Termodinamica Ll Apuntes 21
-
Upload
valentin-vazquez -
Category
Documents
-
view
20 -
download
0
Transcript of Termodinamica Ll Apuntes 21
TERMODINAMICA ll http://www.youtube.com/watch?v=Zq-Ito6gbck 21/enero/2014PV = mRT
PV = nRT
Constante ideal de R del aire: => ideal = 0.082El volumen ideal de 1 mol de cualquier gas es -> V= 22.4 L
R = = = 0.082006 n = No. de moles.
m = masa molar del aire = 28.96 gr/mol.
Peso molecular= 28.96 gr/mol.
1 atm = 760 mm de Hg = 10330 kg/ = 1.01325bar = 101325 Pa = 760 Torr = 14.695948 psi
= 10.332274 mca (metros de columna de agua).PV=mT => = .: M = Pm x mol = m gr
= = = = = 29.2515 = = = 26.4891P= presion barometrica DF = 585 mm de Hg .: = 7951.381579 kg/T= T ambiente= 15 .: 15 + 273.15 K = 288.15 K
V = saln= 6m*6m*2.5m= 90PV= mRT .: m=
m= = = 84.8484 kg de airen= = =2929.8480 molesNo. de Avogadro=(2929.8480 moles)(6.022x )= 1.764354xmol= 0.082PV = nRT .: n= = = 2894.8337 molm= masa ; n= No. de moles; Pmol= peso molecular.
m = n*Pmol =2894.8337 mol * 28.96kg/mol= 83.8343 kg de aire.
22/enero/2013
Isobrico P = cte. J = 427 Q = u + w .:MCp At = MCvAt + =>
Kcal kcal P Qs= MCpAt .: Qs=MCp()
V
Isovolumetrico V= cte. W= 0Q = v P Q= u=MCvAt
=
V
Isotrmico T= cte. Q=w P
= (ln )
= mRT (ln )
=
V
Adiabatico
Q=0 = Q = cte J = 427 u= w
MCvt = =( = ( )K= .: k de aire = 1.41 => k del aire = 1.4
Pm= C = 44 gr/mol
PV = nRT .: R = = = 0.082006 = 0.082R C= = = 19.2528 kg
Se tiene un vehiculo el cual cuenta con unos neumaticos , cuyo volume es de 326 , en cada uno. El cual se encuentra ubicado en la ciudad de Mexico, en donde se tiene una presin baromtrica de 585 mm de Hg. El vehiculo tiene sus neumticos inflados con aire a una presin manomtrica de 2 kg/.
Qu cantidad de nitrgeno (N) se necesitara , si se quiere sustituir el aire por el nitrgeno?
Nota: tmese una temperatura media de 20 , la masa en kg.Si dicho vehiculo se traslada al nivel del mar Cul seria la presin resultante en cada una de las llantas?
Datos : P= = 0.79513815 V= 326 Pabs= Pbaro+Pman .: 0.79513815 + 2=2.79513835P baro= 585mmde Hg = 0.79513815 P man= 2kg/
T= 20 = 293.15 K R = = = 30.25 kgn= = = 1.026 = 1.03 kgm = No. mol x Pmol = 163.32 x 45.98x = 7.5094kgP abs= Pbar+Pman= 1.033 kg + 2kg/ = 3.033 kg/ 23/enero/2013 R = 0.2968 kg/kg K (0.070)(427) kg*m/kcal = 29.89 => 30.29 kg*m/kg K
1 kcal= 40184 kJ = 0.2968 kg/kg K = 0.070 kcal/kg K
P= = = 27773.3547 kg/Pman= Pabs-Pbar =27773.3547 kg/- 10330 kg/ =17443.3547 kg/
Un tanque con 0.708 d volumen contiene aire a una presin absoluta de 1.015 kg y a una temperatura de 21, Cuntas kcalorias se necesitan para calentar el aire hasta 65.5 , y cual seria la presin resultante a dicha temperatura.
