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Unformatted text preview: ERQBLEMAJ
Kuowu a FIND; A rigid tank containing 10 kg of air is being heated. If it is desired to increase the
thperature of the air lOO'C in a period of 1000 s. at what rate must heat be added? SEEEMAILQ. Rigid Tank M =10 kg, AT =100'C,At=1000 sec.
Assume process is reversible. W = 0 QW=%Ei‘ E=U+M‘2v2+Mgz dE=du+ dMV2 + degz ' ﬁ=du dt dt dt dt ’ dt dt For an ideal gas: U = Mcva ; dgtl = Maggi: = 161ch Q = McydI ; For air: cv= 0.7165 kJIkgK Table A7. dt 0 =(—M——1'10H0'17£3 ' 10° = 0.7165 kw ¢=¢=¢=¢=¢=¢= ERQBLEMM w; A system consists of a ﬂywheel with a mass of 70 kg and a mass moment of inertia of 5.096 lsgm2 about its axis of rotation. The temperature of the system remains
constant. FIND ' If there is no heat transfer, how much work is done when the system changes from an
elevation of 1.0 to 11.0 m while its speed of rotation changes from 10 to 5.64
revolutions per second? HE TI ' ASSUMPTIQNS; ERQPERTIES; AEALXSLS:
Thesystemisarotatingmass. QW=AE ; E=U+KE+PE
The process is assumed to be adiabatic. Furthermore, since the system is isothermal we will assume AU = 0. W = AKE + APE _ KE =% for a rotating system where: I = mass moment of inertia = 5.069 kgm2
co = angular rotation = rad/sec. 0)] = 10.0 x 21: rad/see, (02 = 5.64 x 21: rad/sec. AKE = I(mz2  m12)/2 = 5.069 (5.642  102 )(2n)2 l2.0 = 6859.3 J ‘1
APE = Mg(Zz  21) = (70.0)(9.so7)(11.0  1.0) = + 6864.9] w = 6859.3 + 6864.9 . w = 5.60] ==¢==‘=: _. I. RRQBLEMAAE KNQWE ; A pressure cooker with inside volume of 2 liters operates at a pressure of 2 atm. with
water at a quality of 0.5. After operation, the pressure cooker is left aside allowing its
contents to cool. E IED ; If the rate of heat loss is 50 Watts. how long does it take for the pressure inside to drop
to 1 atm. What is the state of the water at this point? Indicate the process on a T—v
diagram. ‘ SQHEMAIIQ
W
PRQPERTIES ;
ANALXSJSJ (v01.)prcssure cooker = 2 liters = 2 X m3 During operation: P1 = 2 atm = 0.2 MPa, x1 = 0.5
/ At 0.2 MPa. speciﬁc volume, v1 = Vf + )6va = 0.001061 + 0.5(0.8857  0.001)
“’ ‘ v1 = 0.443381 m3 [kg mass of water in the pressure cooker :2 x 10'3I0.443381 = 4.511 x 1‘03 kg===°=5= At state (2)  After cooling, Final pressure, P2 = 0.1 MPa ====¢==
Since the pressure cooker vessel is rigid. the total volume remains the same.
Since no water leaves the system, the speciﬁc volume Stays constant. v2 = 1/1 = 0.4434 m3 lkg,
At a pressure of 0.1 MPa, vr= 0.001043 m3 lkg. and v3 = 1.694 m3 [kg
7 Therefore. the water is in the saturated state at (2), v2 = v; + 11201g  w) = xjVg or. 142.: V2 / vs = 0.4434 I 1.694 = 0.262 kg of steam per 1 kg of mixture. °=¢=¢==¢=
Amount of heat'transfer in going from (1) to (2). Q14 = AU + Wm = AEXconst. Vol. process) Therefore. Q12 = (U2  U1)  M(uz'  111) u: = u. + 142 um = 417.36 + 262320887 = 964.6 kJ/kg u: u. + x1 .ufg = 504.5 4L9.5)2025 = 1517 kakg mag = 4.511,; 103  (946.6 — 1517) x 103 = 2573 k]
Total heat loss = 2.513 U. Rate ofHeat Loss = 50 W = 50 1/3
or time taken to cool = 2.573 I 0.05 = 51.46 seconds. (or slightly less than a minute) W KN WN' A system containing 3.0 ltg of air is operated in a cycle consisting of the following
three process:
[2 A constantvolume heat addition: P1 = 0.1 MPa, T1 = 20°C, ansz = 0.2 MPa.
