ch 5 open system

7/27/2019 Ch 5 Open System

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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)



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



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)



( ) 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



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 =



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 =



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.



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



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)



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 .



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



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.



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



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)



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



∑∑

++−

++=−

••••

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)

.

.



−+−+−=−

•••)(

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

−+

−

+−=−

ch 5 open system

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