Problems in Chapter 3 (Fluid Dynamics)
1. An incompressible, inviscid fluid flows steadily
with circular streamlines around a horizontal
bend as shown. The radial variation of the velocity
profile is given by rv = ro vo where vo is the
velocity at the inside of the bend which has
a radius r = ro . Determine the pressure variation
across the bend in terms of vo , ro , ρ , r and po,
where po is the pressure at r = ro . Neglect gravity.
(2)
(1)
Sol) 1. Related with the Bernoulli e.q. across the streamline
2. Choose two points (1) (at r = ro ) & (2) (at r = r ) across the streamline
3. Apply the Bernoulli equation across the streamline (i.e. between (1) & (2))
ro v 2
r v2
po + ρ
dn + γz1 (at r = ro) = p + ρ
dn + γz2 (at r = r)
ro r
ro r
where dn = −dr , z1 = 0 , z 2 = 0 (Horizontal bend)
4. Using the velocity profile, v = (ro vo ) / r
∫
then,
Finally,
∫
po = p − ρ (ro vo ) 2
p = po −
∫
r
ro
1
1
1
1
dr = p + ρ (ro vo ) 2 [ 2 − 2 ]
3
2
r
r
ro
1
1
1
ρ (ro vo ) 2 [ 2 − 2 ]
2
r
ro
or
p = po +
r
1
ρvo 2 [1 − ( o ) 2 ]
r
2
[ANSWER]
2. Water flows from the faucet on the first floor of the building shown
with a maximum velocity of 20 ft/s. For steady inviscid flow,
determine the maximum water velocity from the basement
faucet and from the faucet on the second floor (assume each
floor is 12 ft tall).
Sol) From the simplified diagram as shown,
Bernoulli’s equation along the streamline;
p1 +
1
1
ρv12 + γz1 = p2 + ρv2 2 + γz 2 [Between (1) and (2) (Basement)]
2
2
1
2
= p3 + ρv3 + γz3 [Between (1) and (3) (1st floor)]
2
1
= p4 + ρv4 2 + γz 4 [Between (1) and (4) (2nd floor)]
2
Since all pressures at the faucets (p2, p3, p4) are zero (Free jet)
a) Water velocity at the basement faucet
1
1
p3 + ρv3 2 + γz3 = p2 + ρv2 2 + γz 2
2
2
(4)
16 ft
(3)
4 ft
(1)
8 ft
(2)
v2 =
∴
2⎡
1
⎤
2
2
p3 + ρv3 + γ ( z3 − z 2 ) − p2 ⎥ = v3 + 2 g ( z3 − z 2 )
⎢
ρ⎣
2
⎦
(where v3 = 20 ft/s, and z3 − z2 = 12 ft)
= (20) 2 + 2(32.2)(12) = 34.2 ft/s
[since p2 = p3 = 0]
[ANSWER]
b) Water velocity at the faucet on the second floor
1
1
2
2
p3 + ρv3 + γz3 = p4 + ρv4 + γz 4
2
2
v4 =
2⎡
1
⎤
2
2
p3 + ρv3 + γ ( z3 − z 4 ) − p4 ⎥ = v3 + 2 g ( z3 − z 4 )
⎢
ρ⎣
2
⎦
(where v3 = 20 ft/s, and z3 − z4 = − 12 ft)
[since p4 = p3 = 0]
= (20) 2 + 2(32.2)(−12) = − 372.8 : Imaginary → Unreal physical value
∴ No water at the second floor
[ANSWER]
3. (Bernoulli eq. & Continuity) Water flows into a large tank
at a rate of 0.011 m3/s as shown. The water leaves the tank
through 20 holes in the bottom of the tank, each of which
produces a stream of 10-mm diameter. Determine the
equilibrium height, h, for steady state operation.
(1)
Sol) 1. The equilibrium height h
- Related with the continuity eq.
Q1 (into a tank) = Q2 (leaving through holes)
or
0.011 = 20 (holes) A2V2 = 20
π
4
(2)
2
D2 v2
2. In order to determine v2, the Bernoulli eq. along a streamline
p1 +
0+
1 2
1
ρv1 + γz1 = p2 + ρv2 2 + γz2
2
2
where p1 = p2 = 0 (atmosphere), z1 = h , z 2 = 0 , v1 = 0
1
1
ρ (0) 2 + γ (h) = 0 + ρv2 2 + γ (0)
2
2
&
v 2 = 2 gh
3. Insert v2, determined from the Bernoulli eq. into the continuity eq.
4. (Bernoulli eq. & Continuity) Water flows through a pipe
reducer as shown. The static pressures at (1) and (2) are
measured by the inverted U-tube manometer containing
oil of specific gravity, SG, less than one. Determine
the manometer reading, h.
(B)
(C)
(A)
Sol)
l
1. U-tube manometer reading
From (1) → (A) → (B) → (C) → (2)
p1 − γ ( z 2 − z1 ) − γl − γh + γ oil h + γl = p2
where γ oil = SGγ
or
p1 − p2 = γ ( z 2 − z1 ) + (1 − SG )γh
2. p1 − p2 : To be determined from the Bernoulli eq.
1 2
1
2
p1 + ρv1 + γz1 = p2 + ρv2 + γz2
2
2
1
1
or
p1 − p2 = γ ( z 2 − z1 ) + ρv2 2 − ρv12
2
2
3. By comparing two eqs.,
⎡ ⎛ v ⎞2 ⎤
1
1
2
2
2
(1 − SG )γh = ρ [v2 − v1 ] = ρv2 ⎢1 − ⎜⎜ 1 ⎟⎟ ⎥
2
2
⎢⎣ ⎝ v2 ⎠ ⎥⎦
4. v2 2 − v12 : To be determined from the continuity eq. (Confined flow)
Q1 = A1v1 =
π
4
D12 v1
=
π
4
D2 v2 = A2 v2 = Q2
2
→
2
v1 A2 D2
=
=
v2 A1 D12
5. We need one more condition to determine v2, (e.g. Q = A2 v2 )
5. (Bernoulli eq. & Continuity & Pitot tube)
Two Pitot tubes and two static pressure taps are
placed in the pipe contraction shown. The
flowing fluid is water, and viscous effects are
negligible. Determine the two manometer
readings, h and H.
(C)
(D)
l3
(4)
(3)
(1)
l1
(2)
(B)
Sol) (1) Manometer reading h between (1) & (2)
(A)
From (1) → (A) → (B) → (2)
p1 + γl1 − SGγh − γ (l1 − h) = p2
or
where γ oil = SGγ
(1 − SG )γh = p2 − p1 (Ú Independent to l1)
y p2 − p1 : To be determined from the Bernoulli eq.
1
1
2
2
p1 + ρv1 + γz1 = p2 + ρv2 + γz 2
2
2
where, z1 = 0 , z 2 = 0 , v1 = 0 & v2 = 0 (Stagnation points!!)
∴ p 2 − p1 = 0 and thus
h=0
(2) Manometer reading H between (3) & (4) [ (3) → (C) → (D) → (4)]
p3 − γ (l3 −
or
γ (H −
1
) + γ Air H + γ (l3 − H ) = p4
12
where γ Air : Negligible
1
) = p3 − p4
12
y p3 − p4 : To be determined from the Bernoulli eq. & the continuity eq.
p3 +
1
1
ρv32 + γz3 = p4 + ρv4 2 + γz4
2
2
where, z3 = 3 / 12 ft, z 4 = 2 / 12 ft, v3 = 2 ft/s, v4 =
A3
v3
A2