Chapter 4 Practice Problems

Transcription

1 Chater 4 ractice roblems (4.) (a) the number of microstates is N (b) 3 articles total 3! {, } 3 number microstates of secific arrangement (macrostate)!*! robability = (# microstates of secific arrangement)/(total # of microstates) rob (c ) # microstates. (d) {,} {3,} {4,} {3,3 } 4! 6!*! 5! 0 3!*! 6! 5 4!*! 6! 0 3!*3! macrostate # of microstates* ! * number of microstates = m!(8 m)! total number of microstates is 8 = 56, which is the same as the sum from the table. ortion of microstates (robability) for requested configurations: {5:3} = 56/56 = 0.9 = % {4:4} = 70/56 = 0.73 = 7% {3:5) = % like {5:3} robability of any one of the three most evenly distributed states = % + 7% + % = 7% (e) for 8 article system, tirling s arox will not aly /k = ln({4:4}/{5:3}) = ln (70/56) = 0.3 o accomany Introductory Chemical Engineering hermodynamics J.R. Elliott, C.. Lira, 00-04, all rights reserved. (/6/04) of 7

2 (4.) Initial ( each x reresents 5 molecule) xxxx Chater 4 ractice roblems Final x x x x Create a sace with a three emty boxes for the initial state. he number of molecules is too small to use tirling s aroximation. = 0!/(0!0!0!0!) = = 0!/(5!5!5!5!) = 0*9*8*7*6*5*4*3***0*9*8*7*6/(5*4*3*)^3 = /k = ln(/) = ln( ) = 3.8 (4.3) 5 molecules in 3 boxes, molecules are identical j N!...Eqn. 4.4 m! i ij 5! 9!4!! 5! (5!) k ln.3 (4.4) two dice.?? for going from double sixes to a four and three. k for double sixes, we have robability of /6 for each dice.!!*! 6 6 for one four and one three robability alied for /6 for each one in each dice,! *!*! 6 6 ln ln k (4.5) =?? Assume Nitrogen is an Ideal gas V R. Eqn.. o accomany Introductory Chemical Engineering hermodynamics J.R. Elliott, C.. Lira, 00-04, all rights reserved. (/6/04) of 7

3 3 8.34cm * Ma / mole K Chater 4 ractice roblems *300K 0. 08Ma 3 3L / mole*(000cm /L) imilarly Ma C ln R ln Eqn R C..(ig) 7R * ln 8.34 * ln 30.88J / mole K.07kJ K (4.6) (a) m-balance: dn in = -dn -balance: in d( n) dn in in in in n d dn dn dt dt But hysically, we know that the leaking fluid is at the same state as the fluid in the tank; therefore, the -balance becomes: inside inside ( nd dn) ( dn), and dn dn so 0 from the steam table. tate (Ma) O C (kj/kg) (kj/kg*k) (in) () By interolation, imlies = 0.8 O C (4.7) (a) teady-state flow, = Ws At bar = 0. Ma tart mole basis: x 0.333, x 0.667, adiabatic, C xc xc, C for each is the same anyway. MW xmw xmw 0.333( 6) * 0.66( g / mole) R =.987BU/lbmol-R. W Cd 7 * BU / lbmol & m ton / h 000lb / h. & MW 0.66lb / lbmol 000lb lbmol BU * * h 0.66lb lbmol 6 W,305,000.3*0 BU / h (b)?? of the comressor. R * R o accomany Introductory Chemical Engineering hermodynamics J.R. Elliott, C.. Lira, 00-04, all rights reserved. (/6/04) 3 of 7

4 Chater 4 ractice roblems o find the efficiency of the comressor, = But the enthaly and the internal energy will change which gives a change in the ' W Work.?? W 0 C ln R ln C ln R ln C R C 7 R * 00 *559R 5 35R & C( ) 6.95(35 559) 558BU / lbmol (4.8) Adiabatic, steady-state oen system Q = 0, &(C/R = 7/) ig 65 7R kmole W Cd * (65 300) kJ / kmole * kJ 8kg 300?? K 7 *8.34 C kJ / mol R C * kJ / kmol kJ * 8kg / kmol 7.8% 0.78 (4.9) work required er kg of steam through this comressor? By looking at the steam table in the back of the book (Ma) ( O C) (kj/kg) (kj/kg-k) o accomany Introductory Chemical Engineering hermodynamics J.R. Elliott, C.. Lira, 00-04, all rights reserved. (/6/04) 4 of 7

