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ENGG 3260
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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/s3)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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