difference of the backwork ratio in both the cases is very small.

8



BDA 3043

¹ Steam is the working fluid in an ideal Rankine cycle with superheat and reheat.Steam enters the first-stage turbine at 8.00 MPa, 480 oC and expands to 0.8 MPa.It is then reheated to 440 oC before entering the second-stage turbine, where itexpands to the condenser pressure of 10 kPa. The net power output is 100 MW.Determine:

• The cycle illustrated in p – h diagram:

wt,2

qout

t,1w

boiler

h1 h2

LP tu

rbin

e

h5h6 h4

reheat

superheat

wp

condenser

HP

turb

ine

pum

p

h3

q in,1

q in,2

2

4

3

5

1 6

a) The thermal efficiency of the cycle

η =Net work output

Heat input

=∑woutput −

∑winput∑

qinput

η =|w34|+ |w56| − |w12||q23|+ |q45|

(62)

From the 1. law of T/D applied on a cyclic system:∑energy in =

∑energy out (63)

q23 + q45 + w12 = w34 + w56 − q61

w34 + w56 − w12 = q23 + q45 − q61 (64)

replacing eq. (64) into eq. (62), yields

η = 1− |q61||q23|+ |q45|

(65)

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BDA 3043


E=2

500

E=2

800

E=3

000

E=34

00

sv=0

.005

sv=0

.01

E=3500

4

3

2

1

5

6

You can describe the Rankine cycle using p – h, T – s, or h – s diagram,or you can combine them to obtain more accuracy. Interpolation is oftennecessary if the location point does not lay on a constant line. To calculatethe thermal efficiency of the steam power plant, you can use the previous

10



BDA 3043

method with the data read directly on the diagrams illustrated above. Analternative solution is to use the data taken from the steam table. It will beshown following.

State phase given steam table3 superheated p3 = 8.00 MPa,

t3 = 480 0Cinterpolation

4 p4 = 0.8 MPa5 superheated t5 = 440 0C,

p5 = p4

interpolation

6 p6 = 10 kPa

1 sat. liquid p1 = p6

vf = 0.001010 m3/kg,hf = 191.81 kJ/kg,sf = 0.6492 kJ/kg K

2 liquid

State change Process Meaning3-4 isentropic,work out s3 = s4, wt = w34

4-5 isobar,heat addition p4 = p5, qin = q45

5-6 isentropic,work out s5 = s6, wt = w56

6-1 isobar,heat rejection p6 = p1, qout = q61

1-2 isentropic, compression s1 = s2, wout = w12

2-3 isobar,heat addition p2 = p3, qin = q23

• Interpolation at state 3 , p3 = 8 MPaNumber Temperature Enthalpy Entropy

1 450 3273.3 6.55792 480 hp3 sp3

3 500 3399.5 6.7266

Linear interpolation to determine both the value of h2 = hp3 and s2 = sp3[h3 − h1

t3 − t1

]=

h2 − h1

t2 − t1

=⇒ h2 = h1 +[t2 − t1t3 − t1

]∗ (h3 − h1) (66)

h2 = 3273.3 +[

480− 450500− 450

]∗ (3399.5− 3273.3)

= 3273.3 + 75.72

=⇒ hp3 = 3349.02 kJ/kg (67)

s2 = 6.5579 +[

480− 450500− 450

]∗ (6.7266− 6.5579)

s2 = 6.5579 + 0.10122

=⇒ sp3 = 6.65912 kJ/kg K (68)

11


BDA 3043


• State 4 , p4 = 0.8 MPaSince sp3 = 6.65912 kJ/kg < 6.6616 kJ/kg = sp4,v, so the state 4 is laid inthe liquid+vapor region!.

x =s4 − sfsfg

(69)

=6.65912− 2.0457

4.6160= 0.9994 (70)

h4 = hf + x ∗ hfg (71)

= 720.87 + 0.9994 ∗ 2047.7

= 2, 767.3 kJ/kg (72)

• Interpolation at state 5 , p5 = 0.8 MPa

Number Temperature Enthalpy Entropy1 400 3267.7 7.57352 440 hp5 sp5

3 500 3481.3 7.8692

Linear interpolation to determine both the value of h2 = hp3 and s2 = sp3[h3 − h1

t3 − t1

]=

h2 − h1

t2 − t1

=⇒ h2 = h1 +[t2 − t1t3 − t1

]∗ (h3 − h1) (73)

h2 = 3267.7 +[

440− 400500− 400

]∗ (3481.3− 3267.7)

