Ch 5 Open System
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Transcript of Ch 5 Open System
7/27/2019 Ch 5 Open System
http://slidepdf.com/reader/full/ch-5-open-system 1/17
a) Mass Flow Rate
• Mass flow through a cross-sectional area per unit time iscalled the mass flow rate
∫ =
•
A
ndAV m ρ
where is the velocity normal to the cross-sectional flow area.nV
AV m m ρ =•
• The integration can be performed for one dimensional flow to yield
where
ρ= density of fluid, (kg/m3)V m = mean fluid velocity normal to A (m/s)
A = cross-sectional area normal to flow direction (m2)
5.1 Conservation of mass (pg 220)
7/27/2019 Ch 5 Open System
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b) Volume Flow Rate
• Volume flow through a cross-sectional area per unit time iscalled the volume flow rate
• The mass and volume flow rate are related by
AV dAV V m
A
n == ∫ • (m3/s)
ν ρ
•••
==V
V m
7/27/2019 Ch 5 Open System
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c) Conservation of Mass for Open System or Control Volume
Total massentering thesystem
Total massleaving thesystem
Net changein mass within thesystem
− =
systemout in mmm•••
∆=−
(kg) systemout in mmm ∆=−rate form
(kg/s)
7/27/2019 Ch 5 Open System
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( ) systemout in mmmm 12 −=−∑∑
The mass balance for a control volume can also expressed more explicitly as
dt
dmmm
system
ei=−∑∑
••
where i = inlet, e = exit, 1 = initial state, 2 = final state
and
7/27/2019 Ch 5 Open System
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d) Conservation of Mass for Steady-Flow Processes
The states of the mass streams crossing the control surface or boundary are constant with time. Under these conditions themass and energy content of the control volume are constant with time.
0=∆=•
CV CV m
dt
dm
( ) skg mm out in /∑∑••
=
Total mass entering CV per unit time
Total mass enteringCV per unit time
=
• steady flow
• steady flow (single stream)
21
••
= mm 222111 AV AV ρ ρ =
7/27/2019 Ch 5 Open System
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e) Special Case: Steady Flow of an Incompressible Fluid
The mass flow rate is related to volume flow rate and fluid density by
V m ρ =
• Steady Incompressible Flow :
• Steady Incompressible Flow :(single stream)
∑∑••
= ei V V
21
••
=V V
(m3/s)
2211 AV AV =
7/27/2019 Ch 5 Open System
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Example 5.1Example 5.1
A garden hose attached with a nozzle is used
to fill a 50 L bucket. The inner diameter of
the hose is 2 cm, and it reduces to 1 cm at
the nozzle exit. If it takes 50 s to fill thebucket with water, determine
(a
) The volume and mass flow rates of water through the hose.
(b) The average velocity of water at the nozzle
exit.
7/27/2019 Ch 5 Open System
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5.2 Flow work and the energy of aflowing fluid
The energy required to push the mass into or out of
the control volume is known as the flow work or
flow energy.
• Schematic for flow work
7/27/2019 Ch 5 Open System
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As the fluid upstream pushes mass across the control
surface, work done on that unit of mass is
Pvm
W w
Pmv PV A
A FL FLW
flow
flow
flow
==
====
The term Pv is called the flow work done on
the unit of mass as it crosses the control
surface
(kJ/kg)
1)
2)
7/27/2019 Ch 5 Open System
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a) The total energy of flowing fluid
The total energy carried by a unit of mass as it crosses
the control surface is the sum of the internal energy, flow
work, potential energy, and kinetic energy.
)/(2
)/(2
2
2
kg kJ gz V
h peke Pvu
kg kJ gz V u pekeue
+
+
θ
Here we have used the definition of enthalpy, h = u + Pv .
7/27/2019 Ch 5 Open System
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b) Energy transport by mass
Amount of energy transport:
)(2
2
kJ gz
V
hmm E mass
+θ
Rate of energy transport:
)/(@)(2
2
skJ kW gz V hmm E mass
+•
θ
7/27/2019 Ch 5 Open System
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Example 5.2Example 5.2
Steam is leaving a 4-L pressure cooker whoseSteam is leaving a 4-L pressure cooker whoseoperating pressure is 200 kPa. It is observed that theoperating pressure is 200 kPa. It is observed that theamount of liquid in the cooker has decreased by 0.5 Lamount of liquid in the cooker has decreased by 0.5 L
in 50 min after the steady operating conditions arein 50 min after the steady operating conditions areestablished, and the cross-sectional area of the exitestablished, and the cross-sectional area of the exitopening is 10 mmopening is 10 mm22. Determine. Determine
(a) the mass flow rate of the steam and the exit velocity,(a) the mass flow rate of the steam and the exit velocity,
(b) the total and flow energies of the steam per unit mass(b) the total and flow energies of the steam per unit mass
(c) the rate at which energy leaves the cooker by steam.(c) the rate at which energy leaves the cooker by steam.
7/27/2019 Ch 5 Open System
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5.3 Energy Balances for Steady Flow Systems
• Steady flow process – a process during which a fluid flowsthrough a control volume steadily
• No intensive or extensive properties within the control volume ( the mass, m the volume, V and the total
energy content, E remain constant)
Under steady flow conditions
7/27/2019 Ch 5 Open System
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Mass balance for steady flow process min = mout
Multiple inlets and exits Σ mi
= Σ me
One inlet and one exit m1 = m2 or ρ 1V 1 A1 = ρ 2 V 2 A2
.
.
.
.
. .
Energy Balance for Steady Flow Process
E in - E out = ∆ E system
. . . 0 (steady)
E in = E out . .
Energy balance: (kW)
7/27/2019 Ch 5 Open System
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Energy can be transferred by heat, work and mass only: the generalsteady flow system can also be written as
ee
out
out ii
in
in
mW QmW Q
∑∑
••••••
++=++ θ θ
++= gz
V h
2
2
θ
where
total energy of flowing fluidfor inlet and exit
7/27/2019 Ch 5 Open System
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∑∑
++−
++=−
••••
ii
iiee
ee gz V
hm gz V
hmW Q
22
22
whereQ = heat transferred into the system (heat input)
W = work produced by the system (work output)
.
.
7/27/2019 Ch 5 Open System
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−+−+−=−
•••)(
212
2
1
2
212 z z g
V V hhmW Q
Steady flow for single stream
per unit mass
)(2 12
2
1
2
2
12
z z g V V
hhwq
−+
−
+−=−