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.
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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