School of Engineering, FETBE

EM203/208 (January 2013 Semester) - Assignment 3

1. A saturated liquid–vapor mixture of water at 150 kPa is contained in a well-insulated rigid

tank. Initially, the quality of the mixture is 30%. An electric resistance heater placed in the

tank is now turned on and kept on until only saturated vapor exists in the tank. If the initial

mass of the vapor in the tank is 0.6 kg, determine (a) the final mass of the saturated vapor in

the tank, and (b) entropy change of the steam during this process.

2. Superheated steam enters an adiabatic turbine at a rate of 5 kg/s. The pressure and

temperature of the steam at the turbine inlet is 8 MPa and 550°C. The steam leaves the turbine

at 20 kPa. If the isentropic efficiency of the turbine is 90%, determine (a) the temperature at

the turbine exit and (b) the power output of the turbine.

3. The power output of the turbine is 5 MW. Steam enters the turbine at 10 MPa, 550°C, and

50 m/s and leaves at 20 kPa and 120 m/s with a moisture content of 5 percent. The turbine is

not adequately insulated and it is estimated that heat is lost from the turbine at a rate of 300

kW. Assuming the surroundings to be at 25°C, determine (a) the reversible power output of

the turbine, (b) the exergy destroyed within the turbine, and (c) the second-law efficiency of

the turbine.

4. The pressure ratio of a simple Brayton cycle using air as the working fluid is 10. The

isentropic efficiencies of the compressor and turbine are 80% and 85% respectively.

Assuming a minimum and maximum temperatures in the cycle of 320 and 1180 K

respectively, determine (a) the air temperature at the turbine exit, (b) the net work output, and

(c) the thermal efficiency.

Note: Use Table A-17 to solve this problem.

.

8 MPa, 550°C

20 kPa

.Solution

1.

From the steam tables (Tables A-4 through A-6)

P1=150 kPax1=0.3 }v1=v f +x1 v fg=0.001053+0.3 (1.1594−0.001053 )=0.34856 m3 /kg

s1=sf +x1 sfg=1.4337+0.3 (5.7894 )=3.17052 kJ /kg ∙K

v2=v1

sat. vapour }s2=6.79453 kJ /kg ∙ K (by interpolation, see below )

Interpolation for s2

s2−6.8207

6.7886−6.8207=0.34856−0.37483

0.34261−0.37483→ s2=6.79453 kJ /kg ∙ K

(a) The mass remains the same, so the final mass of the saturated vapor equals the total initial

mass. Now

x1=m1 , g

mt

→ mt=m1 , g

x1

=0.6 kg0.3

=2kg

Thus

m2 , g=mt=2kg

(b) The entropy change is

∆ S=m ( s2−s1 )= (2kg ) (6.79453−3.17052 ) kJ /kg ∙ K=7.248 kJ / K

H2Ox1 = 0.3

P1 = 150 kPa

2. From the steam tables:

P1=8 MPaT1=550° C} h1=3521.8 kJ /kg

s1=6.8800 kJ /kg ∙ K

Above data are obtained from Table A-6

P2=20 kPas2 s=s1

} x2 s=s2 s−s f

sfg

=6.8800−0.83207.0752

=0.8548

h2 s=hf +x2 s h fg=251.42+0.8548 (2357.5 )=2266.6 kJ / kg

Above data are obtained from Table A-5 (see below)

From the isentropic efficiency relation,

ηT=h1−h2 a

h1−h2 s

→ h2 a=h1−ηT (h1−h2 s )=3521.8−0.9 (3521.8−2266.6 )=2392.1 kJ /kg

Thus

P2 a=20 kPah2 a=2392.1 kJ /kg }T 2 a=T sat @20kPa=60.06 °C

since at 20 kPa, hf = 251.42 and hg = 2608.9 so that hf > h2a > hg.