= 3267.7 + 85.44

=⇒ hp5 = 3, 353.14 kJ/kg (74)

s2 = 7.5735 +[

440− 400500− 450

]∗ (7.8692− 7.5735)

s2 = 7.5735 + 0.11828

=⇒ sp5 = 7.6918 kJ/kg K (75)

• At state 6 , p6 = 10 kPa, s6 = s5 = 7.6918 kJ/kg K. At pressure of 10 kPa,the entropy of the fluid at saturated state is of s = 8.1488 kJ/kg K. Sothe state 6 must be must be in the liquid vapor region!.

At p = 10 kPa.sf [kJ/kg K] sfg [kJ/kg K] sg [kJ/kg K]

0.6492 7.4996 8.1488

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BDA 3043

x =s6 − sfsfg

(76)

=7.6918− 0.6492

7.4996= 0.9391 (77)

At p = 10 kPa.hf [kJ/kg K] hfg [kJ/kg K] hg [kJ/kg K]

191.81 2392.1 2583.9

h6 = hf + x ∗ hfg (78)

= 191.81 + 0.9391 ∗ 2392.1

=⇒ h6 = 2, 438.2 kJ/kg (79)

• At state 2 : s2 = s1 = 0.6492 kJ/kg K

w12 = wp = vf (p2 − p1) (80)

= 0.001010 m3/kg ∗ (8000− 10) kPa

= 8.0699m3

kg∗ kPa ∗ N/m2

Pa= 8.0699 kJ/kg (81)

h2 = h1 + w12 (82)

= 191.81 kJ/kg + 8.0699 kJ/kg

=⇒ h2 = 199.88 kJ/kg (83)

We have prepared all the values needed for efficiency:

η = 1− |q61||q23|+ |q45|

q61 = h1 − h6 (84)

= (191.81− 2, 438.2) kJ/kg

= −2, 246.39 kJ/kg (minus sign means heat rejection) (85)

q23 = h3 − h2 (86)

= (3, 349.02 − 199.88) kJ/kg

= 3, 149.14 kJ/kg (87)

q45 = h5 − h4 (88)

= (3, 353.14− 2, 767.3) kJ/kg

= 585.84 kJ/kg (89)

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BDA 3043


η = 1− 2, 246.393, 149.14 + 585.84

η = 0.4 = 40% (90)

b) The mass flow rate of the steam in kg/hr

c) The rate of heat transfer, Qout from the condensing steam as is passes throughthe condenser, in MW.

d) Discuss the effects of reheat on the steam power plant

e) Repeat the analysis with including that the turbine and pump each havethe isentropic efficiency of 85%.

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BDA 3043

º Consider a regenerative steam power plant with one open feedwater heater.Steam enters the turbine at 8.0 MPa, 480 oC and expands to 0.8 MPa, wheresome of the steam is extracted and diverted to the open feedwater heater op-erating at 0.8 MPa.The remaining steam expands through the second-stage turbine to the con-denser pressure of 10 kPa. This portion of the total flow is condensed to saturatedliquid, then pumped to the extraction pressure and introduced into the feedwa-ter heater at 0.8 MPa.The isentropic efficiency of each turbine is 85% and each pump operates isen-tropically. If the net power output of the cycle is 100 MW, determine:


HP

Pump 1

1 Open

feedwater

heater

2

3

5

4

LP

Tu

rbin

es

Co

nd

en

ser

6

Steam generator

Pump 2

(1 −

y)

7

y (1 − y)

E=2

500

E=2

800

E=3

000

E=34

00

sv=0

.005

sv=0

.01

E=3500

3

2

6

1

7

5s5

4

4s

15


BDA 3043


The thermal efficiency of the cycle

ηth =Net work output

Heat input

=∑woutput −

∑winput∑

qinput

ηth =|wt| − |wp||q23|

(91)

Fraction of the total flow at state 4 by y. y = m4/m with the total massm = m4 + m5.Total turbine work per unit of total mass is:

wt =Wt

m=[(h4 − h3) + (1− y)(h5 − h4)

](92)

The pump work per unit of total mass is:

wp =Wp

m=[(h2 − h1) + (1− y)(h7 − h6)

](93)

y =h1 − h7

h4 − h7(94)

• Calculation of the properties at state 3

The specific enthalpy at state 3 is evaluated in the solution to the previ-ous question.

