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Thermodynamics A214 Chapter 5 Summary Notes R59,00
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  3. Stellenbosch University (SUN)
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  5. Thermodynamics 214

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Thermodynamics A214 Chapter 5 Summary Notes
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Handwritten summary notes with examples of chapter 5 in the Thermodynamics - An Engineering Approach Ninth Edition textbook. Includes explanations, annotations, and answered examples from the textbook.

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  • No
  • Chapter 5
  • May 9, 2021
  • 16
  • 2020/2021
  • Summary

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of control
Chapter 5 Mass and
Energy systems
:




P
conservation of Mass :
open systems

Mass conserved
is
always
-




For boundaries Thus
-

open systems ,
mass can cross .
we


need to keep track of the mass flow race .




Mass flow rate in
masse
= =



ime

cross-sectional
area
to


mass flow race in =p UA in = ✓A
T
Lkgls) /^ T
A
density velocity specific
volume

volumetric flow race = j = volume

Time


volumetric flow rate is = VA ← cross-sectional area


( M31 )
s
µ
velocity

in =p j =
I. ← volumetric flow race


✓ ←
specific volume

Inlets
outlets
to t


{ PVA Am
EPVA -
=


I

, flow
For a
steady race : Mass in control volume is constant
-
-




{ pv EPVA
Dig
A = = O


T T
inlet outlet




For flow
a
steady ,
single
inlet / outlet


In cut


PVA =
PVA


flow and flow
incompressible
For inlet /outlet
steady single
a
,




V A = V A liquid p = constant



In one




Example :

37,85 litres
←




I
a)
¥85s
= = 0,757L Is



( (0.757 ) 0 , > 57

¥g )
in =p U
kg Is
= =



→

b) U = VA

0,75>
§ ( ¥1) V ( I4 0,0082 )
=
✗




V 15
.




. .
= ,
I M1S

→

, and of Fluid
Flow work
Energy Flowing
:
a




Flow work - The work required to push mass into or out


of c control volume .




W = PV or w =P ✓

→ t
pressure volume



'
No out
Energy of a e= ut 12 v2 +
gz going
of control
in or

volume
fluid
non -




flowing
of 112 v2 Inlets / outlets
Energy a ② =L -1 +
gZ
fluid
Flowing ✗ T
9,81 vertical

height
Example :




V
a) = 016L V of sale .
liquid @ 1501hPa :[ A -5 ] ✓ = 0,001053


i.
m=¥ =
lm3 = c.
570kg
c. ocio53

in =
E70 2,37×10-4 Is
kg
=



40×60 →

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