(b) There is only one inlet and one exit, and thus m1=m2=m. We take the actual turbine as the

system, which is a control volume since mass crosses the boundary. The energy balance for

this steady-flow system can be expressed in the rate form as

E¿−Eout⏟Rate of net energy transfer

byheat ,work∧mass

= ∆ E system⏟Rate of change∈internal , kinetic

potential ,etc .energies

Thus

E¿¿ Eout

m h1=W out+mh2 a ( sinceQ=∆ KE=∆ PE=0 )

W out=m ( h1−h2 a )Substituting

W out=(5 kg /s ) (3521.8−2392.1 ) kJ /kg=5648.5 kW

0 (steady flow)

3. (a) From the steam tables:

P1=10 MPaT 1=550 °C } h1=3502.0 kJ /kg

s1=6.7585 kJ /kg ∙ K

Above data are obtained from Table A-6

P2=20 kPax2=0.95 } h2=hf +x2h fg=251.42+0.95 (2357.5 )=2491.05 kJ /kg

s2=sf +x2 sfg=0.8320+0.95 (7.0752 )=7.55344 kJ /kg ∙K

Above data are obtained from Table A-5 (see below)

The enthalpy at the dead state is,

T o=25 °Cxo=0 }ho=h f @25 °C=104.83 kJ /kg

The mass flow rate of steam may be determined from an energy balance on the turbine

m(h1+V 1

2

2 )=m(h2+V 2

2

2 )+Qout+W out , actual

m [3502.0 kJ /kg+(50 m /s )2

2 ( 1 kJ /kg

1000 m2/s2 )]=m [2491.05 kJ /kg+(120 m /s )2

2 ( 1 kJ /kg

1000 m2/s2 )]+300 kW +5000 kW

Solving,

m=5.274 kg/ s

The reversible power may be determined from

W rev=m [ (h1−h2 )−T o ( s1−s2)+V 1

2−V 22

2 ]¿(5.274

kgs ) [(3502.0−2491.05 ) kJ

kg−(298 K ) (6.7585−7.55344 ) kJ

kg ∙ K+

(50 m/ s )2−(120 m /s )2

2 ( 1 kJ /kg

1000 m2/ s2 )]Thus

W rev=6822.42 kW =6.822 MW

(b) The exergy destroyed in the turbine is

X dest=W rev−W out ,actual=6.822−5.0=1.822 MW

10 MPa550°C50 m/s

20 kPax = 0.95120 m/s

(c) The second-law efficiency is

η II=W out ,actual

W rev

= 56.822

=0.733∨73.3 %

4. The T-s diagram of the simple Brayton cycle is shown below.

(a) The properties of air are given in Table A-17.

T 1=320 K →h1=320.29kJ/kgPr1=1.7375

Pr 2=P2

P1

Pr 1=(10 ) (1.7375 )=17.375

Interpolating ,17.375−17.3018.36−17.30

=h2 s−617.53

628.07−617.53→ h2 s=618.28

kJkg

ηC=h2 s−h1

h2−h1

=618.28−320.29h2−320.29

→ h2=320.29+ 618.28−320.290.8

=692.78 kJ/kg

T 3=1180 K → h3=1254.34 kJ/kgPr3=222.2

Pr 4=P4

P3

Pr 3=( 110 )(2 22.2 )=2 2. 22

Interpolating ,22 . 2 2−2 1. 862 3 .13−21.86

=h4 s−6 59. 84

670.47−659.84→h4 s=662. 85

kJkg

320 K

1180 K

ηT=h3−h4

h3−h4 s

→h4=h3−ηT ( h3−h4 s )

Thus

h4=1254.34−(0.85 ) (1254.34−662.85 )=751.57 kJ/kg

¿T 4−7 3 0

7 4 0−7 3 0=7 51.5 7−7 45 .6 2

75 6 . 44−745.62→ T 4=7 3 5.5 K

(b )q¿=h3−h2=1254.34−692.78=561.56 kJ/kg

qout=h4−h1=751.5 7−320.29=4 31 .2 8kJ/kg

wnet , out=q¿−qout=561.56−4 31 .28=1 30 .28 kJ/kg

(c )η th=wnet , out

q¿=130 .2 8kJ/kg

561.56 kJ/kg=0.2 3 2∨23 .2%