=⇒ h3 = 3349.02 kJ/kg (95)

=⇒ s3 = 6.65912 kJ/kg K (96)

• Calculation of the properties at state 4

h4 = h3 − wt

|wt| = ηt ∗ |wt,s|= ηt ∗ |h4,s − h3| (97)

At state 4s , p4s = p4 = 0.8 MPas3 = s4s = 6.65912 kJ/kg < 6.6616 kJ/kg = s4s,v, so the state 4s is laid inthe liquid+vapor region!.

x =s4s − sfsfg

(98)

=6.65912− 2.0457

4.6160= 0.9994 (99)

h4s = hf + x ∗ hfg (100)

= 720.87 + 0.9994 ∗ 2047.7

= 2, 767.3 kJ/kg

→ |wt,s| = |2, 767.3− 3349.02| kJ/kg

= 581.72 kJ/kg (101)

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BDA 3043

The value of h4 can be now calculated:

h4 = 3, 349.02− 0.85 ∗ 581.72

= 2854.56 kJ/kg

With this enthalpy value at at the pressure of 0.8 MPa, the state 4 is su-perheated (h4 > 2768.3 kJ/kg), see steam table. p4 = 0.8 MPa

Number Temp. Enthalpy Entropy1 200 2839.8 6.8177- ? h4 s4

3 250 2950.4 7.0402

Linear interpolation is applied to determine s4:[h3 − h1

s3 − s1

]=

h4 − h1

s4 − s1

=⇒ s4 = s1 +[h4 − h1

h3 − h1

]∗ (s3 − s1) (102)

= 6.8177 +[

2854.56− 2839.82950.4− 2839.8

]∗ (7.0402− 6.8177)

=⇒ s4 = 6.8474 kJ/kg K (103)

• Determination of the property at state 6

At this stage the fluid is at saturated with the condenser pressure ofp6 = 10 kPa. The state properties are read at steam table:

vf [m]3/kg hf [kJ/kg] sf [kJ/kg K]0.001010 191.81 0.6492

• Calculation of the properties at state 7 , p7 = 0.8 MPa.

h7 = h6 + wp1

wp1 = vf (p7 − p6)

= 0.001010m3

kg∗ (800− 10) kPa

= 0.7979m3

kg∗ kPa ∗ N/m2

Pa

= 0.7979kNmkg

= 0.7979kJkg

=⇒ h7 = 191.81 + 0.7979

=⇒ h7 = 192.61kJkg

(104)

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BDA 3043


• Determination of the property at state 1

At this stage the fluid is at saturated with the open heater pressure ofp1 = 0.8 MPa. The state properties are read at steam table:

vf [m]3/kg hf [kJ/kg] sf [kJ/kg K]0.001115 720.87 2.0457

• The fraction of the total flow, y:

y =h1 − h7

h4 − h7

=720.87− 192.612854.56− 192.61

=⇒ y = 0.20 (105)

• To determine the efficiency in eq. (91), we need first to calculate boththe pump and turbine works according to eq. (92) and (93).

i. Calculation of the enthalpy at state 2 , p7 = 8 MPa.

h2 = h1 + wp2

wp2 = vf (p2 − p1)

= 0.001115m3

kg∗ (8− 0.8) MPa

= 0.00803m3

kg∗ MPa ∗ N/m2

Pa

= 8.03kJkg

=⇒ h2 = 720.87 + 8.03

=⇒ h2 = 728.9kJkg

(106)

ii. Calculation of the properties at state 5 , p5 = 10 kPa.s5s = s4.

h5 = h4 − wt

= h4 − ηt ∗ |wt,s|= h4 − ηt ∗ (h4 − h5,s) (107)

The state 5s is at the liquid + vapor region! So, we have to calculatex at this pressure.

x =s5s − sf

sfg

=6.8474− 0.6492

7.4996= 0.8265 (108)

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BDA 3043

h5s = hf + x ∗ hfg= 191.81 + 0.8265 ∗ 2392.1

= 2, 168.88 kJ/kg (109)