5 Chater 4 ractice roblems W kJ now find W =?? = 0 (reversible), look in the steam table (@ = 4.0Ma) to find a similar value for = 6.876kJ/kg-K, if this value is not available so find it by interolation ' , ' ' W ' kJ (kj/kg) (kj/kg-k) ' =?? ' = , 67% (4.0)@ =.0 Ma & = 600 O C, = kj/kg, = kJ/kg-K (team table) O C L (kj/kg) va (kj/kg) steam table Interolation steam table Va L q( ) *(44.68) 493k.49J W kJ W??,, ' W ' = 0 ( reversible), look for in the sat d tem. steam table and find by interolation, W ' 408.0kJ , 85% (4.) 0.Ma, at' d 0Ma 00 O C va tate (Ma) ( O C) (kj/kg) (kj/kg-k) o accomany Introductory Chemical Engineering hermodynamics J.R. Elliott, C.. Lira, 00-04, all rights reserved. (/6/04) 5 of 7

6 Chater 4 ractice roblems ' interolation for above table: ' = (interolation) (kj/kg) (kj/kg-k) W kJ mass flow rate = kg/s W 95.35kJ / s 95350watt &watt h W 944.0h & kJ 63.% (4.) Ebal: Δ = W R bal: Δ rev rev C = 0 => => rev = (0+73)*8^(8.34/44) = 506K. W rev = C ( rev - ) = 44*(506-93) = 937J/mol => W act = 937/0.85 = 3.4kJ/mol W act = C ( act - ) = 3400 => act = (3400/44)+93 = 597K (4.5) hrough the valve in in 3Ma O 0. Ma 0 C 383. K (By interolation) Find from steam table = kj/kg At 3Ma table use same value for in to find in By interolation in in = 5.976kJ/kg-K o accomany Introductory Chemical Engineering hermodynamics J.R. Elliott, C.. Lira, 00-04, all rights reserved. (/6/04) 6 of 7

7 Chater 4 ractice roblems he rocess should be irreversible. o find, interolate using temerature at 0. Ma: = 7.48 kj/kg-k, since > in entroy has been generated. he entroy balance is: in in 0 m m gen (4.8) An insulated cylinder is fitted with a freely floating iston,... team 0.5kg, i = 9 bars, q=0.9, goes to satd vaor at 30 bars, W air = -360 kj. he volume change of the air is equal and oosite the volume change of the steam. he volume change of the steam is (using sat roerties at 0.9 Ma) V i steam = 0.5kg (V satl + q (V satv V satl ))m 3 /kg = 0.5( ( )) = m 3 V f steam = 0.5*V satv = 0.5kg*.0667m 3 /kg = m 3 ; V steam = = m 3. V f air = V i air + V air = = 0.54 m 3 Because we are not told the mass and area of the iston, let us consider it massless, so that the initial ressure of the air is 9 bar and the final ressure is 30bar. he air will be treated like an ideal gas. d(nu)= in dn + dwair; integrate term-by-term n f U f n i U i = in (n f n i ) -360 for the air, let R = 0 for 50 bar and 300K, the inlet condition. n i = V/R = 0.9Ma (0.05m 3 )(0 6 cm 3 /m 3 )/ (cm 3 -Ma)/300 = 8.04 moles n f = V/R = 3Ma(0.54m 3 )(0 6 cm 3 /m 3 )/ (cm 3 -Ma)/ f air = 4638/ f air moles Let R =0 at 50 bar and 300K U R = R -(V)=0-R R = J/mol Because U is indeendent of, then U i = U R = J/mol. For air, use Cv =.5R = J/mol-K he energy balance becomes 4638/ f air [0.786( f air -300) ] J= J rial&error f = 97 K; the temerature change is very small for this case. o accomany Introductory Chemical Engineering hermodynamics J.R. Elliott, C.. Lira, 00-04, all rights reserved. (/6/04) 7 of 7