Datos: P Cp= 0.24kcal/kgK
V = 0.0708 Cv=0.17kcal/kgK
Pabs= 1.015 kg
R a = 0.82
= VJ = 427 Q= V Ra= 29.27 kg*m/kg*KV= MCvT
M= = = 0.8655 kg
V = MCv(-) = kcal
V 0.834(0.17)(338.65-294.15) = 6.309 kcal
= = = 11684.7365kg/K = k= 1.4
6.31kcal= = .: (175.07)(6.31)= 0.8346 8471.19 = = 11473.6181 kg/
25/enero/2013
P T 3 adiabatico 2 isotermico 2 1 1 V S Datos:
V = 0.0708 =1.015 kg = 21= 294.15 K
= 65.5 = 338.65 K
= = 0.834 kgRa= 29.27 kg*m/kg*K = = = 11684.7365kg/ ) = ) .: ) = ) => = = = === 481.4965 K ) = ) .: = ) = ) = (0.708)() = 0.2937 = = =0.293843 Energia interna= Q = 0 w= u
w= = = 8642.0468Energia interna = = = 3498 kcal
.: = = 481.49 = 1163.4638 K = = = 1160.12 K
Q = 0 u = = kcal 3 4 Q= + w
=
2
1 = mCv( ) = Q
= (0.834)(0.24)(1163.4-481.49)= 136.4911 kcal = mCv() = (0.834)(0.17)-(1163.4- 481.49)= 96.6911 kcal
= Q-V =P = 40000 = 38.8758 kcal
CICLO OTTO
4 T P 4
3 5 1 3 1
2 2 Vc Va s 1-2 condicion de admisin Vc = Patm manometro
Va = = =1847.2564 Barmetro
= 14 cm
L= 12 cm
Vc= 7 del volumen total
Rc = .: = = 2.05244x Punto Presion kg/ Volumen T Masa Kg
1 = = = M =
2 =+==
3 = == = =
4 ( ) = = +=
5 = = = =
Cc = Qs = Cc ( PCI) Qs = m Cp (-) = m Cp t m Cv tProblema:D= 14 cm P atm = 585 mm HgL = 12 cm = 20
Rc = 10 Gasolina At = 14.83
Rpm = 2500 PCI = 10400 k cal / kg comb
Punto Presion kg/ Volumen T Masa Kg
17951.38 kg/ 2.0525x293.15 1.9020xkg
27951.38 kg/ 2.0525x293.15 1.9020xkg
3199729.6353 kg/ 2.0525x736.351.9020xkg
41.3164 kg/ 2.0525x4853.37 1.9020xkg
552407.50 kg/ 2.0525x1932.16131.9020xkg
27951.38 kg/ 2.0525x - -
17951.38 kg/ 2.0525x - -
=585 mm Hg .: (585 mm Hg x 10330 kg/ ) / (760 mm Hg ) =7951.38 kg/ = 20 + 273.15 = 293.15
= = /Rc-1 => {( x x 0.12) / 4 } / (10-1)= 2.0525x= (7951.38 kg/ 2.0525x) / (29.27 x 293.15 ) = 1.9020x= = 7951.38 kg/ =+ =( 2.0525x)+ {( x x 0.12) / 4 }= 2.0525x==293.15 = (7951.38 kg/ x2.0525x) / (29.27x293.15 ) = 1.9020 x=== 2.0525x = = ( 7951.38 kg/ ) /( 2.0525x / 2.0525x)=199729.6353 kg/ = = (293.15 ) /( 2.0525x / 2.0525x )= 736.35Cc = = 1.9020x / 14.83 = 1.28Qs= Cc x PCI=(1.28)( 10400 k cal / kg comb)= 1.3312 k cal = += (1.3312 k cal) / (1.9020 x x 0.17 kcal/kg K) +736.35=4853.37 ( ) = ( 199729.6353 kg/ ) (4853.37 / 736.35)=1.3164 kg/ ==2.0525x = = (4853.37 ) / (2.0525x /2.0525x)=1932.1613= =2.0525x = = ( 1.9020xkg x 29.27 x 1932.1613 ) / 2.0525x = 52407.50 kg/ = (-)= (7951.38 kg/ ){( 2.0525x) (2.0525x)}=14.68 Kg/m = = =
= - 40698.0323 Kg/m = 0 = = = = -406.56 Kg/m = 0
= (-)= (7951.38 kg/ )( 2.0525x - 2.0525x) = -14.68 Kg/mQs = Cv ( - ) = (1.9020xkg)(0.17)( 4853.37 736.35) =1.3311kcalQp = Cv ( - ) = (1.9020xkg)(0.17)( 293.15 - 1932.1613) = -0.5299 kcal
Qu = Qs+Qp = (1.3311kcal ) + (-0.5299 kcal)= 0.8011 kcal
Wu = Qu x 427 = 342.06 Kg/mWu = w = 14.68 Kg/m - 40698.0323 Kg/m + 0 -406.56 Kg/m +0 -14.68 Kg/m =-41104.5932 Kg/m = Wu x n
12/feb/2013
CICLO DIESEL
Qs T P 3 4 4
5 3 5 1 Qp
2 1 2 Vc Va s
CONDICIONES TERMODINAMICAS. = = ()= ( ) Punto Presion kg/ Volumen T Masa Kg
1 = = = =
2 =+===+
3 = == = =
4 =( ) = +=
5 = ( ) = = =
2 = - -
1 = - -
Problema ciclo diselD = 15.2 cm rpm= 1200 PCI= 10100 kcal
L = 17.8 cm P atm= 1.0330 kg/ t = 14.92Rc = 20 T amb = 30No. cilindro combustible = diesel Punto Presion kg/ Volumen T Masa Kg
110330 kg/ 1.6999 303.15 1.9789 kg
210330 kg/ 3.3998 x 303.153.9579x kg
