Modern Engineering Thermodynamics

306 CHAPTER 9: Second Law Open System Applications
HYDRODYNAMIC FLOW SYSTEMS
Table 9.4 Hydraulic Flow Systems
System
Flow in a straight pipe
Flow through valves, fittings, etc. (“minor losses”)
Flow through sudden contractions or expansions
Flow through a hydraulic jump
Continued
Heat Loss Formula
1(h L ) 2 = f(L/D)(V 2 /2g), where f is the Darcy-Weisbach friction factor
1(h L ) 2 = K M (V 2 /2g), where K M is the minor loss coefficient
[ 1 (h L ) 2 ] contraction = K C (V 2 /2g), where K C is the contraction coefficient and
V is the contraction outlet velocity or expansion inlet velocity. Also
½ 1 ðh L Þ 2
Š expansion
= 1 − D 2 2
1 /D2 2 V 2 /2g
1(h L ) 2 =(y 2 – y 1 ) 3 /(4y 1 y 2 ), where y 2 – y 1 is the jump height
Note: Values for the coefficients f, K M , and K C can be found in standard fluid mechanics textbooks.
Combining the MERB and the MSRB equations for any type of material
produces a general formula for the entropy production rate
inside a steady state, steady flow, single-inlet, single-outlet system
with an isothermal boundary as
_S P = _m s 2 − s 1 + ðg/g c Þ 1 ðh L Þ 2 − u2 + u 1 /Tb
(9.57)
Table 9.4 shows typical head loss formulations for a few common
hydrodynamic flow situations. With these formulae, the entropy
production rates can be calculated for many different hydraulic or
pneumatic flow systems.
The hydraulic jump is a very effective phenomenon for dissipating
energy. It commonly appears at the end of chutes or spillways to dissipate
the kinetic energy of the flow. It is also an effective mixing
process due to the violent dissipative agitation that takes place. The
following example illustrates the entropy production rate in a
hydraulic jump.
8 Recall that these terms mean we assume that the flow is steady state (SS) and
steady flow (SF), that the flow has a single inlet and a single outlet (SI, SO), and
that there is no work (aergonic) or heat transfer (adiabatic).
EXAMPLE 9.10
At a rate of 500. lbm/s, water at 50.0°F flows down a spillway onto a horizontal floor. A hydraulic jump 1.80 ft high
appears at the bottom of the spillway. The jump has an inlet velocity of 8.00 ft/s and an inlet height of 1.00 ft. Determine
the energy dissipation rate and entropy production rate.
Solution
First, draw a sketch of the system (Figure 9.21).
Hydraulic jump
m = 500. lbm/s
1.00 ft
1.80ft
5.14 ft/s
8.00 ft/s
FIGURE 9.21
Example 9.10.
The unknowns are the energy dissipation rate and entropy production rate. The material is liquid water.
From the formula in Table 9.4, we get
and
_m ðg/g c
1ðh L Þ 2
= ð1:80 − 1:00Þ 3 /41:0 ½ ð Þ01:80 ð ÞŠ = 0:0711 ft
Þ 1 ðh L Þ 2 = ð500: lbm/sÞ
32:174 ft/s
2
fð Þ/32:174 ½ lbm.ft/ ðlbf .s 2 ÞŠgð0:0711 ftÞ
= ð35:55 ft.lbf/sÞ/ ð778:17 ft.lbf/BtuÞ = 0:0457 Btu/s