College of Engineering
Spring Semester- 2016
Thermodynamics II - ME 272
Section: MA, MB and MC
Due Day: Thursday of Week4
PROBLEM 1.1 STATEMENT
(25%)
Steam enters a two-stage adiabatic turbine at 8 MPa and 500oC. It expands in the first stage to a
state of 2MPa and 350oC. Steam is then reheated at constant pressure to a temperature of 500oC
before it is routed to the second stage, where it exists at 30 kPa and a quality of 97 %. The work
output of the turbine is 5MW. Assuming the surrounding to be at 25oC, determine the reversible
power output and the rate of the exergy destruction within this turbine.
Note: the surrounding state is dead state.
Fig.1.1 Two Stages Turbine
Solution:
State
P1=8 MPa
T1=500oC
T2=350 oC
P2=2 MPA
P3=2MPa
T3=500oC
P4=30 kPa
x=97%
Phase Description
Superheated vapor
Enthalpy [h] kJ/kg
h1=3399.5
Entropy in [s] kJ/kg.K
s1= 6.7266
Superheated vapor
h2=3137.7
s2= 6.9583
Superheated vapor
h3=3468.3
s3= 7.4337
Sta. liquid Vapor
Mixture
h4=hf + xhfg=2554.51
s4=sf + xsfg=7.56279
State (2): Tgiven is greater than Tsat at P = 2 MPa , [From Table A-5].
Thermodynamics II-ME 272
College of Engineering
Spring Semester- 2016
Thermodynamics II - ME 272
Section: MA, MB and MC
Due Day: Thursday of Week4
State (3): Tgiven is greater than Tsat at P = 2MPa=2000 kPa , [From Table A-5].
The state is superheated.
Thermodynamics II-ME 272
College of Engineering
Spring Semester- 2016
Thermodynamics II - ME 272
Section: MA, MB and MC
Due Day: Thursday of Week4
State [4] Saturated Liquid Vapor Mixture:
h3 = 289.27 + 0.97 × 2335.3 = 2554.51
kJ
kJ
, s3 = 0.9441 + 0.97 × 6.8234 = 7.56279
kg
kg. K
Ẇnet = Ẇu = ṁ[h1 − h2 + h3 − h4 ] → ṁ =
=
Ẇnet
h1 − h2 + h3 − ℎ4
5000
kg
= 4.25
3399.5 − 3137.7 + 3468.3 − 2554.51
s
Ẇrev = Ẇrev = ṁ[[h1 − h2 + h3 − h4 ] − To(s1 − s2 + s3 − s4 )]
= 4.25
× [3399.5 − 3137.7 + 3468.3 − 2554.51 − 298
× (6.7266 − 6.9583 + 7.4337 − 7.56279)] = 5453.19 kW
XDestroyed = Ẇrev − Ẇ u = 5453.51 − 5000 = 453.51 kW
PROBLEM 1.2 STATEMENT
(25%)
In large steam power plants, the feed water is frequently heated in closed feed water heaters, which
are basically heat exchangers, by steam extracted from the turbine at some stage. Steam enters the
feed water heater at 1MPa and 200oC and leaves as staurated liquid at the same pressure. Feed
water enters the heater at 2.5 MPa nad 50oC and leaves 10oC below the exist temperature of the
steam.
Thermodynamics II-ME 272
College of Engineering
Spring Semester- 2016
Thermodynamics II - ME 272
Section: MA, MB and MC
Due Day: Thursday of Week4
Neglect the heat losses from the outer surfaces of the heater, determine (a) the ratio of the mass
flow rates of the extracted steam and the feed water heater and (b) the reversible work for this
process per unit mass of the feed water. Assume the surroundings to be at 25oC.