23 A constanttemperature heat addition.
31 A constantpressure heat rejection. mm (a) Sketch this cycle on a Pv dia mm.
(b) Compute the work on each of5 the three processes 12. 23, and 31.
(c) What is the net cycle work?
(d) What is the net cycle heat transfer?
(e) What is the cycle thermal efﬁciency? ¢=¢=¢=c=¢=¢=
w :1 Pd(Voi.)
0:) W (Vol.) = constant Wm = o =¢=¢=<=<=<=
W Isothermal. Since P(Vol.) = MoRT and T = eonst.
J J _
_ I _ Cd(Vol.) _ _ ((Vol.)3) _ . _ _ _
W23 1 P d(Vol.) —£ ——(V0L) — C 1n (—Volh . C  P (VOL) — M R T W24, = MRT 1n (gig—K) For air R = 0.287 kJ/kgK Table A7, cv= 0.7165
_ MRTl _(3}(01237)(293.15) _ _
( (V05); ——pl — ——(0.1)_(1000) — 2.524 —(Vol.)2 { H
_ P2'V01.h _ . 1ooo2.524 _ _ _ _ MRT3 _ 3 .287 586.291_
T2 — MIR — ——~————(3)_(0.287) — 586.29 K _ T3, (Vol.)3 — ——————P3 — OIHOOO — 5.048
W24, = (3}(0.287}(586.29) in (345% = 349.90 It] ¢==°=°=°=<=
' l
W34 =1 P'd(VOL) = P'((V0].)1  (VOLb)
Emma Constant pressure 3
W34 = (0.1 1000}(2.524  5.048): —252.4 k] ¢=¢=¢=¢=¢=¢=
(C) 2‘” = Wlz + “’23 + W34 = o + 34939  252.4 = + 97.50 1d ¢===<=<=¢=
(d) For a cyclic process 2Q = 2W = 97,50 1;} ¢===c=¢==
= 2 1M _ . . _ _
(e) m to m hqm added Note. heat rs added dunng process (1 2) and (2 3). , 012  Wr—2 = AUrz;
Q12 = AU12 = McVAT = 30.7165(58629293. 15) = 630.1 k] 12:02:55.1: Q23  W2.3 = AU2.3
Since T2 = T1 and AU2_3 = M4:v {r3  T2) AU2_3 = 0.0
_ 97.5100 _ 9 95%
C223 = W23 = 34990 U P (6301+ 349.9) _ ' ¢=¢=¢=c=¢=c= W
N I ' ’ Compute the rate of temperature change of 0.1 kg of air as it is being compressed
adiabatically with a power input to the air of 1.0 kW. Process is adiabatic, Q = 0. Also for a system M = constant. '=dE.=d.U dKEdPE.
Qw dt dt+dt+dt ﬁll: . 1LT.
dt MC"dt W = 1.0 kW, M = 0.1 kg; cy = 0.7165 kJ/kg°C Table A7. For an ideal gas (air) U = Mch,
 (1.0) = (QM(0.716313%; SEEK = 13.96 [US ' :¢=¢=.=¢=¢= EBQBLEM 419
EN!) 5?! E ; Air. assumed to be an ideal gas with constant speciﬁc heats, is compressed in a closed pistoncylinder device in a reversible polytropic process with n = 1.27. The air , . .
temperature before compression is 30 C and after compression is 130'C.  __ \ m2; Compute the heat transfened on the compression process. ' ngEMAIIQ; ? f
ASSQMPTIQNS; . .;
EBQEERTIES; A E A LY§IS .
For reversible polytropic process the work is equal to: w = mm):  pro/0131
 n w = MR{T2  T1!
but P(Vol.) = MRT Thus: 1  n n = 1.27; M = 1 kg; R = 0.287 kJ/kgK Table A7.
