Unformatted text preview: University of California Davis
Department of Mechanical and Aerospace Engineering
ENG 103 Fluid Mechanics
Fall 2011 Professor Kennedy Chapter 1 25 Introduction Homework 1 P1.41
An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is
falling concentrically through a long vertical sleeve of diameter 6.04 cm. The clearance is
filled with SAE 50 oil at 20 C. Estimate the terminal (zero acceleration) fall velocity.
Neglect air drag and assume a linear velocity distribution in the oil. [HINT: You are given
diameters, not radii.]
Solution: From Table A.3 for SAE 50 oil, = 0.86 kg/ms. The clearance is the
difference in radii: 3.02 – 3.0 cm = 0.02 cm = 0.0002 m. At terminal velocity, the
cylinder weight must balance the viscous drag on the cylinder surface:
V
)( DL) , where C clearance rsleeve rcylinder
C
kg
V
or :
30 N
[0.86
)(
)] (0.06 m)(0.40 m)
m s 0.0002 m
Solve for V
0.0925 m / s
Ans.
________________________________________________________________________
W 1.42 wall Awall ( Some experimental values of
T, °K:
, kg/m s: 200
1.50E–5 of helium at 1 atm are as follows: 400
2.43E–5 600
3.20E–5 800
3.88E–5 1000
4.50E–5 1200
5.08E–5 Fit these values to either (a) a powerlaw, or (b) a Sutherland law, Eq. (1.30a,b). Solution: (a) The powerlaw is straightforward: put the values of and T into, say, an
Excel graph, take logarithms, plot them, and make a linear curvefit. The result is:
Powerlaw curvefit: He 1.505E 5 TK
200 K 0.68 Ans. (a) The accuracy is less than 1%. (b) For the Sutherland fit, we can emulate Prob. 1.41 and
perform the leastsquares summation, E
[ i – o (T i /200)1.5(200 S)/(T i S)]2 and 34 Solutions Manual P1.55 Fluid Mechanics, Sixth Edition A block of weight W is being pulled
W over a table by another weight W o , as shown in h Fig. P1.55. Find an algebraic formula for the
steady velocity U of the block if it slides on an
oil film of thickness h and viscosity . The block Wo bottom area A is in contact with the oil.
Neglect the cord weight and the pulley friction. Fig. P1.55 Solution: This problem is a lot easier to solve than to set up and sketch. For steady motion,
Fx,block 0 A U
CW
hapter 2 ( Pressure Distribution in a Fluid
) A Wo
o
h 81 2 W1
Hence P SolveAsmall U
poil for
(40744) o h
222 lbf .
Ans
4A 12
there is no acceleration, and the falling weightP/16 222/16 14 lbfesistance of the oil film:
Therefore the handle force required is F balances the viscous r Ans.
The block weight W has no effect on steady horizontal motion except to smush the oil film. 2.21 In Fig. P2.21 all fluids are at 20 C.
1Gage A or the350 kPa absoviscometer in
.56* F reads coneplate lute. Determine
Fig. P1.56, the angle is very small, and the
(a) the height h in cm; and (b) the reading
gap is filled with test liquid . Assuming a
of gage B in kPa absolute.
linear velocity profile, derive a formula for
the viscosity in terms of the torque M
Solution: Apply
Fig. P1.56
and cone parameters.the hydrostatic formula
from the air to gage A:
Solution: For any radius r R, the liquid gap is h r tan . Thenig. P2.21
F
p A pair
h
r
dr
180000 (9790)h r
d(Torque) dM dA w133100(0.8) 350000 Pa,r, or
2r
r tan
cos
Ans. (a)
R Solve for h 3 6.49 m
2
2
R
3M sin
M
r 2 dr
, or:
Ans.
