fluid_mechanics

5.1 Conservation of Mass—The Continuity Equation 191
SOLUTION
The pumping capacity sought is the volume flowrate delivered by
the fire pump to the hose and nozzle. Since we desire knowledge
about the pump discharge flowrate and we have information
about the nozzle exit flowrate, we link these two flowrates with
the control volume designated with the dashed line in Fig. E5.1b.
This control volume contains, at any instant, water that is within
the hose and nozzle from the pump discharge to the nozzle exit
plane.
Equation 5.5 is applied to the contents of this control volume
to give
0 (flow is steady)
0
r dV nˆ dA 0
(1)
csrV
0t
cv
The time rate of change of the mass of the contents of this control
volume is zero because the flow is steady. Because there is only
one inflow [the pump discharge, section (1)] and one outflow [the
nozzle exit, section (2)], Eq. (1) becomes
so that with m # AV
Because the mass flowrate is equal to the product of fluid density, ,
and volume flowrate, Q (see Eq. 5.6), we obtain from Eq. 2
Liquid flow at low speeds, as in this example, may be considered
incompressible. Therefore
and from Eqs. 3 and 4
2 A 2 V 2 1 A 1 V 1 0
m # 1 m # 2
2 Q 2 1 Q 1
2 1
Q 2 Q 1
(2)
(3)
(4)
(5)
The pumping capacity is equal to the volume flowrate at the nozzle
exit. If, for simplicity, the velocity distribution at the nozzle exit plane,
section (2), is considered uniform (one-dimensional), then from Eq. 5
Q 1 Q 2 V 2 A 2
V 2 4 D 2 2 120 m/s2
0.0251 m 3 /s
2
4 a 40 mm
1000 mm/m b
(Ans)
COMMENT By repeating the calculations for various values
of the nozzle exit diameter, D 2 , the results shown in Fig.
E5.1c are obtained. The flowrate is proportional to the exit area,
which varies as the diameter squared. Hence, if the diameter
were doubled, the flowrate would increase by a factor of four,
provided the exit velocity remained the same.
Q 1 , m3 /s
0.15
0.10
0.05
(40 mm, 0.0251 m 3 /s)
0
0 20 40 60 80 100
F I G U R E E5.1c
D 2 , mm
E XAMPLE 5.2
Conservation of Mass—Steady, Compressible Flow
GIVEN Air flows steadily between two sections in a long,
straight portion of 4-in. inside diameter pipe as indicated in
Fig. E5.2. The uniformly distributed temperature and pressure at
each section are given. The average air velocity 1nonuniform velocity
distribution2 at section 122 is 1000 fts.
FIND Calculate the average air velocity at section 112.
Control volume
SOLUTION F I G U R E E5.2
Flow
Section (1)
p 1 = 100 psia
T 1 = 540 °R
D 1 = D 2 = 4 in.
Pipe
Section (2)
p 2 = 18.4 psia
T 2 = 453 °R
V 2 = 1000 ft/s
The average fluid velocity at any section is that velocity which
yields the section mass flowrate when multiplied by the section
average fluid density and section area 1Eq. 5.72. We relate the
flows at sections 112 and 122 with the control volume designated
with a dashed line in Fig. E5.2.
Equation 5.5 is applied to the contents of this control volume
to obtain
0 1flow is steady2
0
0t r dV cs
rV nˆ dA 0
cv
The time rate of change of the mass of the contents of this control
volume is zero because the flow is steady. The control surface