1. The fluid velocity along the x axis shown in figure
changes from 6 m/s at point A to 18 m/s at point B.
It is also known that the velocity is a linear
function of distance along the streamline.
Determine the acceleration at points A, B, and C.
Assume steady flow.
r
Sol) From the definition of acceleration a ,
r
r
r
r ∂V
a=
+ (V ⋅ ∇)V
∂t
with u = u (x) , v = 0 , and w = 0 (flow along the x axis)
r
∂u
∂u
∂u
) a = ( + u )iˆ = u iˆ
∂t
∂x
∂x
(Steady flow)
Since u is a linear function of x,
u = ax + b
At point A,
u A = a (0) + b = b = 6 (m/s)
At point B,
u B = a (0.1) + b = 0.1a + 6 = 18 (m/s)
where a and b are constant
)
a = 120
Thus, u = 120 x + 6 (m/s)
From the equation of acceleration,
r
∂u
∂
a = u iˆ = (120 x + 6) (120 x + 6)iˆ = 120(120 x + 6)iˆ (m/s2)
∂x
∂x
Finally,
for x A = 0 ,
r
a A = 120(120 ⋅ (0) + 6)iˆ = 720iˆ (m/s2)
for xC = 0.05 m,
r
aC = 120(120 ⋅ (0.05) + 6)iˆ = 1440iˆ (m/s2)
for x B = 0.1 m,
r
aB = 120(120 ⋅ (0.1) + 6)iˆ = 2160iˆ (m/s2)
(ANSWER)
2. A nozzle is designed to accelerate the fluid from V1 to V2 in a linear fashion. That is,
V = ax + b , where a and b are constants. The flow is constant with V1 = 10 m/s at x1 = 0 and
V2 = 25 m/s at x2 = 1 .
(a) Determine the local acceleration, the convective acceleration, and the acceleration at
points (1) and (2).
(b) Repeat Prob. (a) with the assumption that the flow is not steady, but at the time when
V1 = 10 m/s and V2 = 25 m/s, it is known that ∂V1 / ∂t = 20 m/s2 and ∂V2 / ∂t = 60 m/s2.
Sol)
r
a) With u = ax + b , v = 0 , and w = 0 , ( V : Linear velocity along x-axis)
the acceleration can be written as
r
r
r
r
r
r ∂V
r ∂V r
∂V
∂V
∂V
∂u
a=
+ V ⋅ ∇V =
+u
+v
+w
= u iˆ = a x iˆ
∂t
∂t
∂x
∂y
∂z
∂x
r
r
r
∂V ∂V
=
because V : Independent to time and
=0
∂y ∂z
Since V1 = u1 = 10 m/s at x = 0
&
10 = a(0) + b
&
b = 10
V2 = u 2 = 25 m/s at x = 1
&
25 = a (1) + 10 &
a = 15
and
That is, u = 15 x + 10 m/s, so that the acceleration becomes
r
a = (15 x + 10)(15)iˆ
y Local acceleration:
y Convective acceleration:
y Acceleration at (1) & (2):
r
∂V
=0
∂t
r
r
r
r
r
∂V
∂V
∂V
V ⋅ ∇V = u
= (225 x + 150)iˆ m/s2
+w
+v
∂z
∂y
∂x
r
2
at x = 0
a = (150)iˆ m/s
r
at x = 1
(ANSWER)
a = (225 + 150)iˆ = 375iˆ m/s2
b) The acceleration
r ∂u
∂u
a = ( + u )iˆ = a x iˆ
∂x
∂t
∂u ∂
This means,
= (ax + b) ≠ 0 because of unsteadiness & a = a(t ) and b = b(t )
∂t ∂t
At the time ( t = t0 ) when V1 = 10 m/s (at x = 0 ) and V2 = 25 m/s (at x = 1 ),
i.e.
