Impact of a Jet – Conservation of Linear Momentum

Impact of a Jet – Conservation of Linear Momentum
• Measure the force on the
plate Fex due to the jet (jetvane system) via balance of
moment (weight beam).
• Calculate the force on the
plate Fj using the linear
momentum balance.
•Compare measured value Fex
with theoretically derived
value Fj
. Discuss errors causing
the observed difference
between Fex and Fj
S
â‘ 
⓪
z
Use conservation of energy to obtain jet velocity
impinging onto the plate:
hL
g
P V
z
g
P V
z      
2 2
2
1 1
1
2
0 0
0
 
Neglect head loss; P0=P1=0 atmosphere
pressure; z1
-z0=S
V V gS
g
V
S
g
V
2
2 2

2
1 0
2
1
2
0
  
  
Note
V0=Q/A can be obtained from the measurement:
Q: total measured volume in weight tank/duration
1. Calculate the force on the plate by the jet using linear momentum
balance:
F1
minVin

A

A

V

V  APositive
 
A

V

V  A Negative
 
Conservation of linear momentum in z-direction:
  
z
in in
z
out out
z
F m V m V
minV1
 F1
F1
 Fj

,
F1
: force exerted by the plate onto
the jet
Fj
: reactive force; force exert by the
jet onto the plate
 Positive if flow is out of control surface (C.S.); Negative if flow is into C.S.
1 1
 Fj
 m

inV  QV
In a given direction:
net force = net momentum fluxes
Momentum flux = mass flux times velocity
V A
 
mass flux =
 
2. Jet force can be directly measured from the balance of moment of
the weight-beam system:
W
Fsp
a
b
W
Fsp
a
b + x
Fex
x
Fsp
a  W g b
a F g b x Fsp
  0.15(m)

ex
 W  
0.15 m
Note: given in class note W=0.6 Kg
Fex calculated this way can be
compared with that calculated by
linear momentum balance Fj
.
Error Analysis:
Fj=QV1
Error in Fj
can be caused by Q and V1
Fj+ ΔFj =(Q+ΔQ)(V1+ΔV1
)
Fj+ ΔFj =QV1+·Q·ΔV1+·ΔQ·V1+ ·ΔQ·ΔV1
Normalized by Fj
, we obtain:
1
1
V
V
Q
Q
F
F
j
j 




neglect higher
order
0
0.15
W Fj
gx

Error in Fj
can be caused by x and W
  
 W W W
0.15
0.15
W
0.15
W W
Fj
ΔFj
   
 
     

  

x x x
g
g
x
g
x x
0 neglect
 
x
x
x x
g



 
      
W
W
F
ΔF

ΔW W
0.15
F
j
j
j
This relation is linear.
i.e., 2% error in Q and 3% error in V1 make
total error in estimating Fj
to be 5%.
For a circular plate:
A V

A V

A V

S
hL
g
P V
z
g
P V
z      
2 2
2
1 1
1
2
0 0
0
 
V V gS
g
V
S
g
V
2
2 2

2
1 0
2
1
2
0
  
  
z
  F minV1
2mout V1
F1
z
       
F1
1. Energy balance:
2. Momentum balance:
Mass conservation suggest:
min mout

 2 
 F1
 2minV1
 Fj

Force for circular plate is two times
larger than that of flat plate !