( ) Given: In a constant pressure combustor. C4H10 and theoretical air burns at P1 = 0.2 MPa, T1 = 600K. Products exit at P2 = 0.
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1 (SP 9) N-butane (C4H1) i burned with 85 percent theoretical air, and the product of combution, an equilibrium mixture containing only O, CO, CO, H, HO, N, and NO, exit from a combution chamber at K,. MPa. The average pecific heat at contant preure for butane i c = 1.71 / K in the temperature range 3 K 6 K. p ( ) (a) Determine the equilibrium compoition in the product tream neglecting O and NO (thi will require 4 atom balance and 1 equilibrium equation). (b) Now determine the mole fraction of O and NO in the product tream uing two additional equilibrium equation and the mole fraction you determined in part (a). (c) Auming the reactant are premixed and enter the combution chamber at 6 K,. MPa, determine the heat tranfer to or from the combution chamber in per of n-butane. (Hint: the concentration of O and NO will be much lower than the other concentration and do not need to be included in the thermochemitry calculation). Given: In a contant preure combutor C4H1 and theoretical air burn at P1 =. MPa, T1 = 6K Product exit at P =. MPa, T = K Find: (a) y,n, y,co, y,h O, y,co, y,h (b) y,o, y,no (c) Q CV Sytem ketch:
2 Aumption: (1) Ideal gae () Steady tate, uniform flow (3) Contant preure combution (4) Neglect KE, PE (5) W CV = Baic equation: de dt CV ( ) = = Q W CV CV + n h n h k 1, k 1, k, k, k k ( ) ( ) Q = n h n h = H H h T = h + h T K T CV, k, k 1, k 1, k 1 i f,98, i, i 98 k k C D yc y D p = A B ya y B p ref C + D A B Solution: (a) Proce Equation Stoichiometric equation (a toich = x + y 4 ) : C 4 H (O N ) 5H O + 4CO N Reaction equation with 85% theoretical air (a =.85a toich = 5.55) : C 4 H (O N ) bco + cco + dh O + eh + fn + go + hno Neglecting NO and O, the reaction equation become C 4 H (O N ) bco + cco + dh O + eh N Atom balance: C: 4=b+c c=4-b H: 1=d+e e=5-d O: 5.55=b+c+d=4+b+d d=7.5-b
3 N: = Chemical equilibrium Equilibrium equation and contant from Table A-7: CO + H H O + CO K eq = = 4.59 at T = K From our reaction equation, Solve the quadratic equation, It give, K eq = y CO 1 y H O 1 y 1 CO y 1 ( P ) = cd (4 b)(7.5 b) = H P ref be b( b) (4 b)(7.5 b) b( b) = 4.59 b =.5877 and 3.88 Chooe the poitive value. Then, c = 1.418, d = 4.468, and e = Thu, the reaction equation i CH (O N ).5877 CO CO H O H +.774N The equilibrium compoition i.5877 CO CO H O H +.774N At tate, n,tot n,co + n,co + n,h o + n,h + n,n = = / Therefore, mole fraction are y CO = n,co n,tot = =.8691 y CO = n,co = n,tot =.4743 y N = n,n =.774 n,tot =.6967
4 y H = n,h =.5377 n,tot =.186 y H O = n,h O n,tot = = y CO =.8691 y CO =.4743 y N =.6967 y H =.186 y H O = (b) In order to olve equilibrium decompoition for NO and O imultaneouly by adding two additional equilibrium equation, we need to etablih a reaction equation a decribed below,.774n CO CO H O H xco + yco + zo + pn + qno H O H Note: For NO, 1 O + 1 N NO i required. For O, there are two option: CO CO + 1 O and H O H + 1 O. But, we only need one more equilibrium equation a the problem tatement. Then, we have to chooe CO CO + 1 O baed on a higher equilibrium contant: K CO CO+ 1 O = > K H O H + 1 O = Herein, it i aumed that the effect of H O and H i negligible and the gae exit a an inert ga. Atom balance: C: =x + y y = 4 x O: = x + y + z + q z = ( x y q)/ N:.774 =p+q p =.774.5q With equilibrium CO CO + 1 O 1 K eq (T ) = y CO 1 y O ( P ) = =.136 at T = K y 1 CO P ref yz.5 x ( kpa (x + y + z + p + q )11.35 kpa ).5 =.136 yz.5 x(x + y + z + p + q ).5 = =.996
5 With equilibrium 1 O + 1 N NO K eq (T ) = y NO 1 y O 1 1 y N ( P ) = =. P ref q z.5 =. p.5 at T = K Solve 5 equation and 5 unknown (EES code for thi i attached). Then, x =.587 y = z =.8619 p =.77 q =.8463 Therefore, the equilibrium compoition i.587 CO CO O +.77 N NO H O H Conequently, the reaction become C 4 H (O N ).587 CO CO O +.77 N NO H O H The mole fraction for are y CO =.869 y CO =.4746 y N =.6977 y H =.186 y H O =.1499 y NO = y O = EES code "5 equation" y=4-x z=( x-y-q)/ p= *q (y*z^(.5))/x/((x+y+z+p+q )^(.5))=.996 q/(z^(.5))/(p^(.5))=.
