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Unformatted text preview: 57 Air is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined. Assumptions Flow through the nozzle is steady.
Properties The density of air is given to be 2.21 kg/m3 at the \
inlet, and 0.762 kg/m at the exit. V, = 40 m/s AIR
_ 2
Analysis (a) The mass ﬂow rate of air is determined from the A' — 90 cm _> V2 = 180 m/s inlet conditions to be /
ti: = p.A.V. = (2.21kg/m3)(0.009 m2)(40 m/s) = 0.796 kgls (b) There is only one inlet and one exit, and thus Iii, = rhz = rh . Then the exit area of the nozzle is determined to be Iii 0.796 kg/s =—3—= 0.0058 m2 =58 cm2
szz (0.762 kg/m )(180 m/s) rh=p2A2V2 —) A2 = 534 Heat is lost from the steam ﬂowing in a nozzle. The velocity and the volume ﬂow rate at the nozzle exit are to be
determined. Assumptions 1 This is a steadyﬂow process since there is no
change with time. 2 Potential energy change is negligible. 3
There are no work interactions. 4000C STEAM 300°C Analysis We take the steam as the system, which is a control 300 1‘? a —’ 200 kPa
volume since mass crosses the boundary. The energy balance 10 m/s
for this steadyﬂow system can be expressed in the rate form as Q
Energy balance:
Ein  Eon! = AE‘systemﬂo (steady) = 0
Rate 0f “9‘ energy trarislbr Rate of change inintemal. kinetic.
by heat. work. and mess potential. cinemlgies
Ein = Eout
. V12 . V22  . 
m hl+7 =m h2+7 +Qom smce WEAPCEO)
V2 V2
or hl+4=h2 +—Z+Q+'“‘
2 2 m The properties of steam at the inlet and exit are (Table A6) P. = 800 kPa VI = 0.38429 m3/kg
Tl = 400°C h, = 3267.7 kJ/kg
P2 = 200 kPa 02 = 1.31623 m3/kg
T, = 300°C h2 = 3072.1 kJ/kg The mass ﬂow rate of the steam is l l m =—A V =— 0.08m2 lOm/s = 2.082k /s
u, ' ' 0.33429 m3/s( X ) g
Substituting,
2 2
3267.7 kJ/kg + M[LA‘§2] = 3072.1kJ/kg + V—2[L/k2g2] + 25—“
2 1000m /s 2 1000m /s 2.082 kg/s —)V2 = 606 mls The volume flow rate at the exit of the nozzle is ()2 = riwz = (2.082 kg/s)(l.31623 ms/kg) = 2.74 m3Is 548 Saturated R134a vapor is compressed to a speciﬁed state. The power input is given. The exit temperature is to be
determined. Assumptions 1 This is a steadyﬂow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 Heat transfer with the surroundings is negligible. Analysis We take the compressor as the system, which is a control volume since mass crosses the boundary. Noting that one ﬂuid stream enters and leaves the compressor, the energy balance for this steadyﬂow system can be expressed in the
rate form as Ein — Eout = AEsystemao (Steady) = 0
Rate 0‘ “‘31 “HEY trmsfcr Rate of change in internal. kinetic. 700 kPa
by heat, work. and mass potential,etc.cncrgies
E in = Eout
Win + rhhl = rhhz (since Ake E Ape a 0)
Win = 751012  11,)
From R134a tables (Table A12) $ 180 kPa
sat. vap.