=⇒ h5 = h4 − ηt(h4 − wt)

= 2854.56− 0.85 ∗ (2, 854.56− 2, 168.88)

=⇒ h5 = 2271.732 kJ/kg (110)

iii. Total turbine work per unit total mass flow rate, wt:

wt =Wt

m=[(h4 − h3) + (1− y)(h5 − h4)

]= (2854.56− 3349.02) + (1− 0.2) ∗ (2271.732− 2854.56)

= −960.72 kJ/kg (work done by the system) (111)

iv. The pump work per unit of total mass is:

wp =Wp

m=[(h2 − h1) + (1− y)(h7 − h6)

]= 8.03 + (1− 0.2) ∗ 0.7979

= 8.67 kJ/kg (112)

v. Heat supply, q23.

q23 = h3 − h2

= 3349.02− 728.9

= 2620.12 kJ/kg (113)

The efficiency of the cycle is then:

ηth =|wt| − |wp||q23|

=960.72− 8.67

2620.12

=⇒ ηth = 0.363 = 36.3% (114)

b) the mass flow rate of the steam entering the first stage turbine.

Wcycle =∑

Net power output

= |Wt| − |Wp|= m ∗ (|wt| − |wp|)

m =Wcycle

|wt| − |wp|

=100

960.72− 8.67MW

kJ/kg∗ J

Ws

= 0.1050Mgs∗ 1000 k

M∗ 3600 s

hr

=⇒ m = 3.78 ∗ 105 kghr

(115)

19


BDA 3043


» A reheat – regenerative steam power plant operates with two feedwater heaters,a closed feedwater heater, and an open feedwater heater. Steam enters thefirst turbine at 8.0 MPa, 450 °C and expands to 0.8 MPa. The steam is reheatedto 450 °C before entering the second turbine. It expands through the turbine tothe condenser pressure of 10 kPa.Steam is extracted from the first turbine at 2 MPa and fed to the closed feed-water heater. Feedwater leaves the closed heater at 200 °C and 8 MPa, andcondensate exits as saturated liquid at 2 MPa. The condensate is trapped intothe open feedwater heater. Steam extracted from the second turbine at 0.3MPa and also fed into the open feedwater heater, which operates at 0.3 MPa.The stream exiting the open feedwater heater is saturated liquid at 0.3 MPa.The net power output of the cycle is 100 MW. The working fluid experiences noirreversibilities as it passes through the turbines, pumps, condenser, and steamgenerator. Determine:


b) The mass flow rate of the steam entering the first turbine.

Open

feedwater

heater

Pu

mp

1

Pump 29

10

11

8

7

3

y2

Co

nd

en

ser

6

2

1o450 C

12

m.

m2.

13

feedwater

heater

Closed

4

Steam generator

Reheater

Trap

First turbines

Second turbines

1(1 − y )

21

(1 −

y −

y )

5

y1

3m.

m.

m.

2

m5.

6m.

m6.

m.

m.

2

m.

m3.

8 MPa

8 MPa

10 kPa

0.3 MPa

2 MPasat.

liquid

o200 C

0.3 MPa

liquid

sat.

20



BDA 3043

Open

feedwater

heater

Pu

mp

1

Pump 29

10

11

8

3

6

2

1o450 C

12

m.

m2.

13

feedwater

heater

Closed

7

Co

nd

en

ser

o450 C

4

Steam generator

Reheater

Trap

First turbines

Second turbines

5

3m.

m.

m.

2

m5.

6m.

m6.

m.

m.

2

m.

m3.

8 MPa

8 MPa

0.3 MPa

2 MPasat.

liquid

o200 C

0.3 MPa

liquid

sat.

