For isentropic expansion, S’3S2
:=
x’3
S’3Sliq
−
∆Slv
:= x’30.855=
Chapter 8 – Section A – Mathcad Solutions
8.1 With reference to Fig. 8.1, SI units,
At point 2: Table F.2, H23531.5:= S26.9636:=
At point 4: Table F.1, H4209.3:=
At point 1: H1H4
:=
266
(c) The rate of heat addition, Step 1–2:
(d) The rate of heat rejection, Step 3–4:
H3 Hliq x3 Hvap Hliq−()⋅+:= H4 Hliq x4 Hvap Hliq−()⋅+:=
(e) Wdot12 0:= Wdot34 0:=
8.2 mdot 1.0:= (kg/s)
The following property values are found by linear interpolation in Table F.1:
State 1, Sat. Liquid at TH: H1 860.7:= S1 2.3482:= P1 3.533:=
State 2, Sat. Vapor at TH: H2 2792.0:= S2 6.4139:= P2 3.533:=
State 3, Wet Vapor at TC: Hliq 112.5:= Hvap 2550.6:= P3 1616.0:=
State 4, Wet Vapor at TC: Sliq 0.3929:= Svap 8.5200:= P4 1616.0:=
(b) Steps 2–3 and 4–1 (Fig. 8.2) are isentropic, for which S3=S2 and S1=S4.
Thus by Eq. 6.82):
267
8.3 The following vectors contain values for Parts (a) through (f).
Enthalpies and entropies for superheated vapor, Tables F.2 and F.4 @ P2
and T2 (see Fig. 8.4):
3622.7 kJ
kg
⋅
3635.4 kJ
kg
⋅
⎛
⎜
⎜
⎜
⎜
⎞
⎟
⎟
⎟
6.9013 kJ
kg K⋅
⋅
6.9875 kJ
kg K⋅
⋅
⎛
⎜
⎜
⎜
⎜
⎞
⎟
⎟
⎟
Sat. liq. and sat. vap. values from Tables F.2 and F.4 @ P3 = P4:
Hliq
191.832 kJ
kg
⋅
251.453 kJ
kg
⋅
419.064 kJ
kg
⋅
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎞
⎟
⎟
⎟
⎟
⎟
⎟
:= Hvap
2584.8 kJ
kg
⋅
2609.9 kJ
kg
⋅
2676.0 kJ
kg
⋅
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎞
⎟
⎟
⎟
⎟
⎟
⎟
:=
268
0.6493 kJ
kg K⋅
⋅
0.6493 kJ
kg K⋅
⋅
⎛
⎜
⎜
⎜
⎜
⎞
⎟
⎟
⎟
8.1511 kJ
kg K⋅
⋅
8.1511 kJ
kg K⋅
⋅
⎛
⎜
⎜
⎜
⎜
⎞
⎟
⎟
⎟
1.010 cm3
gm
⋅
1.017 cm3
gm
⋅
1.010 cm3
gm
⋅
0.0167 ft3
lbm
⋅
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
0.80
0.75
0.80
0.80
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
0.75
0.75
0.80
0.75
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎠
269
Answers follow:
QdotCQdotHWdot+:=
QdotHH2H1
−mdot⋅
()
→
⎯
⎯⎯⎯⎯⎯
⎯
:=mdot Wdot
Wturbine Wpump
+
→
⎯
⎯⎯⎯⎯⎯
⎯
:=
Wturbine H3H2
−:=H3H2ηturbine H’3H2
−
()
⋅+
⎡
⎣
⎤
⎦
→
⎯
⎯⎯⎯⎯⎯⎯⎯⎯⎯
⎯
:=
H’3Hliq x’3Hvap Hliq
−
()
⋅+
⎡
⎣
⎤
⎦
→
⎯
⎯⎯⎯⎯⎯⎯⎯⎯⎯
⎯
:=x’3
S2Sliq
−
Svap Sliq
−
→
⎯
⎯⎯
⎯
:=S’3S2
=
H1H4Wpump
+:=H4Hliq
:=
Wpump
Vliq P1P4
−
()
⋅
ηpump
→
⎯
⎯⎯⎯⎯
⎯
:=
10 kPa⋅
20 kPa⋅
⎛
⎜
⎜
⎞
⎟
10000 kPa⋅
7000 kPa⋅
⎛
⎜
⎜
⎞
⎟
80
100
⎛
⎜
⎜
⎞
⎟
270
⎜
⎜
⎜
⎜
⎜
⎜
The following vectors give values for temperatures of 450, 550, and 650 degC:
Svap 7.5947 kJ
kg K⋅
⋅:=Sliq 1.0912 kJ
kg K⋅
⋅:=
Hvap 2646.0 kJ
kg
⋅:=Hliq H4
:=
Saturated liquid and vapor at 50 kPa:
P150 kPa⋅:=
H4340.564 kJ
kg
⋅:=V41.030 cm3
gm
⋅:= P43300 kPa⋅:=
Saturated liquid at 50 kPa (point 4)
Subscripts refer to Fig. 8.3.8.4
271
The following vectors give values for pressures of 5000, 7500, and
10000 kPa at 600 degC
⎜
⎜
Wpump V4P4P1
−
()
⋅
⎡
⎣
⎤
⎦
:=
By Eq. (7.24),
Svap 7.7695 kJ
kg K⋅
⋅:=Sliq 0.9441 kJ
kg K⋅
⋅:=
P4
5000
7500
10000
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
kPa⋅:=
Hvap 2625.4 kJ
kg
⋅:=Hliq H4
:=
Saturated liquid and vapor at 30 kPa:
P130 kPa⋅:=H4289.302 kJ
kg
⋅:=V41.022 cm3
gm
⋅:=
Saturated liquid at 30 kPa (point 4)
Subscripts refer to Fig. 8.3.8.5
⎜
⎜
⎜
⎜
ηWturbine Wpump
+
QH
→
⎯
⎯⎯⎯⎯⎯⎯
⎯
:=
QHH2H1
−
()
:=
Wturbine H’3H2
−:=H’3Hliq x’3Hvap Hliq
−
()
⋅+:=
x’3
S’3Sliq
−
Svap Sliq
−
:=S’3S2
:=
272
⎜
⎜
⎜
⎜
S2
7.4898
7.4797
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
kJ
⋅:=H2
3194
3206.8
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
kJ
=W12
572.442−
559.572−
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
kJ
=
H2H1W12
+:=W12 ηH’2H1
−
()
⋅:=η 0.78:=
H’2
3023.9
3032.5
3049.0
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
kJ
⋅:=P2
725
750
800
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎠
kPa⋅:=
The following enthalpies are interpolated in Table F.2 at four values for
intermediate pressure P2:
For sat. liq. and sat. vap. at 20 kPa:
S’2S1
:=S17.2200 kJ
kg K⋅
⋅:=H13766.4 kJ
kg
⋅:=
From Table F.2 at 7000 kPa and 640 degC:8.6
⎜
⎜
Wturbine H’3H2
−:=H’3Hliq x’3Hvap Hliq
−
()
⋅+:=
x’3
S’3Sliq
−
Svap Sliq
−
:=S’3S2
:=
273
W23 ηH’3H2
−
()
⋅:=H’3Hliq x’3Hvap Hliq
−
()
⋅+:=
x’3
S2Sliq
−
Svap Sliq
−
:=W12 H2H1
−:=
The work calculations must be repeated for THIS case:
We can now find the temperature at this state by interplation in Table F.2.
This gives an intermediate steam temperature t2 of 366.6 degC.
H23197.9 kJ
kg
⋅:=linterp P2H2
,765.16 kPa⋅,
()
3197.9 kJ
kg
=
Also needed are values of H2 and S2 at this pressure. Again we do linear
interpolations:
The work difference is essentially linear in P2, and we interpolate linearly to
find the value of P2 for which the work difference is zero:
H’3Hliq x’3Hvap Hliq
−
()
⋅+:=x’3
S2Sliq
−
Svap Sliq
−
:=
where the entropy values are by interpolation in Table F.2 at P2.
