2b m 1b: Sat liq C, h = kj/kg tot 3a: 1 MPa, s = s 3 -> h 3a = kj/kg, T 3b

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1 .6 A upercritical team power plant ha a high preure of 0 Ma and an exit condener temperature of 50 C. he maximum temperature in the boiler i 000 C and the turbine exhaut i aturated vapor here i one open feedwater heater receiving extraction from the turbine at Ma, and it exit i aturated liquid flowing to pump. he ientropic efficiency for the firt ection and the overall turbine are both 88.5%. Find the ratio of the extraction ma flow to total flow into turbine. What i the boiler inlet temperature with and without the feedwater heater? Baically a Rankine Cycle : 50 C,.5 ka, h = 09. kj/kg, = kj/kg K : 0 Ma : 0 Ma, 000 C, h = 55.7 kj/kg, = kj/kg K AC: 50 C, x =, h = 59. kj/kg b a b ac a) C.V. urbine Ideal: S = x S = 0.899, h S = 6.8 kj/kg => w,s = h - h S = 7.86 kj/kg 0 Ma 000 C Ma b a 50 C Actual: w,ac = h - h AC = 96.6 kj/kg, = w,ac /w,s = b) b b m b: Sat liq C, h = 76.8 kj/kg tot a: Ma, = -> h a = 9.09 kj/kg, b a = > w = 05.6 kj/kg b: Ma, w ac = w =.96 kj/kg m w ac = h -h b => h b = 0.7 kj/kg a: w = v ( a - ) kj/kg h a = h + w = 0. kj/kg a C.V. Feedwater Heater: ṁ O h b = ṁ h b + (ṁ O - ṁ )h a ṁ /ṁ O = x = (h b - h a )/(h b - h a ) = 0.78 c) C.V. urbine: (ṁ O ) = (ṁ ) b + (ṁ O - ṁ ) AC W _ = ṁ O h - ṁ h b - (ṁ O - ṁ )h AC = 5 MW = ṁ O w w = h -xh b - (-x)h AC = 8.7 kj/kg => ṁ O =.6 kg/ d) C.V. No FWH, ump Ideal: w = h S - h, S = Steam table h S = 0. kj/kg, S = 5. C FWH, CV:. b = b =.86 kj/kg K => b = 8.9 C

2 Brayton Cycle, Ga urbine.68 Conider an ideal air-tandard Brayton cycle in which the air into the compreor i at 00 ka, 0 C, and the preure ratio acro the compreor i :. he maximum temperature in the cycle i 00 C, and the air flow rate i 0 kg/. Aume contant pecific heat for the air, value from able A.5. Determine the compreor work, the turbine work, and the thermal efficiency of the cycle. Solution: v = 00 ka Compreion ratio = Max temperature = 00 o C ṁ = 0 kg/ he compreion i reverible and adiabatic o contant. From Eq.8. k- k = = 9.() 0.86 = K Energy equation with compreor work in w C = - w = C 0 ( - ) =.00( ) = 0.8 kj/kg he expanion i reverible and adiabatic o contant. From Eq.8. k- = k = = 67.7 K Energy equation with turbine work out w = C 0 ( - ) =.00( ) = 70. kj/kg Scale the work with the ma flow rate Ẇ C = ṁw C = 08 kw, Ẇ = ṁw = 70 kw Energy added by the combution proce q H = C 0 ( - ) =.00( ) = kj/kg H = w NE /q H = ( )/779.5 = 0.509

