CE2134 Hydraulics AY2014 Tutorial 1 Laminar Flows
National University of Singapore Department of Civil & Environmental Engineering CE2134 Hydraulics Academic Year 2014 Tutorial 1 Q&A
Submission date: 13 October 2014 Question 2 for submission •
Question 1. 3 (a) Calculate the Reynolds number for a fluid of density 900 kg/m and viscosity 0.038 2 Ns/m flowing in a 50 mm diameter pipe at the rate of 2.5 litres/s. litres/s. Estimate the critical mean velocity if the pipe is straight, uniform and of moderate roughness. [1508, 1.69 m/s] (Massey) 2
viscosity 0.9 Ns/m ) is pumped at 20 lit/s through (b) Glycerine (relative density 1.26, viscosity 0 straight 100 mm diameter pipe, 45 m long inclined at 15 to the horizontal. The gauge pressure at the lower inlet end of the pipe is 590 kN/m2. Verify that that the flow is laminar and, neglecting ‘end effects’, calculate the pressure at the outlet end of the pipe and the 2 2 average shear stress at the wall. [Re = 357, 116 kN/m , 183.3 N/m ] (Massey) Solution (a) Given the following 3 Fluid density ! = 900 kg/m 2
Fluid viscosity µ = 0.038 Ns/m Pipe diameter D = 50 mm = 0.05 m 3 Discharge Q = 2.5 L/s = 0.0025 m /s
Velocity of flow V
4Q =
D D
!
Reynolds Number Re
2
=
( 4) ( 0.0025) ! ( 0.05) 2
VD ! VD =
=
1.27 m / s
( 900) ( 1.27) ( 0.05)
Rec
=
2000
=
Vc D ! V
V c
=
1.69 m / s
Cross section area of flow
A
1507
!
( 900) ( 0.05)V
C
=
0.038
µ
Critical Velocity
=
0.038
µ
(b) Critical Reynolds No
=
!
(
! 100 / 1000 =
)
2 =
4
0.00785 m
2
CE2134 Hydraulics AY2014 Tutorial 1 Laminar Flows
Reynolds No
(20 / 1000)
Q
Velocity of flow V
=
A
=
! VD
Re
=
=
0.00785
=
2.54 m / s
(1.26 x1000 ) ( 2.54) (0.1)
=
355
0.9
µ
!
Momentum equation: For the same pipeline section and steady flow Rate of momentum of the fluid leaving the outlet = rate of momentum of the fluid at the inlet Using the definition of piezometric pressures,
( p A ) ! ( p A ) ! ( *
0
=
*
i
i
o
o
)
! DL "
( p A ) (590,000)&$ #!(0.1) '
* i
2
=
i
=
%4"
4633.8 N
At the outlet, the pressure has to be determined. From the discharge formula for laminar flow between parallel plates Q
=
!
4 D " dp * %
!
$
128µ #
'
dx & 4
! 20 $ (! ) (0.1) ! dp * $ or ! dp * $ '7333.86 # & ' # & # & " 1000 % " dx % (128) (0.9 ) " dx % =
=
dp* = !7333.86 ( dx ) = ! ( 7333.86) ( 45) = !330023 Pa
po
=
(p
=
p o ! ( 590 k ) + (1260 ) ( 9.8) 45sin150
=
*
o !
*
pi
) (p =
o +
! gzo
) ( pi !
+
! gzi
(
590 k ! 330 k ! 143.8k = 116.2 Pa
)
)
"
Momentum equation for steady state flow 0
( p A) ( p A) ( ) ( *
=
=
=
=
i
*
!
o
!
!
DL )
!
( dp *) A (! DL ) (! ) (330 k ) ( 0.00785) (! ) (0.1) ( 45) (! ) 2.59 k 14.137 (! )
! =
!
!
!
!