P. =180 kPa h, = 242.86 kJ/kg 035 mam“
x, =0 v,=0.11o4m3/kg The mass ﬂow rate is . _ﬂ _ (0.35/60)m3/s m 3 =0.05283 kg/s
vl 0.1104m /kg Substituting for the exit enthalpy,
W. = rh<h2 — h.)
2.35 kW = (0.05283 kg/s)(h2 — 242.86)kJ/kg——>h2 = 287.34 kJ/kg From Table A13, P2 = 700 kPa T = . °
11, =287.34kJ/kg} 48 8 G 567 Steam is throttled from a speciﬁed pressure to a speciﬁed state. The quality at the inlet is to be determined. Assumptions 1 This is a steadyﬂow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 Heat transfer to or from the ﬂuid is negligible. 4 There are no work interactions involved. Analysis There is only one inlet and one exit, and thus :11, = m2 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyﬂow system can be expressed in the
rate form as . . . no“,
Em—Eom=AE,y,,em (”1:0 Em = Em Throttling valve
mh, =th Steam 3 E 100 kPa
—.
hl = h, 2 MP8 120°C Since Q5W=Ake5Ape50.
The enthalpy of steam at the exit is (Table A6), P2 =100kPa h2 = 2716.1 kJ/kg
T2 =120°C The quality of the steam at the inlet is (Table A5) P2 =2000kPa x hz _hf _ 2716.1—908.47 _
hl =h2 =2716.lkJ/kg 576 A hot water stream is mixed with a cold water stream. For a speciﬁed mixture temperature, the mass ﬂow rate of cold water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is wellinsulated so that heat loss to the
surroundings is negligible. 3 Changes in the kinetic and potential energies of ﬂuid streams are negligible. 4 Fluid properties are constant. 5 There are no work interactions. Properties Noting that T < [email protected] 30m = 127.41°C, the water in all three
streams exists as a compressed liquid, which can be approximated as a
saturated liquid at the given temperature. Thus, h] E [email protected] 80°C = 335.02 kJ/kg
hz E [email protected]°c = 83.915 kJ/kg
h; E [email protected]°c = 175.90 kJ/kg Analysis We take the mixing chamber as the system, which is a control
volume. The mass and energy balances for this steadyﬂow system can
be expressed in the rate form as Mass balance:
. . _ . 7I0(steady)_   _ 
min  mout  Amsystem _ 0 ’ ml + m2  ”’3
Energy balance:
‘ ' _ ' $0 (steady) _
Ein  Eout  AEeiystem  0
Rateofnetenergytrmsfer Rat fchn ‘ ‘t Iki ti
e 0 ngeln Ill emn . :1: C.
byhemmork. and mass potential. etc.enagies
Ein = Eon! titlhl + thth = m3h3 (since Q = W =Ake E Ape E 0)
Combining the two relations and solving for r'ng gives ’hlhl + ’hzhz = (’5'! + ”'2 M3 rhz=h'—h’
’13’12 Substituting, the mass ﬂow rate of cold water stream is determined to be (335.02 — 175.90) kJ/kg
— 0.5 k = 0.865 /
(175.90 83.915)kJ/kg( g/s) kg s '5': m2: 1, = 80°C
m1= 0.5 kg/S H20
(P = 250 kPa)
T3 = 42°C T.2 = 20°C /
m2 584 Oil is to be cooled by water in a thinwalled heat exchanger. The rate of heat transfer in the heat exchanger and
the exit temperature of water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is wellinsulated so that heat loss to the
surroundings is negligible and thus heat transfer from the hot ﬂuid is equal to the heat transfer to the cold ﬂuid. 3 Changes
in the kinetic and potential energies of ﬂuid streams are negligible. 4 Fluid properties are constant. Properties The speciﬁc heats of water and oil are given to be 4.18
and 2.20 kJ/kg.°C, respectively. Analysis We take the oil tubes as the system, which is a control 1:23:23]
volume. The energy balance for this steadyﬂow system can be 2 k g/s
expressed in the rate form as Cold
Ein _ Eout = AEsystemﬂo mead” = 0 water
Rateof net ener transﬁr fem ' ' ‘  22 C