T2 = 130“C (403.15 K) ; T1 = 30°C (303.15 K) w = 1_0).0.287(403.15  303.15 = _ 1063 kl r l 1.27 Q  W = AU = Mcv (Tz  T1) for and ideal gas cv =.0.7165 lekg“C Table A7.
Q  (—1063) = (1.0).(0.7165)(403.15  303.15) =  34.65 kllkg ¢=‘=‘=‘=‘=°= W m A certain house requires rate of 12 kW when the outside air is at 10“C and the inside
temperature is 21°C.
m; (a) What is the minimum amount of power required to drive a heat pump to supply this heat at these conditions?
(b) List the factors that would cause the actual power requirement to be greater than this minimum.
SQﬂEMAIIQ;
2W
Reversed Outside Air
Carnot Cycle  10°C
W
W
4515451415151 (a) Minimum input power will result if a Reversed Carnot Cycle is used.
TH = 21°C (294.15 K) ; TL = 10°C (263.15 K) _ QH_T _ . _
031sz 6:7}: , QL_10.74kw ZW= QH  QL = 12.0  10.74 = 1.26 kW ¢=¢=¢=¢==¢= (b) Actual power input will be greater due to heat transfer across ﬁnite temperature differences. i.e. Tcond. > Tm ; Temp. < Toutside ‘=‘=‘=‘=‘=‘= Friction (mechanical and fluid) therefore pressure drops. EgonigEM 412 t 5 W A device removes 10 k] of heat from a lowtemperature reservoir at 5°C and delivers
. heat to a hightemperature reservoir at 25°C. The work into the device is 5 k1. HER; Dentmine the coefﬁcient of performance of this device if it is
a (a) A heat pump.
(b) A refrigerator. ram» =11 = (a) Heat Pump: 2W 5 <=<=¢=4=¢=¢=
BR .1 i = 10 = 2 O (b) Refrigerator: 2W 5 <=¢=¢=¢=¢=¢= W win.— HEW Km It is desired to heat a home using a heat pump. The heating requirement for the house
is 100,000 lehr. It is desired to maintain the inside of the house at 30°C when the outside air temperature is 10“C. FIND ; (a) What is the minimum power requirements to run the heat pump? How does the
power requirements compare with the electrical consumption if resistance
electrical heating were used? (b) If the heat pump is to be driven by a steam power plant (instead of an electric
motor) operating between SOO'C and 30°C, what is the theoretical minimum rate
at which heat must be transferred to the steam power plant in the boiler? How
does this heat transfer rate compare with the amount of heat required to heat the
house by direct gas heating? QH ' 01. = wpump Outside Air
 10°C W (a) Minimum power requirement is that of a reversible heat pump.
QH rQL = TH I TL = 303nm Therefore. QL = 87800 kJ/hr Power requirement: QH  Q = 13200 kJ/hr = 1667 kw ¢=¢=¢==¢=¢= Electrical power requirement for direct heating = 10000013600 227.78 kW ===¢=¢== Power requirement of the reversible heat pump is 7.575 times less than the requirement
for direct heating. (b) Power requirement of HR = OH  Q = 3.664 kW Power Output of steam engine = 3.667 kW mama = for steam engine = (550  30)/(550 +273): 520/823 = 63.184 ‘i‘b‘==¢=¢=== Min Heat added in boiler of steam engine I
OH = 3.667/0.63184 = 5.8 kw =¢=¢=¢=¢=¢= Compared to direct gas heating. the heating requirement is 4.8 time less. W ENS! E E z 100 k] of heat is removed from a system containing 10 kg of steam while the system
undergoes a reversible isothermal process at a temperature of 4(1) K. The heat is transferred from the system to the surroundings which are at a temperature of 300 K.