Fig.1.2 Heat exchanger
Solution:
State
P1=1MPa
T1=200oC
[Hot Stream]
Sat. Liquid
T2=Tsat.=179.88
P2=1 MPa
[ Hot stream]
P3=2.5MPa
T3=50oC
[Cold Stream]
P4=2.5 MPa
T4=T2-10=170oC
[Cold Stream]
Phase Description
Superheated vapor
Enthalpy [h] kJ/kg
h1= 2828.3
Entropy in [s] kJ/kg.K
s1= 6.6956
Sat. Liquid
h2=hf=761.68
s2= 2.1387
Compressed
Liquid
h3=hf@T=50C = 209.34
s3= sf@T=50C=0.7038
Compressed
Liquid
h4=hf=718.33
s4=2.0419
1
3
4
2
ṁ 1 h1 + ṁ 3 h3 = ṁ 2 h2 + ṁ 4 h4
Thermodynamics II-ME 272
College of Engineering
Spring Semester- 2016
Thermodynamics II - ME 272
Section: MA, MB and MC
Due Day: Thursday of Week4
ṁ 1 = ṁ 2 = ṁCold
ṁ 3 = ṁ 4 = ṁHot
mHot h4 − h3 718.33 − 209.34
=
=
= 0.2463
mCold h1 − h2 2828.3 − 761.68
Ẇrev = [(h4 − h3 ) − To (s4 − s3 )] = [(718.33 − 209.34) − 298 × (2.0419 − 0.7038)]
kJ
= 110.23
kg
PROBLEM 1.3 STATEMENT
(25%)
Steam enters an adiabatic turbine at 6 MPa, 600oC, and 80 m/s and leaves at 50 kPa, 100oC, and
140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output
and (b) the second –law efficiency of the turbine. Assume the surrounding to be at 25oC
Fig.1.3 Steam turbine
Solution:
State
P1=6MPa
T1=600oC
[Hot Stream]
T2=100C
P2=50kPa
Phase Description
Superheated vapor
Enthalpy [h] kJ/kg
h1= 3658.8
Entropy in [s] kJ/kg.K
s1= 7.1693
Superheated vapor
h2=2682.4
s2= 7.6953
Thermodynamics II-ME 272
College of Engineering
Spring Semester- 2016
Thermodynamics II - ME 272
Section: MA, MB and MC
Due Day: Thursday of Week4
V12 − V22
]
2000
= 5.155[(3658.8 − 2682.4) − 298(7.1693 − 7.6953) + (802 − 1402 )/2000
= 5807.4 kW
Ẇrev = 𝑚̇[(h1 − h2 ) − To (s1 − s2 ) +
𝑊̇u = 𝑚̇[(3658.8 − 2682.4) + (802 − 1402 )/2000] = 5000
𝑚̇ =
5000
𝑘𝑔
= 5.155
2
2
(3658.8 − 2682.4) + (80 − 140 )/2000
𝑠
𝜂𝐼𝐼 =
𝑊𝑢̇
5000
=
= 0.861
𝑤𝑟𝑒𝑣
̇
5807.4
PROBLEM 1.4 STATEMENT
(25%)
A hot-water stream at 160oF enters an adiabatic mixing chamber with a mass flow rate of 4 Ibm/s,
where it is mixed with a stream of cold water at 70oF. If the mixture leaves the chamber at 110oF,
determine (a) the mass flow rate of the cold water and (b) the exergy destroyed during this adiabatic
mixing process. Assume all the streams are at a pressure of 50 psia and the surroundings are at
75oF.
Ans. [(a) 5.0 Ibm/s, (b) 14.6 Btu/s]
Stream (1) is a hot stream.
Stream (2) is a cold stream.
Energy Balance:
ṁ1 h1 + ṁ2 h2 = ṁ3 h3
Thermodynamics II-ME 272
College of Engineering
Spring Semester- 2016
Thermodynamics II - ME 272
Section: MA, MB and MC
Due Day: Thursday of Week4
Mass Balance:
ṁ1 + ṁ2 = ṁ3
ṁ1 h1 + ṁ2 h2 = (ṁ1 + ṁ2 )h3 ⇾ ṁ2 =
State
Phase Description
P1=50 psai
Compressed liquid
T1=160F
[Hot Stream]
T2=70F
Compressed Liquid
P2=P1
T3=110F
Compressed liquid
P3=P1
To [R] =T[F]+460=75+460=535 R
ṁ1 (h1 − h3 )
128 − 78.02
=4×
= 5.0 Ibm
h3 − h2
78.02 − 38.08
Enthalpy [h] Btu
h1=128
Entropy in [s] btu/Ibm R
s1=0.23136
h2=38.08
s2=0.07459
h3= 78.02
s3=0.14728
Ẇrev = ṁ1 (h1 − To s1 ) + ṁ2 (h2 − To s2 ) − ṁ3 (h3 − To s3 ) = 4 × (128 − 535 × 0.23136) +
5 × (38.08 − 535 × 0.07459) − 9 × (78.02 − 535 × 0.14728) = 14.73 Btu/s
Ẋ destroyed = Ẇrev − Ẇu = Ẇrev = 14.73
Thermodynamics II-ME 272
Btu
, since Ẇu = 0
s