Download Solutions for Section 4.5

Mechanics of Fluids, 4th Edition
Chapter 4: The Integral Forms of the
Fundamental Laws
Solutions for Section 4.5
1.
Air moves through the duct shown in the problem. The work-rate terms that must be
accounted for include:
& and pn
&
ˆ ⋅ VdA and W
(C) W
S
∫
shear
&
&
The belt inputs energy in the form of W
shear . The fan inputs energy as WS . And,
there would be a pressure difference between the inlet and outlet requiring flow
ˆ ⋅ VdA.
work, i.e., ∫ pn
2. The owner of a cabin that is located on a small stream is thinking of placing a dam across the
stream to create a possible drop of 120 cm. The stream is measured to have approximate
dimensions of 180 cm by 10 cm and a leaf on the water’s surface is observed to travel 10 m in
8 s. If an 80% efficient turbine is used, how much power can the owner expect to generate?
(B) 2 kW
The flow rate is calculated, based on the estimates with the assumption of uniform
flow, to be
Q = AV = (1.8 × 0.1) ×
10
= 0.225 m3 /s
8
The energy equation, Eq. 4.5.24, with p1 = p2 = 0 and negligible kinetic energy
change, provides
p1
p
V22
V12
+
+ 2 + z2 or HT = 1.2 m
+ 1.2 = HT +
2g
2g
γ
γ
Equation (4.5.25) predicts that the energy output may be
& = γ QH η = 9810 × 0.225 ×1.2 × 0.8 = 2120 W
W
T
T T
31
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Mechanics of Fluids, 4th Edition
Chapter 4: The Integral Forms of the
Fundamental Laws
3. The hydroturbine shown in the problem accepts water at 4 MPa and exits to the atmosphere. If
the flow rate is 2 m3/s, the maximum power output is nearest:
(A) 80 MW
First, find the velocities: V1 = 2/π × 0.2 2 = 15.9 m/s, V2 = 2/π × 0.32 = 7.07 m/s. Now,
apply the energy equation, Eq. 4.5.24, assuming negligible change in elevation:
15.92 40 × 106
7.07 2 p2
+
+ z1 = HT +
+
+ z2 or HT = 4090 m
2g
9810
2g
γ
The power is then
& = γ QH = 9810 × 2 × 4090 = 80.2 ×106 W
W
T
T
4. In the setup shown in the problem, if water is flowing, the pressure gage should read:
(D) 940 kPa
Position a and b on the low point of the mercury, as in Example 4.8. Then the
manometer demands that (review Example 3.5)
p a = pb
0.2 × (13.6 × 9810) + z1γ + p1 = 0.2 × 9810 + z2γ + γ
V22
+ p2
2g
since the pressure p2 at the exit is zero gage and the two elevations cancel out. The
energy equation between the inlet and the outlet, neglecting any losses since it’s a
contraction, provides
p1
γ
+
p2 V22
V12
+ γ z1 =
+
+ γ z2
2g
γ
2g
or
p1
γ
+
V12 V22
=
2g 2g
Subtract this energy equation from the manometer equation and obtain
V12
= 0.2 × 12.6.
2g
∴ V1 = 7.03 m/s
Continuity requires V2 = V1 A1 /A2 = 7.03 × 102 /4 2 = 43.9 m/s . The energy equation
(or Bernoulli’s equation) then gives the pressure as
p1
7.032
43.92
+
=
.
9810 2 × 9.81 2 × 9.81
∴ p1 = 939 000 Pa
32
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.