274
For a single isentropic expansion from the initial pressure to the
final pressure, which yields a wet exhaust:
x’3
S1Sliq
−
Svap Sliq
−
:= H’3Hliq x’3Hvap Hliq
−
()
⋅+:=
Whence the overall efficiency is:
275
Exhaust is wet: for sat. liq. & vap.:
S’4S2
:=
Isentropic expansion to 20 kPa:
By interpolation at 350 kPa and this entropy,
S’3S2
:=S27.0311 kJ
kg K⋅
⋅:=H23439.3 kJ
kg
⋅:=
From Table F.2 for steam at 4500 kPa and 500 degC:
8.7
276
P1P6
:=β 1
Vsat.liq
1.083 1.063−
20
⎛
⎜
⎝
⎞
⎠
⋅cm3
gm K⋅
⋅:=
Also by approximation, the definition of the volume expansivity yields:
Vsat.liq 1.073 cm3
gm
⋅:=Psat 294.26 kPa⋅:=Hsat.liq 558.5 kJ
kg
⋅:=
At this temperature, 132.87 degC, interpolation in Table F.1 gives:
We need the enthalpy of compressed liquid at point 1, where the pressure is
4500 kPa and the temperature is:
(degC)
t7138.87:=H7584.270 kJ
kg
⋅:=
For sat. liq. at 350 kPa (Table F.2):
H6H5Wpump
+:=Wpump
V5P6P5
−
()
⋅
η
:=
P64500 kPa⋅:=P520 kPa⋅:=V51.017 cm3
gm
⋅:=H5Hliq
:=
H’4Hliq x’4Hvap Hliq
−
()
⋅+:=x’4
S’4Sliq
−
:=
277
8.8 Refer to figure in preceding problem.
Although entropy values are not needed for most points in the process, they are
recorded here for future use in Problem 15.8.
From Table F.4 for steam at 650(psia) & 900 degF:
H21461.2 BTU
lbm
⋅:= S21.6671 BTU
lbmrankine⋅
⋅:= S’3S2
:=
By interpolation at 50(psia) and this entropy,
By Eq. (7.25),
H1Hsat.liq Vsat.liq 1βT1
⋅−
()
⋅P1Psat
−
()
⋅+:= H1561.305 kJ
kg
=
By an energy balance on the feedwater heater:
Work in 2nd section of turbine:
WII 1kg⋅mass−()H
4H3
−
()
⋅:= WII 307.567−kJ=
P51 psi⋅:= H5Hliq
:= V50.0161 ft3
lbm
⋅:=
For sat. liq. at 50(psia) (Table F.4):
H7250.21 BTU
lbm
⋅:= t7281.01:= S70.4112 BTU
lbmrankine⋅
⋅:=
We need the enthalpy of compressed liquid at point 1, where the pressure is
650(psia) and the temperature is
Isentropic expansion to 1(psia): S’4S2
:=
Exhaust is wet: for sat. liq. & vap.:
Hliq 69.73 BTU
lbm
⋅:= Hvap 1105.8 BTU
lbm
⋅:=
Sliq 0.1326 BTU
lbmrankine⋅
⋅:= Svap 1.9781 BTU
lbmrankine⋅
⋅:=
x’4
S’4Sliq
−
Svap Sliq
−
:= H’4Hliq x’4Hvap Hliq
−
()
⋅+:=
x’40.831=H’4931.204 BTU
lbm
=
279
WII 158.051−BTU=WII 1lb
m
⋅mass−
()
H4H3
−
()
⋅:=
Work in 2nd section of turbine:
H1257.6 BTU
lbm
=H1Hsat.liq Vsat.liq 1βT1
⋅−
()
⋅P1Psat
−
()
⋅+:=
By Eq. (7.25) and (7.26),
β4.95 10 5−
×1
rankine
=
P1P6
:=β 1
Vsat.liq
0.01726 0.01709−
20
⎛
⎜
⎝
⎞
⎠
⋅ft3
lbmrankine⋅
⋅:=
The definition of the volume expansivity yields:
At this temperature, 270.01 degF, interpolation in Table F.3 gives:
280
WIηH’3H2
−
()
⋅:=η 0.80:=H’33142.6 kJ
kg
⋅:=
By double interpolation in Table F.2,
S’3S2
:=
At point 3 the pressure must be such that the steam has a condensation
temperature in feedwater heater I of 195 degC, 5 deg higher than the
temperature of the feed water to the boiler at point 1. Its saturation pressure,
corresponding to 195 degC, from Table F.1, is 1399.0 kPa. The steam at point 3
is superheated vapor at this pressure, and if expansion from P2 to P3 is
isentropic,
8.9
281