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

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