Exercices Lecture 8
• The boat is moving and consequently the control
volume must be moving with it.
• The flow entering into the control volume is the
flow passing by the engine (the pump). In a
relative reference this flow is entering with the
speed of the boat and is leaving with the speed of
the jet.
 V Q
j
kV
V 
V 
Q 
  Q 2
 4k V jQ
2
2k

 Q 


 2k 
2

V jQ
k
2
   V Q    V j Q  0
Units of k?
2k
Q
   V Q  F  kV
k  
  V Q 
V 
2

  V  AV 
V 
2
  A 
Rationale:
• Momentum flux variation balances applied forces;
• Mass is conserved;
• Energy is not conserved: Bernoulli equation is not applicable.
2

 D1
 U1

4

2


D 2
  U2


4



  Q


The pressure is assumed uniform at
the entrance, both on the wall and
in the open section.
D1
2
U 2  U1
D2
2


 U 2 Q   U 1 Q   PA n 1  PA n 2
2
2
D1 
 D1
 U 1 2  U 1
4
D2 
P1
P1
 P2 D 2
2
2


 D1
   U 1  U 1


4


2

D 2
  P1  P2 

4

2

 2
D
2
1
 D1
   U 1  1 
2 
D2 

 P2     U 1
2
2
2

D1 
D1 
A1 
2 A1




1




U
1

1
2 
2 

A2 
A 2 
D2 
D2 
Rationale:
• Mass is conserved;
• Momentum flux changes balances the applied force.
• There is no momentum flux change and thus the
forces balance
2

 D1
 U1

4

2


D 2
  U2


4



  U1  U 2




  U 2 Q    U 1 Q   PA n 1  PA n 2  F
F  P1  P2 
D
4
2
 145 * 10 *
3
 0 .1
4
2
 1 . 14 * 10 N  114 kg
3
Are data of exercise
P3.85 plausible?
Rationale:
• Between the orifice (section O) and section 2 one
can apply the same rationale as in P3.59
• Between section 1 and the orifice the figure suggests
that there are no eddies and thus no energy
dissipation and consequently the Bernoulli Equation
could be applied.
• We can thus relate the pressure at section 1 and
section 2 combining a energy budget between 1 and
O and a momentum budget between O and 2.
2

DO
UO

4

2


D 2
  U2


4



  Q


D2
2
UO  U2
DO
2


 U 2 Q   U O Q   PA n O  PA n 2
PO
 P2 
2


D
2
2

  U 2  1 
2 
DO 

1
1


2 
2 
U    P 
U 
P 
2
2

1

O
PO  P1 
P1
1
2
 U
2
1
 U
2
O

4
2


 D 1  
A1 
1
2

 P1 
 U  1  2   P1 
 U 1 1  
 

2
2
D
A
O 
 O  


1
2
1
 P2   P1  PO   PO  P2 
4


 D 1  
D 
1
2
2
 

  U 2  1 
 U 1 1  
 


2
D
D
 O  



2
2
2
O
P1
 P2 
4

2





D
1
D
2
2
1   1  

  U 2   1 

2
D  
2 
D O 

 O   


500 gal / min  0 . 0315 m
3
/ s
U  4m / s
D2
2
D
 2 . 78
2
O
 D1

D
 O
P1




4
 7 . 72
1


3
2
1  7 . 72   82 KPa
 P2   10 * 4 *  1  2 . 78  
2


This is about one half of the pressure
drop suggested in the text of the exercise.
Two hypotheses can be considered:
1) The data is wrong.
2) The data is ok and the stream lines
are not well represented. If they were
as represented in the Figure, then the
effective orifice diameter is smaller
than that considered in the problem
(there is a vena contracta)
Could we calculate the flow discharge using the
diaphragm of P3.85 and measuring the pressure drop
between entrance and the orifice?
1
1


2 
2 
U    P 
U 
P 
2
2

1

O
PO  P1 
U 
Yes, we could, but if there is a vena contracta one will get an
excessive discharge because the real pressure drop is higher
than that assumed in this calculation.
The discharge computed using this equation is called the
ideal discharge. The real discharge is this multiplied by a
coefficient smaller than unity.
Q 
1
2
 U
P1
2
1
 U
2
O

2

A1 
 P1 
 U  1  2 
2
AO 

 PO 
4




1 
D1
 
 1  
2 
D O  



D
4
P1
2
 PO 
4




D
1

 
 1  
2 
D O  



1
1
2
1