3 .76 Repeat roblem.7, but include a regenerator with 75% efficiency in the cycle. A large tationary Brayton cycle ga-turbine power plant deliver a power output of 00 MW to an electric generator. he minimum temperature in the cycle i 00 K, and the maximum temperature i 600 K. he minimum preure in the cycle i 00 ka, and the compreor preure ratio i to. Calculate the power output of the turbine. What fraction of the turbine output i required to drive the compreor? What i the thermal efficiency of the cycle? Solution: Both compreor and turbine are reverible and adiabatic o contant, Eq.8. relate then to auming contant heat capacity. k- Compreor: = ( / ) k = 00() 0.86 = 68. K w C = h - h = C 0 ( - ) =.00 ( ) = 9.5 kj/kg k- urbine = = ( / ) k = 600 (/) 0.86 = 75. K w = h h = C 0 ( ) =.00 ( ) = 85. kj/kg w NE = = 5.7 kj/kg ṁ = Ẇ NE /w NE = /5.7 = 95. kg/ Ẇ = ṁw = = 66. MW w C /w = 9.5/85. = 0.99 x x' = 00 ka For the regenerator REG = 0.75 = h X - h h X' - h = X - X = X = 7.7 K urbine and compreor work not affected by regenerator. Combutor need to add le energy with the regenerator a q H = C 0 ( - X ) =.00( ) = kj/kg H = w NE /q H = 5.7/879.8 = 0.58

4 .78 A two-tage compreor in a ga turbine bring atmopheric air at 00 ka, 7 o C to 500 ka, then cool it in an intercooler to 7 o C at contant. he econd tage bring the air to 000 ka. Aume both tage are adiabatic and reverible. Find the combined pecific work to the compreor tage. Compare that to the pecific work for the cae of no intercooler (i.e. one compreor from 00 to 000 ka). Solution: C.V. Stage : => Reverible and adiabatic give contant which from Eq.8. give: = ( / ) (k-)/k = 90 (500/00) = 59. K w cin = C ( - ) =.00(59. 90) = 87.0 kj/kg C.V. Stage : => Reverible and adiabatic give contant which from Eq.8. give: = ( / ) (k-)/k = 00 (000/500) = 65.7 K w cin = C ( - ) =.00( ) = kj/kg w tot = w c + w c = = 5 kj/kg he intercooler reduce the work for tage a i lower and o i pecific volume. C.V. One compreor => 5 Reverible and adiabatic give contant which from Eq.8. give: 5 = ( 5 / ) (k-)/k = 90 (000/00) = K w in = C ( 5 - ) =.00( ) = 7 kj/kg 5 v ka 500 ka 00 ka he reduction in work due to the intercooler i haded in the -v diagram.

5 .9 A gaoline engine ha a volumetric compreion ratio of 9. he tate before compreion i 90 K, 90 ka, and the peak cycle temperature i 800 K. Find the preure after expanion, the cycle net work and the cycle efficiency uing propertie from able A.5. Compreion to : = From Eq.8. and Eq.8. = (v /v ) k- = = 698. K = (v /v ) k = = ka Combution to at contant volume: v = v q H = u u = C v ( ) = 0.77 ( ) = kj/kg Expanion to : = ( / ) = (800 / 698.) = ka = From Eq.8. and Eq.8. = (v /v ) k- = 800 (/9) 0. = 77. K = ( / )(v /v ) = (77./800) (/9) = ka Find now the net work w = u - u = C v( - ) = 0.77( ) = -9.8 kj/kg w = u - u = C v( - ) = 0.77( ) = 75.7 kj/kg Net work and overall efficiency w NE = w + w = = 6.9 kj/kg = w NE /q H = 6.9/ = Comment: We could have found from Eq..8 and then w NE = q H. v v

6 .09 At the beginning of compreion in a dieel cycle = 00 K, = 00 ka and after combution (heat addition) i complete = 500 K and = 7.0 Ma. Find the compreion ratio, the thermal efficiency and the mean effective preure. Solution: Standard Dieel cycle. See -v and - diagram for tate number. Compreion proce (ientropic) from Eq = = 7000 ka => v / v = ( / ) / k = (7000 / 00) 0.7 =.67 = ( / ) (k-) / k = 00(7000 / 00) = 88. K Expanion proce (ientropic) firt get the volume ratio v / v = / = 500 / 88. =.8 v / v = v / v = (v / v )( v / v ) =.67 /.8 = 7 he exhaut temperature follow from Eq.8. = (v / v ) k- = (500 / 7) 0. = K q L = C vo ( - ) = 0.77( ) = 78.5 kj/kg q H = h - h C po ( - ) =.00( ) = 67 kj/kg Overall performance = - q L / q H = / 67 = w net = q net = q H - q L = = 95.5 kj/kg v max = v = R / = / 00 = 0.05 m /kg v min = v max / (v / v ) = 0.05 /.67 = 0.0 m /kg meff = w net v max v min = 95.5 / ( ) = 997 ka v v Remark: hi i a too low compreion ratio for a practical dieel cycle.

