Transcript Fluids - Eastern Illinois University

Fluids
Your Comments
Everything is going great.
It seems like there is homework due and we haven’t gone over any problems like
it yet, but you do a good job of going over problems like that and help us with
the homework.
Just am curious how im doing in the class with regard to test grades
I’m very far behind with my homework
I like working problems in class.
The assignments and preflights seem a little tedious when I have a ton of stuff to do.
Tough stuff
Keep smilin
This class is very interesting, we do fun hands on stuff, but all the equations make
it so much less fun
Woohoo class!
We’re almost done! 
I would appreciate more examples I feel a little confused when doing homeworks
Comments Continued
One thing in particular... Due to the load of work we have in this
class, and the pace that we continuously work at, it would be nice to
know where we're at as far as grades go. Being a student in college,
I TOTALLY understand what it feels like to be bogged down, and
completely swamped. Next time around for this class, instead of
assigning so much "busy work" that highly influences our grade,
maybe drop it down a notch so we can know what we're getting in
the class on a more consistent basis. Less to grade usually means
faster and more efficient grading right? For me, its just hard to
judge how I'm doing right now because there aren't any grades up,
and the semester is creeping up quick. PLUS if the grades are
coming back faster, not only do i know if i get the points or not, i
can get feedback as to whether or not I'm doing the work right. I
do the homework how I think it should be done, but I'm not sure
whether its right or not. Thanks for putting up with us!
Pascal’s Principle
• A change in pressure at any point in a
confined fluid is transmitted everywhere
in the fluid.
• Hydraulic Lift
DP1 = DP2
F1/A1 = F2 / A2
F1 = F2 (A1/A2)
• Compare the work done by F1 with the
work done by F2
A) W1 > W2
B) W1 = W
C) W1 < W2
= F2 (A / A ) d
W = F d cos q
W1 = F1 d1
= F2 (A1 / A2) d1
2
1
2
= F2 V1 / A2
= F2 d2 = W2
1
Dam ACT
B
A
A
Two dams of equal height prevent water
from entering the basin. Compare the net
force due to the water on the two dams.
A) FA > FB
B) FA=FB
C) FA< FB
F = P A, and pressure is rgh. Same pressure, same
area same force even though more water in B!
Pressure and Depth
Barometer: a way to measure
atmospheric pressure
For non-moving fluids, pressure depends only on
depth.
p1=0
p2 = p1 + rgh
Patm - 0 = rgh
p2=patm
Measure h, determine patm
example--Mercury
r = 13,600 kg/m3
patm = 1.05 x 105 Pa
 h = 0.757 m = 757 mm = 29.80” (for 1 atm)
h
Preflight
Is it possible to stand on the roof of a five story
(50 foot) tall house and drink, using a straw, from
a glass on the ground?
Pa
p=0
81% 1. No
Pa  rgh
h
rg
p
a
h
19% 2. Yes
Even if the person were able to remove all the air from the straw, the height
to which the outside air pressure moves the water up the straw would not
be high enough for the person to drink the water.
Why not? A straw is a straw.
I never heard of it not working.
h 

1.01 x 105 N
1000 kg
m3

m2
 9.8 m
 10.3m
s2
It would be much more difficult since
atmospheric pressure is no
longer pushing the fluid upwards.
Archimedes’ Principle
• Determine force of fluid on immersed
cube
– Draw FBD
• FB = F2 – F1
•
= P2 A – P1 A
•
= (P2 – P1)A
•
=rgdA
•
=rgV
• Buoyant force is weight of displaced
fluid!
Archimedes Example
A cube of plastic 4.0 cm on a side with
density = 0.8 g/cm3 is floating in the water.
When a 9 gram coin is placed on the block,
how much does it sink below the water
surface?
SF=ma
Fb – Mg – mg = 0
r g Vdisp = (M+m) g
Vdisp = (M+m) / r
h A = (M+m) / r
h = (M + m)/ (r A)
= (51.2+9)/(1 x 4 x 4) = 3.76 cm
Fb
Mg m
g
h
M = rplastic Vcube
= 4x4x4x0.8
= 51.2 g
Archimedes’ Principle
• Buoyant Force (FB)
–
–
–
–
–
weight of fluid displaced
FB = rfluidVoldisplaced g
Fg = mg = robject Volobject g
object sinks if robject > rfluid
object floats if robject < rfluid
• If object floats…
– FB = Fg
– Therefore: rfluid g Voldispl. = robject g Volobject
– Therefore: Voldispl./Volobject = robject / rfluid
Preflight 2
Which weighs more:
1. A large bathtub filled to the brim with water.
2. A large bathtub filled to the brim with water with a
battle-ship floating in it.
Tub of water + ship
CORRECT
3. They will weigh the same.
Tub of water
Weight of ship = Buoyant force =
Weight of displaced water
Overflowed water
Continuity of Fluid Flow
• Watch “plug” of fluid moving through the narrow part of the tube (A1)
•Time for “plug” to pass point Dt = x1 / v1
• Mass of fluid in “plug”
m1 = r Vol1 =r A1 x1 or m1 = rA1v1Dt
• Watch “plug” of fluid moving through the wide part of the tube (A2)
•Time for “plug” to pass point Dt = x2 / v2
• Mass of fluid in “plug”
m2 = r Vol2 =r A2 x2 or m2 = rA2v2Dt
• Continuity Equation says m1 = m2 fluid isn’t building up or disappearing
•A1 v1 = A2 v2
Faucet Preflight
A stream of water gets narrower as it
falls from a faucet (try it & see).
Explain this phenomenon using the
V1
equation of continuity
A1
A2
V2
As the the water falls, its velocity is increasing. Since the continuity equations
states that if density doesn't change...Area1*velocity1=Area2*velocity2. From
this equation, we can say that as the velocity of the water increases, its area is
going to decrease
I tried to do this experiment and my roommate yelled at me for
raising the water bill.
Adhesion and cohesion…
Fluid Flow Concepts
r
A1 P1 v1
A2 P2
v2
• Mass flow rate: rAv (kg/s)
• Volume flow rate: Av (m3/s)
• Continuity: rA1 v1 = rA2 v2
i.e., mass flow rate the same everywhere
e.g., flow of river
Bernoulli’s Equation
• Consider tube where both Area, height
change.
Note:
– W = DK + DU
(P1-P2) V = ½ m (v22 – v12) + mg(y2-y1)
(P1-P2) V = ½ rV (v22 – v12) + rVg(y2-y1)
P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22
W=F d
= PA d
=PV
Lift a House
Calculate the net lift on a 15 m x 15 m
house when a 30 m/s wind (1.29 kg/m3)
blows over the top.
P1+rgy1 + ½ rv12 = P2+rgy2 + ½rv22
P1 – P2 = ½ r (v22 – v12)
= ½ r (v22 – v12)
= ½ (1.29) (302) N / m2
= 581 N/ m2
F = PA
= 581 N/ m2 (15 m)(15 m) = 131,000 N
= 29,000 pounds! (note roof weighs 15,000 lbs)
Fluid Flow Summary
r
A1 P1 v1
•
•
•
•
A2 P2
v2
Mass flow rate: rAv (kg/s)
Volume flow rate: Av (m3/s)
Continuity: rA1 v1 = rA2 v2
Bernoulli: P1 + 1/2 rv12 + rgh1 = P2 + 1/2 rv22 + rgh2