Impact of a Jet â€“ Conservation of Linear Momentum

â€¢ Measure the force on the

plate Fex due to the jet (jetvane system) via balance of

moment (weight beam).

â€¢ Calculate the force on the

plate Fj using the linear

momentum balance.

â€¢Compare measured value Fex

with theoretically derived

value Fj

. Discuss errors causing

the observed difference

between Fex and Fj

S

â‘

â“ª

z

Use conservation of energy to obtain jet velocity

impinging onto the plate:

hL

g

P V

z

g

P V

z ï€« ï€« ï€½ ï€« ï€« ï€«

2 2

2

1 1

1

2

0 0

0

ï§ ï§

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

ïƒž ï€½ ï€

ïƒž ï€½ ï€«

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

ï€¦

A

ï²

A

ï²

V

ï¶

V ïƒ— AïƒžPositive

ï¶ ï¶

A

ï²

V

ï¶

V ïƒ— Aïƒž Negative

ï¶ ï¶

Conservation of linear momentum in z-direction:

ïƒ¥ ï€½ïƒ¥ ï€ïƒ¥

z

in in

z

out out

z

F mï€¦ V mï€¦ V

ï€minV1

ï€½ F1

F1

ï€½ ï€Fj

ï€¦

,

F1

: force exerted by the plate onto

the jet

Fj

: reactive force; force exert by the

jet onto the plate

ïƒž Positive if flow is out of control surface (C.S.); Negative if flow is into C.S.

1 1

ïœ Fj

ï€½ m

ï€¦

inV ï€½ ï²QV

In a given direction:

net force = net momentum fluxes

Momentum flux = mass flux times velocity

V A

ï¶ ï¶

mass flux =

ï² ï‚·

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

ïƒ—a ï€½ Wïƒ— g ïƒ—b

a F g ï€¨b xï€© Fsp

ïƒ— ï€« 0.15(m)

ïƒ—

ex

ï€½ Wïƒ— ïƒ— ï€«

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=ï²QV1

Error in Fj

can be caused by Q and V1

Fj+ Î”Fj =ï²(Q+Î”Q)(V1+Î”V1

)

Fj+ Î”Fj =ï²QV1+ï²Â·QÂ·Î”V1+ï²Â·Î”QÂ·V1+ ï²Â·Î”QÂ·Î”V1

Normalized by Fj

, we obtain:

1

1

V

V

Q

Q

F

F

j

j ï„

ï€«

ï„

ï€½

ï„

neglect higher

order

0

0.15

W Fj

gx

ï€½

Error in Fj

can be caused by x and W

ï€¨ ï€©ï€¨ ï€©

ï€¨ W W Wï€©

0.15

0.15

W

0.15

W W

Fj

Î”Fj

ï„ ï„ ï„ ï„

ï„ ï„

ï€« ïƒ— ï€« ïƒ— ï€« ïƒ—

ï€½

ï€½ ï€« ï€«

ï€«

x x x

g

g

x

g

x x

0 neglect

ï€¨ ï€©

x

x

x x

g

ï„

ï€«

ï„

ïƒž ï€½

ïœ ï„ ï€½ ïƒ— ï€« ï„ ïƒ—

W

W

F

Î”F

Î”W 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

ï€«

A V

ï€«

A V

ï€

S

hL

g

P V

z

g

P V

z ï€« ï€« ï€½ ï€« ï€« ï€«

2 2

2

1 1

1

2

0 0

0

ï§ ï§

V V gS

g

V

S

g

V

2

2 2

2

1 0

2

1

2

0

ïƒž ï€½ ï€

ïƒž ï€½ ï€«

z

ï€¨ ï€© F minV1

2mout V1

F1

z

ïƒ¥ ï€½ ï€ ï€¦ ï€« ï€¦ ï€ ï€½

F1

1. Energy balance:

2. Momentum balance:

Mass conservation suggest:

min mout

ï€¦

ï€½ 2 ï€¦

ïƒž F1

ï€½ ï€2minV1

ï€½ ï€Fj

ï€¦

Force for circular plate is two times

larger than that of flat plate !