Solutions Manual • Fluid Mechanics, Fifth Edition
814
Solution: For gasoline at 20°C, take ρ ≈ 680 kg/m3 ≈ 1.32 slug/ft3. Compute the power
æ 400 ö
1.32(32.2) ç
H
è 449 ÷ø
ft⋅lbf ρgQH
P = 20 × 550 = 11000
, solve H ≈ 232 ft Ans. (a)
=
=
s
0.80
η
Then ∆p = ρ gH = 1.32(32.2)(232) = 9870 psf ÷ 144 ≈ 69 psi
Ans. (b)
11.14 A pump delivers gasoline at 20°C and 12 m3/h. At the inlet, p1 = 100 kPa, z1 = 1 m,
and V1 = 2 m/s. At the exit p2 = 500 kPa, z2 = 4 m, and V2 = 3 m/s. How much power is
required if the motor efficiency is 75%?
Solution: For gasoline, take ρ g ≈ 680(9.81) = 6671 N/m3. Compute head and power:
p2 V22
p
V2
500000
(3)2
100000
(2)2
+
+ z 2 − 1 − 1 − z1 =
+
+4−
−
− 1,
ρ g 2g
ρ g 2g
6671
2(9.81)
6671 2(9.81)
æ 12 ö
6671 ç
(63.2)
è 3600 ÷ø
ρgQH
≈ 1870 W Ans.
or: H ≈ 63.2 m, Power =
=
η
0.75
H=
11.15 A lawn sprinkler can be used as a
simple turbine. As shown in Fig. P11.15,
flow enters normal to the paper in the center
and splits evenly into Q/2 and Vrel leaving
each nozzle. The arms rotate at angular
velocity ω and do work on a shaft. Draw the
velocity diagram for this turbine. Neglecting
friction, find an expression for the power
delivered to the shaft. Find the rotation rate
for which the power is a maximum.
Solution: Utilizing the velocity diagram at
right, we apply the Euler turbine formula:
P = ρ Q(u2 Vt2 − u1Vt1 ) = ρ Q[u(W − u) − 0]
or: P = ρ Qω R( Vrel − ω R)
Ans.
V
dP
= ρQ(Vrel − 2u) = 0 if ω = rel Ans.
2R
du
2
where Pmax = ρQu(2u − u) = ρQ(ω R)
Fig. P11.15