Combustion Chamber. (0.1 MPa)

Transcription

1 ME 354 Tutial #10 Winte 001 Reacting Mixtues Pblem 1: Detemine the mle actins the pducts cmbustin when ctane, C 8 18, is buned with 00% theetical ai. Als, detemine the dew-pint tempeatue the pducts i the pessue is 0.1 MPa. Step 1: Daw a diagam t epesent the sstem C 8 18 Ai ( 00% Theetical) Cmbustin Chambe (0.1 MPa) C N Step : State u assumptins Assumptins: 1) Cmbustin is cmplete ) Cmbustin gases ae ideal gases 3) Stead peating cnditins exist Step 3: Calculatins Bee we stat the slutin t this pblem let us examine the geneal cases cmplete cmbustin t develp a pcedue balancing the eactin equatins. 1) Geneal Case 100% Theetical Ai and Cmplete Cmbustin C x + v N ) vc C + v + v N ( N whee the v i s ae the stichimetic ceicients. Peming a mass balance n each the elements i.e. equiing that the ttal mass mle numbe each ELEMENT in the pducts shuld be equal t that in the eactants. Mass Balance C s v C x s v s v N s v v x C + + N v 3.76v 4 1

2 ) Geneal Case Excess Ai and Cmplete Cmbustin C x + v 1+ a)( N ) vc C + v + av + v N ( N Let (1+a) the pecent theetical ai, whee a is the pecentage excess ai. F example, 300% theetical ai (00% excess ai), (1+a)3 s a. Mass Balance C s v C x s v v s ( 1+ a) v v av v x C N s v.76(1 + a) v N 3 4 In the pblem we ae tld ctane, C 8 18, is cmbusted with 00% theetical ai ( 100% excess ai) s a 1. Using the mulas develped the geneal case with excess ai: C s x 8 v C 18 s v 9 18 s v x N s v 3.76(1 + a) v 3.76()(1.5) 94 N Theee, the balanced eactin equatin is C 5 + N ( N ) 8C The ttal mles the pducts, N pducts, is N pducts ( )13.5 The mle actins, i, each cmpnent ae: C N C /N pducts (8/13.5) 6.47% N /N pducts (9/13.5) 7.9% Answe Pblem#1 a) N /N pducts (1.5/13.5) 10.1% N N /N pducts (94/13.5) 76.1% Recall the dew-pint tempeatue the pducts is the tempeatue at which the wate vap in the pducts stats t cndense as the pducts ae cled. This is the satuated tempeatue cespnding t the pessue (i.e. the patial pessue in the mixtue). Since we have assumed that the cmbustin

3 gases behave as ideal gases, we can detemine the patial pessue m the ttal pessue multiplied b the mle actin,. P P(0.079)(0.1 [MPa]) 7.9 kpa Fm Table A-4, inteplating in between 35 C and 40 C T dp T dp 39.7 C Answe Pblem#1 b) Pblem #: A small gas tubine uses C 8 18 (l), with 400% theetical ai. The ai and ente at 5 C and the pducts cmbustin leave at 900K. The utput the engine and the cnsumptin ae measued, and it is und that the speciic cnsumptin is 0.5 kg/s pe megawatt utput. Detemine the heat tanse m the engine pe kml. Assume cmplete cmbustin. Step 1: Daw a diagam t epesent the sstem Q C Ai C C N 900K (400% Theetical) Cmbustin Chambe Step : State u assumptins Assumptins: 1) Cmbustin is cmplete ) Cmbustin gases ae ideal gases 3) ke, pe 0 4) stead peating cnditins exist Step 3: Calculatins Using the pcedues develped in Pblem #1, we can balance the eactin equatin 400% theetical ai (a3). C s x 8 v 18 s v 9 18 s v x

4 N s v 3.76(1 + a) v 3.76(4)(1.5) 188 N C ( N ) 8C N 18 In de t detemine the heat tanse m the sstem we need pem a ist law analsis n the cmbustin chambe. Fm Cengel and Bles (14-9) n p777, the ist law cmbustin analsis (n a pe mle basis ) is as shwn in Eq1. Q in + Win + N ( h + h h ) Qut + Wut + N p ( h + h h ) p (Eq1) We will assume the heat tanse and wk ae ut the sstem s the Q in & W in tems ae ze. We will ist detemine the apppiate enthalpies the eactants. Nte: the cicle supescipt is used t dente that the ppet value is eeenced t a 5 C, 1atm eeence state. Reactants (C 8 18,, and N ) Enthalp matin, h F stable elements and N, the enthalp matin is ze h 0 N h 0 We can ind the enthalp matin C 8 18 (l), m Table A-6 h kj/kml Ideal Gas Enthalp elative t eeence state, h h In the h h tem, the h tem is the ideal gas enthalp at the tempeatue inteest and h is the ideal gas enthalp at the eeence tempeatue. B subtacting h m h we ae, in eect, eeencing the ideal gas enthalp at the tempeatue inteest t an ideal gas enthalp ze at the eeence state. B ding this all the eactants and pducts we ensue that all the enthalp values ae eeenced t the same eeence state (5 C, 1atm) and ae thus cmpaable. Since the eactants ae alead at the eeence state, eactants. ( h h ) ( h h ) N ( h h ) C h h 0 all the 4

5 Pducts (C,,, and N ) Enthalp matin, h F stable elements and N, the enthalp matin is ze h 0 N h 0 We can ind the enthalp matins C and (g), m Table A kj/kml h C h 4180 kj/kml Ideal Gas Enthalp elative t eeence state, h h Since the pducts ae at 900K, we must detemine h at 900K and subtact the h at the eeence state 5 C/98K (this is h ) t ensue u enthalpies ae all calculated with espect t the same eeence state. F C, kj/kml (Table A-0) 9364 kj/kml (Table A-0) h h@ 98K ( h h ) C 8041 kj/kml F (g), 3188 kj/kml (Table A-3) 9904 kj/kml (Table A-3) h h@ 98K ( h h ) F, 194 kj/kml 798 kj/kml (Table A-19) h h@ 98K 868 kj/kml (Table A-19) ( h h ) F N, 1946 kj/kml 6890 kj/kml (Table A-18) h h@ 98K 8669 kj/kml (Table A-18) 5

6 ( h h ) N 181 kj/kml We ae als given the speciic cnsumptin is 0.5 kg/s pe megawatt utput the engine. We can use this inmatin t ind the W ut as shwn belw. MJ 1 s 114.4[ kg ] MJ kj Wut kg 1[ kml ] kml kml 0.5 s Nte: The mla mass C 8 18 was detemined m the mla mass C (1.01kg/kml) and (1.008kg/kml) M 8(1.01)+18(1.008) 114.4kg /kml. Eq1 can be eaanged t islate Q ut as shwn in Eq. ut N h + h ) Wut Q ( N ( h + h) p p (Eq) Substituting in the values detemined abve int Eq we detemine the heat tanse m the engine. Nte: we can detemine the numbe mles each pduct using u balanced eactin equatin: N C 8, N 9, N 37.5, and N N 188. Q ut kj ( 49950) kml kj (456960) kml 8( ) + 9( (1946) + 188(181) 194) Q ut ( ) kj/kml Q ut kj/kml Answe Pblem #) 6