I D ' (a) What is the speciﬁc entropy change of the system? (b) What is the entropy change of the surroundings? (c) What is the entropy change of the universe? _ SQEEMMLCJ
Surroundings
@ 300 K
\{7
ASSQMP ! IQNS; 5
EKQEEELIES;
A NALYSIS ;
M = 10 kg
System undergoes a reversible, isothermal process at 400 K
Note: sts = ' qurr
(3) ASsys = sts 1' Tsys = 400/400 = 0.25 kJ/K;
As = AS/M= 0.25/10 = 0.025 kJ/kg‘K ¢=¢=¢=c=¢=¢=
(b) ASSDIT = qurr I Tsurr = HOD/300 = +0.333 kJ/K =¢=¢=¢=¢=¢=
(c) ASuniv = ASsys + Assurr = 0.25 + 0.333 = + 0.033 leK <=<=<=<=¢=¢= PBQELEM 423 MN; The entr0py of a system containing 5.0 kg of helium is decreased by 6.5 kaK while the
system undergoes a constantpressure quasi equilibrium process. The initial
temperature of the helium is 100°C. IND: (a) What is the ﬁnal temperature?
(b) What is the heat transfer for this process?
SCHEMAIIQ:
ASSQMPTIQES:
ERQEEBIIES:
ANALYSIS;
M = 5.0 kg ; T1 = 100°C (373.15 K), As = 6.5 kJ/K For Helium: c = 5.1926 lekgK. cv =3.1156 kJ/kgK. R = 2.077 kJIkgK Table A7.
(a) As = c In (T 1)  Roln (P2fP1) For a constant pressure process In (Pg/P1) = 0 Therelpore; as = M'As = Mcp ln (T2IT1)=(5.0)(5.1926)oln (rzrrl) =6.5 1n ('I'z/T1)= 0.2504; T2/T1= 0.7735 T2 = 0.7735(373JS) = 290.5 K ¢=¢=¢=¢=<=¢=
(b) w = P’dWOIi = P’((V01}2  (Vol)1) = P2(Vol. )2  Pl(Vol.)1 since P = P1 = P2 but P.(Vol.) = MRT ; W = MRTz  MRT1 = MFR(T2  T1)
W = (5.0)'(2.077)(290.5  373.15) =  858.3 k]
AU = MCV AT = (5.0)(3.1156)(290.5  373.15): —1287.5 Q  W = AU; Q  (8583) = 4287.5 ; Q = 2145 k] ¢=¢=¢=¢=¢=<= W
IND' State whether the following processes are reversible, irreversible, or impossible, for
liquid water (assumed incompressible) undergoing an adiabatic process. (a) ti;  ul is positive. (b) u2  u1 = 0. (c) u2  u1 = is negative.
W
AW
PBQEERIIES;
A EALYS IS ;
Tds = du + Pdv
Since the water is assumed to be incompressible. dv = 0
T'ds = du For an adiabatic process ds 2 0
<=<=<=¢=¢=¢= (a) If du is positive, process is irreversible
(b) If du is zero, process is reversible ==¢====
(c) If du is negative, impossible =‘=‘=‘=‘=‘=
W
WN' Saturated water vapor at P = 0.40 MPa is expanded reversibly and adiabatggaﬂy in a
pistoncylinder device to a pressure of P = 0.1 MPa. F D: (a) Sketch the process on a Ts diagram. Show the saturated vapor line.
(b) What is the quality of the water at state 2?
(c) How much work is done?
SCHEMATIQ:
A P '
PRQPERTIES:
A E AL YS IS ;
P1: 0.40 MPa, x1 = 1.0, P2 = 0.10 MPa
(3) For an adiabatic. reversible process: s1 = 52
S <=¢=¢=¢=<=¢=
(b) P1 = 0.40 MPa, x1 = 1.0: from Table (A1.2) 31 = 6.8959 kJ/kgK, u] = 2553.6 lekg
P2 = 0.10 MPa : from Table (ﬁt1.2) Sf = 1.3206 lekgK, uf = 417.36 kakg
53 = 7.3594 kJ/kgK. ug = 2506.1 lekg
(1  x)sr + X’Sg =s ‘ ‘ 
(1  x)(l.3296) + x(7.3594) = 6.8959 therefore it = 0.923 ‘=‘=<=¢=¢=¢=
(c) qw=Au=u2u1,butq=0 ‘ U2 = (1  X)'Uf + xug = (l « 0.923)(417.63) + (0.923)(2506.1) = 2345.3 kJ/kg
 w = 2345.3  2553.6, therefole w = 2033 kmcg ¢=¢=¢=¢=¢=¢= W Kw; A steam power plant is assumed to work on a Carnot cycle with water/steam as the
working substance (Fig. P4—15). The reversible isothermal heat addition takes place in
a boiler at a pressure of 20 atm. and, during this process. saturated liquid water is
converted to saturated water vapor. The reversible isothermal heat rejection takes place in a condenser at a pressure of 1 atm. (a) Calculate the heat transfer on each process in the Cycle? (b) What IS the net work output of the engine?
