= 0, no work has been added

Transcription

1 hater Practice Problems (P1) Enery Balance: Eqn 7 d u z u z u z m U m m Q & + + & lets lets E + S (P) closed system, no mass flow m & m& 0, valve stuck revent any steam from o m& 0, no chane volume occurs (no exansion or contraction) E 0, no work has been added or removed, and there is no um nor turbe S 0 Inore KE and PE d the fal enery balance becomes ( mu ) Q Before the valve is stuck, there is steam esca and therefore, there is an enthaly for the vaor leav, no exansion / contraction work, no shaft work, no ketic or otential enery occurred d enery balance will be ( mu ) m& + Q, note m & m& (P3) (Solution for as furnace) Assume the house is air-tiht Furnace has been heat the home steadily, the system (furnace) is a steady-state oen system, d u z m U m& m& + + 0, Furnace is fixed size, no exansion/contraction work occurs onsider just the hot side of the heat exchaner loss m& m& Q Furnace Fuel and air (ha 14 will consider reactions such as ombustion ases combustion, but the E-balance doesn t need to be written differently!) Note: if the small shaft work from blower is cluded, then the boundary cludes the house air and of the furnace, and the boundary cludes both sides of the heat To accomany Introductory hemical Eneer Thermodynamics JR Elliott, T Lira, 001, all rihts reserved (7/31/001) 1 of 8 ouse

2 hater Practice Problems exchaner For this boundary, all heat transfer is with the furnace, and there is no Q term, but there are four mass flow terms (fuel/air, combustion as, house air, house air return) and a small (neliible comared to other terms) s term (P4) (Solution for as furnace) Assume house is air-tiht, m m 0 System: the house and all contents, consider just the house side of heat exchaner closed, unsteady-state system there is heat a and heat lost the house No exansion / contraction work nor ketic, otential enery chane The ternal enery of the system could chane with resect to time deend on heat a and heat loss, or could be zero if they are balanced If the furnace is roerly sized, there will be a net a temerature of the house! Fuel and air loss Furnace ouse ombustion ases d ( mu ) Q + Q a loss, where Q loss is neative Note: if blower is cluded with system, then you also should be clear whether the boundary cludes the entire heat exchaner ( which case the Q a term is relaced with enthaly flow terms), or just the house side of the heat exchaner In either event, the s term will be small relative to other terms sce there is so much heat com from furnace (P6) System: the bulb and its contents d System is closed steady state system, ( ) 0 is left side of equation There is no work act on the bulb and no mass flow or U 0 Photons loss onsider hotons as heat transfer by radiation net 0 hotons and heat loss from bulb,, bulb To accomany Introductory hemical Eneer Thermodynamics JR Elliott, T Lira, 001, all rihts reserved (7/31/001) of 8

3 hater Practice Problems (P7) System: the sunbather at 1:00 noon (oen system) No work has been done, no ketic or otential eneries No mass flow, but ersiration loss, m & m& Unsteady state, still ett hot onsider hotons to be heat transfer by radiation by us the equation 64 (text book) and dro all the zero values, enery balance will be: d ( m body U body body V ) m& + Q& (steam tables could be used to estimate the enthaly of the evaorat ersiration if the T is known or estimated) (P8) System: the balloon and its contents Unsteady state system, oen system as No heat a or loss No shaft work Mass chan with resect to time, as it ets flat, m & m& Volume of the balloon is chan the work of exansion / contraction occurs Ketic enery effected by chan the velocity of the balloon No otential enery chane d mu balloon u dm enery balance is: mu & + (P9) (a) Potential enery (PE), PΕ m** h, ( 1 J 1Nm ) where m mass, ravity, h heiht or distance that the body moved by P E 1000Nm P E m h / c, h m / 1k *98066N / k h m c This is why otential enery is often ored when thermal chanes are resent the heiht chane must be very lare to be sificant (b) Ketic enery otential enery K E body 1 1/ 1 c K E *1( km / Ns ) *1000Nm, 1 mu u m k u 447m / s This is why ketic enery is often ored when thermal chanes are resent the velocity chane must be very lare to be sificant m& E Q hotons To accomany Introductory hemical Eneer Thermodynamics JR Elliott, T Lira, 001, all rihts reserved (7/31/001) 3 of 8

4 hater Practice Problems (P10) losed system, unsteady-state, 1 block, water, 3 tank Enery balance U0 m1* U1+ m* U + m3* U 0 T ( U vdt ), T1 U v T 1 3 Q 0 E 0 (T 400)*00* *4000*(T-300)+500*0380*(T-300) 0 Solve for T, T K U BLOK 00*038*( ) -7566J U ATER 4000*4184*( ) 74809J (P11) Unsteady-state, closed system, Enery balance with block at 50m dee the water m1* U1+ m* U + m3* U +m4(k)* z(m)*/c(n/k) 0 (1J1km /s ) 00*38*(T-400)-0*50* *4184*(T-300)+500*38*(T-300) 0 T U BLOK (P1) a Unsteady-state closed system U as nv T5mole*00*5cal/mol-K 5000cal; with as AND vessel U n as v T + m v T m v T (015) cal vessel as vessel vessel E 0( riid) U Q b U Q + U Q Q m T E E 5mole*00*7cal/mol-K 7000cal (P13) Unsteady state, closed system Assume Ideal Gas v + R, and 7R/ (diatomic) 7R/ R v; Q U + ( no work ) Q U 0785* ( ) Q 079J/mol v 5R/ 5*8314/ 0785J/-k To accomany Introductory hemical Eneer Thermodynamics JR Elliott, T Lira, 001, all rihts reserved (7/31/001) 4 of 8