6 (c) Uing Table A-3 and A-5, Reactant T = 6 K, p = bar h c 1 1 f,98, C H 4 1 = 15,51 = 1.71 c = = 99.4 K K K p, C H p, C H h, C ( ) ( ) ( ) ( ) 4H 6 K h 1, C 98 4H K = c 1 p, C H T = 1 K K h, C ( 6 ) 4H K h ( ) 1, C4H 98 K = 3, 1 hc ( ) ( ) ( ) 4H 6 K = h 1 f,98, C4H + h 1, C 6 4H K h 1, C 98 95, 49 4H K = 1 ho ( 6 K ) = h f,98,o + h ( ) ( ),O 6 K h,o 98 K = 17,99 8, 68 = 9, 47 hn ( 6 K ) = h f,98, N + h ( ) ( ), N 6 K h, N 98 K = 17,563 8, 669 = 8,894 H 1 = n 1,k h 1,k = (1 ) ( 95,49 ) + (5.55 ) (9,47 C k 4 H 1 ) O + (.774 ) (8,894 ) N = 14,36
7 Product T = K, p = bar 1 hco ( K ) = h ( ) ( ) f,98, CO + h, CO K h, CO 98 K = 393,5 + 1,84 9,364= 3, 8 hco ( K ) = hf,98, CO + h, CO ( K ) h, CO ( 98 K ) = 11, , 48 8, 669 = 5379 hh ( ) O K = hf,98, H ( ) ( ) O + h, HO K h, H 98 41,8 8,593 9, O K = + = hh ( K ) = h f,98, H + h ( ) ( ), H K h, H 98 K = 5,97 hn ( K ) = h f,98, N + h ( ) ( ), N K h, N 98 K = 64,81 8, 669 = 56,14 H = n,k k h,k = (.5877 ) ( 3,8 ) CO + (1.418 ) ( 53,79 ) CO + (4.468 ) ( 169,13 ) H O + (.774 ) (56,14 ) N = 417,635 Q CV = H H 1 = 417,635 14,36 = 557,995 + (.5377 ) (5,97 ) H Thu, of n butane 1 Q CV 1 of n butane per econd in the reaction = 557,995 = 557,995 of n butane
8 (SP3 & SP31 Canceled) (SP3) A imple ideal Rankine cycle with water a the working fluid operate between the preure limit of 15 MPa in the boiler and 1 kpa in the condener. Saturated team enter the turbine. a. Determine the work produced by the turbine per unit ma, the heat tranferred in the boiler per unit ma, and the thermal efficiency of the cycle. b. Now conider that irreveribilitie in the turbine caue the team quality at the outlet of the turbine to be.7. Determine the ientropic efficiency of the turbine and the thermal efficiency of the cycle. Given: A Rankine cycle with water operate at Pcond = 1 kpa, Pboiler = 15 MPa the team quality at the turbine inlet i 1. Find: (a) W turbine/m, Q bolier/m, and η th (b) η ien, η th,irrev Sytem ketch: None Aumption: 1. Steady tate, teady flow. KE and PE negligible 3. Internally reverible pump 4. No work at the boiler and the condener 5. Adiabatic at the turbine and the pump 6. Water enter the pump a a aturated liquid 7. Negligible change of pecific volume in the pump
9 Baic equation: de cv dt = = Q cv W cv + m i (h i + v i i + gz i) m e (h e + v e e + gz e) Solution: (a) T- diagram for the ideal Rankine cycle i Q in/m Q out/m State 1-: pump State -3: boiler State 3-4: turbine State 4-1: condener For bolier and condenor: Q cv/m + h in h out = For turbine and pump: W cv/m + h in h out = Steam propertie can be found from the team table, Table A- and A-3 h 1 = h f,1kpa = v 1 = v f,1kpa =.143 m3 Work done by an internally reverible adiabatic pump (refer Chapter 8..): W pump m = v 1 (P P 1 ) =.143 m3 (15, 1)kPa = h = h 1 + W pump m = = 433.