(c) What is the thermal efﬁciency of the engine?
Compare this value with the theoretical maximum value.
some:
as
ASSLIMEUQNSI
BBQ BERTIES ; From Table A l .2 @ 2.0 MPa;
S3  s3 = 6.3409 kJ/kgK,
T2 = T3 = 212.43 'C = 485.57 K
From Table Al.2 @ 0.1 MPa;
s3 = 7.3595 kakgK,
T1: T4 = 99.63 'C = 372.78 K
ANALXSLS;
(a) 912 = 0
q2_3 = T; (S3  52) = 485.570.8935) = 1890.57 KJ/kg
%4=0
(14.1 = '1‘; (s1  34) = 372.78(3.8935) = 1451.42KJlkg
(b) Wm = qnet = 1890.57  1451.42 = 439.15 KJlkg
(C) Tlmenmt = Wm I q; = 439.15/1890.57 = 23.23 % 11c = (TH  TL)” H = (485.57  372.78)I485.57 = 23.23 % $2 = sr= 2.4474 kJ/kgK.
ng = 5:: 1.3026 kakgK.
5:3 = 6.0568 kJ/kgK, ¢==¢=¢=¢=€= cccccc ¢=¢=¢=<=<=<= ¢=¢=<=<=¢=<= <=¢=¢=¢=¢=¢= <=<=¢=<=<=¢= <=<=¢=¢=<=¢= W
w A system contains 0.1 kg of steam at a pressure of 1.0 MPa and a temperature of 250C. lt expands adiabatically to a pressure of 0. 15 MPa while producing 26 1d of
work. KIND; (a) What is the actual quality of state 2?
(b) Calculate the adiabagp expansion efﬁciency deﬁned as:
r2. actual w 12 ﬂvmibla Idllbllic where the ideal work assumes the expansion
process is both reversible and adiabatic from the initial state to the final
pressure. (c) Sketch the actual and ideal process on a Ts diagram (show the saturated liquid and
saturated vapor lines). naa= W
PER ' AEALYEIS;
M = 0.10 kg, P1 = 1.0 MPa, T1 = 250°C (523.15 K), P2 = 0.15 MPa (3) QW=Au=M(u2u1) W=M'(ut112)
P1 = 1. MPa, T1 = 250'C: from Table A1.2; u] = 2709.9 kJ/kg, S] = 6.9247 kJ/kgK
26.0 = 0.10(2709.9  u2) therefore u2 = 2709.9  260 = 2449.9 lekg P2 = 0.15 MPa: from Table A1.2; [If = 466.94 kJ/kg. ug = 2519.7 kJ/kg
U2 =(1  x)uf + xug : 2449.9 = (l  x)(466.94) + x(2519.7)
Therefore x = 0.966 ‘=‘=‘=‘=¢=¢= (b) If the process is adiabatic and reversible: 51 = 52 = 6.9247 lekgK
P2 = 0.15 MPa: from Table A1.2; Sf: 1.4336 kI/kgK, $3 = 7.2233 lekgK
6.9247 = (1  x')~(l.4336) + x'(7.2233) ; x' = 0.9484
u2‘ = (1  0.9484)(466.94) + (0.9484)(2519.7) = 2413.8 lekg
Q  W' = M(uz  u] ) = 0.10(2413.8  2709.9)
W’ = 29.61 H (reversible, adiabatic work) 1mg = $37 = = 0.878 = 87.8 % ¢=<=<=¢=<=<= (c) <=<=<=¢=¢=¢= E Bﬁtlgg ...
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This note was uploaded on 05/11/2008 for the course EGN 3358 taught by Professor Sleiti during the Fall '07 term at University of Central Florida.
 Fall '07
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