3.16 An incompressible fluid flows past
an impermeable flat plate, as in Fig. P3.16,
with a uniform inlet profile u = U and a
o
cubic
polynomial
exit
profile
⎛ 3η − η 3 ⎞
y
≈
u
U
⎟ where η =
o⎜
Fig. P3.16
δ
⎝ 2 ⎠
Compute the volume flow Q across the top surface of the control volume.
Solution:
For the given control volume and incompressible flow, we obtain
δ
0 = Q top + Q right − Q left
δ
⎛ 3 y y3 ⎞
= Q + ∫ Uo ⎜
−
b dy − ∫ U o b dy
⎝ 2δ 2δ 3 ⎟⎠
0
0
5
3
= Q + U o bδ − U o bδ , solve for Q = U o bδ
8
8
Ans.
According to Torricelli’s theorem, the
velocity of a fluid draining from a hole in a
tank is V ≈ (2gh)1/2, where h is the depth of
water above the hole, as in Fig. P3.28. Let
the hole have area Ao and the cylindrical
tank have bottom area Ab. Derive a formula
for the time to drain the tank from an initial
depth ho.
3.28
Solution: For a control volume around the tank,
d ⎡
ρ dv ⎤⎦ + m out = 0
dt ⎣ ∫
ρ Ab
Fig. P3.28
0
∫
ho
dh
h
t
=∫
0
dh
= − m out = − ρ Ao 2gh
dt
Ao 2 g
dt;
Ab
t=
Ab
Ao
ho
2g
Ans.
3.54 For the pipe-flow reducing section
of Fig. P3.54, D1 = 8 cm, D2 = 5 cm, and p2
= 1 atm. All fluids are at 20°C. If V1 = 5
m/s and the manometer reading is h = 58
cm, estimate the total horizontal force
resisted by the flange bolts.
Fig. P3.54
Solution:
Let the CV cut through the bolts and through section 2. For the given
manometer reading, we may compute the upstream pressure:
p1 − p2 = (γ merc − γ water )h = (132800 − 9790)(0.58 m) ≈ 71300 Pa (gage)
Now apply conservation of mass to determine the exit velocity:
Q1 = Q2 , or (5 m/s)(π /4)(0.08 m)2 = V2 (π /4)(0.05)2 , solve for V2 ≈ 12.8 m/s
Finally, write the balance of horizontal forces:
2 − V1 ),
∑ Fx = −Fbolts + p1,gage A1 = m(V
or: Fbolts = (71300)
π
4
(0.08)2 − (998)
π
4
(0.08)2 (5.0)[12.8 − 5.0] ≈ 163 N
Ans.
3.55 In Fig. P3.55 the jet strikes a vane
which moves to the right at constant velocity
Vc on a frictionless cart. Compute (a) the force
Fx required to restrain the cart and (b) the
power P delivered to the cart. Also find the
cart velocity for which (c) the force Fx is a
maximum and (d) the power P is a maximum.
Fig. P3.55
Solution: Let the CV surround the vane and cart and move to the right at cart speed.
The jet strikes the vane at relative speed Vj − Vc. The cart does not accelerate, so the
horizontal force balance is
∑ Fx = −Fx = [ ρ A j (Vj − Vc )](Vj − Vc ) cosθ − ρ A j (Vj − Vc )2
or: Fx = ρ A j (Vj − Vc )2 (1 − cosθ ) Ans. (a)
The power delivered is P = Vc Fx = ρ A j Vc (Vj − Vc )2 (1 − cosθ ) Ans. (b)
The maximum force occurs when the cart is fixed, or: Vc = 0 Ans. (c)
The maximum power occurs when dP/dVc = 0, or: Vc = Vj /3 Ans. (d)
3.170 If losses are neglected in Fig.
P3.170, for what water level h will the flow
begin to form vapor cavities at the throat of
the nozzle?
Solution: Applying Bernoulli from (a) to (2)
gives Torricelli’s relation: V2 = √(2gh). Also,
V1 = V2 (D2 /D1 )2 = V2 (8/5)2 = 2.56V2
Fig. P3.170
Vapor bubbles form when p1 reaches the vapor pressure at 30°C, pvap ≈ 4242 Pa (from
Table A.5), while ρ ≈ 996 kg/m3 at 30°C (Table A.1). Apply Bernoulli between 1 and 2:
p1
ρ
+
V 21
2
+ gz1 ≈
p2
ρ
+
V 22
2
+ gz 2 , or:
4242 (2.56V2 )2
100000 V 22
+
+0≈
+
+0
996
2
996
2
Solve for V 22 = 34.62 = 2gh, or h = 34.62/[2(9.81)] ≈ 1.76 m
Ans.
In the spillway flow of Fig. P3.176,
the flow is assumed uniform and
hydrostatic at sections 1 and 2. If losses are
neglected, compute (a) V2 and (b) the force
per unit width of the water on the spillway.
3.176
Solution:
For mass conservation,
V2 = V1h1 /h 2 =
Fig. P3.176
(a) Now apply Bernoulli from 1 to 2:
p1
γ
+
V12
2g
5.0
V1 = 7.14V1
0.7
Solve for V 21 =
+ h1 ≈
p2
γ
+
V 22
2g
+ h 2 ; or: 0 +
V12
2g
+ 5.0 ≈ 0 +
(7.14V1 )2
+ 0.7
2g
2(9.81)(5.0 − 0.7)
m
m
, or V1 = 1.30
, V2 = 7.14V1 = 9.28
2
s
s
[(7.14) − 1]
Ans. (a)
(b) To find the force on the spillway (F ←), put a CV around sections 1 and 2 to obtain
∑ Fx = − F +
γ
2
h12 −
γ
2
2 − V1 ), or, using the given data,
h 22 = m(V
1
N
F = (9790)[(5.0)2 − (0.7)2 ] − 998[(1.30)(5.0)](9.28 − 1.30) ≈ 68300
m
2
Ans. (b)
3.178 For the water channel flow of Fig.
P3.178, h1 = 0.45 ft, H = 2.2 ft, and V1 =
16 ft/s. Neglecting losses and assuming
uniform flow at sections 1 and 2, find the
downstream depth h2. Show that two realistic
solutions are possible.
The analysis is quite similar to
Prob. 3.177 - continuity + Bernoulli:
Solution:
V2 = V1
h1 16(0.45)
=
;
h2
h2
Fig. P3.178
V12
V2
V12
(7.2/h 2 )2
+ h1 = 2 + h 2 + H =
+ 0.45 =
+ h 2 + 2.2
2g
2g
2(32.2)
2(32.2)
Combine into a cubic equation: h 32 − 2.225 h 22 + 0.805 = 0 . The three roots are:
h 2 = −0.540 ft (impossible); h 2 = +2.03 ft (subcritical);
h 2 = +0.735 ft (supercritical) Ans.