5 hater Practice Problems (P14) Steady state, oen system 660 o F o T( o ) (T( o F) 3)*5/9 Throuh the valve 0 m & m& From steam tables (back of the book) Interolate for o to fd J/ Outlet at 1 atm Use sat T table to fd P1014 MPa (close enouh to 1 atm) is less than satv 676 J/, so the let is two-hase va 56 J/ and fd satl 419 J/ va q + SatL q*(56) q 0954 (P16) Steady state, oen system 0 m& m& + Q a) P1 15 MPa, T1 600 o kj/k (Steam Table) P 10MPa, T 700 o eat Q 1-87 kj/k b) for i take lowest P at 600 o and 001MPa i 3706 J/ ( at 600 o and 01MPa) 3870 kj/k (Steam Table) R * T i (35831J / 3706J / ) *18 / mole 8314J / mole K *( ) K NOTE: (answer tyed wron the book rts 1-3) To accomany Introductory hemical Eneer Thermodynamics JR Elliott, T Lira, 001, all rihts reserved (7/31/001) 5 of 8

6 hater Practice Problems (P17) There are several ways to look at this roblem, and some choices of system boundries are more difficult than others One method is resented (a) System: iston, donote with subscrit closed system, no Q (nelect) no flow (riid body) no chane of ternal enery (iston doesn t chane T) u m Z + m S, surface forces cause movement; treat this as s 01m 3 i P 10bar 0 1m area Now we need to determe s The work done by the as (subscrit ) and the atmoshere (subscrit a) will be equal to the work done on the iston The as and atmoshere are closed systems that chane size System: as E, P dv () System: atmoshere Note : dv d V a E, a Pa dv a Pa dv (3) omb work teraction at boundary s, E, E, a hich results the work equation m Z + m u S, n ( RT / V dv P dv S, ) a P dv P dv a (4) V 05 S, (n RT1* ln - 01*(05-01)) (P1V1* ln - (01*015))(5) V S, (10*01*ln5-015)*1E6cm 3 /m J From E Bal mu / c S, - m z/ c 7669J - 700*981*(5m-1m) 66386J(6) u 1376m/s To accomany Introductory hemical Eneer Thermodynamics JR Elliott, T Lira, 001, all rihts reserved (7/31/001) 6 of 8

7 hater Practice Problems (b) Free fliht, system: Piston, closed system, constant T Equation (1) alies where S, 0 (no surface forces act on iston, because we will ore air resistance) The ketic enery will o to zero at the to of fliht The itial ketic enery from art (a) u m z + m 700(981) z 66386J z 9 6m (not count 5 meters of tube) 0(7) (c) P V constant, sce isothermal, P P1V1/V 4 bar (d) V 588, V V lu to (5) S, (10*01*ln588-01*( ))*1E6 cm 3 /m 3 S,P J Us (6) mu / c *981(588 1) J lu to (7) 700 *981* z 94840J, z 138m, free, fliht (e) the iston will accelerate when the uword force (Fu) > down ward force (Fdown) The iston will decelerate when Fu< Fdown The maximum exit velocity will be obtaed if the iston leaves the cylder at the condition Fu Fdown F F u down P A P A + a m m P A Pa A + (8) P can be related to V sce as is isothermal, PV constant, PV 1 1 P (9) V Substitut (9) to (8), and sert the values (1E 6Pa) ( V V )(01m ) (01E6Pa)(01m ) 700* To accomany Introductory hemical Eneer Thermodynamics JR Elliott, T Lira, 001, all rihts reserved (7/31/001) 7 of 8

8 V V 1 593, V 0593m Answer will be almost the same as (d) Plu to (5) hater Practice Problems S, (10*01*ln593-01*( ))*1E6 cm 3 /m 3 S, J Us (6), mu / c *981(593 1) J sub Into (7) 700 *981* z 94846J, z 138m, free, fliht 3 (P18) 1 m V1 Initial and fal conditions, P1 50 si 647 sia 0446MPa ball: 3*54cm/76 cm dia cylder volume: V100cm*314*(76/) 45581cm 3 ball mass: 015lbm*454k/lbm057k ; exit velocity: 40mi/hr *580ft/mi*03048m/ft*1hr/3600sec 1788 m/s Enery Balance: see E Bal From (P17), excet otential enery is not cluded sce iston is horizontal, ball relaces the iston of (P17) adat (4), (5) from (P17) S,ball 0057/*(1788) 911 J m u / c S,ball 0446*V1*ln(V/V1) - 01*(45581-V1) 911J Us solver V1 387cm 3 not count Vball To accomany Introductory hemical Eneer Thermodynamics JR Elliott, T Lira, 001, all rihts reserved (7/31/001) 8 of 8