7 .6 he effect of a number of open feedwater heater on the thermal efficiency of an ideal cycle i to be tudied. Steam leave the team generator at 0 Ma, 600 C, and the cycle ha a condener preure of 0 ka. Determine the thermal efficiency for each of the following cae. A: No feedwater heater. B: One feedwater heater operating at Ma. C: wo feedwater heater, one operating at Ma and the other at 0. Ma. a) no feed water heater w = vd 0.000(0000-0) = 0. kj/kg h = h + w = =.0 = = = x x = h = = w = h - h = = 77.9 kj/kg w N = w - w = = 57.7 q H = h - h = = 5.6 H = w N q H = = 0.8 S. GEN. URBINE. COND. 0 Ma o 600 C 0 ka b) one feedwater heater w = 0.000(000-0) =.0 kj/kg h = h + w = = 9.8 w = ) =. kj/kg h = h + w = = = 5 = =.87 + x 6.78 S. GEN. 5 6 HR. URBINE. COND. 7

8 x 6 = h 6 = = 7. CV: heater cont: m = m 6 + m =.0 kg t law: m 6 h 6 + m h = m h m 6 = = Ma 5 o 600 C Ma 6 0 ka 7 m = 0.776, h 7 = ( = h of part a) ) CV: turbine w = (h 5 - h 6 ) + m (h 6 - h 7 ) = ( ) ( ) = 5.5 kj/kg CV: pump w = m w + m w = 0.776(.0) + (.) =. kj/kg w N = = 0. kj/kg CV: team generator q H = h 5 - h = = 75. kj/kg H = w N /q H = 0./75. = 0.7 c) two feedwater heater w = (00-0) = 0. kj/kg h = w + h = = 9.0 w = (000-00) =.0 kj/kg h = h + w = = S. GEN. 6 H HR URBINE. L HR 9 0 COND.

9 w 56 = 0.007( ) = 0.7 kj/kg h 6 = h 5 + w 56 = = = 7 = = 9. o C at 8 = Ma h 8 = = 8 = =.50 + x o 600 C 7 Ma Ma 0. Ma 0 ka x 9 = => h 9 = = 6.8 kj/kg CV: high preure heater cont: m 5 = m + m 8 =.0 kg ; t law: m 5 h 5 = m h + m 8 h m 8 = = 0.00 m = CV: low preure heater cont: m 9 + m = m = m ; t law: m 9 h 9 + m h = m h ( ) m 9 = = m = = CV: turbine w = (h 7 - h 8 ) + ( - m 8 )(h 8 - h 9 ) + ( - m 8 - m 9 )(h 9 - h 0 ) = ( ) ( ) ( ) = 8.0 kj/kg CV: pump w = m w + m w + m 5 w 56 = 0.687(0.) (.0) + (0.7) =. kj/kg w N = =.8 kj/kg CV: team generator q H = h 7 - h 6 = = kj/kg H = w N /q H =.8/508.5 = 0.88

