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Unformatted text preview: Chapter 4 Practice Problems
(P4.1) COP = coef. of performance = QC WS ,net
Using state numbers of Fig 4.94.10. PH plot will look like Fig 4.10 of pg 151. Use PH
chart pg 653 and table pg 654:
state 2 is satV at –40C ! H2 = 372 kJ/kg (chart)
state 3, outlet of the reversible compressor is found by following the isentropic
line to 40C, where H3’ = 438 kJ/kg.
state 4 is satL at H4 = 256 kJ/kg (table)
state 1, H1 = H4 QC = (H 2 − H 1 ) = 372 − 256 = 116kJ / kg
WS = (H3 – H2) = 438 – 372 = 66 kJ/kg
&
Q
116
⇒ COP = C =
= 1.76
&
66
W
S QC = COP (WS) = 1.76 (9000 J/day) = 16 kJ/day
(P4.2) Methane,. See Fig 4.94.10 pg 1501.Methane chart pg 651.
Tevap = −280 O F , P = 0.032MPa
⇒ H 2satV = 339 Btu / lbm
S 2satV = 2.35Btu / lb.O F
Pressure of 3’ isn’t given, however, it will be the same as the condenser outlet (P4) since
the condenser will be considered to be isobaric. State 4 will be satL at 40F, so P4 = P3’ =
2 MPa. Follow isentropic line from the compressor inlet state to 2MPa. At P3’=2MPa,
and S3’ = 2.35 Btu/lbmF, H3’ = 485 Btu/lbm, and T3’ = 40F. At the outlet of the
condenser, the fluid will be saturated liquid at 2 MPa,
T4 = −160 O F , P4 = 2MPa
⇒ H 4satL = 218Btu / lbm
satliq S4 1.54 Btu / lb.O F State
1
2 Sat'd vap
3'
4 COP = O T( F) O P(MPa) H(Btu/lb) S(Btu/lb* F)
280
0.032
218
280
0.032
339
2.35
40
485
2
2.35
160
2
218
1.54 QC
H 2 − H1
H − H 4 339 − 218
=
=2
=
′
′
H 3 − H 2 WS ,net H 3 − H 2 485 − 339 ⇒ COP = 0.83
(P4.3) Rankine Cycle, Fig 4.3 Pg. 143
temperatures 200 O C & 99.6 O C , Sat vapor from Turbine outlet can be found in the sat
Pressure table p642.
To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/24/2001) 1 of 2 Chapter 4 Practice Problems
The turbine is assumed to be adiabatic and reversible since no other specifications are
given. The inlet state will then have the same entropy as the outlet. Since the
boiler/superheater outlet is to operate at T3 = 200 C, and S3 = 7.3592, we must hunt in the
superheated tables to find this combination. This will occur between 0.2 and 0.3 MPa.
Interpolating, the value of H can be found. Values are tabulated below:
O State
TC
P(MPa) H(kJ/kg) S(kJ/kg*K) V(m3/kg)
4’ sat vap
99.6
0.098
2675
7.359
5 sat liq.
99.6
0.098
417
1.3028 0.001043
3
200
0.271
2867
7.3592
6
417.18 ′
WS′,turbine = (H 4 − H 3 ) = 2675 − 2867 = −192kJ / kg
∆H pump = WS , pump = ∫ VdP = V∆P 1043cm 3 * (0.271 − 0.098)
1J
kJ
*3
*3
kg
cm * MPa 10 J
= 0.180kJ / kg. ⇒ ∆H pump = ⇒ ∆H pump
⇒ WS′,net = −192 + 0.180 = −191.82kJ / kg
&η = − WS′,net
Qboiler = − (− 191.82 ) − (− 191.82 )
=
(H 3 − H 6 ) (2867 − H 6 ) ⇒ H 6 = 417 + 0.18 = 417.18kJ / kg − (− 191.82 )
= 0.0783
(2867 − 417.18)
⇒ η = 7.83%
⇒η = To accompany Introductory Chemical Engineering Thermodynamics
© J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/24/2001) 2 of 2 ...
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This note was uploaded on 01/31/2010 for the course CHEM 2106 taught by Professor Liz during the Spring '08 term at University of Minnesota Crookston.
 Spring '08
 LIZ

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