MEB2043 FLUID MECHANICS I

JANUARY 2021 SEMESTER

ASSIGNMENT 4

(FINITE CONTROL VOLUME ANALYSIS)

NAME : INTAN NUR HASLINDA BINTI ROSLAN

STUDENT ID : 18001912

PREPARED FOR : DR TUAN MOHAMMAD YUSOFF SHAH TUAN

DATE OF SUBMISSION : 8th MARCH 2021

INTAN NUR HAS LINDA BINTI ROS LAN 18001912 ME Group 2

Tutorial 6

① ②

It for Jc using
ri SA ideal gas law
0 St + pv O
-
-
-
,

¥
,

⇐ = -
in

in =p AV

A, =
Az

For steady flow between sections 11 ) and 121,

in a
= in , Pi T2
Va = -
Vi
A, V, Pz T,
Ps As Vz =p .

Vz =
I At V, -
( t) =
( 77 kPa ) ( 240 K )
( 205 m/s ,
Pz Az ( 45 kPa ) ( 268 K )

=
314 Mls
 ① ③

¥ for p ft t
Jc ,
pv - ri SA -
- o

steady flow

{ s
p V
- ri SA = O

water and alcohol can be considered incompressible

relative to each other

Q3 = Qi t Qz in =p Q

in 3 = mi , + in ,

P3 ( Qi t Q2 ) =p Q , ,
t
p , Q2
3) 1000 Co l ) 80010.3 )
pz ( O
. I t O .
= -
t

' }
P3
=
850 kg Im
For steady flow,

in , t in 2 = in 3

pi Qi t
p - Q2 =p 303 Cl )
-

since water and oil are incompressible ,

Q, t Q2 =
Q3 -
12 )

Q2)
pi Qi tpa Q2 =p, ( Q ,
t

Q, t SG , Q2 = SG 3 ( Q , t Q 3)

Q1 ( I -
SG 3 )
Q, =

SG z -
SG ,

( l Mls ) ( l -
0.95 )
=

0.95 -
0.90
3
=
I m Is
 (1) →
(3) →

(2) →

For steady incompressible flow through control volume,

Q, t Q2 =
Q3

V, A , t Q2 = V3 A 3

Q2 =
Vs Az -
Vi Al
=
( 6 Mls ) ( O 075 m2 ) -
-
130 Mls ) ( 0.01 m2 )
3
=
0.15 m Is

in liters per second

Q2
-
-

(0.15 MI ) ( 1000,2mi}eI)
=
150 liters Is
INTAN NUR HAS LINDA BINTI ROS LAN 18001912 ME Group 2

Tutorial 7

(2)

(1)

Assume flow from Cl ) to 12 ) and use the

energy equation ( Eq 5- 84)
.

to get for the contents of the control volume shown :

't hi!
"

Pg +
12g 2, =

Pj +
Yg t 2, + he

Thus ,

he =

Pg -

Pg t 2 ,
-

22

= 3 m -
I m -
1.5 m

= 0.5M

and since he > O

,
the assumed direction of flow is correct .
 (1)

(2)

For the control volume shown , application of the energy equation ( Eg 5.847 yields .

O
'>

Pg t
Igi t 2, =

Pj +
I'g t 2, t hs -
he 111

However V?
he = 1.2 12 )
2g

hs =
hp =
20 -
2000 Q2 ( 3)

Since Q =
Vz Az
,
from eq .
12 )

"

j} ( AQ )
'
he = (4)

Combining egs . 117,13) and 141 ,

" "

Lg ( AQ ) 13g ( AQ )
'
t 22 = 2, t Zo -
2000 Q -

(5)
 ( 2g Azz )
' '
Q2 t
z
g
jazz t 2000 = 2, -
22 t 20

Therefore ,

12g
2, -
22 t 20

Q
a' zgiiaI.pt
=

+ 2000
gig
.

1219.8111%04071212
G t 20

.gg/iico.gnyzt
=

+ 2000
2cg

=
0 0521
. m3 Is
 control
volume
↳

For the water flowing through the control volume , the x and y direction components
of the linear momentum are

Vs p Vz A 2 t Vz COS Op V 3 A 3=0 Cl )

and
-

Vip Vi A, t Vs sin Op Vs A 3=0 121

From the conservation of mass principle ,

p Vi A , p Vz Az t
p V3 A3 0 13 )
- - =

combining egg . Ill and 121 ,

'"
Hit Ii )
"

!! !
'
" Ai
tano = = = = O 3086
-

1041212 )
'

Vi Ar (6) 21
IT

'
tan ( o 3086 )
-

O = .

=
17 150
-
Combining eqs . Ill and 131 ,

Vip Az t Vz cos 0 ( p V A , tpvz Az )

.
= O

>
V2 A2
✓3 =

cos 0 ( Vi Ai t Vz A 2)

2
Vz 'd ,
=

'
0 ( Vi di )
'
cos t Vzdz

(6) 210.1212
=

( cos 17.150 ) ( ( 4) ( O 1) ' . t (6) CO 1212 ]

-

=
4 29 . m/s

To determine the loss of available energy associated with the flow through this
control volume we obtain by applying the energy equation ( Eg 5.641 .

'

( ) mis ( vis VI ) in
'

( ie Y ) in W
Y
'
-

,
t
,
-

.
t t t 3=0 141

The conservation of mass equation Eq ,

13) can also be written as

-
in , t in z t in 3=0 15 )

combining egs . 141 and 151 ,

f ! )
'

Ivi )
'

(
" " "
ni , his -
vi. It in luv z Wal ,
-
-
- in , tins co ,
Rate of available energy loss

Vi
( -132 )
Vi
( )
V'
rate of loss =
p Vi A, + p Va Az

))
'

Id ( Ii )
" "
I
- "
air .
=
. v. +
;

( 101441/4 42.292 ) 121461/62 I. ))

' '

99491
( " - -
29
= co . + co .

= 558 69 .
N -
m/s