3sin
Then, with h known,sin can evaluate the pressure at gage B: R 3
we 0
2
pB 180000 + 9790(6.49 0.80) = 251000 Pa 251 kPa Ans. (b) 2.22 The fuel gage for an auto gas tank
reads proportional to the bottom gage
pressure as in Fig. P2.22. If the tank
accidentally contains 2 cm of water plus
gasoline, how many centimeters “h” of air
remain when the gage reads “full” in error?
Fig. P2.22 2.45 Determine the gage pressure at point A in Fig. P2.45, in pascals. Is it higher or lower
than Patmosphere ?
Solution: Take
9790 N m3 for water and 133100 N m3 for mercury. Write the
hydrostatic formula between the atmosphere and point A:
patm (0.85)(9790)(0.4 m)
(133100)(0.15 m) (12)(0.30 m)
(9790)(0.45 m) p A , Chapter 2 (10000)(15) (288.2h 2 )[(h/2) csc 60 MB 109 Pressure Distribution in a Fluid
(h/6) csc 60 ] 4898(7.5cos60 ) 0, 3 F
or: 110.9h 150000 18369,ig.hP2.45
(131631/110.9)1/3 10.6 ft
or: p A patm 12200 Pa 12200 Pa (vacuum) Ans. Ans. 2.63 The tank in Fig. P2.63 has a 4cmdiameter plug which will pop out if the
hydrostatic force on it reaches 25 N. For
20 C fluids, what will be the reading h on
the manometer when this happens?
Solution: The water depth when the plug
pops out is
F 25 N h CG A or h CG (9790)h CG (0.04)2
4 Fig. P2.63 2.032 m It makes little numerical difference, but the mercurywater interface is a little deeper than
this, by the amount (0.02 sin 50 ) of plugdepth, plus 2 cm of tube length. Thus
patm (9790)(2.032 0.02 sin 50
or: h 0.152 m 0.02) (133100)h
Ans. patm , P2.97 The contractor ran out of gunite mixture and finished the deep corner, of a
5mwide swimming pool, with a quartercircle 2m water piece of PVC pipe, labeled AB in A Fig. P2.97. Compute the (a) horizontal and 1m Fig. P2.97 B (b) vertical water forces on the curved panel AB. Solution: For water take = 9790 N/m3. (a) The horizontal force relates to the vertical
projection of the curved panel AB:
FH , AB hCG Aprojected N (9790 m3 )(2.5 m)[(1m)(5m)] 122, 000 N Ans.(a) (b) The vertical force is the weight of water above panel AB:
FV (9790 168 N
m 3 )[(2m)(1m) Solutions Manual 4 (1 m) 2 ](5 m) 136, 000 N Ans.(b) Fluid Mechanics, Fifth Edition 2.147 The tank of water in Fig. P2.147
accelerates uniformly by rolling without
friction down the 30 inclined plane. What
is the angle of the free surface? Can you
explain this interesting result?
Solution: If frictionless, F W sin
ma
along the incline and thus a g sin 30 0.5g.
Thus tan ax
g az Fig. P2.147 0.5g cos30
; solve for
g 0.5g sin 30 30 ! Ans. The free surface aligns itself exactly parallel with the 30 incline. P2.148 A child is holding a string onto which is attached a heliumfilled balloon. (a)
The child is standing still and suddenly accelerates forward. In a frame of reference
moving with the child, which way will the balloon tilt, forward or backward? Explain.
(b) The child is now sitting in a car that is stopped at a red light. The heliumfilled
balloon is not in contact with any part of the car (seats, ceiling, etc.) but is held in place
by the string, which is held by the child. All the windows in the car are closed. When
the traffic light turns green, the car accelerates forward. In a frame of reference moving
with the car and child, which way will the balloon tilt, forward or backward? Explain.
(c) Purchase or borrow a heliumfilled balloon. Conduct a scientific experiment to see if Q. 8 ...
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This note was uploaded on 12/18/2011 for the course ENG 103 taught by Professor Chattot during the Fall '08 term at UC Davis.
 Fall '08
 Chattot
 Fluid Mechanics, Meter

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