fluid_mechanics

5.3 First Law of Thermodynamics — The Energy Equation 233
where the mass flowrate, m # , is (from Eq. 5.6)
m # AV D 2 2
4 V 2
For fluid density, , we use 1.23 kg/m 3 (standard air) and, thus,
from Eqs. 4 and 5 we obtain
W # shaft
net in
w shaft
net in 1rpD 2 242V 2
10.4 kW231000 1Nm21skW24
11.23 kgm 3 231p210.6 m2 2 44112 ms2
(5)
or
From Eqs. 2, 3, and 6 we obtain
w shaft 95.8 Nm/kg
net in
72.0 Nm/kg
95.8 Nm/kg 0.752
(6)
(Ans)
COMMENT Note that only 75% of the power that was delivered
to the air resulted in useful effects, and, thus, 25% of the
shaft power is lost to air friction.
F l u i d s i n t h e N e w s
Curtain of air An air curtain is produced by blowing air through
a long rectangular nozzle to produce a high-velocity sheet of air,
or a “curtain of air.” This air curtain is typically directed over a
doorway or opening as a replacement for a conventional door.
The air curtain can be used for such things as keeping warm air
from infiltrating dedicated cold spaces, preventing dust and other
contaminates from entering a clean environment, and even just
keeping insects out of the workplace, still allowing people to enter
or exit. A disadvantage over conventional doors is the added
power requirements to operate the air curtain, although the advantages
can outweigh the disadvantage for various industrial
applications. New applications for current air curtain designs
continue to be developed. For example, the use of air curtains as
a means of road tunnel fire security is currently being investigated.
In such an application, the air curtain would act to isolate
a portion of the tunnel where fire has broken out and not allow
smoke and fumes to infiltrate the entire tunnel system. (See
Problem 5.123.)
If Eq. 5.82, which involves energy per unit mass, is multiplied by fluid density, r, we obtain
V5.14 Water plant
aerator
p out rV 2 out
2
gz out p in rV 2 in
2 gz in rw shaft r1loss2
net in
(5.83)
where g rg is the specific weight of the fluid. Equation 5.83 involves energy per unit volume and
the units involved are identical with those used for pressure 1ft # lbft 3 lbft 2 or N # mm 3 Nm 2 2.
If Eq. 5.82 is divided by the acceleration of gravity, g, we get
p out
(5.84)
g V 2 out
2g z out p in
g V 2 in
2g z in h s h L
where
The energy equation
written in
terms of energy
per unit weight
involves heads.
(5.85)
is the shaft work head and h L lossg is the head loss. Equation 5.84 involves energy per unit weight
1ft # lblb ft or N # mN m2. In Section 3.7, we introduced the notion of “head,” which is energy
per unit weight. Units of length 1for example, ft, m2 are used to quantify the amount of head involved.
If a turbine is in the control volume, h s is negative because it is associated with shaft work out of
the control volume. For a pump in the control volume, h s is positive because it is associated with
shaft work into the control volume.
We can define a total head, H, as follows
Then Eq. 5.84 can be expressed as
h s w shaft net ing
W # shaft
net in
m # g
H p g V 2
2g z
W # shaft
net in
gQ
H out H in h s h L