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Assignment__10 - Prob 10.1 PROBLEM 9.7 PROBLEM STATEMENT...

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Prob 10.1 PROBLEM 9.7 PROBLEM STATEMENT: Air enters the compressor of an ideal gas turbine engine (ideal air standard Brayton cycle) at 25 o C and 0.1 MPa. The pressure ratio is 6:1, and the maximum temperature in the cycle is 700 o C. Determine the net-work (kJ/ kg) and the cycle thermal efficiency. DIAGRAM DEFINING SYSTEM AND PROCESS: GIVEN: State 1: T 1 = 25 o C, P 1 = 0.1 MPa, P 2 :P 1 =6:1 State 3: T 3 = 700 o C FIND: w NET and η BRAYTON ASSUMPTIONS: IG, but will not assume constant specific heats; SFSS, NKEPE, no P in combustor or cooler, compressor and turbine adiabatic and reversible. GOVERNING RELATIONS : 1. First law for given assumptions: , , , , ( ) ( ) IN ab OUT ab IN ab OUT ab b a q q w w h h - + - = - 2. Isentropic process relation for P o functions: o b b o a a P P P P = PROPERTIES: From Table 5s: o o 1 3 P (25 C) 1.3175 and P (700 C) 103.75 ° = ° = QUANTITATIVE SOLUTION: We know that P 2 =P 3 and P 4 =P 1 . From the isentropic process relationships: ( 29 o 2 P 1.3715 0.6 0.1 8.229 = = T 2s =220.41 o C from Table 5s ( 29 o 4 P 103.75 0.1 0.6 17.29 = = . 4s T 331 4 C = ° from Table 5s Also from Table 5s: h 1 = 299.03 kJ/kg, h 2 = 497.33 kJ/kg ( 29 ( 29 o o o o 2 1 2 1 4 3 4 3 P P P P and P P P P = = 2 3 1 4 w NET P 1 =0.1 atm T 1 =25 o C Turbine Comp. Combustor T 3 =700 o C Cooler P 2 =6P 1 q OUT q IN
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Prob 10.1 h 3 = 1016.6 kJ/kg, and h 4 = 614.91 kJ/kg ( 29 ( 29 ( 29 ( 29 ( 29 NET T C 3 4 2 1 w = w - w = h - h - h - h = 1016.6 - 614.91 - 497.33 - 299.03 kJ/kg 203.4kJ/kg = ( 29 ( 29 ( 29 ( 29 NET NET BRAYTON IN 3 2 203.4 kJ/kg w w = = = q h - h 1016.6 - 497.33 kJ/kg η DISCUSSION OF RESULTS: Using the P o functions and the enthalpy values from the ideal gas tables accounts for variation in specific heat with temperature and hence gives somewhat more accurate results than assuming constant specific heats. BRAYTON 39.2% η =
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Prob 10.1 PROBLEM 9.8 PROBLEM STATEMENT: The schematic diagram of an actual gas turbine is shown in the figure. Determine for both the ideal (constant properties) and temperature dependent cases: a). the compressor efficiency, η COMP b). net work, w NET (kJ/kg) c). thermal efficiency, η BRAYTON DIAGRAM DEFINING SYSTEM AND PROCESS: GIVEN: State 1: T 1 = 28 o C = 301K; P 1 = 0.1 MPa State 2: T 2 = 227 o C = 500K; P 2 = 0.5 MPa State 3: T 3 = 871 o C = 1144K State 4: T 4 = 482 o C = 755K q loss = 30 kJ/kg FIND: A) for the case of constant properties and B) for the case of temperature dependent properties a) the compressor efficiency, η COMP b) net work, w NET (kJ/kg) c) thermal efficiency, η BRAYTON ASSUMPTIONS: SFSS, NKEPE, IG (air), no P in combustor GOVERNING RELATIONS: 1. First Law for a control volume: IN,ab IN,ab OUT,ab OUT,ab b a q w q w h h + - - = - 2. Isentropic process, ideal gas with constant specific heats: - = k 1 k b b,s a a P T T P 3. Isentropic process, nonconstant specific heats: = o b b o a a P P P P PROPERTY DATA: Evaluate constant specific heats at the inlet temperature of 28 ° C. From Table 5s: 2 3 1 4 q loss =30 kJ/kg W NET P 1 =0.1 MPa T 1 =28 o C P 2 =0.5 MPa T 2 =227 o C T 3 =871 o C T 4 =482 o C Turbine Comp.
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