10 .66 A jet ejector, a device with no moving part, function a the equivalent of a coupled turbine-compreor unit (ee roblem 9.8 and 9.90). hu, the turbinecompreor in the dual-loop cycle of Fig..09 could be replaced by a jet ejector. he primary tream of the jet ejector enter from the boiler, the econdary tream enter from the evaporator, and the dicharge flow to the condener. Alternatively, a jet ejector may be ued with water a the working fluid. he purpoe of the device i to chill water, uually for an air-conditioning ytem. In thi application the phyical etup i a hown in Fig..6. Uing the data given on the diagram, evaluate the performance of thi cycle in term of the ratio Q L /Q H. a. Aume an ideal cycle. b. Aume an ejector efficiency of 0% (ee roblem 9.90). VA o 50 C. Q H BOIL. H. JE EJEC. COND. 0 o C o C CHILL. 8 QL VA 0 o C FLASH CH. L. LIQ 0 o C (from mixing tream & 9). = = 5 = 8 = 9 = 0 = G 0 o C =.6 ka = = G 50 o C = 75.8 ka, ,0 6 = 7 = 0 o C = 50 o C = 0 o C 9 = 0 o C Aume 5 = 0 ' = 6 = 7 = G 0 o C =.76 ka ' Cont: ṁ + ṁ 9 = ṁ 5 + ṁ 0, ṁ 5 = ṁ 6 = ṁ 7 + ṁ ṁ 7 = ṁ 8 = ṁ 9, ṁ 0 = ṁ = ṁ, ṁ = ṁ a) ṁ + ṁ = ṁ ; ideal jet ejector = & = (' & ' at = ) then, ṁ (h - h ) = ṁ (h - h )

11 From = = x 8.06; x = h = = kj/kg From = = = C, h = 70. kj/kg ṁ /ṁ = =.5677 Alo h = 5.79 kj/kg, h 7 =.0 kj/kg, h 9 = 8.96 kj/kg Mixing of tream & 9 5 & 0: (ṁ + ṁ )h + ṁ 7 h 9 = (ṁ 7 + ṁ + ṁ )h 5 = 0 Flah chamber (ince h 6 = h 5 ) : (ṁ 7 +ṁ )h 5 = 0 = ṁ h + ṁ 7 h uing the primary tream ṁ = kg/: ṁ = (ṁ )h 5 & (ṁ )h 5 = ṁ 7.0 Solving, ṁ 7 = 0.67 & h 5 = 8.88 kj/kg L pump: -w L = 0.000( ) = 0.00 kj/kg h 8 = h 7 - w L = =.0 kj/kg Chiller: Q. L = ṁ 7 (h 9 -h 8 ) = 0.67( ) = 8500 kw (for ṁ = ) H pump: -w H = ( ) = 0.7 kj/kg h = h 0 - w H = = 85.5 kj/kg Boiler: Q. = ṁ (h - h ) = ( ) = 66. kw Q. L /Q. = 8500/66. =.9 H b) Jet eject. eff. = (ṁ /ṁ ) AC /(ṁ /ṁ ) IDEAL = 0.0 (ṁ /ṁ ) AC = = 0.75 uing ṁ = kg/: ṁ = (ṁ )h 5 & (ṁ )h 5 = ṁ 7.0 Solving, ṁ 7 = 9.76 & h 5 = h 0 = kj/kg

12 hen, Q. = 9.76( ) = 668 kw L h = = 86.6 kj/kg Q. = ( ) = 660. kw H & Q. L /Q. = 668/660. = 0.67 H

14 EXAMLE E9-9 A combined ga turbine-team power plant ha a net power output of 500 MW. Air enter the compreor of the ga turbine at 00 ka, 00 K, and ha a compreion ratio of and an ientropic efficiency of 85%. he turbine ha an ientropic efficiency of 90%, inlet condition of 00 ka and 00 K, and an exit preure of 00 ka. he air from the turbine exhaut pae through a heat exchanger and exit at 00 K. On the team turbine ide, team at 8 Ma, 00 o C enter the turbine, which ha an ientropic efficiency of 85%, and expand to the condener preure of 8 ka. Saturated water at 8 ka i circulated back to the heat exchanger by a pump with an ientropic efficiency of 80%. Determine (a) the ratio of ma flow rate in the two cycle, (b) the ma flow rate of air, and (c) the thermal efficiency. (d) What-if-Scenario: What would the thermal efficiency be if the turbine inlet temperature increaed to 600 K? [Manual Solution] [ES Solution]

SOLUTION MANUAL CHAPTER 12

SOLUTION MANUAL CHAPTER 12 SOLUION MANUAL CHAPER CONEN SUBSECION PROB NO. In-ext Concept Quetion a-g Concept problem - Brayton cycle, ga turbine - Regenerator, Intercooler, nonideal cycle 5-9 Ericon cycle 0- Jet engine cycle -5