Modern Engineering Thermodynamics

674 CHAPTER 16: Compressible Fluid Flow
where ∑ net F ext g c is the net sum of all the external forces acting on the system. Note that, for a closed system,
_m = 0andm closed system = constant. Then, Eq. (16.32) reduces to Newton’s second law for a fixed mass closed
system:
d
dt ðmVÞ closed
system
= m dV
= ma
dt closed closed
system system
= ∑ F ext g c
net
Also, the external forces are normally divided into two categories: surface forces, such as pressure and contact
forces, and body forces, such as gravity and magnetic forces. That is,
∑F ext = ∑F surface +∑F body
For a steady state, steady flow, single-inlet, single-outlet open system, Eq. (16.32) further reduces to
Linear momentum rate balance (SS, SF, SI, SO):
∑F ext = _mðV out − V in Þ/g c (16.33)
EXAMPLE 16.10
An air jet is used to levitate a 5.00 × 10 –3 kg sheet of paper, as shown in Figure 16.20. The diameter of the jet at the nozzle
exit is 3.00 × 10 –3 m and it is at atmospheric pressure at 20.0°C. Determine the velocity of the jet.
Solution
W
Assume the flow is steady state and steady flow. Then, the y
Paper
component of Eq. (16.33) is
F y = −W = _mðV out − V in Þ y
/g c
where W is the weight of the paper, given by
W = mg/g c = ð5:00 × 10 −3 kgÞð9:81 m=s 2 Þ/ð1Þ = 0:0491 N
y
0.003 m in diameter
and
ðV out Þ y
= 0
Nozzle
Also, _m = ρAV = ρπ D 2 V/4, where
Then,
So,
ρ =
p
RT = 101:3 × 10 3 kg/ðm.s 2 Þ
½286 m 2 /ðs 2 . = 1:21 kg/m3
KÞŠð293:15 KÞ
W = _mðV in Þ y
/g c = ρAV 2 in /g c = ρπ D 2 V 2 in /4g c
FIGURE 16.20
Example 16.10.
x
Air flow
V in =
4g " # 1/2
cW
4ð1Þð0:0491 kg.m=s 2 1/2
Þ
=
ρπ D 2 ð1:21 kg=m 3 ÞðπÞð3:00 × 10 − 3 mÞ 2 = 75:7m=s
Exercises
25. The nozzle diameter in Example 16.10 is to be reduced from 3.00 × 10 –3 m to 1.50 × 10 –3 m. Determine the velocity of
the resulting jet required to levitate the sheet of paper, assuming all the other variables remain unchanged. Answer:
V jet = 151 m/s.
26. Suppose the sheet of paper in Example 16.10 is replaced by a sheet of cardboard with a mass of 5 × 10 –2 kg. Determine
the velocity of the jet required to levitate the cardboard, assuming all the other variables remain unchanged. Answer:
V jet = 239 m/s.
27. If the air jet in Example 16.10 is at 0.00°C rather than 20.0°C, determine the velocity of the jet required to levitate the
sheet of paper assuming all the other variables remain unchanged. Answer: V jet = 73.1 m/s.