0.183k Pa
"
CE2134 Hydraulics AY2014 Tutorial 1 Laminar Flows
Question 2* (a) A viscous fluid of constant mass density ! and coefficient of viscosity µ falls through
the small gap between two vertical plates spaced 2h apart as shown in the figure below. The flow is due entirely to gravity. Determine the fully developed velocity profile between the plates. [ w
=
! g
(h
2
! x 2
) / 2µ ]
3
2
(b) Fluid of density 1260 kg/m and viscosity 0.9 Ns/m passes between two infinite parallel plates, 2 cm separation. Calculate the flow rate.
Solution
Taking coordinates with x- horizontal as shown and z vertically downwards from the middle of the two plates consider the small elemental volume shown.
(
)
p*dx ! p* + dp* dx +
(
! +
!dp*dx + d ! dz = 0 d !
=
dp*
dx dz d " dw %
dp
$ µ ' = * dx # dx & dz µ
d 2 w dx
2
=
d dz
( p ! " gz )
=
)
d ! dz ! ! dz = 0
!! g
CE2134 Hydraulics AY2014 Tutorial 1 Laminar Flows
We note that under normal situations where the coordinate z is taken to be positive upwards, p
=
p + ! gz .
p
=
p ! ! gz .
*
*
However, when we set the z coordinate to be positive downwards, then Also note that the velocity w is taken as positive along the z-coordinate.
There are no slip conditions at the walls
µ w
w
=
=
!
! g
2
2
+ Ax + B
0 at x =
0
1 2 1 =
2 1 =
2
±
! gh
=
0
µ w
x
2
! gh
! g
(h
h
+ A
2
2
h
+ B
2
! A
h 2
! x 2
+ B
)
(b) The discharge through the plates can be found as follows. dQ
=
wdx
h
Q
=
h
"
w dx
!h
! g
" 2µ (h
2
! x
2
) dx
!h
&h %h x ! ( 2µ $ 3' !h
! g # =
=
2
x
3
! g 2 =
h
! g
3 =
2µ 3
h
3
3µ
Substitution gives,
(1260) (9.8) ! 1 $ # & 3(0.9) " 100 %
3
Q
=
=
4573
lit / sec /m width
CE2134 Hydraulics AY2014 Tutorial 1 Laminar Flows
Question 3.
(a) A thin film of oil, thickness h and viscosity that the velocity profile is given by u
! g
=
(h 2µ
2
µ ,
flows down an incline plane. Show
! y 2 ) sin"
where u is the local velocity at a depth y below the free surface, to the horizontal and ! is the fluid density. (Douglas et al)
! is
the plate inclination
3
(b) A film of fluid, density 2400 kg/m , flows down a vertical plate with a free surface velocity of 0.75 m/s. If the film is 20 mm thick, determine the fluid viscosity. [6.28 2 Ns/m ] (Douglas et al) Solution
Referring to the figure above, s is the coordinate from the free surface. Consider the control volume as shown. The flow is parallel to the slope and we use the piezometric pressure which is constant across the parallel flow, i.e. in the s direction. The direction of the shear stresses on this control volume is as shown and, based on convention pertaining to the s-coordinate, it is positive at the level s and directed in the downslope x-direction. The component of the weight of the fluid in the control volume down the slope is dW sin ! , where ! is the inclination of the slope to the horizontal. When using the piezometric head concept, this weight component is not used in the momentum equation. Applying the momentum equation for a steady flow, 0
= =
p*s ! ( p* + dp* ) ( s) + (! ) dx
! ( dp* ) ( s) + (! ) ( dx )
Dividing through by (dx )
0
=
" dp % !s $ ' + ! # dx & *
CE2134 Hydraulics AY2014 Tutorial 1 Laminar Flows
'! dp µ = s )# ds (" dx du
*
$* ' d ( p + ! gz ) * dz &, = ) , s = s! g %+ ( dx dx +
=
s! g (- sin! )
since dz decreases as dx increases
Note that with a free surface and a thin film, the pressure gradient is zero
( dp / dx
=
)
0 and
gravity is the only driving potential. The above equation also implies that at the free surface (where s=0), the shear stress is zero, and it increases linearly to the sloping wall. Integrating µ u
At
s
=
,
z u
=
=
! # g sin "
s
2
2
+
B
0
( " g sin ! ) z
2
B
=
µ u
1 µ
=
2u
( ! g sin! ) ( z ! s 2
1 =
(
)
2 0.75 =
2
2
! ! g sin! $ # & ( z ' s " 2 % 2
=
)
( 2400) (9.8) sin 90
6.28 Nm / s
2
(
0
" 20 %2 $ ' # 1000 &
2
)
(
CE2134 Hydraulics AY2014 Tutorial 1 Laminar Flows
Question 4. The radial clearance between a hydraulic plunger and the cylinder wall is 0.15 mm, the length of the plunger 0.25 m and the diameter 150 mm. Calculate the leakage rate past the plunger at an instant when the pressure differential between the two ends of the 2 plunger is 15 m of water. Viscosity of hydraulic fluids is 0.9 Ns/m [0.0052 litres/min] (Douglas et al). Solution For laminar flow in a annular space which is very small such that the approximate equation can be used.