2 MPa

0.8

MP

a

0.3

MP

a

10 kPa

10 kPa

sat.liquid

a) The thermal efficiency is calculated as follows:

ηth =Net work output

Heat input=

Net power outputHeat flow input

=∑Woutput −

∑Winput∑

Qinput=|Wt| − |Wp||Qin|

=|Wt/m| − |Wp/m|

|Qin/m|(116)

Turbine power output:

Wt = Wt1 + Wt2

= W12 + W23 + W45 + W56

= mw12 + m3w23 + m3w45 + m6w56

wt ≡Wt

m= w12 +

m3

mw23 +

m3

mw45 +

m6

mw56 (117)

Pump power input:

Wp = Wp1 + Wp2

= m6wp1 + mwp2

wp ≡Wp

m=

m6

mw78 + w9−10

Heat flow input:

Qin = Qsteam gen. + Qreheater

= mqsteam gen. + m3qreheater

qin ≡Qin

m= q11-1 +

m3

mq34 (118)

The thermal efficiency is then:

=⇒ ηth =|wt| − |wp||qin|

(119)

21


BDA 3043


• Mass balance on turbines control volume (thick dashed lines):

m = m2 + m3

= m2 + m5 + m6

1 =m2

m+m5

m+m6

m(120)

fractions of the total flow are defined: y1 =m2

m, and y2 =

m5

m

→ m6

m= 1− y1 − y2 (121)

• Energy balance on the control volume of the Closed feedwater heater:

0 = H2 − H12 + H10 − H11

= m2h2 − m2h12 + mh10 − mh11

= m2(h2 − h12) + m(h10 − h11)

m2(h2 − h12) = m(h11 − h10)

m2

m≡ y1 =

h11 − h10

h2 − h12(122)

• Energy balance on the control volume of the Open feedwater heater:

0 = H5 + H8 + H13 − H9

= m5h5 + m6h8 + m2h13 − mh9

=m5

mh5 +

m6

mh8 +

m2

mh13 − h9

= y2h5 + (1− y1 − y2)h8 + y1h13 − h9

= y2[h5 − h8] + y1[h13 − h8] + [h8 − h9]

y2[h5 − h8] = y1[h8 − h13] + [h9 − h8]

y2 =h9 − h8 + y1(h8 − h13)

h5 − h8(123)

Calculation of the state properties at every location:

• State 1 . Reading at the steam table:

p[bar] t [°C] h[kJ/kg] s[kJ/kg K] Phase

80 450 3273.3 6.558 superheated

• State 2 . 1 – 2 is isentropic expansion. Reading at the steam table withp2 = 2 MPa = 20 bar:

22



BDA 3043

No. t [°C] h[kJ/kg] s[kJ/kg K]

1 250 2903.2 6.548

2 ? h2 6.558

3 300 3024.2 6.768

Linear interpolation to get the value of h4:

h2 = h1 +s2 − s1s3 − s1

∗ (h3 − h1)

= 2903.2 +6.558− 6.5486.768− 6.548

∗ (3024.2− 2903.2)

=⇒ h2 = 2, 908.70 kJ/kg (124)

1 4

5

8 MPa

450

10 kPa

3

67

2 MPa

2

12

0.3 MPa

0.8 MPa

9 13

s

Tem

pera

tur

8

20011

10

• State 3 . 1 – 3 is isentropic expansion (s1 = s3 = 6.558 kJ/kg K). Readingat the steam table with p3 = 0.8 MPa = 8 bar:

p[MPa] t [°C] Enthalpy[kJ/kg] Entropy[kJ/kg K]

hf hfg hg sf sfg sg

0.8 170.41 720.9 2047.4 2768.3 2.0457 4.6159 6.6616

Steam quality, x at state 3:

x =s3 − sfsfg

x =6.558− 2.0457

4.6159= 0.9776 (125)

h3 = hf + x ∗ hfg= 720.9 + 0.9776 ∗ 2047.4

=⇒ h3 = 2, 722.3478 kJ/kg (126)

23


BDA 3043


• State 4 . Reading at the steam table:

p[bar] t [°C] h[kJ/kg] s[kJ/kg K] Phase

8 450 3373.9 7.726 superheated

• State 5 . Reading at the pressure of p5 = 3 bar in the steam table:

No. t [°C] h[kJ/kg] s[kJ/kg K]

1 300 3069.6 7.704

5 ? h5 7.726

3 350 3172.0 7.875

Linear interpolation to get the value of h5:

h5 = h1 +s5 − s1s3 − s1

∗ (h3 − h1)

= 3069.6 +7.726− 7.7047.875− 7.704

∗ (3172.0− 3069.6)

=⇒ h5 = 3, 082.77 kJ/kg (127)

