Class Problem.pdf

7 Pages
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Department
Engineering
Course Code
ENGG 3260
Professor
Linda Gerber

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Description
Class Problem Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions. Properties From the steam tables (Table A-6) P 15 MPa  v1= 0.06857 m /kg o  90 kJ/s T1=500 C  h1=3433.8 kJ/kg and 1 Steam 2 P 2 2MPa v = 0.15120m /kg  2 T2= 400 C  h 2= 3247.6kJ/kg Analysis (a) There is only one inlet and one exit, and thus m = m = m& . 1 2 The mass flow rate of steam is 1 1 −4 2 m = V 1 1 3 (80 m/s)(50×10 m ) = 5.833 kg/s v1 0.06857m /kg (b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as E − E & = ∆E 0 (steady)= 0 1 424 3ut 144y4244 3 Rate of net energy Rate of change in internal, kinetic, by heat, work, and maspotential, etc. energies E = E & in out m h + V 2/2) = Q +m h + V 2/2) (sinceW ≅ ∆pe ≅ 0) 1 1 out 2 2  V 2 − V 2  − Q&out= m h 2 − h1+ 2 1   2  Substituting, the exit velocity of the steam is determined to be  2 2  − 90kJ/s = 5.833kg/s 3247.6 − 3433.8+ V2− (80m/s)  1kJ/kg   2 1000m /s 2     It yields V 2 589.9 m/s (c) The exit area of the nozzle is determined from 3 & 1 mv 2 (.833 kg/s ).1512 m /kg ) −4 2 m = v V2A 2 → A = 2 V = 589.9 m/s =15.0×10 m 2 2 Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 3 Properties The gas constant of air is 0.287 kPa.m /kg.K (Table A-1). The enthalpies are (Table A-17) T1= 27°C = 300 K → h1= 300.19 kJ/kg T2= 42°C = 315 K → h2= 315.27 kJ/kg Analysis (a) There is only one inlet and one exit, and thus m 1 m 2 m &. We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as E inE&out = ∆Esystem0 (steady)= 0 1 424 3 144 4244 3 Rby heat, work, and massof change in internal, kinetic, 18 kJ/s potential, etc. energies Ein E&out 2 2 1 AIR 2 m h 1+ V 1 /2) =Q&out+ m h2 + V 2/2) (sinceW ≅ ∆pe ≅ 0) 2 2 &  V 2 − V1  − Qout= m  2 − h1+   2  Substituting, the exit velocity of the air is determined to be 2 2  V2 − (220 m/s)  1kJ/kg  − 18 kJ/s= (2.5 kg/s3)5.27 −00.19 +  2 2   2 1000 m /s  It yields V 2 62.0 m/s (b) The exit pressure of air is determined from the conservation of mass and the ideal gas relations, 2 m = 1 A V  → v = A 2 2 = 0.04m )62m/s )= 0.992m /kg v2 2 2 2 m 2.5kg/s and RT (0.287kPa⋅m /kg ⋅K (315K ) P2 2= RT 2 → P =2 2 = = 91.1kPa v 2 0.992m /kg Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A-4 through 6) P1= 10 MPa   v1= 0.02975 m /kg o  P = 10 MPa T1= 450 C  h1= 3240.9 kJ/kg 1 T1= 450°C and V1= 80 m/s P2=10 kPa  h 2 h +fx h 2 fg=191.83+ 0.92× 2392.8 = 2393.2kJ/kg x2= 0.92  ·STEAM Analysis (a) The change in kinetic energy is determined from m = 12 kg/s W· V 2 −V 2 (50m/s)2 − (80m/s)2 1kJ/kg  ∆ ke= 2 1 =   = −1.95kJ/kg 2 2  1000m /s 2   P = 10 kPa (b) There is only one inlet and one exit, and thus m 1 m =2m &. We take 2 x2= 0.92 the turbine as the system, which is a control volume since mass crosses V2= 50 m/s the boundary. The energy balance for this steady-flow system can be expressed in the rate form as E − E& = ∆E& 0 (steady) = 0 1424 3ut 144sy4244 43 Rate of net energy Rate of change in internal, kinetic, by heat, work, and masspotential, etc. energies Ein E &out m h +V 2/ 2) W& + m h + V 2/2) (sinceQ ≅ ∆pe ≅ 0) 1 1 out
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