3684766.2 kg/ 1.6999 1004.77 3.9579x kg
4684766.2 kg/ 5.6342 3684.0 3.9579x kg
555293.33 kg/ )3.3998 x 1795.0277 3.9579x kg
210330 kg/ 3.3998 x - -
110330 kg/ 1.6999 - -
= 1.0330 kg/ x 10000 = 10330 kg/ = (0.152)(0.178) / 4 = 3.2299x = = => = 3.2299x /20 - 1 =1.6999 = 30 + 273.15 = 303.15 = == = 1.9789 kg =10330 kg/ = =303.15=+=(1.6999 ) +( 3.2299x ) = 3.3998 x
= = = 3.9579x kg = = = 684766.2 kg/ ===1.6999 == 3.9579x kg = = (303.15) / (1.6999 / 3.3998 x ) = 1004.77 Ma = -= Cc = = = 2.52 Qs= Cc x PCI = 2.52 x 10100 kcal = 2.5453kcal = += 1004.77 + = 3684.0 =( )= 1.6999 ( )= 5.6342 = 684766.2 kg/ ==3.9579x kg==3.9579x kg== 3.3998 x = ( ) = (684766.2 kg/ ) ( ) = 55293.33 kg/ ) = = (3684.0 ) / ( 3.3998 x / 5.6342 ) = 1795.0277 = =10330 kg/ = =3.3998 x = =10330 kg/ = =1.6999 Qs = Cp ( - )= (3.9579x kg) (0.24)( 3684.0 - 1004.77 )= 2.6510 kcal
Qp= Cv ( - )= 3.9579x kg x 0.17 (303.15 - 1795.0277 ) = -1.0037 kcalQu = Qs + Qp = 2.6510 kcal-1.0037 kcal = 1.6473kcal
Wu = Qu J =1.6473kcal x 427 = 703.39 kg m
= = = 7033.9 kg m / s = 1-{ ()k(-) =-5595915.886 = = = 0.6213 26 / Febrero/2013Ley general de un gas ideal:
PV = nRT
R = = 0.082 condiciones ideales para la R airePV = nRT
Ejemplo: n = => n = = 582.4039 mol
P = 14 atm m = Ma x No. mol .: gr/mol x molGas= C ( C = 12; 0 = 16 x 2) = 44gr/mol m =44gr/mol x 1 mol = 44gr=0.044kgV = 1 = 1000 l m= 582.4039 mol x 0.044kg = 25.6257 kgT = 20 (20 +273.15 = 293.15 Esto ya es especifico , con esto se puede calcular la R de cualquier gas, solo cambiando la masa atmica del gas ( m )
R C = = =19.2528 R C = ( 19.2528 ) / ( 427J) = 0.045kcal/kg K Ciclo compresor 6 de marzo del 2013
Motor: conjunto de dispositivos con movimiento relativo entre 1 o mas de sus elementos que sirve para transformar una energa X en energa mecnica.
Maquina: conjunto de dispositivos que interrelacionados entre si transforman un X tipo de energa en otro tipo de energa.
Maquinas motrices: son aquellas en las cuales se suministra una energa X y se obtiene energa mecnica ( motores).Maquinas conducidas son: aquellas en las cuales se les suministra una energa X ( puede ser energa mecnica y se obtiene otro tipo de energa).
P Pd 3 2 Vd=volumen desplazado .: Vc= volumen de cmara .: x Vd Va= - Datos: Pa 4 1 = 10% = 2 pulgadas =5.08 cm = 0.0508 m Va V T = 20 => 20+273.15= 293.15 K P baro = 7951.38 kg/ Vc Vd = 3 atm L = 3.5 pulg = 8.89 cm =0.0889 m PuntoVolumen Presin kg/Temperatura K Masa kg
11.9819x 7951.38 kg/ 293.15 K1.8366xkg
27.5035x 30990 kg/432.3316 K1.8774x
31.8018x 30990 kg/432.3316 K4.4125 xkg
44.7603 7951.38 kg/
293.15 K4.4125 xkg
= Vc + Vd .: ( x Vd )+( L )=(0.1 x Vd ) +( 0.0889) = (0.1 * 1.8018xx) + ( 1.8018x ) = 1.9819x Vc = 1.8018x = 1.9819x = 7951.38 kg/ = 20 +273.15= 293.15 KR = 29.27 kg*m/kg*K = = 1.8366xkg = 3atm .: ( (3atm) =30990 kg/()=( ).: ()= => = ()= () = 1.9819x ()= 7.5035x = ( ) => = = = 432.3316 K = =1.8774x = =30990 kg/ = = 1.8018x = 432.3316 K = = = 4.4125 xkg ==7951.38 kg/ =293.15 K ==4.4125 xkg
= ( )= (1.8018x ) ( ) =4.7603 Gasto masico m =Masa descargada por ciclo
Mad = - Wc = W = W1-2+ W2-3+W3-4+W4-5
= = = = -)Hoja1
Columna1PVTMColumna2
puntokg/m^2 Kkg
11.0150.708294.150.834
211684.73650.708338.650.834
340019.540.2937481.490.834
440019.540.7081160.690.834
v