10 P 3 = 15, kpa, x 3 = 1 h 3 = = K P 4 = 1 kpa, 4 = 4 x 4 = 4 f fg = = / h 4 = h f + x 4 h fg = = Thu, W turbine/m = h 3 h 4 = = / Q bolier/m = h 3 h = = / Q condenor/m = h 4 h 1 = = / The thermal efficiency of the cycle i η th = 1 Q out Q in = 1 Q condenor/m Q bolier/m = =.3139 (b) T- diagram for the Rankine cycle with irreveribilitie at the turbine i Q in/m Q out/m State 1-: pump State -3: boiler State 3-4: turbine State 4-1: condener Except propertie at tate 4, all propertie are identical with part (a).
11 h 1 = h = 433. h 3 = 61.5 h 4 = Becaue the team quality at the outlet of the turbine i.7 P 4 = 1 kpa, x 4 = 1 h 4 = h f + x 4 h fg = = / The ientropic efficiency of the turbine i η ien = h 3 h 4 h 3 h 4 = =.876 Newly calculated Q condenor/m, Q condenor m = h 4 h 1 = = The thermal efficiency of the cycle with irrerveribilitie at the turbine i η th,irrev = 1 Q out Q in = Q condenor/m Q bolier/m = =.741
12 (SP33) A team power plant operate on an ideal reheat Rankine cycle between the preure limit of 15 MPa and 1 kpa. The ma flow rate of team through the cycle i 1 /. Steam enter both tage of the turbine at 5 C. If the moiture content of the team at the exit of the low-preure turbine i not to exceed 1 percent, a. Determine the preure at which reheating take place. b. Determine the total rate of heat input in the boiler. c. Determine the thermal efficiency of the cycle. d. Plot the cycle on a T- diagram with repect to aturation line. Given: An ideal reheat Rankine cycle with water operate at Pmin = 1 kpa, Pmax = 15 MPa Tturb1,inlet =Tturb,inlet = 5 m = 1 / The quality at the exit of the low-preure turbine i targeted.9 Find: (a) Preheat (b) Q in (c) η th (d) T- diagram Sytem ketch: None Aumption: 1. Steady tate, teady flow. KE and PE negligible 3. Internally reverible pump
13 4. No work at the boiler and the condener 5. Adiabatic at the turbine and the pump 6. Water enter the pump a a aturated liquid 7. Negligible change of pecific volume in the pump Baic equation: de cv dt = = Q cv W cv + m i (h i + v i i + gz i) m e (h e + v e e + gz e) Solution: T- diagram for the ideal reheat Rankine cycle i State 1-: pump 15 MPa State -3: boiler State 3-4: 1 t turbine State 4-5: reheat State 5-6: nd turbine State 6-1: condener (a) Steam propertie can be found from the team table, Table A-, A-3, and A-4 h 1 = h f,1kpa = v 1 = v f,1kpa =.11 m3 Work done by an internally reverible adiabatic pump (refer Chapter 8..): W pump m = v 1 (P P 1 ) =.11 m3 h = h 1 + W pump m = = 6.97 (15, 1)kPa = 15.14
14 P3 = 15, kpa, T3 = 5 h3 = / 3 = /K (need double interpolation from the table) From the condition, the quality at the exit of the low-preure turbine i targeted.9, we et x 6 =.9 P6 = 1 kpa h 6 = h f + x 6 h fg = = = f + x 6 fg = ( ) = 7.41 K T5 = 5, 5 = = MPa P 5 MPa 3MPa P 5 =.16 MPa(the reheat preure) = h 5 h = h 5 = / P4 =.16 MPa, 4 = 3 h 4 = 8.55 h 4 = (need double interpolation from the table) h 4 = (b) The rate of heat input i Q in = m [(h 3 h ) + (h 5 h 4 )] = 1 = 45,4.8 [( ) + ( )] (c)
15 The thermal efficiency i determined from Q in and Q out. Q out = m (h 6 h 1 ) = 1 ( ) = 5,84. Thu, η th = 1 Q out Q in (d) = 1 5,84. =.46 or 4.6 % 45,4.8 It i drawn at the initial part of the olution.
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