• State 6 . 4 – 6 is isentropic expansion (s6 = s4 = 7.726 kJ/kg K). Readingat the steam table with p3 = 10 kPa = 0.1 bar:

p[kPa] t [°C] Enthalpy[kJ/kg] Entropy[kJ/kg K]

hf hfg hg sf sfg sg

10 45.81 191.8 2392.1 2583.9 0.6492 7.4996 8.1488

Steam quality, x at state 6:

x =s6 − sfsfg

x =7.726− 0.6492

7.4996= 0.9436 (128)

h6 = hf + x ∗ hfg= 191.8 + 0.9436 ∗ 2392.1

=⇒ h6 = 2, 449.04 kJ/kg (129)

• State 7 . Reading at the steam table with p7 = 10 kPa = 0.1 bar, see thevalues in the state 6 above. The state 7 is saturated liquid.vf = 0.00101 m3/kg, h7 = 191.8 kJ/kg.

• State 8 . State change 7 – 8 is an isentropic compression in the liquidphase with the specific pump work:

wp1 = vf (p8 − p7)

= 0.00101m3

kg∗ (300− 10) kPa

N/m2

Pawp1 = 0.2929 kJ/kg (130)

24



BDA 3043

The specific enthalpy at the state 8 can be then calculated:

h8 = wp1 + h7

=⇒ h8 = 0.2929 + 191.8 = 192.09 kJ/kg (131)

• State 9 . It is saturated liquid at pressure of p9 = 0.3 MPa = 3 bar.

p[MPa] v[f ] Enthalpy[kJ/kg] Entropy[kJ/kg K]

hf hfg hg sf sfg sg

0.3 0.00107 561.4 2163.5 2724.9 1.6717 5.3199 6.9916

→ h9 = 561.4 kJ/kg• State 10 . State change 9 – 10 is an isentropic compression in the liquid

phase with the specific pump work:

wp2 = vf (p10 − p9)

= 0.00107m3

kg∗ (8000− 300) kPa

N/m2

Pawp2 = 8.239 kJ/kg (132)

The specific enthalpy at the state 8 can be then calculated:

h10 = wp2 + h9

=⇒ h10 = 8.239 + 561.4 = 569.639 kJ/kg (133)

• State 11 . It is a compressed liquid water at 8 MPa and 200 °C. The en-thalpy value at this state can be interpolated in the compressed liquidwater table.

p[MPa] h[kJ/kg]5 853.688 h11

10 855.80

h11 = h5 +h10 − h5p10 − p5

∗ (p8 − p5)

= 853.68 +855.80− 853.68

10− 5∗ (8− 5)

=⇒ h11 = 854.95 kJ/kg (134)

• State 12 . It is a saturated liquid state at 2 MPa. From the steam table,we read h12 = 908.47 kJ/kg.

• State 13 . 12 – 13 is a throttling process or isenthalpic process, so h13 =s12 = 908.47 kJ/kg.

25


BDA 3043


• Calculation of the fraction of the total flow y1 in eq. (122)

y1 =h11 − h10

h2 − h12

=854.95− 569.6392, 908.70− 908.47

= 0.143 (135)

• Calculation of the fraction of the total flow y2 in eq. (123)

y2 =h9 − h8 + y1(h8 − h13)

h5 − h8

= 0.092 (136)

• Turbine power output:

wt = w12 +m3

mw23 +

m3

mw45 +

m6

mw56

m3

m= 1− y1 = 0.857

m6

m= 1− y1 − y2 = 0.765

→ wt = (h2 − h1) + 0.857 ∗ [(h3 − h2) + (h5 − h4)] + 0.765(h6 − h5)

= −364.6 + 0.857 ∗ (−477.48) + 0.765 ∗ (−633.73)

→ wt = −1258.75 kJ/kg (137)

• Pump power input:

wp =m6

mw78 + w9−10

= (1− y1 − y2) ∗ (h8 − h7) + (h10 − h9)

= 0.765 ∗ (192.09− 191.80) + (569.64− 561.4)

= 8.46 kJ/kg (138)

• Heat flow input:

qin = q11-1 +m3

mq34

= (h1 − h11) + (1− y1) ∗ (h4 − h3)

= (3, 273.30− 854.95) + 0.857 ∗ (3, 373.90− 2, 722.35)

= 2976.97 kJ/kg (139)

• The thermal efficiency of the cycle is then:


=1258.75− 8.46

2976.97=

1250.292976.97

= 0.42

=⇒ ηth = 42% (140)

26



BDA 3043

b) The mass flow rate of the steam entering the first turbine.