Dh 3 ' dp $ ! % " 12µ & dx # (
Q
=
=
(
( )(0.150)(0.15 / 1000)3 (15)(1000) g ! 12(0.9) 0.25 (
=
)
8.66 10 !5 litres / s
=
(
=
8.66(10 !8 ) m 3 / s
)
5.2 10 !3 litres / min
This is actually a problem related to laminar flow in an annular space with the flow initiated by the pressure difference. We have
& , )# * '! $ 2 2 2 * ( R - R1 ) '! dp ) $ . , Q= * - ' $ R42 - R14 - * 2 '! 8µ + dx ( , ) R 2 * log * ' '! $ e* '' * $% + R1 ( (!" + Here, R1
=
& 150 # $ ! = 75 mm = 0.075 m % 2 "
R2
=
(75
+
)
0.15 mm
=
0.07515 m
! ( +$ 2 2 2 * ( 0.07515 ' 0.075 ) -& ! (15) (1000 ) ( 9.8) $ # ! 4 4 # -& Q = # & (0.07515) - ( 0.075) - * ( + 0.07515 8 ( 0.9 ) " 0.25 * log -& %# - -& * e* #" ) 0.075 , ,% ) ! 2.2522 ) (10'5 ) $ ( ! ' 7 & ( 58800 )#(2.5388) (10 ) ' 7.2 0.001998 &% #" =
=
(
)
8.667 10'8 m 3 / s
=
0.0052 litres / s
.
CE2134 Hydraulics AY2014 Tutorial 1 Laminar Flows
Question 5. A steel sphere, 1.5 mm diameter and of mass 13.7 mg, falls steadily in oil 3 through a vertical distance of 500 mm in 56 s. The oil has a density of 950 kg/m and is contained in a drum so large that ‘wall effects’ are negligible. [Assume that the motion is such that the inertia forces on the fluid particles may be neglected in comparison with the shearing forces due to viscosity. At such low Reynolds number, the resisting viscous forces is given by 3! µ uD , where u is the steady velocity, D is the diameter of the sphere
and µ is the viscosity of the oil]. What is the viscosity of the oil? Verify any assumption made. [Hint: Consider buoyancy force] Solution Assume a totally laminar condition during the passage of the sphere through the oil. Then check if the Reynolds number is small for the flow to be laminar.
Steady laminar mo tion : No net force on sphere mg
=
3! µ uD
+
!
!"#
downward body force
upward viscous resis tan ce
(
! " 3 $ D & " % 6 ! # "# $
! g #
upward buoyancy
! 0.5 $ 3 ! '3 & (1.5) (10 ) + ( 950) ( 9.8) (0.0015 ) " 56 % 6
)
13.7 10 '6 ( 9.8) = 3! µ #
(
134.2 = 0.000126 µ + 1.645 10 '5
) (
µ = 0.934 Pa. s
Reynolds number=
! uD
=
( 950) ( 0.5/56 ) ( 1.5/1000 )
µ
=0.01362
0.934
