# hw11 - Prob 10.3 PROBLEM 9.8 PROBLEM STATEMENT: The...

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Prob 10.3 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: P 1 =0.1 MPa T 1 =28 o C Combustor Comp. Turbine T 3 =871 o C P 2 =0.5 MPa T 2 =227 o C 4 T 4 =482 o C q loss =30 kJ/kg 1 3 2 q IN W NET 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: qw IN,ab OUT,ab OUT,ab b a q w h h + −− = 2. Isentropic process, ideal gas with constant specific heats:  =   k1 k b b,s a a P P TT 3. Isentropic process, nonconstant specific heats: = o bb o a a PP P P

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Prob 10.3 PROPERTY DATA: Evaluate constant specific heats at the inlet temperature of 28 ° C. From Table 5s: k air = 1.4 and c P,air =1.005 kJ/kgK QUANTITATIVE SOLUTION: A) For the case of constant properties  == =   k1 1.4 1 k 1.4 2 2,s 1 1 P 0.5MPa 3 0 1 K 4 7 7 P0 . 1 M P a TT K −− = = = η IN,COMP,s 2,s 1 P 2,s 1 2,s 1 COMP I N , C O M P , a 21 P wh h c ( T T ) T T 477K 301K w h h c (T T ) T T 500K 301K η= COMP 88.3% Finding the various works and heat transfers: =−− = =− OUT,TURB 3 4 loss P 3 4 loss w ( h h)q c( T T )q kJ kJ 1.005 (1144K 755K) 30 361kJ kgK kg = =−= −= IN,COMP 2 1 P 2 1 kJ w (h h ) c (T T ) 1.005 (500K 301 kgK = K) 200kJ () − = = IN 3 2 p 3 2 kJ kJ q h h c (T T ) 1.005 1144K 500K 647 kgK kg = = NET OUT,TURB IN,COMP NET kJ kJ kJ w w w 361 200 w 161 kg kg kg = ηη NET BRAYTON BRAYTON IN w 161kJ/kg 24.9% q6 4 7 k J / k g B) For the case of temperature dependent properties Interpolating Table 5s to get: P 0 1 = 1.43; h 1 = 302.05 kJ/kg; h 2 = 504.12 kJ/kg; h 3 = 1213.6 kJ/kg; h 4 = 773.99 kJ/kg Using the relationship: = oo 2 1 P 0.5MPa P 1.43 7.15, . 1 M P a P and interpolation: h 2,s = 479.01 kJ/kg. = η = η IN,COMP,s 2,s 1 COMP COMP IN,COMP,a 2 1 h (479.01 302.05)kJ/kg 87.6% w h h (504.12 finding the various works and heat transfers:
Prob 10.3 = =−− = OUT,TURB 3 4 loss kJ kJ w (h h ) q (1213.6 773.99) 30 409 kg kg =−= = IN,COMP 2 1 kJ kJ w (h h ) (504.12 302.05) 202 kg kg () = IN 3 2 kJ kJ q h h 1213.6 504.12 709.5 kg kg =− = = NET OUT,TURB IN,COMP NET kJ kJ kJ w w

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## This homework help was uploaded on 03/19/2008 for the course ME 326 taught by Professor Schmidt during the Spring '07 term at University of Texas.

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hw11 - Prob 10.3 PROBLEM 9.8 PROBLEM STATEMENT: The...

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