Wcycle = m(|wt| − |wp|) (141)

m =Wcycle

|wt| − |wp|

=100

1250.29∗ MW

kJ/kg∗ J

Ws

= 0.08Mgs∗ 1000k

M∗ 3600 s

hr

=⇒ m = 2.88 ∗ 105 kghr

(142)

9

11

10

13

1

3

4

6

2

8 MPa

2 MPa

0.8 MPa

10 kPa

0.3 MPa

7

5

8

12

27


BDA 3043


¼ The schematic of a single-flash geothermal power plant with state numbers isgiven in the figure. Geothermal resource exists as saturated liquid at 230 °C. Thegeothermal liquid is withdrawn from the production well at rate of 230 kg/s, andis flashed to a pressure of 500 kPa by an essentially isenthalpic flashing processwhere resulting vapor is separated from the liquid in a separator and directedto the turbine.The steam leaves the turbine at 10 kPa with a moisture content of 10 percentand enters the condenser where it is condensed and routed to a reinjectionwell along with liquid coming off separator.

3

2

1

6

4

5

Turbine

Se

pa

rato

r

Condenser

Fla

sh c

ha

mb

er

Reinjection wellProduction well

4

3

5

1

26

10 kPa

s

2.797 MPa

4s

Tem

pera

tur

230

0.5 MPa

flashing

a) The mass flow rate of steam through the turbineFrom the saturated water table, we get the following state properties:

Temp [°C] psat [kPa]Enthalpy [kJ/kg]

hf hfg hg

230 2,797.1 990.14 1812.8 2802.9

28



BDA 3043

State 1 is at saturated liquid.

h1 = 990.14 kJ/kg

Process 1 – 2 is isenthalpic process to pressure of 500 kPa.

psat [kPa] Temp [°C]Enthalpy [kJ/kg] Entropy [kJ/kg K]

hf hfg hg sf sfg sg

500 151.83 640.09 2108.0 2748.1 1.8604 4.9603 6.8207

h1 = h2 = hf + xhfg (143)

x =h2 − hfhfg

=990.14− 640.09

2, 108.0x = 0.1661 (144)

m3 = x ∗ m1 (145)

= 0.1661 ∗ 230 kg/s

=⇒ m3 = 38.2 kg/s

b) the isentropic efficiency of the turbine Process 3 – 4 is isentropic expansionto pressure of 10 kPa.

psat [kPa] Temp [°C]Enthalpy [kJ/kg] Entropy [kJ/kg K]

hf hfg hg sf sfg sg

10 45.81 191.81 2392.1 2583.9 0.6492 7.4996 8.1488

s3 = s2,g = s4,s = 6.8207 kJ/kg K. Calculation of the steam quality at 10 kPa:

x =s3 − sfsfg

=6.8207− 0.6492

7.4996x = 0.8229 (146)

The isentropic enthalpy of the steam leaving the turbine:

h4s = hf + xhfg

= 191.81 + 0.8229 ∗ 2392.1

= 2160.27 kJ/kg (147)

29


BDA 3043


The actual enthalpy, h4, is calculated at the steam quality of x = 0.9 (Mois-ture content 10 % means the steam quality is 90 %).

h4 = hf + xhfg

= 191.81 + 0.9 ∗ 2392.1

= 2344.7 kJ/kg (148)

The turbine efficiency is calculated as follows:

ηt =|h4 − h3||h4s − h3|

=|2344.7− 2748.1||2160.27− 2748.1|

=403.4587.83

=⇒ ηt = 0.686 = 68.6%

c) the power output of the turbine

Wt = m3 ∗ wt= m ∗ (h4 − h3)

= 38.2kgs∗ (−403.4)

kJkg

= −15, 409.88 kW (minus sign means power output)

=⇒ Wt = 15.41 MW

d) the thermal efficiency of the plant

ηth =|Wt||Qin|

Energy input comes from the production well:

Qin = m1 ∗ h1

= 230 kg/s ∗ 990.14 kJ/kg

= 227.73 MW

The thermal efficiency is then:

ηth =15.41227.73

= 0.068

=⇒ ηth = 6.8%

30

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