{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MAE204_F11_HW6_Solutions

MAE204_F11_HW6_Solutions - 5-7 Air is accelerated in a...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 5-7 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 = 5-34 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 steady-flow 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 steady-flow 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 A-6) 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 5-48 Saturated R-134a vapor is compressed to a specified state. The power input is given. The exit temperature is to be determined. Assumptions 1 This is a steady-flow 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 steady-flow 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 A-12) $ 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 A-13, P2 = 700 kPa T = . ° 11, =287.34kJ/kg} 48 8 G 5-67 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 steady-flow 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 steady-flow 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 A-6), P2 =100kPa h2 = 2716.1 kJ/kg T2 =120°C The quality of the steam at the inlet is (Table A-5) P2 =2000kPa x hz _hf _ 2716.1—908.47 _ hl =h2 =2716.lkJ/kg 5-76 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 well-insulated 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 steady-flow 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 5-84 Oil is to be cooled by water in a thin-walled 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 well-insulated 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 steady-flow 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) 5-98 Air at a specified rate is heated by an electrical heater. The current is to be determined. Assumptions 1 This is a steady-flow 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 A-l). The constant pressure specific heat of air at room temperature is c,, = 1.005 kJ/kg-°C (Table A-2a). Analysis We take the pipe in which the air is heated as the system, which is a control volume. The energy balance for this steady-flow 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)(30-15)K 1000v1 " v ' 110 v 1kJ/s J: 49.7 Amperes 5-135 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 uniform-flow 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 A-2). 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 uniform-flow 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 . 4-k) (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 ...
View Full Document

{[ snackBarMessage ]}