V1 = u1 = a (t 0 )(0) + b(t 0 ) = 10
&
b(t0 ) = 10
and
V2 = u2 = a (t0 )(1) + 10 = 25
&
a (t0 ) = 15
Also
∂V1
= 20 m/s2
∂t
at x = 0 and t = t0
(Local acceleration, ANSWER)
∂V2
= 60 m/s2
∂t
at x = 1 and t = t0
(Local acceleration, ANSWER)
In addition,
y Convective acceleration:
At x = 0 and t = t0
At x = 1 and t = t0
r
r
∂u
V ⋅ ∇V = u1 1 = 15[15(0) + 10]iˆ = 150iˆ m/s2
∂x
r
r
∂u2
V ⋅ ∇V = u 2
= 15[15(1) + 10]iˆ = 375iˆ m/s2
∂x
(ANSWER)
(ANSWER)
y Fluid acceleration:
At x = 0 and t = t0
At x = 1 and t = t0
∂u ⎞
⎛ ∂u
a = ⎜ 1 + u1 1 ⎟iˆ = (20 + 150)iˆ = 170iˆ m/s2
∂x ⎠
⎝ ∂t
r ⎛ ∂u
∂u ⎞
a = ⎜ 2 + u2 2 ⎟iˆ = (60 + 375)iˆ = 435iˆ m/s2
∂x ⎠
⎝ ∂t
(ANSWER)
(ANSWER)
3. Water flows steadily through the funnel shown in figure.
Throughout most of the funnel the flow is approximately
radial (along rays from O) with a velocity of V = c / r 2 ,
where r is the radial coordinate and c is a constant. If the
velocity is 0.4 m/s when r = 0.1 m, determine the
acceleration at points A and B.
Sol) From the equation of acceleration in the streamline coordinates,
r
a = a n nˆ + a s sˆ
V2
where a n =
= 0 since R = ∞ (For the straight streamline)
R
c
∂V
∂V
= −V
where V = 2 and sˆ = −rˆ
as = V
∂s
∂r
r
Since V = 0.4 m/s when r = 0.1 m,
c = Vr 2 = (0.4)(0.1) 2 = 4 × 10−3 m3/s
)
V =
4 × 10−3
r2
m/s
Thus,
c ∂ c
c
2c
2c 2
as = −( 2 ) ( 2 ) = −( 2 )(− 3 ) = 5 m/s2
r ∂r r
r
r
r
At point A,
At point B,
as =
as =
2(4 × 10−3 ) 2
(0.1)
5
2(4 × 10−3 ) 2
(0.1167)5
= 3.20 m/s2
= 1.48 m/s
0.12 m
2
(ANSWER)
0.1 m
A
0.06 m
B
0.1 m
rB = (0.1) 2 + (0.06) 2 = 0.1167 m
O
4. Air flows from a pipe into the region between a circular disk and a cone as shown in figure.
The fluid velocity in the gap between the disk and the cone is closely approximated by
V = V0 R 2 / r 2 , where R is the radius of the disk, r is the radial coordinate, and V0 is the fluid
velocity at the edge of the disk. Determine the acceleration for r = 0.5 and 2 ft if V0 = 5 ft/s
and R = 2 ft.
Sol) From the equation of acceleration in the streamline coordinates,
r
a = a n nˆ + a s sˆ
where a n =
V2
= 0 since R = ∞ (For the straight streamline)
R
∂V
∂V
= −V
as = V
∂s
∂r
where V =
V0 R 2
r2
and sˆ = − rˆ
Thus,
V R2 ∂ V R2
V R2
2V R 2
2V 2 R 4
ft/s2
as = −( 0 2 ) ( 0 2 ) = −( 0 2 )(− 03 ) = 0 5
∂r r
r
r
r
r
=
2(5) 2 (2) 4
r
5
=
800
At point r = 0.5 ft,
At point r = 2 ft,
r
5
ft/s2
as =
as =
800
= 25,600 ft/s2
(0.5)
5
800
= 25 ft/s2
(2)
5
(ANSWER)
5. Water is squirted from a syringe with a speed of V = 5 m/s by pushing in the plunger with a
speed of V p = 0.03 m/s as shown in figure. The surface of the deforming control volume
consists of the sides and end of the cylinder and the end of the plunger. The system consists of
the water in the syringe at t = 0 when the plunger is at section (1) as shown. Make a sketch to
indicate the control surface and the system when t = 0.5 .
Sol) During the t = 0.5 s time interval the plunger moves l p = V pδt = (0.03)(0.5) = 0.015 m and
the water initially at the exit moves lw = Vδt = (5)(0.5) = 2.5 m. The corresponding control
surfaces and the systems at t = 0 and t = 0.5 s shown in the figure below.
0.065
0.08 m
CV at t = 0
CV at t = 0.5 s
System at t = 0.5 s