Study Guides (248,644)
ENGG 3260 (5)

# Class Problem.pdf

7 Pages
115 Views

Department
Engineering
Course Code
ENGG 3260
Professor
Linda Gerber

This preview shows pages 1 and half of page 2. Sign up to view the full 7 pages of the document.
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
More Less

Only pages 1 and half of page 2 are available for preview. Some parts have been intentionally blurred.

Unlock Document

Unlock to view full version

Unlock Document
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.