by heat, work.3% mass Ra‘eﬁmgﬁf chfglgmm’ 1.5 kg/ S
Em = Em l 40°C mh. = Q0‘“ + mhz (since Ake a Ape a 0)
Q... = 429,0“.  T2)
Then the rate of heat transfer from the oil becomes
Q = [rhcpmn  7;!!!)10“ = (2 kg/s)(2.2 kJ/kg.°C)(150°C  40°C) = 484 kW
Noting that the heat lost by the oil is gained by the water, the outlet temperature of the water is determined from  . ' 434 kJ/s
Q = lmcp(Tour ‘ Tinnwata __)Tout = Tin + Q = 22°C , + —o = 992°C
mwﬂcp (1.5kg/s)(4.18kJ/kg. C) 598 Air at a speciﬁed rate is heated by an electrical heater. The current is to be determined. Assumptions 1 This is a steadyﬂow process since there is no change with time. 2 Kinetic and potential energy changes are
negligible. 3 The heat losses from the air is negligible. Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table Al). The constant pressure speciﬁc heat of air at room
temperature is c,, = 1.005 kJ/kg°C (Table A2a). Analysis We take the pipe in which the air is heated as the system, which is a control volume. The energy balance for this
steadyﬂow system can be expressed in the rate form as Em —Eom = AEsystemm (5mm : 0
\_.__‘,_J —’___4 '
Rateofnet energy mfer Rnteof changein intemaLkinetic, W 
by heat. MIKE!!! "13$ potential. etc. energies e.tn
E. = E — t
. . "‘ _°“‘ 100 kPa, 15°C Air 100 kPa
mhl +I’Vejn =mh2 0.3 [113/8 ﬂ 30°C
Wejn = ri1(h2 —hl)
VI =n'1cp(T2 —T.)
The inlet speciﬁc volume and the mass ﬂow rate of air are
RT . . 3 
V] =_1 =W = 0.8266m3/kg
Pl 100 kPa
' 3
,5, = ﬂ = ﬂ = 0.3629 kg/s v, 0.8266 m3/kg Substituting into the energy balance equation and solving for the current gives I _ rhcp(T2  T1) _ (0.3629 kg/s)(1.005 kJ/kg K)(3015)K 1000v1
" v ' 110 v 1kJ/s J: 49.7 Amperes 5135 An insulated rigid tank initially contains helium gas at high pressure. A valve is opened, and half of the mass of
helium is allowed to escape. The ﬁnal temperature and pressure in the tank are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it
can be analyzed as a uniformﬂow process by using constant average properties for the helium leaving the tank. 2 Kinetic
and potential energies are negligible. 3 There are no work interactions involved. 4 The tank is insulated and thus heat
transfer is negligible. 5 Helium is an ideal gas with constant speciﬁc heats. Properties The speciﬁc heat ratio of helium is k =l.667 (Table A2). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the
microscopic energies of ﬂowing and nonflowing ﬂuids are represented by enthalpy h and internal energy u, respectively,
the mass and energy balances for this uniformﬂow system can be expressed as Mass balance: min  mom = Amsys‘em —> me = ml  m2 m2 =‘é'm1 (given)——> me =m2 =1lrm1 Energy balance: Ein _ Eout = AEsystem h——~——l ~—.,———l
Net Cher “if“ Change in internal, kinetic,
by heat. wot .and mm potmtial. etc. energies mehe =m2u2 mlul (since W E Q‘=' keg pea 0) Note that the state and thus the enthalpy of helium leaving the tank is changing during this process. But for simplicity, we
assume constant properties for the exiting steam at the average values. Combining the mass and energy balances: 0 = Elmlhe +%mlu2 m,ul Dividingby ml/Z 0=he+u2 2ul or 0=cp T' :T2 +ch2 2€le
Dividingbycu: 0=k(T+T2)+2T2—4T, sincek=cp/cv . 4k) (4—1.667)
Sol f T: T=( T=——403K=257K mg m 2 2 (2+k) ' (2+l.667)( ) The ﬁnal pressure in the tank is  "'sz P, =1E(3ooo kPa)=956 Rh
2 403 ...
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 Fall '08
 Errington

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